Finite time calculation in QED [Archive]

Igor Khavkine

Mar30-05, 08:59 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-03-30, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n>> I don\'t have time to do this calculation myself. But, if you are\n>> willing, I can tell you exactly what to do.\n>\n> Sure, I would like to know that. I don\'t need detailed calculations.\n> It would be enough if I see that the step is doable in principle.\n> Below I repeat the 8-step procedure for your convenience.\n> I have a problem understanding how your Step 1 is going to yield\n> any numerical result. The state exp(iH(t - -oo))|0> is an infinite\n> linear combination of multi-bare-particle states. Are you going to\n> write this linear combination in an explicit form? Are you using\n> original QED Hamiltonian (with physical masses and charges) or the\n> Hamiltonian with counterterms?\n\nUnderstanding things in principle can be a very ephemeral. Understanding\nitself is of little value. One must be able to do something with it. I\nwould strongly recommend performing at least one calculation of this\nsort. The one I have in mind would be of special interest to you. You\'ve\noften complained that you have not seen a definitive calculation of the\nspeed of interaction in QED. Now is your chance to actually do one.\n\nThe model itself is not QED but should have all the important features.\nConsider a complex scalar field phi. It is minimally coupled to gauge\nfield A. The Lagrangian density is\n\nL_1 = -|Dphi|^2 - m^2 |phi|^2 - V|phi|^2 - (1/4) tr(F^2).\n\nHere D = @ - ieA, the covariant derivative. F is the field strength for\nA. V is the external potential. There is a preferred frame in which the\npotential V = V(x) is stationary and is two infinite square wells,\none [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].\n\nThe only reason I think a scalar field is easier to handle is because in\nthe case of Dirac fields, the boundary conditions for the infinite\nsquare well are not so simple.\n\nIn the infinite past, a particle is placed in each well. At any finite\ntime t the properties of this tate can be examined, for example the\nexpectation vale for the position of the particle in each well. At t=0 a\ntime dependent perturbation can be added to V which sets one of the\nparticles into motion. The effects on the other particle can be\ncalculated.\n\nNow, lets take baby steps.\n\nIn the free field case (e=0), what is are the equations of motion for phi?\nSolve them. Quantize the fields.\n\nWhat is the Hamiltonian? Diagonalize it. Write down and solve the\nHeisenberg equations of motion for the fields.\n\nWhat is the position operator? Put a particle in one of the boxes. What\nis the position expectation value for this state?\n\nWrite the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear\ntime independent) part and H_1 the interaction part (the rest). Switch\nto the interaction picture. Write down the equations of motion for the\nfield operators and the states.\n\nOnce you get these done, we can talk more.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-03-30, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

>> I don't have time to do this calculation myself. But, if you are
>> willing, I can tell you exactly what to do.
>
> Sure, I would like to know that. I don't need detailed calculations.
> It would be enough if I see that the step is doable in principle.
> Below I repeat the 8-step procedure for your convenience.
> I have a problem understanding how your Step 1 is going to yield
> any numerical result. The state \exp(iH(t - -oo))|0> is an infinite
> linear combination of multi-bare-particle states. Are you going to
> write this linear combination in an explicit form? Are you using
> original QED Hamiltonian (with physical masses and charges) or the
> Hamiltonian with counterterms?

Understanding things in principle can be a very ephemeral. Understanding
itself is of little value. One must be able to do something with it. I
would strongly recommend performing at least one calculation of this
sort. The one I have in mind would be of special interest to you. You've
often complained that you have not seen a definitive calculation of the
speed of interaction in QED. Now is your chance to actually do one.

The model itself is not QED but should have all the important features.
Consider a complex scalar field \phi. It is minimally coupled to gauge
field A. The Lagrangian density is

L_1 = -|Dphi|^2 - m^2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F^2).

Here D = @ - ieA, the covariant derivative. F is the field strength for
A. V is the external potential. There is a preferred frame in which the
potential V = V(x) is stationary and is two infinite square wells,
one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].

The only reason I think a scalar field is easier to handle is because in
the case of Dirac fields, the boundary conditions for the infinite
square well are not so simple.

In the infinite past, a particle is placed in each well. At any finite
time t the properties of this tate can be examined, for example the
expectation vale for the position of the particle in each well. At t=0 a
time dependent perturbation can be added to V which sets one of the
particles into motion. The effects on the other particle can be
calculated.

Now, lets take baby steps.

In the free field case (e=0), what is are the equations of motion for \phi?
Solve them. Quantize the fields.

What is the Hamiltonian? Diagonalize it. Write down and solve the
Heisenberg equations of motion for the fields.

What is the position operator? Put a particle in one of the boxes. What
is the position expectation value for this state?

Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear
time independent) part and H_1 the interaction part (the rest). Switch
to the interaction picture. Write down the equations of motion for the
field operators and the states.

Once you get these done, we can talk more.

Igor

Eugene Stefanovich

Mar30-05, 12:52 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n> On 2005-03-30, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>>Igor Khavkine wrote:\n>\n>\n>>>I don\'t have time to do this calculation myself. But, if you are\n>>>willing, I can tell you exactly what to do.\n>>\n>>Sure, I would like to know that. I don\'t need detailed calculations.\n>>It would be enough if I see that the step is doable in principle.\n>>Below I repeat the 8-step procedure for your convenience.\n>>I have a problem understanding how your Step 1 is going to yield\n>>any numerical result. The state exp(iH(t - -oo))|0> is an infinite\n>>linear combination of multi-bare-particle states. Are you going to\n>>write this linear combination in an explicit form? Are you using\n>>original QED Hamiltonian (with physical masses and charges) or the\n>>Hamiltonian with counterterms?\n>\n>\n> Understanding things in principle can be a very ephemeral. Understanding\n> itself is of little value. One must be able to do something with it. I\n> would strongly recommend performing at least one calculation of this\n> sort. The one I have in mind would be of special interest to you. You\'ve\n> often complained that you have not seen a definitive calculation of the\n> speed of interaction in QED. Now is your chance to actually do one.\n>\n> The model itself is not QED but should have all the important features.\n> Consider a complex scalar field phi. It is minimally coupled to gauge\n> field A. The Lagrangian density is\n>\n> L_1 = -|Dphi|^2 - m^2 |phi|^2 - V|phi|^2 - (1/4) tr(F^2).\n>\n> Here D = @ - ieA, the covariant derivative. F is the field strength for\n> A. V is the external potential. There is a preferred frame in which the\n> potential V = V(x) is stationary and is two infinite square wells,\n> one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].\n>\n> The only reason I think a scalar field is easier to handle is because in\n> the case of Dirac fields, the boundary conditions for the infinite\n> square well are not so simple.\n>\n> In the infinite past, a particle is placed in each well. At any finite\n> time t the properties of this tate can be examined, for example the\n> expectation vale for the position of the particle in each well. At t=0 a\n> time dependent perturbation can be added to V which sets one of the\n> particles into motion. The effects on the other particle can be\n> calculated.\n>\n> Now, lets take baby steps.\n>\n> In the free field case (e=0), what is are the equations of motion for phi?\n> Solve them. Quantize the fields.\n>\n> What is the Hamiltonian? Diagonalize it. Write down and solve the\n> Heisenberg equations of motion for the fields.\n>\n> What is the position operator? Put a particle in one of the boxes. What\n> is the position expectation value for this state?\n>\n> Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear\n> time independent) part and H_1 the interaction part (the rest). Switch\n> to the interaction picture. Write down the equations of motion for the\n> field operators and the states.\n>\n> Once you get these done, we can talk more.\n\nThe point is that I already did this calculation. You can consider my\nentire book as a detailed solution of this particular problem: what\nis the dynamics of interaction between charged particles? I considered\nfermions instead of scalar particles and I did not put them in the\ninfinite well, as you suggest, but nevertheless I made all the above\nsteps in the book. And I\'ve got an answer: the interaction acts\ninstantaneously, i.e., the force acting on particle B depends on\nthe position and velocity of particle A at the same time instant.\nIf you ask me to repeat this calculation for scalar fields, I\'ll\nfollow exactly the same route as I did for QED. I am pretty sure that\nI\'ll get the same answer: there is no retardation. I am so sure\nabout that without doing actual calculation, because retardation\nof the direct interaction between dressed particles would contradict the\nlaw of conservation of momentum.\n\nIf I understand you correctly, neither you nor Arnold Neumaier are\nsaying that what I did is wrong. You said that particle dynamics can\nbe also evaluated by more traditional methods of QFT (closed-path,\nDirac-Fock, wavefunction renormalization, 8-step formula,...). I may\ntrust you on that. However, you also claim that thus obtained\ninteractions are bound to be retarded. That\'s where I start to worry.\nBecause that\'s not what my approach shows. This means that my approach\nis not equivalent to yours. Then which one is correct? I opened all my\ncards: all derivations of interparticle instantaneous potentials are\npresented in the book. Now it is your turn: show how you calculate\ndynamics and where the retardation comes from.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On 2005-03-30, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>>Igor Khavkine wrote:
>
>
>>>I don't have time to do this calculation myself. But, if you are
>>>willing, I can tell you exactly what to do.
>>
>>Sure, I would like to know that. I don't need detailed calculations.
>>It would be enough if I see that the step is doable in principle.
>>Below I repeat the 8-step procedure for your convenience.
>>I have a problem understanding how your Step 1 is going to yield
>>any numerical result. The state \exp(iH(t - -oo))|0> is an infinite
>>linear combination of multi-bare-particle states. Are you going to
>>write this linear combination in an explicit form? Are you using
>>original QED Hamiltonian (with physical masses and charges) or the
>>Hamiltonian with counterterms?
>
>
> Understanding things in principle can be a very ephemeral. Understanding
> itself is of little value. One must be able to do something with it. I
> would strongly recommend performing at least one calculation of this
> sort. The one I have in mind would be of special interest to you. You've
> often complained that you have not seen a definitive calculation of the
> speed of interaction in QED. Now is your chance to actually do one.
>
> The model itself is not QED but should have all the important features.
> Consider a complex scalar field \phi. It is minimally coupled to gauge
> field A. The Lagrangian density is
>
> L_1 = -|Dphi|^2 - m^2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F^2).
>
> Here D = @ - ieA, the covariant derivative. F is the field strength for
> A. V is the external potential. There is a preferred frame in which the
> potential V = V(x) is stationary and is two infinite square wells,
> one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].
>
> The only reason I think a scalar field is easier to handle is because in
> the case of Dirac fields, the boundary conditions for the infinite
> square well are not so simple.
>
> In the infinite past, a particle is placed in each well. At any finite
> time t the properties of this tate can be examined, for example the
> expectation vale for the position of the particle in each well. At t=0 a
> time dependent perturbation can be added to V which sets one of the
> particles into motion. The effects on the other particle can be
> calculated.
>
> Now, lets take baby steps.
>
> In the free field case (e=0), what is are the equations of motion for \phi?
> Solve them. Quantize the fields.
>
> What is the Hamiltonian? Diagonalize it. Write down and solve the
> Heisenberg equations of motion for the fields.
>
> What is the position operator? Put a particle in one of the boxes. What
> is the position expectation value for this state?
>
> Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear
> time independent) part and H_1 the interaction part (the rest). Switch
> to the interaction picture. Write down the equations of motion for the
> field operators and the states.
>
> Once you get these done, we can talk more.

The point is that I already did this calculation. You can consider my
entire book as a detailed solution of this particular problem: what
is the dynamics of interaction between charged particles? I considered
fermions instead of scalar particles and I did not put them in the
infinite well, as you suggest, but nevertheless I made all the above
steps in the book. And I've got an answer: the interaction acts
instantaneously, i.e., the force acting on particle B depends on
the position and velocity of particle A at the same time instant.
If you ask me to repeat this calculation for scalar fields, I'll
follow exactly the same route as I did for QED. I am pretty sure that
I'll get the same answer: there is no retardation. I am so sure
about that without doing actual calculation, because retardation
of the direct interaction between dressed particles would contradict the
law of conservation of momentum.

If I understand you correctly, neither you nor Arnold Neumaier are
saying that what I did is wrong. You said that particle dynamics can
be also evaluated by more traditional methods of QFT (closed-path,
Dirac-Fock, wavefunction renormalization, 8-step formula,...). I may
trust you on that. However, you also claim that thus obtained
interactions are bound to be retarded. That's where I start to worry.
Because that's not what my approach shows. This means that my approach
is not equivalent to yours. Then which one is correct? I opened all my
cards: all derivations of interparticle instantaneous potentials are
presented in the book. Now it is your turn: show how you calculate
dynamics and where the retardation comes from.

Eugene Stefanovich.

Igor Khavkine

Mar30-05, 01:06 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <424AED06.6020800@synopsys.com>, Eugene Stefanovich wrote:\n>\n>\n> Igor Khavkine wrote:\n>> On 2005-03-30, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>>\n>>>Igor Khavkine wrote:\n>>\n>>\n>>>>I don\'t have time to do this calculation myself. But, if you are\n>>>>willing, I can tell you exactly what to do.\n>>>\n>>>Sure, I would like to know that. I don\'t need detailed calculations.\n>>>It would be enough if I see that the step is doable in principle.\n>>>Below I repeat the 8-step procedure for your convenience.\n>>>I have a problem understanding how your Step 1 is going to yield\n>>>any numerical result. The state exp(iH(t - -oo))|0> is an infinite\n>>>linear combination of multi-bare-particle states. Are you going to\n>>>write this linear combination in an explicit form? Are you using\n>>>original QED Hamiltonian (with physical masses and charges) or the\n>>>Hamiltonian with counterterms?\n>>\n>>\n>> Understanding things in principle can be a very ephemeral. Understanding\n>> itself is of little value. One must be able to do something with it. I\n>> would strongly recommend performing at least one calculation of this\n>> sort. The one I have in mind would be of special interest to you. You\'ve\n>> often complained that you have not seen a definitive calculation of the\n>> speed of interaction in QED. Now is your chance to actually do one.\n>>\n>> The model itself is not QED but should have all the important features.\n>> Consider a complex scalar field phi. It is minimally coupled to gauge\n>> field A. The Lagrangian density is\n>>\n>> L_1 = -|Dphi|^2 - m^2 |phi|^2 - V|phi|^2 - (1/4) tr(F^2).\n>>\n>> Here D = @ - ieA, the covariant derivative. F is the field strength for\n>> A. V is the external potential. There is a preferred frame in which the\n>> potential V = V(x) is stationary and is two infinite square wells,\n>> one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].\n>>\n>> The only reason I think a scalar field is easier to handle is because in\n>> the case of Dirac fields, the boundary conditions for the infinite\n>> square well are not so simple.\n>>\n>> In the infinite past, a particle is placed in each well. At any finite\n>> time t the properties of this tate can be examined, for example the\n>> expectation vale for the position of the particle in each well. At t=0 a\n>> time dependent perturbation can be added to V which sets one of the\n>> particles into motion. The effects on the other particle can be\n>> calculated.\n>>\n>> Now, lets take baby steps.\n>>\n>> In the free field case (e=0), what is are the equations of motion for phi?\n>> Solve them. Quantize the fields.\n>>\n>> What is the Hamiltonian? Diagonalize it. Write down and solve the\n>> Heisenberg equations of motion for the fields.\n>>\n>> What is the position operator? Put a particle in one of the boxes. What\n>> is the position expectation value for this state?\n>>\n>> Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear\n>> time independent) part and H_1 the interaction part (the rest). Switch\n>> to the interaction picture. Write down the equations of motion for the\n>> field operators and the states.\n>>\n>> Once you get these done, we can talk more.\n\n> The point is that I already did this calculation. You can consider my\n> entire book as a detailed solution of this particular problem: what\n> is the dynamics of interaction between charged particles? I considered\n> fermions instead of scalar particles and I did not put them in the\n> infinite well, as you suggest, but nevertheless I made all the above\n> steps in the book. And I\'ve got an answer: the interaction acts\n> instantaneously, i.e., the force acting on particle B depends on\n> the position and velocity of particle A at the same time instant.\n> If you ask me to repeat this calculation for scalar fields, I\'ll\n> follow exactly the same route as I did for QED. I am pretty sure that\n> I\'ll get the same answer: there is no retardation. I am so sure\n> about that without doing actual calculation, because retardation\n> of the direct interaction between dressed particles would contradict the\n> law of conservation of momentum.\n\nI\'m not asking you to do this calculation the same way you did in your\nbook. I\'m asking you to do it in the way that I outline. And neither am\nI asking you calculate the force or the potential due to either\nparticle. The goal of the calculation is to evaluate the expectation\nvalue of the position of a particle in one of the wells. I predict that\nyou will not get the answer that you expect.\n\n> If I understand you correctly, neither you nor Arnold Neumaier are\n> saying that what I did is wrong. You said that particle dynamics can\n> be also evaluated by more traditional methods of QFT (closed-path,\n> Dirac-Fock, wavefunction renormalization, 8-step formula,...). I may\n> trust you on that. However, you also claim that thus obtained\n> interactions are bound to be retarded. That\'s where I start to worry.\n> Because that\'s not what my approach shows. This means that my approach\n> is not equivalent to yours. Then which one is correct? I opened all my\n> cards: all derivations of interparticle instantaneous potentials are\n> presented in the book. Now it is your turn: show how you calculate\n> dynamics and where the retardation comes from.\n\nNo need to just trust, now is your chance to find out for sure. And I\'m\nafraid the turn is still yours. No-one except yourself has a steak in,\nand hence has a reason to dedicate time to, finding where your approach\nand the standard one cease to be equivalent. Now is your chance to find\nout for sure. More talk without any calculations will lead nowhere.\n\nIgor\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <424AED06.6020800@synopsys.com>, Eugene Stefanovich wrote:
>
>
> Igor Khavkine wrote:
>> On 2005-03-30, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>>
>>>Igor Khavkine wrote:
>>
>>
>>>>I don't have time to do this calculation myself. But, if you are
>>>>willing, I can tell you exactly what to do.
>>>
>>>Sure, I would like to know that. I don't need detailed calculations.
>>>It would be enough if I see that the step is doable in principle.
>>>Below I repeat the 8-step procedure for your convenience.
>>>I have a problem understanding how your Step 1 is going to yield
>>>any numerical result. The state \exp(iH(t - -oo))|0> is an infinite
>>>linear combination of multi-bare-particle states. Are you going to
>>>write this linear combination in an explicit form? Are you using
>>>original QED Hamiltonian (with physical masses and charges) or the
>>>Hamiltonian with counterterms?
>>
>>
>> Understanding things in principle can be a very ephemeral. Understanding
>> itself is of little value. One must be able to do something with it. I
>> would strongly recommend performing at least one calculation of this
>> sort. The one I have in mind would be of special interest to you. You've
>> often complained that you have not seen a definitive calculation of the
>> speed of interaction in QED. Now is your chance to actually do one.
>>
>> The model itself is not QED but should have all the important features.
>> Consider a complex scalar field \phi. It is minimally coupled to gauge
>> field A. The Lagrangian density is
>>
>> L_1 = -|Dphi|^2 - m^2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F^2).
>>
>> Here D = @ - ieA, the covariant derivative. F is the field strength for
>> A. V is the external potential. There is a preferred frame in which the
>> potential V = V(x) is stationary and is two infinite square wells,
>> one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].
>>
>> The only reason I think a scalar field is easier to handle is because in
>> the case of Dirac fields, the boundary conditions for the infinite
>> square well are not so simple.
>>
>> In the infinite past, a particle is placed in each well. At any finite
>> time t the properties of this tate can be examined, for example the
>> expectation vale for the position of the particle in each well. At t=0 a
>> time dependent perturbation can be added to V which sets one of the
>> particles into motion. The effects on the other particle can be
>> calculated.
>>
>> Now, lets take baby steps.
>>
>> In the free field case (e=0), what is are the equations of motion for \phi?
>> Solve them. Quantize the fields.
>>
>> What is the Hamiltonian? Diagonalize it. Write down and solve the
>> Heisenberg equations of motion for the fields.
>>
>> What is the position operator? Put a particle in one of the boxes. What
>> is the position expectation value for this state?
>>
>> Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear
>> time independent) part and H_1 the interaction part (the rest). Switch
>> to the interaction picture. Write down the equations of motion for the
>> field operators and the states.
>>
>> Once you get these done, we can talk more.

> The point is that I already did this calculation. You can consider my
> entire book as a detailed solution of this particular problem: what
> is the dynamics of interaction between charged particles? I considered
> fermions instead of scalar particles and I did not put them in the
> infinite well, as you suggest, but nevertheless I made all the above
> steps in the book. And I've got an answer: the interaction acts
> instantaneously, i.e., the force acting on particle B depends on
> the position and velocity of particle A at the same time instant.
> If you ask me to repeat this calculation for scalar fields, I'll
> follow exactly the same route as I did for QED. I am pretty sure that
> I'll get the same answer: there is no retardation. I am so sure
> about that without doing actual calculation, because retardation
> of the direct interaction between dressed particles would contradict the
> law of conservation of momentum.

I'm not asking you to do this calculation the same way you did in your
book. I'm asking you to do it in the way that I outline. And neither am
I asking you calculate the force or the potential due to either
particle. The goal of the calculation is to evaluate the expectation
value of the position of a particle in one of the wells. I predict that
you will not get the answer that you expect.

> If I understand you correctly, neither you nor Arnold Neumaier are
> saying that what I did is wrong. You said that particle dynamics can
> be also evaluated by more traditional methods of QFT (closed-path,
> Dirac-Fock, wavefunction renormalization, 8-step formula,...). I may
> trust you on that. However, you also claim that thus obtained
> interactions are bound to be retarded. That's where I start to worry.
> Because that's not what my approach shows. This means that my approach
> is not equivalent to yours. Then which one is correct? I opened all my
> cards: all derivations of interparticle instantaneous potentials are
> presented in the book. Now it is your turn: show how you calculate
> dynamics and where the retardation comes from.

No need to just trust, now is your chance to find out for sure. And I'm
afraid the turn is still yours. No-one except yourself has a steak in,
and hence has a reason to dedicate time to, finding where your approach
and the standard one cease to be equivalent. Now is your chance to find
out for sure. More talk without any calculations will lead nowhere.

Igor

Eugene Stefanovich

Mar31-05, 12:53 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote:\n\n>>>\n>>>The model itself is not QED but should have all the important features.\n>>>Consider a complex scalar field phi. It is minimally coupled to gauge\n>>>field A. The Lagrangian density is\n>>>\n>>> L_1 = -|Dphi|^2 - m^2 |phi|^2 - V|phi|^2 - (1/4) tr(F^2).\n>>>\n>>>Here D = @ - ieA, the covariant derivative. F is the field strength for\n>>>A. V is the external potential. There is a preferred frame in which the\n>>>potential V = V(x) is stationary and is two infinite square wells,\n>>>one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].\n>>>\n>>>The only reason I think a scalar field is easier to handle is because in\n>>>the case of Dirac fields, the boundary conditions for the infinite\n>>>square well are not so simple.\n>>>\n>>>In the infinite past, a particle is placed in each well. At any finite\n>>>time t the properties of this tate can be examined, for example the\n>>>expectation vale for the position of the particle in each well. At t=0 a\n>>>time dependent perturbation can be added to V which sets one of the\n>>>particles into motion. The effects on the other particle can be\n>>>calculated.\n>>>\n>>>Now, lets take baby steps.\n>>>\n>>>In the free field case (e=0), what is are the equations of motion for phi?\n>>>Solve them. Quantize the fields.\n>>>\n>>>What is the Hamiltonian? Diagonalize it. Write down and solve the\n>>>Heisenberg equations of motion for the fields.\n>>>\n>>>What is the position operator? Put a particle in one of the boxes. What\n>>>is the position expectation value for this state?\n>>>\n>>>Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear\n>>>time independent) part and H_1 the interaction part (the rest). Switch\n>>>to the interaction picture. Write down the equations of motion for the\n>>>field operators and the states.\n>>>\n>>>Once you get these done, we can talk more.\n>>\n>\n>>The point is that I already did this calculation. You can consider my\n>>entire book as a detailed solution of this particular problem: what\n>>is the dynamics of interaction between charged particles? I considered\n>>fermions instead of scalar particles and I did not put them in the\n>>infinite well, as you suggest, but nevertheless I made all the above\n>>steps in the book. And I\'ve got an answer: the interaction acts\n>>instantaneously, i.e., the force acting on particle B depends on\n>>the position and velocity of particle A at the same time instant.\n>>If you ask me to repeat this calculation for scalar fields, I\'ll\n>>follow exactly the same route as I did for QED. I am pretty sure that\n>>I\'ll get the same answer: there is no retardation. I am so sure\n>>about that without doing actual calculation, because retardation\n>>of the direct interaction between dressed particles would contradict the\n>>law of conservation of momentum.\n>\n>\n> I\'m not asking you to do this calculation the same way you did in your\n> book. I\'m asking you to do it in the way that I outline.\n> And neither am\n> I asking you calculate the force or the potential due to either\n> particle. The goal of the calculation is to evaluate the expectation\n> value of the position of a particle in one of the wells. I predict that\n> you will not get the answer that you expect.\n>\n>\n>>If I understand you correctly, neither you nor Arnold Neumaier are\n>>saying that what I did is wrong. You said that particle dynamics can\n>>be also evaluated by more traditional methods of QFT (closed-path,\n>>Dirac-Fock, wavefunction renormalization, 8-step formula,...). I may\n>>trust you on that. However, you also claim that thus obtained\n>>interactions are bound to be retarded. That\'s where I start to worry.\n>>Because that\'s not what my approach shows. This means that my approach\n>>is not equivalent to yours. Then which one is correct? I opened all my\n>>cards: all derivations of interparticle instantaneous potentials are\n>>presented in the book. Now it is your turn: show how you calculate\n>>dynamics and where the retardation comes from.\n>\n>\n> No need to just trust, now is your chance to find out for sure. And I\'m\n> afraid the turn is still yours. No-one except yourself has a steak in,\n> and hence has a reason to dedicate time to, finding where your approach\n> and the standard one cease to be equivalent. Now is your chance to find\n> out for sure. More talk without any calculations will lead nowhere.\n\nThen, I afraid we\'ll be stuck at the same place. The whole point of\nwriting this book was that I did not understand how dynamical problems\ncan be solved "in the way that you outline". If I follow that way, then\nI would expect that the Hamiltonian after renormalization has infinite\nmatrix elements and bare particles are not stable wrt to the time\nevolution. I have no idea how your approach is going to deal with it. I\nsuspect that the "standard" way of doing things is simply wrong. That\'s\nmy conclusion. If you disagree with me, then the logic of discussion\nsuggests that you either indicate the place where I made a mistake or\nyou make your own calculation. If you don\'t have a stake in it and not\ngoing to dedicate time to it, then my conclusion remains unchallenged.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:

>>>
>>>The model itself is not QED but should have all the important features.
>>>Consider a complex scalar field \phi. It is minimally coupled to gauge
>>>field A. The Lagrangian density is
>>>
>>> L_1 = -|Dphi|^2 - m^2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F^2).
>>>
>>>Here D = @ - ieA, the covariant derivative. F is the field strength for
>>>A. V is the external potential. There is a preferred frame in which the
>>>potential V = V(x) is stationary and is two infinite square wells,
>>>one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].
>>>
>>>The only reason I think a scalar field is easier to handle is because in
>>>the case of Dirac fields, the boundary conditions for the infinite
>>>square well are not so simple.
>>>
>>>In the infinite past, a particle is placed in each well. At any finite
>>>time t the properties of this tate can be examined, for example the
>>>expectation vale for the position of the particle in each well. At t=0 a
>>>time dependent perturbation can be added to V which sets one of the
>>>particles into motion. The effects on the other particle can be
>>>calculated.
>>>
>>>Now, lets take baby steps.
>>>
>>>In the free field case (e=0), what is are the equations of motion for \phi?
>>>Solve them. Quantize the fields.
>>>
>>>What is the Hamiltonian? Diagonalize it. Write down and solve the
>>>Heisenberg equations of motion for the fields.
>>>
>>>What is the position operator? Put a particle in one of the boxes. What
>>>is the position expectation value for this state?
>>>
>>>Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear
>>>time independent) part and H_1 the interaction part (the rest). Switch
>>>to the interaction picture. Write down the equations of motion for the
>>>field operators and the states.
>>>
>>>Once you get these done, we can talk more.
>>
>
>>The point is that I already did this calculation. You can consider my
>>entire book as a detailed solution of this particular problem: what
>>is the dynamics of interaction between charged particles? I considered
>>fermions instead of scalar particles and I did not put them in the
>>infinite well, as you suggest, but nevertheless I made all the above
>>steps in the book. And I've got an answer: the interaction acts
>>instantaneously, i.e., the force acting on particle B depends on
>>the position and velocity of particle A at the same time instant.
>>If you ask me to repeat this calculation for scalar fields, I'll
>>follow exactly the same route as I did for QED. I am pretty sure that
>>I'll get the same answer: there is no retardation. I am so sure
>>about that without doing actual calculation, because retardation
>>of the direct interaction between dressed particles would contradict the
>>law of conservation of momentum.
>
>
> I'm not asking you to do this calculation the same way you did in your
> book. I'm asking you to do it in the way that I outline.
> And neither am
> I asking you calculate the force or the potential due to either
> particle. The goal of the calculation is to evaluate the expectation
> value of the position of a particle in one of the wells. I predict that
> you will not get the answer that you expect.
>
>
>>If I understand you correctly, neither you nor Arnold Neumaier are
>>saying that what I did is wrong. You said that particle dynamics can
>>be also evaluated by more traditional methods of QFT (closed-path,
>>Dirac-Fock, wavefunction renormalization, 8-step formula,...). I may
>>trust you on that. However, you also claim that thus obtained
>>interactions are bound to be retarded. That's where I start to worry.
>>Because that's not what my approach shows. This means that my approach
>>is not equivalent to yours. Then which one is correct? I opened all my
>>cards: all derivations of interparticle instantaneous potentials are
>>presented in the book. Now it is your turn: show how you calculate
>>dynamics and where the retardation comes from.
>
>
> No need to just trust, now is your chance to find out for sure. And I'm
> afraid the turn is still yours. No-one except yourself has a steak in,
> and hence has a reason to dedicate time to, finding where your approach
> and the standard one cease to be equivalent. Now is your chance to find
> out for sure. More talk without any calculations will lead nowhere.

Then, I afraid we'll be stuck at the same place. The whole point of
writing this book was that I did not understand how dynamical problems
can be solved "in the way that you outline". If I follow that way, then
I would expect that the Hamiltonian after renormalization has infinite
matrix elements and bare particles are not stable wrt to the time
evolution. I have no idea how your approach is going to deal with it. I
suspect that the "standard" way of doing things is simply wrong. That's
my conclusion. If you disagree with me, then the logic of discussion
suggests that you either indicate the place where I made a mistake or
you make your own calculation. If you don't have a stake in it and not
going to dedicate time to it, then my conclusion remains unchallenged.

Eugene Stefanovich.

Arnold Neumaier

Mar31-05, 09:42 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> If you ask me to repeat this calculation for scalar fields, I\'ll\n> follow exactly the same route as I did for QED. I am pretty sure that\n> I\'ll get the same answer: there is no retardation.\n\nEven if there is no retardation in a fixed loop approximation,\nthe number of derivatives involverd increases with the loop number,\nand in the limit you have fullblown retardation...\n\n\n> If I understand you correctly, neither you nor Arnold Neumaier are\n> saying that what I did is wrong.\n\nWe both say that your interpretation of what you did is misleading.\nYou just get an instantaneous approximation to a retarded interaction.\nIt is trivial that retarded interactions can be approimated by\ninstantaneous ones. Thus as long as you don\'t take the limit,\nyour calculations don\'t say anything about your claim.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> If you ask me to repeat this calculation for scalar fields, I'll
> follow exactly the same route as I did for QED. I am pretty sure that
> I'll get the same answer: there is no retardation.

Even if there is no retardation in a fixed loop approximation,
the number of derivatives involverd increases with the loop number,
and in the limit you have fullblown retardation...

> If I understand you correctly, neither you nor Arnold Neumaier are
> saying that what I did is wrong.

We both say that your interpretation of what you did is misleading.
You just get an instantaneous approximation to a retarded interaction.
It is trivial that retarded interactions can be approimated by
instantaneous ones. Thus as long as you don't take the limit,
your calculations don't say anything about your claim.

Arnold Neumaier

Eugene Stefanovich

Mar31-05, 07:13 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>\n>> If you ask me to repeat this calculation for scalar fields, I\'ll\n>> follow exactly the same route as I did for QED. I am pretty sure that\n>> I\'ll get the same answer: there is no retardation.\n>\n>\n> Even if there is no retardation in a fixed loop approximation,\n> the number of derivatives involverd increases with the loop number,\n> and in the limit you have fullblown retardation...\n\nI am not sure what derivatives you are talking about. Let\'s say we are\ninterested in electron-electron interaction. We neglect photon emission\n(i.e., bremsstrahlung interactions) or pair creation.\nThen we are confined to the 2-electron sector of the Fock space, and\nthe only interaction term relevant in this case is a^*a^*aa.\nThere are contributions to this interaction from all even perturbation\norders: 2nd, 4th, etc. However all of them are instantaneous:\nthe interaction operators can be rewritten as functions of particle\npositions and momenta, and the force on particle B depends on\ninstantaneous values of observables of particle A. I do not see how\na sum (even infinite) of instantaneous interactions can become retarded.\n\nMoreover, the retardation is forbidden by the law of conservation of\nmomentum. If interaction is retarded and the momentum is conserved at\nall times, then the transferred momentum must be stored in some material\nform while it transfers from one particle to another.\nSince in the dressed particle approach the virtual particles are\neliminated, this "material form" can correspond only to real photons.\nSo, the retarded part of interaction is described by the exchange of\nreal photons. This exchange does explain such effects as radio\ntransmissions, but it cannot explain the direct Coulomb and magnetic\n(e.g., static) forces between charged particles. These forces\nare described by instantaneous potentials.\n\nEugene Stefanovich\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>
>> If you ask me to repeat this calculation for scalar fields, I'll
>> follow exactly the same route as I did for QED. I am pretty sure that
>> I'll get the same answer: there is no retardation.
>
>
> Even if there is no retardation in a fixed loop approximation,
> the number of derivatives involverd increases with the loop number,
> and in the limit you have fullblown retardation...

I am not sure what derivatives you are talking about. Let's say we are
interested in electron-electron interaction. We neglect photon emission
(i.e., bremsstrahlung interactions) or pair creation.
Then we are confined to the 2-electron sector of the Fock space, and
the only interaction term relevant in this case is a^*a^*aa.
There are contributions to this interaction from all even perturbation
orders: 2nd, 4th, etc. However all of them are instantaneous:
the interaction operators can be rewritten as functions of particle
positions and momenta, and the force on particle B depends on
instantaneous values of observables of particle A. I do not see how
a sum (even infinite) of instantaneous interactions can become retarded.

Moreover, the retardation is forbidden by the law of conservation of
momentum. If interaction is retarded and the momentum is conserved at
all times, then the transferred momentum must be stored in some material
form while it transfers from one particle to another.
Since in the dressed particle approach the virtual particles are
eliminated, this "material form" can correspond only to real photons.
So, the retarded part of interaction is described by the exchange of
real photons. This exchange does explain such effects as radio
transmissions, but it cannot explain the direct Coulomb and magnetic
(e.g., static) forces between charged particles. These forces
are described by instantaneous potentials.

Eugene Stefanovich

Arnold Neumaier

Apr1-05, 01:07 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> The whole point of\n> writing this book was that I did not understand how dynamical problems\n> can be solved "in the way that you outline".\n\nSo your book is a document of your ignorance of the state of the art.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> The whole point of
> writing this book was that I did not understand how dynamical problems
> can be solved "in the way that you outline".

So your book is a document of your ignorance of the state of the art.

Arnold Neumaier

Igor Khavkine

Apr1-05, 01:07 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-03-31, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> Then, I afraid we\'ll be stuck at the same place. The whole point of\n> writing this book was that I did not understand how dynamical problems\n> can be solved "in the way that you outline". If I follow that way, then\n> I would expect that the Hamiltonian after renormalization has infinite\n> matrix elements and bare particles are not stable wrt to the time\n> evolution. I have no idea how your approach is going to deal with it. I\n> suspect that the "standard" way of doing things is simply wrong. That\'s\n> my conclusion. If you disagree with me, then the logic of discussion\n> suggests that you either indicate the place where I made a mistake or\n> you make your own calculation. If you don\'t have a stake in it and not\n> going to dedicate time to it, then my conclusion remains unchallenged.\n\nNeither of your expectations has any bearing on the outcome of the\ncalculation. Just because you did not understand how such calculations\ncould be done does not mean they are impossible. Your suspicion is\nunfounded unless you actually verify it. Pointing out superficial\ndifficulties is not sufficient.\n\nI strongly recommend starting this calculation and I offer to clarify\nwhatever steps cause problems. It will be up to you, however, to relate\nto compare with your own work.\n\nThe logic of discussion also indicates that results of the standard\ntheory stay unchallenged as well. Which is fine for most physicists.\nWhich in turn indicates that your work remains unnoticed.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-03-31, Eugene Stefanovich <eugenev@synopsys.com> wrote:

> Then, I afraid we'll be stuck at the same place. The whole point of
> writing this book was that I did not understand how dynamical problems
> can be solved "in the way that you outline". If I follow that way, then
> I would expect that the Hamiltonian after renormalization has infinite
> matrix elements and bare particles are not stable wrt to the time
> evolution. I have no idea how your approach is going to deal with it. I
> suspect that the "standard" way of doing things is simply wrong. That's
> my conclusion. If you disagree with me, then the logic of discussion
> suggests that you either indicate the place where I made a mistake or
> you make your own calculation. If you don't have a stake in it and not
> going to dedicate time to it, then my conclusion remains unchallenged.

Neither of your expectations has any bearing on the outcome of the
calculation. Just because you did not understand how such calculations
could be done does not mean they are impossible. Your suspicion is
unfounded unless you actually verify it. Pointing out superficial
difficulties is not sufficient.

I strongly recommend starting this calculation and I offer to clarify
whatever steps cause problems. It will be up to you, however, to relate
to compare with your own work.

The logic of discussion also indicates that results of the standard
theory stay unchallenged as well. Which is fine for most physicists.
Which in turn indicates that your work remains unnoticed.

Igor

Eugene Stefanovich

Apr1-05, 03:21 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n\n>\n> I strongly recommend starting this calculation and I offer to clarify\n> whatever steps cause problems. It will be up to you, however, to relate\n> to compare with your own work.\n\nOK, let\'s do it your way.\nI remind you the model you suggested:\n\n> The model itself is not QED but should have all the important features.\n> Consider a complex scalar field phi. It is minimally coupled to gauge\n> field A. The Lagrangian density is\n\n> L_1 = -|Dphi|^2 - m2 |phi|^2 - V|phi|^2 - (1/4) tr(F2).\n\n> Here D = @ - ieA, the covariant derivative. F is the field strength for\n> A. V is the external potential. There is a preferred frame in which the\n> potential V = V(x) is stationary and is two infinite square wells,\n> one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].\n\n> The only reason I think a scalar field is easier to handle is because in\n> the case of Dirac fields, the boundary conditions for the infinite\n> square well are not so simple.\n\n> In the infinite past, a particle is placed in each well. At any finite\n> time t the properties of this tate can be examined, for example the\n> expectation vale for the position of the particle in each well. At t=0 a\n> time dependent perturbation can be added to V which sets one of the\n> particles into motion. The effects on the other particle can be\n> calculated.\n\nI have a few comments about the formulation of this problem.\nFirst, introducing internal potentials looks rather artificial.\nI would prefer to stay closer to reality as much as possible and\nconsider just two interacting particle without any extra stuff.\nThe decision about retarded/instantaneous nature of interaction can be\nmade by examining the dependence of the interparticle force on the\nparticles\' observables. If you throw away the wells you\'ll also\nsave yourself the aggravation with boundary conditions.\nThen, there is no any advantage in using the scalar field, we can work\nin QED just as well. Moreover, the Hamiltonian in QED is already\navailable. Now you write\n\n> Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear\n> time independent) part and H_1 the interaction part (the rest).\n\nDone.\n\n> Switch\n> to the interaction picture.\n\nWould you allow me to stay in the Schroedinger picture?\nI would like to see how some simple states, like vacuum or one-particle\nstates develop in time. Before attempting to solve a 2-particle problem\nwe should make sure that 0-particle and 1-particle systems make sense.\n\n> Write down the equations of motion for the\n> field operators and the states.\n\nThe interaction part H_1 contains trilinear terms like a^*c^*a and a^*ac\nWhen I evaluate the time evolution of a one-electron state I obtain\n\nexp(iHt) a^* |0> = exp[i(H_0 + H_1)t] a^* |0>\n= exp[i(H_0 + a^*c^*a + a^*ac + ...)t] a^* |0>\n= (1 + iH_0t + ia^*c^*a t+ ia^*ac t+ ...) a^* |0>\n= a^* |0> + it a^*c^* |0> + ...\n\nI see that my electron "created" a photon out of nothing even in the\nlowest (1st) perturbation order. The higher orders are even worse.\nI beg for your help here! The problem is the same in the scalar field\nmodel that you suggested, because its Hamiltonian also contains unphys\nterms that I consider inappropriate.\n\nThe only consistent solution of this problem that I know of is\nto apply a unitary transformation to H_0 + H_1 that eliminates\nthe unphys terms from there. I am open for other suggestions.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:

>
> I strongly recommend starting this calculation and I offer to clarify
> whatever steps cause problems. It will be up to you, however, to relate
> to compare with your own work.

OK, let's do it your way.
I remind you the model you suggested:

> The model itself is not QED but should have all the important features.
> Consider a complex scalar field \phi. It is minimally coupled to gauge
> field A. The Lagrangian density is

> L_1 = -|Dphi|^2 - m2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F2).

> Here D = @ - ieA, the covariant derivative. F is the field strength for
> A. V is the external potential. There is a preferred frame in which the
> potential V = V(x) is stationary and is two infinite square wells,
> one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].

> The only reason I think a scalar field is easier to handle is because in
> the case of Dirac fields, the boundary conditions for the infinite
> square well are not so simple.

> In the infinite past, a particle is placed in each well. At any finite
> time t the properties of this tate can be examined, for example the
> expectation vale for the position of the particle in each well. At t=0 a
> time dependent perturbation can be added to V which sets one of the
> particles into motion. The effects on the other particle can be
> calculated.

I have a few comments about the formulation of this problem.
First, introducing internal potentials looks rather artificial.
I would prefer to stay closer to reality as much as possible and
consider just two interacting particle without any extra stuff.
The decision about retarded/instantaneous nature of interaction can be
made by examining the dependence of the interparticle force on the
particles' observables. If you throw away the wells you'll also
save yourself the aggravation with boundary conditions.
Then, there is no any advantage in using the scalar field, we can work
in QED just as well. Moreover, the Hamiltonian in QED is already
available. Now you write

> Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear
> time independent) part and H_1 the interaction part (the rest).

Done.

> Switch
> to the interaction picture.

Would you allow me to stay in the Schroedinger picture?
I would like to see how some simple states, like vacuum or one-particle
states develop in time. Before attempting to solve a 2-particle problem
we should make sure that 0-particle and 1-particle systems make sense.

> Write down the equations of motion for the
> field operators and the states.

The interaction part H_1 contains trilinear terms like a^*c^*a and a^*ac
When I evaluate the time evolution of a one-electron state I obtain

\exp(iHt) a^* |0> = \exp[i(H_0 + H_1)t] a^* |0>= \exp[i(H_0 + a^*c^*a + a^*ac + ...)t] a^* |0>= (1 + iH_0t + ia^*c^*a t+ ia^*ac t+ ...) a^* |0>= a^* |0> + it a^*c^* |0> + ...

I see that my electron "created" a photon out of nothing even in the
lowest (1st) perturbation order. The higher orders are even worse.
I beg for your help here! The problem is the same in the scalar field
model that you suggested, because its Hamiltonian also contains unphys
terms that I consider inappropriate.

The only consistent solution of this problem that I know of is
to apply a unitary transformation to H_0 + H_1 that eliminates
the unphys terms from there. I am open for other suggestions.

Eugene Stefanovich.

Igor Khavkine

Apr1-05, 05:18 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-01, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n>>\n>> I strongly recommend starting this calculation and I offer to clarify\n>> whatever steps cause problems. It will be up to you, however, to relate\n>> to compare with your own work.\n>\n> OK, let\'s do it your way.\n> I remind you the model you suggested:\n>\n> > The model itself is not QED but should have all the important features.\n> > Consider a complex scalar field phi. It is minimally coupled to gauge\n> > field A. The Lagrangian density is\n>\n> > L_1 = -|Dphi|^2 - m2 |phi|^2 - V|phi|^2 - (1/4) tr(F2).\n>\n> > Here D = @ - ieA, the covariant derivative. F is the field strength for\n> > A. V is the external potential. There is a preferred frame in which the\n> > potential V = V(x) is stationary and is two infinite square wells,\n> > one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].\n>\n> > The only reason I think a scalar field is easier to handle is because in\n> > the case of Dirac fields, the boundary conditions for the infinite\n> > square well are not so simple.\n>\n> > In the infinite past, a particle is placed in each well. At any finite\n> > time t the properties of this tate can be examined, for example the\n> > expectation vale for the position of the particle in each well. At t=0 a\n> > time dependent perturbation can be added to V which sets one of the\n> > particles into motion. The effects on the other particle can be\n> > calculated.\n>\n> I have a few comments about the formulation of this problem.\n> First, introducing internal potentials looks rather artificial.\n> I would prefer to stay closer to reality as much as possible and\n> consider just two interacting particle without any extra stuff.\n> The decision about retarded/instantaneous nature of interaction can be\n> made by examining the dependence of the interparticle force on the\n> particles\' observables. If you throw away the wells you\'ll also\n> save yourself the aggravation with boundary conditions.\n> Then, there is no any advantage in using the scalar field, we can work\n> in QED just as well. Moreover, the Hamiltonian in QED is already\n> available.\n\nI do not trust conclusions about the speed of the interaction just by\nlooking at the form of what appears to the force or the potential. I\nwant to calculate something that is concievably measurable. That is why\nI chose the expectation value of the position operator. My choice of\nexternal potential mimics the design of a controlled experiment for such\na measurement.\n\nA priori, we know neither what the vacuum, the 1-particle, or the\n2-particle states of the interacting theory look like. We only know that\nin the distant past they can be approximated by corresponding free\nstates. If we do not introduce a localising potential wells, any state\nwe start with in the distant past will result in scattering by the time\nwe get to t=0. In other words, if we start with a two particle state in\nthe distant past, we have no idea what kind of state we\'ll get at t=0.\nWith the potential wells we do.\n\n> Now you write\n> > Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear\n> > time independent) part and H_1 the interaction part (the rest).\n>\n> Done.\n>\n> > Switch\n> > to the interaction picture.\n>\n> Would you allow me to stay in the Schroedinger picture?\n> I would like to see how some simple states, like vacuum or one-particle\n> states develop in time. Before attempting to solve a 2-particle problem\n> we should make sure that 0-particle and 1-particle systems make sense.\n\nThe separation H = H_0 + H_1 into free and interaction parts is\narbitrary. If you choose H_0 = 0 and H = H_1 you recover the\nSchroedinger picrure. The difference will appear in the expression for\nthe propagators and the iteraction vertices. If you choose H_0 = 0, then\nall your bare propagators will have value 1. And in addition to the\ninteraction vertices you expect, you\'ll also see vertices like\n----o---- representing the mass term and ----x---- representing the\nkinetic term. I will give all instructions in the interaction picture\nwhere H_0 is taken to be the free scalar field and EM field\nHamiltonians. If you wish you can always compare with the Schroedinger\npicture, the map between them is known.\n\n> > Write down the equations of motion for the\n> > field operators and the states.\n>\n> The interaction part H_1 contains trilinear terms like a^*c^*a and a^*ac\n> When I evaluate the time evolution of a one-electron state I obtain\n>\n> exp(iHt) a^* |0> = exp[i(H_0 + H_1)t] a^* |0>\n> = exp[i(H_0 + a^*c^*a + a^*ac + ...)t] a^* |0>\n> = (1 + iH_0t + ia^*c^*a t+ ia^*ac t+ ...) a^* |0>\n> = a^* |0> + it a^*c^* |0> + ...\n>\n> I see that my electron "created" a photon out of nothing even in the\n> lowest (1st) perturbation order. The higher orders are even worse.\n> I beg for your help here! The problem is the same in the scalar field\n> model that you suggested, because its Hamiltonian also contains unphys\n> terms that I consider inappropriate.\n\nFirst, you did not write down the equations of motion in the interaction\npicture as I asked. Second, your solution is not correct, the\nHamiltonian will have a time dependent potential that will kick one of\nthe particles into motion. In the same vein, in the interaction picture\nall particle operators will have time dependence given to them by H_0.\nThird, as has been explained several times, the expression you wrote\ndown has zero physical meaning. You are still far from anything that can\nbe directly physically interpreted.\n\nYou still have the following steps to complete:\n* Write down the equations of motion of the field operators and the\nstates.\n* Diagonalize H_0.\n* Solve the equations of motion for the field operators phi and A.\n\nWorking knowledge of quantum mechanics as well as patience are\nprerequisites for this calculation.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-01, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:
>>
>> I strongly recommend starting this calculation and I offer to clarify
>> whatever steps cause problems. It will be up to you, however, to relate
>> to compare with your own work.
>
> OK, let's do it your way.
> I remind you the model you suggested:
>
> > The model itself is not QED but should have all the important features.
> > Consider a complex scalar field \phi. It is minimally coupled to gauge
> > field A. The Lagrangian density is
>
> > L_1 = -|Dphi|^2 - m2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F2).
>
> > Here D = @ - ieA, the covariant derivative. F is the field strength for
> > A. V is the external potential. There is a preferred frame in which the
> > potential V = V(x) is stationary and is two infinite square wells,
> > one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].
>
> > The only reason I think a scalar field is easier to handle is because in
> > the case of Dirac fields, the boundary conditions for the infinite
> > square well are not so simple.
>
> > In the infinite past, a particle is placed in each well. At any finite
> > time t the properties of this tate can be examined, for example the
> > expectation vale for the position of the particle in each well. At t=0 a
> > time dependent perturbation can be added to V which sets one of the
> > particles into motion. The effects on the other particle can be
> > calculated.
>
> I have a few comments about the formulation of this problem.
> First, introducing internal potentials looks rather artificial.
> I would prefer to stay closer to reality as much as possible and
> consider just two interacting particle without any extra stuff.
> The decision about retarded/instantaneous nature of interaction can be
> made by examining the dependence of the interparticle force on the
> particles' observables. If you throw away the wells you'll also
> save yourself the aggravation with boundary conditions.
> Then, there is no any advantage in using the scalar field, we can work
> in QED just as well. Moreover, the Hamiltonian in QED is already
> available.

I do not trust conclusions about the speed of the interaction just by
looking at the form of what appears to the force or the potential. I
want to calculate something that is concievably measurable. That is why
I chose the expectation value of the position operator. My choice of
external potential mimics the design of a controlled experiment for such
a measurement.

A priori, we know neither what the vacuum, the 1-particle, or the
2-particle states of the interacting theory look like. We only know that
in the distant past they can be approximated by corresponding free
states. If we do not introduce a localising potential wells, any state
we start with in the distant past will result in scattering by the time
we get to t=0. In other words, if we start with a two particle state in
the distant past, we have no idea what kind of state we'll get at t=0.
With the potential wells we do.

> Now you write
> > Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear
> > time independent) part and H_1 the interaction part (the rest).
>
> Done.
>
> > Switch
> > to the interaction picture.
>
> Would you allow me to stay in the Schroedinger picture?
> I would like to see how some simple states, like vacuum or one-particle
> states develop in time. Before attempting to solve a 2-particle problem
> we should make sure that 0-particle and 1-particle systems make sense.

The separation H = H_0 + H_1 into free and interaction parts is
arbitrary. If you choose H_0 = and H = H_1 you recover the
Schroedinger picrure. The difference will appear in the expression for
the propagators and the iteraction vertices. If you choose H_0 = 0, then
all your bare propagators will have value 1. And in addition to the
interaction vertices you expect, you'll also see vertices like
----o---- representing the mass term and ----x---- representing the
kinetic term. I will give all instructions in the interaction picture
where H_0 is taken to be the free scalar field and EM field
Hamiltonians. If you wish you can always compare with the Schroedinger
picture, the map between them is known.

> > Write down the equations of motion for the
> > field operators and the states.
>
> The interaction part H_1 contains trilinear terms like a^*c^*a and a^*ac
> When I evaluate the time evolution of a one-electron state I obtain
>
> \exp(iHt) a^* |0> = \exp[i(H_0 + H_1)t] a^* |0>
> = \exp[i(H_0 + a^*c^*a + a^*ac + ...)t] a^* |0>
> = (1 + iH_0t + ia^*c^*a t+ ia^*ac t+ ...) a^* |0>
> = a^* |0> + it a^*c^* |0> + ...
>
> I see that my electron "created" a photon out of nothing even in the
> lowest (1st) perturbation order. The higher orders are even worse.
> I beg for your help here! The problem is the same in the scalar field
> model that you suggested, because its Hamiltonian also contains unphys
> terms that I consider inappropriate.

First, you did not write down the equations of motion in the interaction
picture as I asked. Second, your solution is not correct, the
Hamiltonian will have a time dependent potential that will kick one of
the particles into motion. In the same vein, in the interaction picture
all particle operators will have time dependence given to them by H_0.
Third, as has been explained several times, the expression you wrote
down has zero physical meaning. You are still far from anything that can
be directly physically interpreted.

You still have the following steps to complete:
* Write down the equations of motion of the field operators and the
states.
* Diagonalize H_0.
* Solve the equations of motion for the field operators \phi and A.

Working knowledge of quantum mechanics as well as patience are
prerequisites for this calculation.

Igor

Arnold Neumaier

Apr2-05, 02:42 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> I do not see how\n> a sum (even infinite) of instantaneous interactions can become retarded.\n\nIn spite of having it had explained several times in previous threads.\n\nI am tired of preaching deaf ears. Instead of trying to learn\nyou are just trying to defend yourself.\n\nUnderstandable but unproductive.\n\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> I do not see how
> a sum (even infinite) of instantaneous interactions can become retarded.

In spite of having it had explained several times in previous threads.

I am tired of preaching deaf ears. Instead of trying to learn
you are just trying to defend yourself.

Understandable but unproductive.

Arnold Neumaier

Eugene Stefanovich

Apr2-05, 02:42 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>The whole point of\n>>writing this book was that I did not understand how dynamical problems\n>>can be solved "in the way that you outline".\n>\n>\n> So your book is a document of your ignorance of the state of the art.\n\nI would agree with you if my book contained just complaints.\nHowever, I indentify the problem and propose a solution that really\nworks.\n\nOn the other hand I don\'t see that you even understand what the problem\nis. So far I haven\'t heard from you any coherent proposal how\nare you going to calculate both S-matrix and time evolution with the\nsame Hamiltonian. You gave some vague references to the "wavefunction\nrenormalization". Could you be more specific, please?\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>The whole point of
>>writing this book was that I did not understand how dynamical problems
>>can be solved "in the way that you outline".
>
>
> So your book is a document of your ignorance of the state of the art.

I would agree with you if my book contained just complaints.
However, I indentify the problem and propose a solution that really
works.

On the other hand I don't see that you even understand what the problem
is. So far I haven't heard from you any coherent proposal how
are you going to calculate both S-matrix and time evolution with the
same Hamiltonian. You gave some vague references to the "wavefunction
renormalization". Could you be more specific, please?

Eugene Stefanovich.

Eugene Stefanovich

Apr2-05, 02:43 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote:\n\n>>OK, let\'s do it your way.\n>>I remind you the model you suggested:\n>>\n>>\n>>>The model itself is not QED but should have all the important features.\n>>>Consider a complex scalar field phi. It is minimally coupled to gauge\n>>>field A. The Lagrangian density is\n>>\n>>> L_1 = -|Dphi|^2 - m2 |phi|^2 - V|phi|^2 - (1/4) tr(F2).\n>>\n>>>Here D = @ - ieA, the covariant derivative. F is the field strength for\n>>>A. V is the external potential. There is a preferred frame in which the\n>>>potential V = V(x) is stationary and is two infinite square wells,\n>>>one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].\n>>\n>>>The only reason I think a scalar field is easier to handle is because in\n>>>the case of Dirac fields, the boundary conditions for the infinite\n>>>square well are not so simple.\n>>\n>>>In the infinite past, a particle is placed in each well. At any finite\n>>>time t the properties of this tate can be examined, for example the\n>>>expectation vale for the position of the particle in each well. At t=0 a\n>>>time dependent perturbation can be added to V which sets one of the\n>>>particles into motion. The effects on the other particle can be\n>>>calculated.\n>>\n>>I have a few comments about the formulation of this problem.\n>>First, introducing internal potentials looks rather artificial.\n>>I would prefer to stay closer to reality as much as possible and\n>>consider just two interacting particle without any extra stuff.\n>>The decision about retarded/instantaneous nature of interaction can be\n>>made by examining the dependence of the interparticle force on the\n>>particles\' observables. If you throw away the wells you\'ll also\n>>save yourself the aggravation with boundary conditions.\n>>Then, there is no any advantage in using the scalar field, we can work\n>>in QED just as well. Moreover, the Hamiltonian in QED is already\n>>available.\n>\n>\n> I do not trust conclusions about the speed of the interaction just by\n> looking at the form of what appears to the force or the potential. I\n> want to calculate something that is concievably measurable. That is why\n> I chose the expectation value of the position operator. My choice of\n> external potential mimics the design of a controlled experiment for such\n> a measurement.\n\nI suggest to simplify the problem and instead of 2 particles consider\njust one. We\'ll return to the 2-particle case later. So, let us forget\nabout the external potential for the moment.\n\n>\n> A priori, we know neither what the vacuum, the 1-particle, or the\n> 2-particle states of the interacting theory look like. We only know that\n> in the distant past they can be approximated by corresponding free\n> states. If we do not introduce a localising potential wells, any state\n> we start with in the distant past will result in scattering by the time\n> we get to t=0. In other words, if we start with a two particle state in\n> the distant past, we have no idea what kind of state we\'ll get at t=0.\n> With the potential wells we do.\n\nAre you saying that 0-particle and 1-particle states will also result\nin some "scattering" at time t=0? They certainly will if you use your\nHamiltonian, but it doesn\'t make any sense. I don\'t know much about\nexperiment, but I know enough to say that no scattering occurs in real\n0- and 1-particle states. To me this indicates one of the two\npossibilities:\n1) the things you call "particles" have no relationship to the real\nparticles we observe in nature\n2) the thing you call "Hamiltonian" has no relationship to the real\nquantum mechanical time evolution operator\nDo you have any other explanation?\n\n>\n>\n>>Now you write\n>>\n>>>Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear\n>>>time independent) part and H_1 the interaction part (the rest).\n>>\n>>Done.\n>>\n>>\n>>>Switch\n>>>to the interaction picture.\n>>\n>>Would you allow me to stay in the Schroedinger picture?\n>>I would like to see how some simple states, like vacuum or one-particle\n>>states develop in time. Before attempting to solve a 2-particle problem\n>>we should make sure that 0-particle and 1-particle systems make sense.\n>\n>\n> The separation H = H_0 + H_1 into free and interaction parts is\n> arbitrary.\n\nI don\'t think that separation is arbitrary. I would prefer the\nseparation which has clear physical meaning: H_0 is Hamiltonian of\nfree particles, e.g. for massive particles\n\nH_0 = \\int dp sqrt(p^2 + m^2) a^*(p)a(p)\n\nwhere m is the experimentally measured particle mass. The rest is\ninteraction. In this case, if you drop the interaction term you\'ll\nobtain the system of free particles moving freely without interaction,\nas it should be. All other choices are artificial.\n\n\n\n> If you choose H_0 = 0 and H = H_1 you recover the\n> Schroedinger picrure. The difference will appear in the expression for\n> the propagators and the iteraction vertices. If you choose H_0 = 0, then\n> all your bare propagators will have value 1. And in addition to the\n> interaction vertices you expect, you\'ll also see vertices like\n> ----o---- representing the mass term and ----x---- representing the\n> kinetic term. I will give all instructions in the interaction picture\n> where H_0 is taken to be the free scalar field and EM field\n> Hamiltonians. If you wish you can always compare with the Schroedinger\n> picture, the map between them is known.\n\nI am afraid you are trying to make simple things complex.\nFirst, there is no use of propagators when we are trying to calculate\nthe time evolution. Propagators are useful for the S-matrix calculation.\nSecond, the interaction picture could be convenient for the S-matrix\ncalculations, but it becomes rather confusing in the case of time\nevolution: both states and operators become time-dependent. Let us\nkeep simple things simple: use the Schroedinger picture in which\nH_0 and H_1 are time-independent and try to calculate the time\ndependence of some state vector, e.g., a^*|0>\n\n\n\n>\n>\n>>>Write down the equations of motion for the\n>>>field operators and the states.\n>>\n>>The interaction part H_1 contains trilinear terms like a^*c^*a and a^*ac\n>>When I evaluate the time evolution of a one-electron state I obtain\n>>\n>>exp(iHt) a^* |0> = exp[i(H_0 + H_1)t] a^* |0>\n>> = exp[i(H_0 + a^*c^*a + a^*ac + ...)t] a^* |0>\n>> = (1 + iH_0t + ia^*c^*a t+ ia^*ac t+ ...) a^* |0>\n>> = a^* |0> + it a^*c^* |0> + ...\n>>\n>>I see that my electron "created" a photon out of nothing even in the\n>>lowest (1st) perturbation order. The higher orders are even worse.\n>>I beg for your help here! The problem is the same in the scalar field\n>>model that you suggested, because its Hamiltonian also contains unphys\n>>terms that I consider inappropriate.\n>\n>\n> First, you did not write down the equations of motion in the interaction\n> picture as I asked.\n\nEquations of motion for what? Fields? I do not care about fields.\nI have a Hamiltonian and I have an initial state. My "working knowledge\nof quantum mechanics" tells me that this is all I need to compute the\ntime evolution of the state.\n\n> Second, your solution is not correct, the\n> Hamiltonian will have a time dependent potential that will kick one of\n> the particles into motion.\n\nRemember, we are talking about a single particle. Let\'s try something\nsimple for starters.\n\n> In the same vein, in the interaction picture\n> all particle operators will have time dependence given to them by H_0.\n\nAgain, I am working in the Schroedinger picture. Please forgive me this\nweakness, but I prefer having my operators time-independent, then it is\neasier to figure out what is going on.\n\n> Third, as has been explained several times, the expression you wrote\n> down has zero physical meaning.\n> You are still far from anything that can\n> be directly physically interpreted.\n\n\nIf you don\'t like my formula for the time evolution of a 1-particle\nstate, then what is your formula?\n\n\n>\n> You still have the following steps to complete:\n> * Write down the equations of motion of the field operators\n\nI don\'t need those, as I said above.\n\n> and the states.\n\nAre you talking about the non-stationary Schroedinger equation?\n\n-i d/dt \\Psi(t) = H \\Psi(t)\n\nI wrote the solution of this equation for the 1-particle state above.\nDid I make any mistake?\n\n> * Diagonalize H_0.\n\nEigenvectors of H_0 are |0>, a^*|0>, a^*c^*|0>, ...\nThe eigenvalues are sums of energies of created particles.\n\n> * Solve the equations of motion for the field operators phi and A.\n\nI don\'t need those, as I said above.\n\n>\n> Working knowledge of quantum mechanics as well as patience are\n> prerequisites for this calculation.\n\nAgreed.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:

>>OK, let's do it your way.
>>I remind you the model you suggested:
>>
>>
>>>The model itself is not QED but should have all the important features.
>>>Consider a complex scalar field \phi. It is minimally coupled to gauge
>>>field A. The Lagrangian density is
>>
>>> L_1 = -|Dphi|^2 - m2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F2).
>>
>>>Here D = @ - ieA, the covariant derivative. F is the field strength for
>>>A. V is the external potential. There is a preferred frame in which the
>>>potential V = V(x) is stationary and is two infinite square wells,
>>>one [-L-w,w]x[0,L]x[0,L] and the other [w,w+L]x[0,L]x[0,L].
>>
>>>The only reason I think a scalar field is easier to handle is because in
>>>the case of Dirac fields, the boundary conditions for the infinite
>>>square well are not so simple.
>>
>>>In the infinite past, a particle is placed in each well. At any finite
>>>time t the properties of this tate can be examined, for example the
>>>expectation vale for the position of the particle in each well. At t=0 a
>>>time dependent perturbation can be added to V which sets one of the
>>>particles into motion. The effects on the other particle can be
>>>calculated.
>>
>>I have a few comments about the formulation of this problem.
>>First, introducing internal potentials looks rather artificial.
>>I would prefer to stay closer to reality as much as possible and
>>consider just two interacting particle without any extra stuff.
>>The decision about retarded/instantaneous nature of interaction can be
>>made by examining the dependence of the interparticle force on the
>>particles' observables. If you throw away the wells you'll also
>>save yourself the aggravation with boundary conditions.
>>Then, there is no any advantage in using the scalar field, we can work
>>in QED just as well. Moreover, the Hamiltonian in QED is already
>>available.
>
>
> I do not trust conclusions about the speed of the interaction just by
> looking at the form of what appears to the force or the potential. I
> want to calculate something that is concievably measurable. That is why
> I chose the expectation value of the position operator. My choice of
> external potential mimics the design of a controlled experiment for such
> a measurement.

I suggest to simplify the problem and instead of 2 particles consider
just one. We'll return to the 2-particle case later. So, let us forget
about the external potential for the moment.

>
> A priori, we know neither what the vacuum, the 1-particle, or the
> 2-particle states of the interacting theory look like. We only know that
> in the distant past they can be approximated by corresponding free
> states. If we do not introduce a localising potential wells, any state
> we start with in the distant past will result in scattering by the time
> we get to t=0. In other words, if we start with a two particle state in
> the distant past, we have no idea what kind of state we'll get at t=0.
> With the potential wells we do.

Are you saying that 0-particle and 1-particle states will also result
in some "scattering" at time t=0? They certainly will if you use your
Hamiltonian, but it doesn't make any sense. I don't know much about
experiment, but I know enough to say that no scattering occurs in real
0- and 1-particle states. To me this indicates one of the two
possibilities:
1) the things you call "particles" have no relationship to the real
particles we observe in nature
2) the thing you call "Hamiltonian" has no relationship to the real
quantum mechanical time evolution operator
Do you have any other explanation?

>
>
>>Now you write
>>
>>>Write the Hamiltonian as H = H_0 + H_1, where H_0 is the free (bilinear
>>>time independent) part and H_1 the interaction part (the rest).
>>
>>Done.
>>
>>
>>>Switch
>>>to the interaction picture.
>>
>>Would you allow me to stay in the Schroedinger picture?
>>I would like to see how some simple states, like vacuum or one-particle
>>states develop in time. Before attempting to solve a 2-particle problem
>>we should make sure that 0-particle and 1-particle systems make sense.
>
>
> The separation H = H_0 + H_1 into free and interaction parts is
> arbitrary.

I don't think that separation is arbitrary. I would prefer the
separation which has clear physical meaning: H_0 is Hamiltonian of
free particles, e.g. for massive particles

H_0 = \int dp \sqrt(p^2 + m^2) a^*(p)a(p)

where m is the experimentally measured particle mass. The rest is
interaction. In this case, if you drop the interaction term you'll
obtain the system of free particles moving freely without interaction,
as it should be. All other choices are artificial.

> If you choose H_0 = and H = H_1 you recover the
> Schroedinger picrure. The difference will appear in the expression for
> the propagators and the iteraction vertices. If you choose H_0 = 0, then
> all your bare propagators will have value 1. And in addition to the
> interaction vertices you expect, you'll also see vertices like
> ----o---- representing the mass term and ----x---- representing the
> kinetic term. I will give all instructions in the interaction picture
> where H_0 is taken to be the free scalar field and EM field
> Hamiltonians. If you wish you can always compare with the Schroedinger
> picture, the map between them is known.

I am afraid you are trying to make simple things complex.
First, there is no use of propagators when we are trying to calculate
the time evolution. Propagators are useful for the S-matrix calculation.
Second, the interaction picture could be convenient for the S-matrix
calculations, but it becomes rather confusing in the case of time
evolution: both states and operators become time-dependent. Let us
keep simple things simple: use the Schroedinger picture in which
H_0 and H_1 are time-independent and try to calculate the time
dependence of some state vector, e.g., a^*|0>

>
>
>>>Write down the equations of motion for the
>>>field operators and the states.
>>
>>The interaction part H_1 contains trilinear terms like a^*c^*a and a^*ac
>>When I evaluate the time evolution of a one-electron state I obtain
>>
>>\exp(iHt) a^* |0> = \exp[i(H_0 + H_1)t] a^* |0>>> = \exp[i(H_0 + a^*c^*a + a^*ac + ...)t] a^* |0>>> = (1 + iH_0t + ia^*c^*a t+ ia^*ac t+ ...) a^* |0>>> = a^* |0> + it a^*c^* |0> + ...
>>
>>I see that my electron "created" a photon out of nothing even in the
>>lowest (1st) perturbation order. The higher orders are even worse.
>>I beg for your help here! The problem is the same in the scalar field
>>model that you suggested, because its Hamiltonian also contains unphys
>>terms that I consider inappropriate.
>
>
> First, you did not write down the equations of motion in the interaction
> picture as I asked.

Equations of motion for what? Fields? I do not care about fields.
I have a Hamiltonian and I have an initial state. My "working knowledge
of quantum mechanics" tells me that this is all I need to compute the
time evolution of the state.

> Second, your solution is not correct, the
> Hamiltonian will have a time dependent potential that will kick one of
> the particles into motion.

Remember, we are talking about a single particle. Let's try something
simple for starters.

> In the same vein, in the interaction picture
> all particle operators will have time dependence given to them by H_0.

Again, I am working in the Schroedinger picture. Please forgive me this
weakness, but I prefer having my operators time-independent, then it is
easier to figure out what is going on.

> Third, as has been explained several times, the expression you wrote
> down has zero physical meaning.
> You are still far from anything that can
> be directly physically interpreted.

If you don't like my formula for the time evolution of a 1-particle
state, then what is your formula?

>
> You still have the following steps to complete:
> * Write down the equations of motion of the field operators

I don't need those, as I said above.

> and the states.

Are you talking about the non-stationary Schroedinger equation?

-i d/dt \Psi(t) = H \Psi(t)

I wrote the solution of this equation for the 1-particle state above.
Did I make any mistake?

> * Diagonalize H_0.

Eigenvectors of H_0 are |0>, a^*|0>, a^*c^*|0>, ...
The eigenvalues are sums of energies of created particles.

> * Solve the equations of motion for the field operators \phi and A.

I don't need those, as I said above.

>
> Working knowledge of quantum mechanics as well as patience are
> prerequisites for this calculation.

Agreed.

Eugene Stefanovich.

Igor Khavkine

Apr3-05, 04:45 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-02, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n>> Working knowledge of quantum mechanics as well as patience are\n>> prerequisites for this calculation.\n>\n> Agreed.\n\nMy goal in this thread is to show you how a finite time calculation is\ndone in standard theory because by your own account you\'ve never seen\nsuch a calculation. On your own time you can convert these calculations\nto your favorite framework. I do not wish to discuss that. This means\nthat I expect you to be patient, to follow instructions, and to reserve\nphysical interpretation until the end. If you can\'t agree, then I\'m not\ninclined to continue the discussion. I can explain the reasoning behind\neach step, but I will not compare and contrast them to your favorite\napproach.\n\n> I suggest to simplify the problem and instead of 2 particles consider\n> just one. We\'ll return to the 2-particle case later. So, let us forget\n> about the external potential for the moment.\n\nNo. The number of particles is immaterial. The external potential is\nessential for the experimental setup.\n\n> Are you saying that 0-particle and 1-particle states will also result\n> in some "scattering" at time t=0? They certainly will if you use your\n> Hamiltonian, but it doesn\'t make any sense. I don\'t know much about\n> experiment, but I know enough to say that no scattering occurs in real\n> 0- and 1-particle states. To me this indicates one of the two\n> possibilities:\n> 1) the things you call "particles" have no relationship to the real\n> particles we observe in nature\n\nBingo! That\'s why they are called *bare* particles.\n\n>> The separation H = H_0 + H_1 into free and interaction parts is\n>> arbitrary.\n>\n> I don\'t think that separation is arbitrary. I would prefer the\n> separation which has clear physical meaning: H_0 is Hamiltonian of\n> free particles, e.g. for massive particles\n>\n> H_0 = \\int dp sqrt(p^2 + m^2) a^*(p)a(p)\n>\n> where m is the experimentally measured particle mass. The rest is\n> interaction. In this case, if you drop the interaction term you\'ll\n> obtain the system of free particles moving freely without interaction,\n> as it should be. All other choices are artificial.\n\nChoice is choice, whenever it exists one is allowed to exercise it.\nYour H_0 is a good start. However, you are not using the external\npotential and you are missing the E&M Hamiltonian. Whatever your\nopinions about it\'s existence, it is part of the model. Also, leave m as\na parameter. Physical interpretation comes in the end.\n\n> I am afraid you are trying to make simple things complex.\n> First, there is no use of propagators when we are trying to calculate\n> the time evolution. Propagators are useful for the S-matrix calculation.\n> Second, the interaction picture could be convenient for the S-matrix\n> calculations, but it becomes rather confusing in the case of time\n> evolution: both states and operators become time-dependent. Let us\n> keep simple things simple: use the Schroedinger picture in which\n> H_0 and H_1 are time-independent and try to calculate the time\n> dependence of some state vector, e.g., a^*|0>\n\nYou are wrong, and if you are patient enough you will see why. The\nperturbative calculation will be expanded in Feynman diagrams and will\nmake use of propagators.\n\n>> First, you did not write down the equations of motion in the interaction\n>> picture as I asked.\n>\n> Equations of motion for what? Fields? I do not care about fields.\n> I have a Hamiltonian and I have an initial state. My "working knowledge\n> of quantum mechanics" tells me that this is all I need to compute the\n> time evolution of the state.\n\nWhether you care for fields or not is immaterial, they are part of the\nmodel. Your working knowledge of quantum mechanics should tell you that\nyou need states *and* operators. In QFT the fields are operators out of\nwhich all observables are built. You call them "particle operators", but\nthat does not change that everyone esle calls them "fields" or "field\noperators".\n\nField operators are not some mysterious quantities that spring out of\nthe vacuum. They appear naturally when you try to work with quantum\nmechanics with an indeterminate number of particles. See Sections 59 and\n65 of Dirac\'s _Principles of Quantum Mechanics_.\n\n[...some irrelevant objections snipped...]\n\n>> You still have the following steps to complete:\n>> * Write down the equations of motion of the field operators\n>\n> I don\'t need those, as I said above.\n\nIn ordinary quantum mechanics you need the position and momentum\noperators x and p. In QFT you need field operators, \'nuf said.\n\n>> and the states.\n>\n> Are you talking about the non-stationary Schroedinger equation?\n>\n> -i d/dt \\Psi(t) = H \\Psi(t)\n>\n> I wrote the solution of this equation for the 1-particle state above.\n> Did I make any mistake?\n\nFirst, you have a typo: -i -> i. All equations of motion in quantum\nmechanics stem from the Schroedinger equation. I am asking you to write\ndown the equations of motion for the states and operators in the\ninteraction picture. This is no more complicated than what\'s done in\nSection 5.5 of Sakurai.\n\n>> * Diagonalize H_0.\n>\n> Eigenvectors of H_0 are |0>, a^*|0>, a^*c^*|0>, ...\n> The eigenvalues are sums of energies of created particles.\n\nClose, but it\'s time to start getting specific. In QFT,\n"diagonalization" of a bilinear Hamiltonian refers to writing it as\n\nH_0 = sum_n E_n b^+_n b_n\n\nwhere [b_n,b^+_m] = delta_{n,m}. Your job is to find the E_n and how the\nb_n are related to the field operators phi(x). This involves solving the\nKlein-Gordon equation in the external potential. When there is no\nexternal potential, the label n represents momentum and the b_n are the\nwhat you call "particle operators".\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-02, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

>> Working knowledge of quantum mechanics as well as patience are
>> prerequisites for this calculation.
>
> Agreed.

My goal in this thread is to show you how a finite time calculation is
done in standard theory because by your own account you've never seen
such a calculation. On your own time you can convert these calculations
to your favorite framework. I do not wish to discuss that. This means
that I expect you to be patient, to follow instructions, and to reserve
physical interpretation until the end. If you can't agree, then I'm not
inclined to continue the discussion. I can explain the reasoning behind
each step, but I will not compare and contrast them to your favorite
approach.

> I suggest to simplify the problem and instead of 2 particles consider
> just one. We'll return to the 2-particle case later. So, let us forget
> about the external potential for the moment.

No. The number of particles is immaterial. The external potential is
essential for the experimental setup.

> Are you saying that 0-particle and 1-particle states will also result
> in some "scattering" at time t=0? They certainly will if you use your
> Hamiltonian, but it doesn't make any sense. I don't know much about
> experiment, but I know enough to say that no scattering occurs in real
> 0- and 1-particle states. To me this indicates one of the two
> possibilities:
> 1) the things you call "particles" have no relationship to the real
> particles we observe in nature

Bingo! That's why they are called *bare* particles.

>> The separation H = H_0 + H_1 into free and interaction parts is
>> arbitrary.
>
> I don't think that separation is arbitrary. I would prefer the
> separation which has clear physical meaning: H_0 is Hamiltonian of
> free particles, e.g. for massive particles
>
> H_0 = \int dp \sqrt(p^2 + m^2) a^*(p)a(p)
>
> where m is the experimentally measured particle mass. The rest is
> interaction. In this case, if you drop the interaction term you'll
> obtain the system of free particles moving freely without interaction,
> as it should be. All other choices are artificial.

Choice is choice, whenever it exists one is allowed to exercise it.
Your H_0 is a good start. However, you are not using the external
potential and you are missing the E&M Hamiltonian. Whatever your
opinions about it's existence, it is part of the model. Also, leave m as
a parameter. Physical interpretation comes in the end.

> I am afraid you are trying to make simple things complex.
> First, there is no use of propagators when we are trying to calculate
> the time evolution. Propagators are useful for the S-matrix calculation.
> Second, the interaction picture could be convenient for the S-matrix
> calculations, but it becomes rather confusing in the case of time
> evolution: both states and operators become time-dependent. Let us
> keep simple things simple: use the Schroedinger picture in which
> H_0 and H_1 are time-independent and try to calculate the time
> dependence of some state vector, e.g., a^*|0>

You are wrong, and if you are patient enough you will see why. The
perturbative calculation will be expanded in Feynman diagrams and will
make use of propagators.

>> First, you did not write down the equations of motion in the interaction
>> picture as I asked.
>
> Equations of motion for what? Fields? I do not care about fields.
> I have a Hamiltonian and I have an initial state. My "working knowledge
> of quantum mechanics" tells me that this is all I need to compute the
> time evolution of the state.

Whether you care for fields or not is immaterial, they are part of the
model. Your working knowledge of quantum mechanics should tell you that
you need states *and* operators. In QFT the fields are operators out of
which all observables are built. You call them "particle operators", but
that does not change that everyone esle calls them "fields" or "field
operators".

Field operators are not some mysterious quantities that spring out of
the vacuum. They appear naturally when you try to work with quantum
mechanics with an indeterminate number of particles. See Sections 59 and
65 of Dirac's _Principles of Quantum Mechanics_.

[...some irrelevant objections snipped...]

>> You still have the following steps to complete:
>> * Write down the equations of motion of the field operators
>
> I don't need those, as I said above.

In ordinary quantum mechanics you need the position and momentum
operators x and p. In QFT you need field operators, 'nuf said.

>> and the states.
>
> Are you talking about the non-stationary Schroedinger equation?
>
> -i d/dt \Psi(t) = H \Psi(t)
>
> I wrote the solution of this equation for the 1-particle state above.
> Did I make any mistake?

First, you have a typo: -i -> i. All equations of motion in quantum
mechanics stem from the Schroedinger equation. I am asking you to write
down the equations of motion for the states and operators in the
interaction picture. This is no more complicated than what's done in
Section 5.5 of Sakurai.

>> *[/itex] Diagonalize H_0.
>
> Eigenvectors of H_0 are |0>, a^*|0>, a^*c^*|0>, ...
> The eigenvalues are sums of energies of created particles.

Close, but it's time to start getting specific. In QFT,
"diagonalization" of a bilinear Hamiltonian refers to writing it as

[itex]H_0 = sum_n E_n b^+_n b_n

where [b_n,b^+_m] = \delta_{n,m}. Your job is to find the E_n and how the
b_n are related to the field operators \phi(x). This involves solving the
Klein-Gordon equation in the external potential. When there is no
external potential, the label n represents momentum and the b_n are the
what you call "particle operators".

Igor

Eugene Stefanovich

Apr3-05, 04:46 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>I do not see how\n>>a sum (even infinite) of instantaneous interactions can become retarded.\n>\n>\n> In spite of having it had explained several times in previous threads.\n>\n> I am tired of preaching deaf ears. Instead of trying to learn\n> you are just trying to defend yourself.\n>\n> Understandable but unproductive.\n\nProbably your explanation was not convincing enough.\n\nI say that in each perturbation order (the same is true for the\nsum of all perturbation orders) the dressed interaction operator\nin the 2-electron sector of the Fock space has the form\n\nV = \\int dp dq dk f(p,q,k) a^*(p-k)a^*(q+k)a(p)a(q)\n\nwhere f(p,q,k) is a rotationally invariant function of its arguments.\nThis operator can be exactly expressed through operators of\nmomentum p and q and position r_1 and r_2 of the two particles:\n\nV(p, q, r_1-r_2)\n\nThis is a general expression for an instantaneous momentum- and\ndistance-dependent potential. The arguments are presented in subsections\n9.2.7 and 12.2.3 of the book. I would appreciate if you attack these\narguments rather than attacking me personally.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>I do not see how
>>a sum (even infinite) of instantaneous interactions can become retarded.
>
>
> In spite of having it had explained several times in previous threads.
>
> I am tired of preaching deaf ears. Instead of trying to learn
> you are just trying to defend yourself.
>
> Understandable but unproductive.

Probably your explanation was not convincing enough.

I say that in each perturbation order (the same is true for the
sum of all perturbation orders) the dressed interaction operator
in the 2-electron sector of the Fock space has the form

V = \int dp dq dk f(p,q,k) a^*(p-k)a^*(q+k)a(p)a(q)

where f(p,q,k) is a rotationally invariant function of its arguments.
This operator can be exactly expressed through operators of
momentum p and q and position r_1 and r_2 of the two particles:

V(p, q, r_1-r_2)

This is a general expression for an instantaneous momentum- and
distance-dependent potential. The arguments are presented in subsections
9.2.7 and 12.2.3 of the book. I would appreciate if you attack these
arguments rather than attacking me personally.

Eugene Stefanovich.

Eugene Stefanovich

Apr3-05, 04:46 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nEugene Stefanovich wrote:\n\n>\n>>In the same vein, in the interaction picture\n>>all particle operators will have time dependence given to them by H_0.\n>\n>\n> Again, I am working in the Schroedinger picture. Please forgive me this\n> weakness, but I prefer having my operators time-independent, then it is\n> easier to figure out what is going on.\n>\n\n\nOoops, sorry, my mistake. In the Schroedinger picture creation and\nannihilation operators must have the same transformation laws as states,\ne.g.,\n\na(t) = exp(iHt) a exp(-iHt)\n\nwhere H is the full interacting Hamiltonian, and a is annihilation\noperator. This does not change anything in my other arguments, though.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

>
>>In the same vein, in the interaction picture
>>all particle operators will have time dependence given to them by H_0.
>
>
> Again, I am working in the Schroedinger picture. Please forgive me this
> weakness, but I prefer having my operators time-independent, then it is
> easier to figure out what is going on.
>

Ooops, sorry, my mistake. In the Schroedinger picture creation and
annihilation operators must have the same transformation laws as states,
e.g.,

a(t) = \exp(iHt) a \exp(-iHt)

where H is the full interacting Hamiltonian, and a is annihilation
operator. This does not change anything in my other arguments, though.

Eugene Stefanovich.

Igor Khavkine

Apr4-05, 06:40 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <424F9DD2.4020801@synopsys.com>, Eugene Stefanovich wrote:\n\n> Ooops, sorry, my mistake. In the Schroedinger picture creation and\n> annihilation operators must have the same transformation laws as states,\n> e.g.,\n>\n> a(t) = exp(iHt) a exp(-iHt)\n>\n> where H is the full interacting Hamiltonian, and a is annihilation\n> operator. This does not change anything in my other arguments, though.\n\nNo. Creation/annihilation operators are operators and have the same time\ndependence as other operators. Namely, they are time independent in the\nSchroedinger picture. You must have been confused with the density\nmatrix.\n\nIgor\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <424F9DD2.4020801@synopsys.com>, Eugene Stefanovich wrote:

> Ooops, sorry, my mistake. In the Schroedinger picture creation and
> annihilation operators must have the same transformation laws as states,
> e.g.,
>
> a(t) = \exp(iHt) a \exp(-iHt)
>
> where H is the full interacting Hamiltonian, and a is annihilation
> operator. This does not change anything in my other arguments, though.

No. Creation/annihilation operators are operators and have the same time
dependence as other operators. Namely, they are time independent in the
Schroedinger picture. You must have been confused with the density
matrix.

Igor

Eugene Stefanovich

Apr5-05, 04:45 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote:\n> On 2005-04-02, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>>Igor Khavkine wrote:\n>\n>\n>>>Working knowledge of quantum mechanics as well as patience are\n>>>prerequisites for this calculation.\n>>\n>>Agreed.\n>\n>\n> My goal in this thread is to show you how a finite time calculation is\n> done in standard theory because by your own account you\'ve never seen\n> such a calculation. On your own time you can convert these calculations\n> to your favorite framework. I do not wish to discuss that. This means\n> that I expect you to be patient, to follow instructions, and to reserve\n> physical interpretation until the end. If you can\'t agree, then I\'m not\n> inclined to continue the discussion. I can explain the reasoning behind\n> each step, but I will not compare and contrast them to your favorite\n> approach.\n\nOK, I will hold my scepticism for a while and let you develop your\napproach if you promise to "explain the reasoning behind each step".\n\n>\n>\n>>I suggest to simplify the problem and instead of 2 particles consider\n>>just one. We\'ll return to the 2-particle case later. So, let us forget\n>>about the external potential for the moment.\n>\n>\n> No. The number of particles is immaterial.\n\nIf the number is immaterial, then I would prefer to talk about just one\nparticle.\n\n> The external potential is\n> essential for the experimental setup.\n\nI am not sure what kind of potential you\'ll introduce\nlater for describing the experiment, but I am becoming suspicious. The\nexternal potential is definitely an approximation. Isn\'t it dangerous\nto introduce approximations at such an early stage?\n\n>\n>\n>>Are you saying that 0-particle and 1-particle states will also result\n>>in some "scattering" at time t=0? They certainly will if you use your\n>>Hamiltonian, but it doesn\'t make any sense. I don\'t know much about\n>>experiment, but I know enough to say that no scattering occurs in real\n>>0- and 1-particle states. To me this indicates one of the two\n>>possibilities:\n>>1) the things you call "particles" have no relationship to the real\n>>particles we observe in nature\n>\n>\n> Bingo! That\'s why they are called *bare* particles.\n>\n>\n>>>The separation H = H_0 + H_1 into free and interaction parts is\n>>>arbitrary.\n>>\n>>I don\'t think that separation is arbitrary. I would prefer the\n>>separation which has clear physical meaning: H_0 is Hamiltonian of\n>>free particles, e.g. for massive particles\n>>\n>>H_0 = \\int dp sqrt(p^2 + m^2) a^*(p)a(p)\n>>\n>>where m is the experimentally measured particle mass. The rest is\n>>interaction. In this case, if you drop the interaction term you\'ll\n>>obtain the system of free particles moving freely without interaction,\n>>as it should be. All other choices are artificial.\n>\n>\n> Choice is choice, whenever it exists one is allowed to exercise it.\n> Your H_0 is a good start. However, you are not using the external\n> potential and you are missing the E&M Hamiltonian.\n\nI only wrote the part of H_0 relevant to massive particles. In the\npresence of photons you should also add the corresponding part\n\nH_0 = ... + \\int dp |p| c^*(p)c(p)\n\n> Whatever your\n> opinions about it\'s existence, it is part of the model. Also, leave m as\n> a parameter. Physical interpretation comes in the end.\n\nOK, I am still following you.\n\n>\n>>>First, you did not write down the equations of motion in the interaction\n>>>picture as I asked.\n>>\n>>Equations of motion for what? Fields? I do not care about fields.\n>>I have a Hamiltonian and I have an initial state. My "working knowledge\n>>of quantum mechanics" tells me that this is all I need to compute the\n>>time evolution of the state.\n>\n>\n> Whether you care for fields or not is immaterial, they are part of the\n> model. Your working knowledge of quantum mechanics should tell you that\n> you need states *and* operators. In QFT the fields are operators out of\n> which all observables are built. You call them "particle operators", but\n> that does not change that everyone esle calls them "fields" or "field\n> operators".\n\nI agree that creation and annihilation operators can be expressed\nthrough fields and vice versa. Operators of observables can be\nexpressed through both particle operators and fields. As long as you do\nnot assign any physical meaning to particle operators and fields, I\nam with you.\n\n>\n> Field operators are not some mysterious quantities that spring out of\n> the vacuum. They appear naturally when you try to work with quantum\n> mechanics with an indeterminate number of particles. See Sections 59 and\n> 65 of Dirac\'s _Principles of Quantum Mechanics_.\n>\n> [...some irrelevant objections snipped...]\n>\n>\n>>>You still have the following steps to complete:\n>>>* Write down the equations of motion of the field operators\n>>\n>>I don\'t need those, as I said above.\n>\n>\n> In ordinary quantum mechanics you need the position and momentum\n> operators x and p. In QFT you need field operators, \'nuf said.\n>\n>\n>>>and the states.\n>>\n>>Are you talking about the non-stationary Schroedinger equation?\n>>\n>>-i d/dt \\Psi(t) = H \\Psi(t)\n>>\n>>I wrote the solution of this equation for the 1-particle state above.\n>>Did I make any mistake?\n>\n>\n> First, you have a typo: -i -> i.\n\nI disagree with you here. In my book I went through a great pain to\nselect all signs consistently. However, this sign does not have any\nrelevance to our discussion, and I do not mind to change it\n\ni d/dt \\Psi(t) = H \\Psi(t)\n\n> All equations of motion in quantum\n> mechanics stem from the Schroedinger equation. I am asking you to write\n> down the equations of motion for the states and operators in the\n> interaction picture. This is no more complicated than what\'s done in\n> Section 5.5 of Sakurai.\n\nI don\'t have Sakurai book. Do you want me to write smthng like\nMaxwell\'s equations and Klein-Gordon equations in the external\npotential?\n\n>\n>\n>>>* Diagonalize H_0.\n>>\n>>Eigenvectors of H_0 are |0>, a^*|0>, a^*c^*|0>, ...\n>>The eigenvalues are sums of energies of created particles.\n>\n>\n> Close, but it\'s time to start getting specific. In QFT,\n> "diagonalization" of a bilinear Hamiltonian refers to writing it as\n>\n> H_0 = sum_n E_n b^+_n b_n\n>\n> where [b_n,b^+_m] = delta_{n,m}. Your job is to find the E_n and how the\n> b_n are related to the field operators phi(x). This involves solving the\n> Klein-Gordon equation in the external potential. When there is no\n> external potential, the label n represents momentum and the b_n are the\n> what you call "particle operators".\n\nAgreed, without the external potential you should get H_0 as I wrote it\nabove.\n\nH_0 = \\int dp sqrt(p^2 + m^2) a^*(p)a(p) + \\int dp |p| c^*(p)c(p)\n\nOK, we get the free particle Hamiltonian, what\'s next?\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On 2005-04-02, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>>Igor Khavkine wrote:
>
>
>>>Working knowledge of quantum mechanics as well as patience are
>>>prerequisites for this calculation.
>>
>>Agreed.
>
>
> My goal in this thread is to show you how a finite time calculation is
> done in standard theory because by your own account you've never seen
> such a calculation. On your own time you can convert these calculations
> to your favorite framework. I do not wish to discuss that. This means
> that I expect you to be patient, to follow instructions, and to reserve
> physical interpretation until the end. If you can't agree, then I'm not
> inclined to continue the discussion. I can explain the reasoning behind
> each step, but I will not compare and contrast them to your favorite
> approach.

OK, I will hold my scepticism for a while and let you develop your
approach if you promise to "explain the reasoning behind each step".

>
>
>>I suggest to simplify the problem and instead of 2 particles consider
>>just one. We'll return to the 2-particle case later. So, let us forget
>>about the external potential for the moment.
>
>
> No. The number of particles is immaterial.

If the number is immaterial, then I would prefer to talk about just one
particle.

> The external potential is
> essential for the experimental setup.

I am not sure what kind of potential you'll introduce
later for describing the experiment, but I am becoming suspicious. The
external potential is definitely an approximation. Isn't it dangerous
to introduce approximations at such an early stage?

>
>
>>Are you saying that 0-particle and 1-particle states will also result
>>in some "scattering" at time t=0? They certainly will if you use your
>>Hamiltonian, but it doesn't make any sense. I don't know much about
>>experiment, but I know enough to say that no scattering occurs in real
>>0- and 1-particle states. To me this indicates one of the two
>>possibilities:
>>1) the things you call "particles" have no relationship to the real
>>particles we observe in nature
>
>
> Bingo! That's why they are called *bare* particles.
>
>
>>>The separation H = H_0 + H_1 into free and interaction parts is
>>>arbitrary.
>>
>>I don't think that separation is arbitrary. I would prefer the
>>separation which has clear physical meaning: H_0 is Hamiltonian of
>>free particles, e.g. for massive particles
>>
>>H_0 = \int dp \sqrt(p^2 + m^2) a^*(p)a(p)
>>
>>where m is the experimentally measured particle mass. The rest is
>>interaction. In this case, if you drop the interaction term you'll
>>obtain the system of free particles moving freely without interaction,
>>as it should be. All other choices are artificial.
>
>
> Choice is choice, whenever it exists one is allowed to exercise it.
> Your H_0 is a good start. However, you are not using the external
> potential and you are missing the E&M Hamiltonian.

I only wrote the part of H_0 relevant to massive particles. In the
presence of photons you should also add the corresponding part

H_0 = .[/itex].. + \int dp |p| c^*(p)c(p)

> Whatever your
> opinions about it's existence, it is part of the model. Also, leave m as
> a parameter. Physical interpretation comes in the end.

OK, I am still following you.

>
>>>First, you did not write down the equations of motion in the interaction
>>>picture as I asked.
>>
>>Equations of motion for what? Fields? I do not care about fields.
>>I have a Hamiltonian and I have an initial state. My "working knowledge
>>of quantum mechanics" tells me that this is all I need to compute the
>>time evolution of the state.
>
>
> Whether you care for fields or not is immaterial, they are part of the
> model. Your working knowledge of quantum mechanics should tell you that
> you need states *and* operators. In QFT the fields are operators out of
> which all observables are built. You call them "particle operators", but
> that does not change that everyone esle calls them "fields" or "field
> operators".

I agree that creation and annihilation operators can be expressed
through fields and vice versa. Operators of observables can be
expressed through both particle operators and fields. As long as you do
not assign any physical meaning to particle operators and fields, I
am with you.

>
> Field operators are not some mysterious quantities that spring out of
> the vacuum. They appear naturally when you try to work with quantum
> mechanics with an indeterminate number of particles. See Sections 59 and
> 65 of Dirac's _Principles of Quantum Mechanics_.
>
> [...some irrelevant objections snipped...]
>
>
>>>You still have the following steps to complete:
>>>* Write down the equations of motion of the field operators
>>
>>I don't need those, as I said above.
>
>
> In ordinary quantum mechanics you need the position and momentum
> operators x and p. In QFT you need field operators, 'nuf said.
>
>
>>>and the states.
>>
>>Are you talking about the non-stationary Schroedinger equation?
>>
>>-i d/dt \Psi(t) = H \Psi(t)
>>
>>I wrote the solution of this equation for the 1-particle state above.
>>Did I make any mistake?
>
>
> First, you have a typo: -i -> i.

I disagree with you here. In my book I went through a great pain to
select all signs consistently. However, this sign does not have any
relevance to our discussion, and I do not mind to change it

i d/dt \Psi(t) = H \Psi(t)

> All equations of motion in quantum
> mechanics stem from the Schroedinger equation. I am asking you to write
> down the equations of motion for the states and operators in the
> interaction picture. This is no more complicated than what's done in
> Section 5.5 of Sakurai.

I don't have Sakurai book. Do you want me to write smthng like
Maxwell's equations and Klein-Gordon equations in the external
potential?

>
>
>>>* Diagonalize H_0.
>>
>>Eigenvectors of H_0 are [itex]|0>, a^*|0>, a^*c^*|0>, ...
>>The eigenvalues are sums of energies of created particles.
>
>
> Close, but it's time to start getting specific. In QFT,
> "diagonalization" of a bilinear Hamiltonian refers to writing it as
>
> H_0 = sum_n E_n b^+_n b_n
>
> where [b_n,b^+_m] = \delta_{n,m}. Your job is to find the E_n and how the
> b_n are related to the field operators \phi(x). This involves solving the
> Klein-Gordon equation in the external potential. When there is no
> external potential, the label n represents momentum and the b_n are the
> what you call "particle operators".

Agreed, without the external potential you should get H_0 as I wrote it
above.

H_0 = \int dp \sqrt(p^2 + m^2) a^*(p)a(p) + \int dp |p| c^*(p)c(p)

OK, we get the free particle Hamiltonian, what's next?

Eugene Stefanovich.

Eugene Stefanovich

Apr5-05, 04:45 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote:\n> In article <424F9DD2.4020801@synopsys.com>, Eugene Stefanovich wrote:\n>\n>\n>>Ooops, sorry, my mistake. In the Schroedinger picture creation and\n>>annihilation operators must have the same transformation laws as states,\n>>e.g.,\n>>\n>>a(t) = exp(iHt) a exp(-iHt)\n>>\n>>where H is the full interacting Hamiltonian, and a is annihilation\n>>operator. This does not change anything in my other arguments, though.\n>\n>\n> No. Creation/annihilation operators are operators and have the same time\n> dependence as other operators. Namely, they are time independent in the\n> Schroedinger picture. You must have been confused with the density\n> matrix.\n>\n\nI see what you are saying, but I think the time dependence of particle\noperators can be defined both ways. All depends on definition. My\ndefinition is that a^*(p,t) is operator creating a one-particle state\nwith momentum p out of vacuum at time t. With this definition, in the\nSchroedinger picture, we should have\n\na^*(p,t) = exp(iHt) a^*(p,0) exp(-iHt)\n\nActing by this operator on the time-dependent vacuum state |0, t>\nwe obtain the correct one-particle state at time t\n\n|p, t> = a^*(p,t) |0, t>\n= exp(iHt) a^*(p,0) exp(-iHt) exp(iHt) |0, 0>\n= exp(iHt) a^*(p,0) |0, 0>\n= exp(iHt) |p, 0>\n\nIn order to define time-independent operators of observables\nwe can use particle operators at time t=0. E.g., the operator of\ntotal momentum is\n\nP = \\int dp p a^*(p,0) a(p,0)\n\nOn the other hand, you may choose to work with time-independent\nparticle operators a^*(p) and a(p) in the Schroedinger picture.\nThen, to create the one-particle state at time t you should first\nbring vacuum back to time 0, apply a^*(p), and then time translate the\nobtained vector to time t\n\n|p,t> = exp(iHt) a^*(p) exp(-iHt) |0, t>\n= exp(iHt) a^*(p) |0, 0>\n= exp(iHt) |p, 0>\n\nAs you see, the result is the same, but the 2nd definition looks\nmore cumbersome: creation of the 1-particle state at time t requires\napplication of 3 operators exp(iHt), a^*(p), and exp(-iHt).\n\nOperators a^*(p) and a^*(p,t) have no physical significance.\nTheir only purpose is to simplify calculations with states and\noperators of observables which do have physical significance.\nWe can choose creation and annihilation operators as we see convenient\nto make mathematical manipulations as simple as possible.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> In article <424F9DD2.4020801@synopsys.com>, Eugene Stefanovich wrote:
>
>
>>Ooops, sorry, my mistake. In the Schroedinger picture creation and
>>annihilation operators must have the same transformation laws as states,
>>e.g.,
>>
>>a(t) = \exp(iHt) a \exp(-iHt)
>>
>>where H is the full interacting Hamiltonian, and a is annihilation
>>operator. This does not change anything in my other arguments, though.
>
>
> No. Creation/annihilation operators are operators and have the same time
> dependence as other operators. Namely, they are time independent in the
> Schroedinger picture. You must have been confused with the density
> matrix.
>

I see what you are saying, but I think the time dependence of particle
operators can be defined both ways. All depends on definition. My
definition is that a^*(p,t) is operator creating a one-particle state
with momentum p out of vacuum at time t. With this definition, in the
Schroedinger picture, we should have

a^*(p,t) = \exp(iHt) a^*(p,0) \exp(-iHt)

Acting by this operator on the time-dependent vacuum state |0, t>
we obtain the correct one-particle state at time t

|p, t> = a^*(p,t) |0, t>= \exp(iHt) a^*(p,0) \exp(-iHt) \exp(iHt) |0, 0>= \exp(iHt) a^*(p,0) |0, 0>= \exp(iHt) |p, 0>

In order to define time-independent operators of observables
we can use particle operators at time t=0. E.g., the operator of
total momentum is

P = \int dp p a^*(p,0) a(p,0)

On the other hand, you may choose to work with time-independent
particle operators a^*(p) and a(p) in the Schroedinger picture.
Then, to create the one-particle state at time t you should first
bring vacuum back to time 0, apply a^*(p), and then time translate the
obtained vector to time t

|p,t> = \exp(iHt) a^*(p) \exp(-iHt) |0, t>= \exp(iHt) a^*(p) |0, 0>= \exp(iHt) |p, 0>

As you see, the result is the same, but the 2nd definition looks
more cumbersome: creation of the 1-particle state at time t requires
application of 3 operators \exp(iHt), a^*(p), and \exp(-iHt).

Operators a^*(p) and a^*(p,t) have no physical significance.
Their only purpose is to simplify calculations with states and
operators of observables which do have physical significance.
We can choose creation and annihilation operators as we see convenient
to make mathematical manipulations as simple as possible.

Eugene Stefanovich.

Igor Khavkine

Apr6-05, 03:23 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-05, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n> If the number is immaterial, then I would prefer to talk about just one\n> particle.\n\nNo. You need two particles to setup an interaction experiment. But we\nare still a long way from there. Right now you must work with an\narbitrary number of particles to setup the framework.\n\n>> The external potential is\n>> essential for the experimental setup.\n>\n> I am not sure what kind of potential you\'ll introduce\n> later for describing the experiment, but I am becoming suspicious. The\n> external potential is definitely an approximation. Isn\'t it dangerous\n> to introduce approximations at such an early stage?\n\nExternal potential == localization. I will say no more.\n\n>> All equations of motion in quantum\n>> mechanics stem from the Schroedinger equation. I am asking you to write\n>> down the equations of motion for the states and operators in the\n>> interaction picture. This is no more complicated than what\'s done in\n>> Section 5.5 of Sakurai.\n>\n> I don\'t have Sakurai book. Do you want me to write smthng like\n> Maxwell\'s equations and Klein-Gordon equations in the external\n> potential?\n\nNo. Let\'s take this a bit slower, because I still haven\'t seen you write\ndown the correct Hamiltonian. Consider the Lagrangian density\n\nL_1 = -|Dphi|^2 - m^2 |phi|^2 - V|phi|^2 - (1/4) tr(F^2),\n\nwhere phi is a complex scalar field, F is the field strength for the\n4-vector potential A and the covariant derivative is D = @ - ieA. What\nis the Hamiltonian density? Write everything in position space. I want\nto see what your expression is. Feel free to use compact notation as I\ndid.\n\nWhat are the classical equations of motion for the free fields. Solve\nthem and expand the fields in eigenmodes (momentum modes for A and\ndiscrete modes for phi). Re-express the Hamiltonian in terms of the mode\ncoefficients.\n\nStandard reference for all of this is the last chapter of Goldstein.\n\nHamiltonian first, more later. This has to be done before a proper\nquantized theory is constructed.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-05, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

> If the number is immaterial, then I would prefer to talk about just one
> particle.

No. You need two particles to setup an interaction experiment. But we
are still a long way from there. Right now you must work with an
arbitrary number of particles to setup the framework.

>> The external potential is
>> essential for the experimental setup.
>
> I am not sure what kind of potential you'll introduce
> later for describing the experiment, but I am becoming suspicious. The
> external potential is definitely an approximation. Isn't it dangerous
> to introduce approximations at such an early stage?

External potential == localization. I will say no more.

>> All equations of motion in quantum
>> mechanics stem from the Schroedinger equation. I am asking you to write
>> down the equations of motion for the states and operators in the
>> interaction picture. This is no more complicated than what's done in
>> Section 5.5 of Sakurai.
>
> I don't have Sakurai book. Do you want me to write smthng like
> Maxwell's equations and Klein-Gordon equations in the external
> potential?

No. Let's take this a bit slower, because I still haven't seen you write
down the correct Hamiltonian. Consider the Lagrangian density

L_1 = -|Dphi|^2 - m^2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F^2),

where \phi is a complex scalar field, F is the field strength for the
4-vector potential A and the covariant derivative is D = @ - ieA. What
is the Hamiltonian density? Write everything in position space. I want
to see what your expression is. Feel free to use compact notation as I
did.

What are the classical equations of motion for the free fields. Solve
them and expand the fields in eigenmodes (momentum modes for A and
discrete modes for \phi). Re-express the Hamiltonian in terms of the mode
coefficients.

Standard reference for all of this is the last chapter of Goldstein.

Hamiltonian first, more later. This has to be done before a proper
quantized theory is constructed.

Igor

Eugene Stefanovich

Apr6-05, 11:52 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n\n>>>All equations of motion in quantum\n>>>mechanics stem from the Schroedinger equation. I am asking you to write\n>>>down the equations of motion for the states and operators in the\n>>>interaction picture. This is no more complicated than what\'s done in\n>>>Section 5.5 of Sakurai.\n>>\n>>I don\'t have Sakurai book. Do you want me to write smthng like\n>>Maxwell\'s equations and Klein-Gordon equations in the external\n>>potential?\n>\n>\n> No. Let\'s take this a bit slower, because I still haven\'t seen you write\n> down the correct Hamiltonian. Consider the Lagrangian density\n>\n> L_1 = -|Dphi|^2 - m^2 |phi|^2 - V|phi|^2 - (1/4) tr(F^2),\n>\n> where phi is a complex scalar field, F is the field strength for the\n> 4-vector potential A and the covariant derivative is D = @ - ieA. What\n> is the Hamiltonian density? Write everything in position space. I want\n> to see what your expression is. Feel free to use compact notation as I\n> did.\n>\n> What are the classical equations of motion for the free fields. Solve\n> them and expand the fields in eigenmodes (momentum modes for A and\n> discrete modes for phi). Re-express the Hamiltonian in terms of the mode\n> coefficients.\n>\n> Standard reference for all of this is the last chapter of Goldstein.\n>\n> Hamiltonian first, more later. This has to be done before a proper\n> quantized theory is constructed.\n\nCan we do it another way? You write the Hamiltonian and we will discuss\nthe physics of it. I must admit that I do not understand the procedure\nof "field quantization", i.e., the way the Hamiltonian is derived from\nthe Lagrangian in quantum field theories. That\'s not because I am lazy\nand don\'t want to read numerous good books written about the subject.\nI read enough of those books, and still the entire procedure does not\nmake sense to me. I admire the final result of\nthis procedure, e.g., in the case of QED the quantization results in\na relativistically invariant Hamiltonian that (after renormalization)\nleads to the most accurate S-matrix. However, there are so many\nhandwavings and magic spells on the path of quantization, that\nI consider it witchcraft. I wouldn\'t try it myself.\n\nSo, I suggest that you write the Hamiltonian and I will trust you that\nthis is the correct Hamiltonian of the theory. Then we\'ll discuss\nphysics stemming from this Hamiltonian. In any event, we cannot discuss\nquantum physics until the Hamiltonian is written in an explicit form.\nOn the other hand, after the Hamiltonian is written, all physics is in\nour hands.\n\nEugene Stefanovich.\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:

>>>All equations of motion in quantum
>>>mechanics stem from the Schroedinger equation. I am asking you to write
>>>down the equations of motion for the states and operators in the
>>>interaction picture. This is no more complicated than what's done in
>>>Section 5.5 of Sakurai.
>>
>>I don't have Sakurai book. Do you want me to write smthng like
>>Maxwell's equations and Klein-Gordon equations in the external
>>potential?
>
>
> No. Let's take this a bit slower, because I still haven't seen you write
> down the correct Hamiltonian. Consider the Lagrangian density
>
> L_1 = -|Dphi|^2 - m^2 |\phi|^2 - V|\phi|^2 - (1/4) tr(F^2),
>
> where \phi is a complex scalar field, F is the field strength for the
> 4-vector potential A and the covariant derivative is D = @ - ieA. What
> is the Hamiltonian density? Write everything in position space. I want
> to see what your expression is. Feel free to use compact notation as I
> did.
>
> What are the classical equations of motion for the free fields. Solve
> them and expand the fields in eigenmodes (momentum modes for A and
> discrete modes for \phi). Re-express the Hamiltonian in terms of the mode
> coefficients.
>
> Standard reference for all of this is the last chapter of Goldstein.
>
> Hamiltonian first, more later. This has to be done before a proper
> quantized theory is constructed.

Can we do it another way? You write the Hamiltonian and we will discuss
the physics of it. I must admit that I do not understand the procedure
of "field quantization", i.e., the way the Hamiltonian is derived from
the Lagrangian in quantum field theories. That's not because I am lazy
and don't want to read numerous good books written about the subject.
I read enough of those books, and still the entire procedure does not
make sense to me. I admire the final result of
this procedure, e.g., in the case of QED the quantization results in
a relativistically invariant Hamiltonian that (after renormalization)
leads to the most accurate S-matrix. However, there are so many
handwavings and magic spells on the path of quantization, that
I consider it witchcraft. I wouldn't try it myself.

So, I suggest that you write the Hamiltonian and I will trust you that
this is the correct Hamiltonian of the theory. Then we'll discuss
physics stemming from this Hamiltonian. In any event, we cannot discuss
quantum physics until the Hamiltonian is written in an explicit form.
On the other hand, after the Hamiltonian is written, all physics is in
our hands.

Eugene Stefanovich.

Igor Khavkine

Apr6-05, 12:16 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In article <425410FB.90308@synopsys.com>, Eugene Stefanovich wrote:\n> Igor Khavkine wrote:\n\n>> Standard reference for all of this is the last chapter of Goldstein.\n>>\n>> Hamiltonian first, more later. This has to be done before a proper\n>> quantized theory is constructed.\n>\n> Can we do it another way? You write the Hamiltonian and we will discuss\n> the physics of it. I must admit that I do not understand the procedure\n> of "field quantization", i.e., the way the Hamiltonian is derived from\n> the Lagrangian in quantum field theories. That\'s not because I am lazy\n> and don\'t want to read numerous good books written about the subject.\n> I read enough of those books, and still the entire procedure does not\n> make sense to me. I admire the final result of\n> this procedure, e.g., in the case of QED the quantization results in\n> a relativistically invariant Hamiltonian that (after renormalization)\n> leads to the most accurate S-matrix. However, there are so many\n> handwavings and magic spells on the path of quantization, that\n> I consider it witchcraft. I wouldn\'t try it myself.\n\nSo far we have not even gotten to quantization, since you haven\'t even\nwritten down the classical Hamiltonian. Do you also consider Hamiltonian\nmechanics witchcraft? Is writing deriving the simple harmonic oscillator\nHamiltonian from its Lagrangian also a magic spell? I am sorry that you\nlack intuition for some of the steps behind quantization, but that\nshould not stop you from following the algorithm that I am willing to\noutline.\n\n> So, I suggest that you write the Hamiltonian and I will trust you that\n> this is the correct Hamiltonian of the theory. Then we\'ll discuss\n> physics stemming from this Hamiltonian. In any event, we cannot discuss\n> quantum physics until the Hamiltonian is written in an explicit form.\n> On the other hand, after the Hamiltonian is written, all physics is in\n> our hands.\n\nWriting down the classical Hamiltonian and solving the classical\nequations of motion is the first and one of the simplest steps of this\ncalculation. I have no intention of doing this work for you. If you are\nnot willing to take at least this step forward, then we have no more to\ndiscuss.\n\nArnold Neumaier suggested that you take a year to study the state of the\nart in constructive field theories. I suggest you take at least as much\ntime to study the very basics of classical and quantum field theory, and\nmore importantly do the exercises and the calculations outlined in the\nstandard texts instead of throwing your hands up at the horrifying\nphilosophical implications and giving up. Progress is 1 percent\ninspiration, 99 percent perspiration.\n\nIgor\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In article <425410FB.90308@synopsys.com>, Eugene Stefanovich wrote:
> Igor Khavkine wrote:

>> Standard reference for all of this is the last chapter of Goldstein.
>>
>> Hamiltonian first, more later. This has to be done before a proper
>> quantized theory is constructed.
>
> Can we do it another way? You write the Hamiltonian and we will discuss
> the physics of it. I must admit that I do not understand the procedure
> of "field quantization", i.e., the way the Hamiltonian is derived from
> the Lagrangian in quantum field theories. That's not because I am lazy
> and don't want to read numerous good books written about the subject.
> I read enough of those books, and still the entire procedure does not
> make sense to me. I admire the final result of
> this procedure, e.g., in the case of QED the quantization results in
> a relativistically invariant Hamiltonian that (after renormalization)
> leads to the most accurate S-matrix. However, there are so many
> handwavings and magic spells on the path of quantization, that
> I consider it witchcraft. I wouldn't try it myself.

So far we have not even gotten to quantization, since you haven't even
written down the classical Hamiltonian. Do you also consider Hamiltonian
mechanics witchcraft? Is writing deriving the simple harmonic oscillator
Hamiltonian from its Lagrangian also a magic spell? I am sorry that you
lack intuition for some of the steps behind quantization, but that
should not stop you from following the algorithm that I am willing to
outline.

> So, I suggest that you write the Hamiltonian and I will trust you that
> this is the correct Hamiltonian of the theory. Then we'll discuss
> physics stemming from this Hamiltonian. In any event, we cannot discuss
> quantum physics until the Hamiltonian is written in an explicit form.
> On the other hand, after the Hamiltonian is written, all physics is in
> our hands.

Writing down the classical Hamiltonian and solving the classical
equations of motion is the first and one of the simplest steps of this
calculation. I have no intention of doing this work for you. If you are
not willing to take at least this step forward, then we have no more to
discuss.

Arnold Neumaier suggested that you take a year to study the state of the
art in constructive field theories. I suggest you take at least as much
time to study the very basics of classical and quantum field theory, and
more importantly do the exercises and the calculations outlined in the
standard texts instead of throwing your hands up at the horrifying
philosophical implications and giving up. Progress is 1 percent
inspiration, 99 percent perspiration.

Igor

Eugene Stefanovich

Apr6-05, 04:34 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n\n>\n>>>Standard reference for all of this is the last chapter of Goldstein.\n>>>\n>>>Hamiltonian first, more later. This has to be done before a proper\n>>>quantized theory is constructed.\n>>\n>>Can we do it another way? You write the Hamiltonian and we will discuss\n>>the physics of it. I must admit that I do not understand the procedure\n>>of "field quantization", i.e., the way the Hamiltonian is derived from\n>>the Lagrangian in quantum field theories. That\'s not because I am lazy\n>>and don\'t want to read numerous good books written about the subject.\n>>I read enough of those books, and still the entire procedure does not\n>>make sense to me. I admire the final result of\n>>this procedure, e.g., in the case of QED the quantization results in\n>>a relativistically invariant Hamiltonian that (after renormalization)\n>>leads to the most accurate S-matrix. However, there are so many\n>>handwavings and magic spells on the path of quantization, that\n>>I consider it witchcraft. I wouldn\'t try it myself.\n>\n>\n> So far we have not even gotten to quantization, since you haven\'t even\n> written down the classical Hamiltonian. Do you also consider Hamiltonian\n> mechanics witchcraft? Is writing deriving the simple harmonic oscillator\n> Hamiltonian from its Lagrangian also a magic spell? I am sorry that you\n> lack intuition for some of the steps behind quantization, but that\n> should not stop you from following the algorithm that I am willing to\n> outline.\n\nI would prefer to start with the quantum Hamiltonian expressed in\nterms of creation and annihilation operators. If you want to express\nthe Hamiltonian in terms of quantum fields, that is fine too.\nI can always go from fields to particle operators. I just want to avoid\nany reference to classical Lagrangians and Hamiltonians. Otherwise\nI am going to ask you many questions which are not relevant to the topic\nof this discussion. If you agree that having a Hamiltonian operator is\nsufficient for describing the time evolution in quantum mechanics or\nQFT, then let us start with the Hamiltonian. I trust that you can\nderive fine quantum Hamiltonians using your quantization procedures.\nLet us just skip this step. Please write the quantum Hamiltonian, and\nwe\'ll start talking.\n\nIf you are saying that\ncalculations of time evolution in QM (or QFT) requires something in\naddition to the Hamiltonian, then I disagree with you strongly.\n\n>\n>\n>>So, I suggest that you write the Hamiltonian and I will trust you that\n>>this is the correct Hamiltonian of the theory. Then we\'ll discuss\n>>physics stemming from this Hamiltonian. In any event, we cannot discuss\n>>quantum physics until the Hamiltonian is written in an explicit form.\n>>On the other hand, after the Hamiltonian is written, all physics is in\n>>our hands.\n>\n>\n> Writing down the classical Hamiltonian and solving the classical\n> equations of motion is the first and one of the simplest steps of this\n> calculation. I have no intention of doing this work for you. If you are\n> not willing to take at least this step forward, then we have no more to\n> discuss.\n\nAre you saying that we cannot solve quantum dynamical problem without\nfirst doing classical calculation? This sounds very suspicious. I always\nthought that classical theory is just an approximation to the quantum\none. So, let us start with the exact approach from the beginning and\nskip the unnecessary classical steps.\n\n>\n> Arnold Neumaier suggested that you take a year to study the state of the\n> art in constructive field theories. I suggest you take at least as much\n> time to study the very basics of classical and quantum field theory, and\n> more importantly do the exercises and the calculations outlined in the\n> standard texts instead of throwing your hands up at the horrifying\n> philosophical implications and giving up. Progress is 1 percent\n> inspiration, 99 percent perspiration.\n\nThe title of this thread is "Finite time calculation in QED".\nI suggest we take the Hamiltonian of QED (or the theory of charged\nscalar particles) and perform this calculation.\nIf you want to discuss the derivation of the QED Hamiltonian by the\nquantization of Maxwell\'s theory you can open a new thread, but I am not\ngoing to participate there.\n\nEugene Stefanovich.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:

>
>>>Standard reference for all of this is the last chapter of Goldstein.
>>>
>>>Hamiltonian first, more later. This has to be done before a proper
>>>quantized theory is constructed.
>>
>>Can we do it another way? You write the Hamiltonian and we will discuss
>>the physics of it. I must admit that I do not understand the procedure
>>of "field quantization", i.e., the way the Hamiltonian is derived from
>>the Lagrangian in quantum field theories. That's not because I am lazy
>>and don't want to read numerous good books written about the subject.
>>I read enough of those books, and still the entire procedure does not
>>make sense to me. I admire the final result of
>>this procedure, e.g., in the case of QED the quantization results in
>>a relativistically invariant Hamiltonian that (after renormalization)
>>leads to the most accurate S-matrix. However, there are so many
>>handwavings and magic spells on the path of quantization, that
>>I consider it witchcraft. I wouldn't try it myself.
>
>
> So far we have not even gotten to quantization, since you haven't even
> written down the classical Hamiltonian. Do you also consider Hamiltonian
> mechanics witchcraft? Is writing deriving the simple harmonic oscillator
> Hamiltonian from its Lagrangian also a magic spell? I am sorry that you
> lack intuition for some of the steps behind quantization, but that
> should not stop you from following the algorithm that I am willing to
> outline.

I would prefer to start with the quantum Hamiltonian expressed in
terms of creation and annihilation operators. If you want to express
the Hamiltonian in terms of quantum fields, that is fine too.
I can always go from fields to particle operators. I just want to avoid
any reference to classical Lagrangians and Hamiltonians. Otherwise
I am going to ask you many questions which are not relevant to the topic
of this discussion. If you agree that having a Hamiltonian operator is
sufficient for describing the time evolution in quantum mechanics or
QFT, then let us start with the Hamiltonian. I trust that you can
derive fine quantum Hamiltonians using your quantization procedures.
Let us just skip this step. Please write the quantum Hamiltonian, and
we'll start talking.

If you are saying that
calculations of time evolution in QM (or QFT) requires something in
addition to the Hamiltonian, then I disagree with you strongly.

>
>
>>So, I suggest that you write the Hamiltonian and I will trust you that
>>this is the correct Hamiltonian of the theory. Then we'll discuss
>>physics stemming from this Hamiltonian. In any event, we cannot discuss
>>quantum physics until the Hamiltonian is written in an explicit form.
>>On the other hand, after the Hamiltonian is written, all physics is in
>>our hands.
>
>
> Writing down the classical Hamiltonian and solving the classical
> equations of motion is the first and one of the simplest steps of this
> calculation. I have no intention of doing this work for you. If you are
> not willing to take at least this step forward, then we have no more to
> discuss.

Are you saying that we cannot solve quantum dynamical problem without
first doing classical calculation? This sounds very suspicious. I always
thought that classical theory is just an approximation to the quantum
one. So, let us start with the exact approach from the beginning and
skip the unnecessary classical steps.

>
> Arnold Neumaier suggested that you take a year to study the state of the
> art in constructive field theories. I suggest you take at least as much
> time to study the very basics of classical and quantum field theory, and
> more importantly do the exercises and the calculations outlined in the
> standard texts instead of throwing your hands up at the horrifying
> philosophical implications and giving up. Progress is 1 percent
> inspiration, 99 percent perspiration.

The title of this thread is "Finite time calculation in QED".
I suggest we take the Hamiltonian of QED (or the theory of charged
scalar particles) and perform this calculation.
If you want to discuss the derivation of the QED Hamiltonian by the
quantization of Maxwell's theory you can open a new thread, but I am not
going to participate there.

Eugene Stefanovich.

Igor Khavkine

Apr6-05, 08:41 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-06, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> I would prefer to start with the quantum Hamiltonian expressed in\n> terms of creation and annihilation operators. If you want to express\n> the Hamiltonian in terms of quantum fields, that is fine too.\n> I can always go from fields to particle operators. I just want to avoid\n> any reference to classical Lagrangians and Hamiltonians. Otherwise\n> I am going to ask you many questions which are not relevant to the topic\n> of this discussion. If you agree that having a Hamiltonian operator is\n> sufficient for describing the time evolution in quantum mechanics or\n> QFT, then let us start with the Hamiltonian. I trust that you can\n> derive fine quantum Hamiltonians using your quantization procedures.\n> Let us just skip this step. Please write the quantum Hamiltonian, and\n> we\'ll start talking.\n\nUnfortunately, quantum Hamiltonians don\'t grow on trees and they have to\nsatisfy the correspondence principle. They must reproduce the intended\nclassical system in the hbar -> 0 limit. However, by the law of\nconservation of suffering, before we can construct a quantum system that\nreproduces a given classical one, we must know what the solutions of the\nclassical system look like. Quantum systems are more general than\nclassical ones, but one cannot expect to do the general case before\nsolving the simpler one. So far you have not asked any such questions,\nbut I would consider them on topic and try to answer to the best of my\nability.\n\n> If you are saying that\n> calculations of time evolution in QM (or QFT) requires something in\n> addition to the Hamiltonian, then I disagree with you strongly.\n\nNo, we only need the Hamiltonian, but before the right Hamiltonian is\nwritten down, one must solve the classical equations of motion. Just\nlike before constructing a second quantized theory, one must describe\nthe states of the first quantized one. The standard form of the QED\nHamiltonian in the momentum representation is useless here since\nmomentum states are no longer free eigenstates, due to the external\npotential.\n\n> Are you saying that we cannot solve quantum dynamical problem without\n> first doing classical calculation? This sounds very suspicious. I always\n> thought that classical theory is just an approximation to the quantum\n> one. So, let us start with the exact approach from the beginning and\n> skip the unnecessary classical steps.\n\nThe classical steps are quite necessary. As I\'ve already said, without\nthem you don\'t even know which Hamiltonian to work with. Besides, the\nGreen functions of the classical equations of motion will give you the\npropagators that you\'ll use in evaluating Feynman diagrams.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-06, Eugene Stefanovich <eugenev@synopsys.com> wrote:

> I would prefer to start with the quantum Hamiltonian expressed in
> terms of creation and annihilation operators. If you want to express
> the Hamiltonian in terms of quantum fields, that is fine too.
> I can always go from fields to particle operators. I just want to avoid
> any reference to classical Lagrangians and Hamiltonians. Otherwise
> I am going to ask you many questions which are not relevant to the topic
> of this discussion. If you agree that having a Hamiltonian operator is
> sufficient for describing the time evolution in quantum mechanics or
> QFT, then let us start with the Hamiltonian. I trust that you can
> derive fine quantum Hamiltonians using your quantization procedures.
> Let us just skip this step. Please write the quantum Hamiltonian, and
> we'll start talking.

Unfortunately, quantum Hamiltonians don't grow on trees and they have to
satisfy the correspondence principle. They must reproduce the intended
classical system in the \hbar -> limit. However, by the law of
conservation of suffering, before we can construct a quantum system that
reproduces a given classical one, we must know what the solutions of the
classical system look like. Quantum systems are more general than
classical ones, but one cannot expect to do the general case before
solving the simpler one. So far you have not asked any such questions,
but I would consider them on topic and try to answer to the best of my
ability.

> If you are saying that
> calculations of time evolution in QM (or QFT) requires something in
> addition to the Hamiltonian, then I disagree with you strongly.

No, we only need the Hamiltonian, but before the right Hamiltonian is
written down, one must solve the classical equations of motion. Just
like before constructing a second quantized theory, one must describe
the states of the first quantized one. The standard form of the QED
Hamiltonian in the momentum representation is useless here since
momentum states are no longer free eigenstates, due to the external
potential.

> Are you saying that we cannot solve quantum dynamical problem without
> first doing classical calculation? This sounds very suspicious. I always
> thought that classical theory is just an approximation to the quantum
> one. So, let us start with the exact approach from the beginning and
> skip the unnecessary classical steps.

The classical steps are quite necessary. As I've already said, without
them you don't even know which Hamiltonian to work with. Besides, the
Green functions of the classical equations of motion will give you the
propagators that you'll use in evaluating Feynman diagrams.

Igor

Eugene Stefanovich

Apr8-05, 02:21 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote:\n> On 2005-04-06, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>\n>>I would prefer to start with the quantum Hamiltonian expressed in\n>>terms of creation and annihilation operators. If you want to express\n>>the Hamiltonian in terms of quantum fields, that is fine too.\n>>I can always go from fields to particle operators. I just want to avoid\n>>any reference to classical Lagrangians and Hamiltonians. Otherwise\n>>I am going to ask you many questions which are not relevant to the topic\n>>of this discussion. If you agree that having a Hamiltonian operator is\n>>sufficient for describing the time evolution in quantum mechanics or\n>>QFT, then let us start with the Hamiltonian. I trust that you can\n>>derive fine quantum Hamiltonians using your quantization procedures.\n>>Let us just skip this step. Please write the quantum Hamiltonian, and\n>>we\'ll start talking.\n>\n>\n> Unfortunately, quantum Hamiltonians don\'t grow on trees and they have to\n> satisfy the correspondence principle. They must reproduce the intended\n> classical system in the hbar -> 0 limit. However, by the law of\n> conservation of suffering, before we can construct a quantum system that\n> reproduces a given classical one, we must know what the solutions of the\n> classical system look like. Quantum systems are more general than\n> classical ones, but one cannot expect to do the general case before\n> solving the simpler one. So far you have not asked any such questions,\n> but I would consider them on topic and try to answer to the best of my\n> ability.\n>\n>\n>>If you are saying that\n>>calculations of time evolution in QM (or QFT) requires something in\n>>addition to the Hamiltonian, then I disagree with you strongly.\n>\n>\n> No, we only need the Hamiltonian, but before the right Hamiltonian is\n> written down, one must solve the classical equations of motion. Just\n> like before constructing a second quantized theory, one must describe\n> the states of the first quantized one. The standard form of the QED\n> Hamiltonian in the momentum representation is useless here since\n> momentum states are no longer free eigenstates, due to the external\n> potential.\n>\n>\n>>Are you saying that we cannot solve quantum dynamical problem without\n>>first doing classical calculation? This sounds very suspicious. I always\n>>thought that classical theory is just an approximation to the quantum\n>>one. So, let us start with the exact approach from the beginning and\n>>skip the unnecessary classical steps.\n>\n>\n> The classical steps are quite necessary. As I\'ve already said, without\n> them you don\'t even know which Hamiltonian to work with. Besides, the\n> Green functions of the classical equations of motion will give you the\n> propagators that you\'ll use in evaluating Feynman diagrams.\n\nThis thread is not about teaching me how to do canonical quantization.\nIf you insist on doing that, then we both wasting our time.\nI see that you do not want to write down an explicit expression for the\nquantum Hamiltonian of your theory. This just confirms my suspicion\nthat your theory (just as QED) does not have a well-defined Hamiltonian\nH that can be simply inserted in formula exp(iHt) and directly used for\ncalculations of the time evolution.\n\nIf I understand your correcly, there is a lot of extra stuff required in\nyour approach to the time evolution: external potential, classical\nequations of motion, Green functions, propagators, to name a few.\nWe haven\'t started to talk about regulators, counterterms and other fun\nthings yet. I conclude that all these smoke and mirrors are just hiding\nthe simple fact that you are not doing quantum mechanics. You are doing\nsomething else. For me, the rules of quantum mechanics are rigid and\nsimple: there should be a well-defined Hamilton operator in the Hilbert\nspace, and the time evolution of any state |Psi> should be expressed\nby formula exp(iHt)|Psi>. If your theory does not have that, then\nit is not quantum mechanics.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On 2005-04-06, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>
>>I would prefer to start with the quantum Hamiltonian expressed in
>>terms of creation and annihilation operators. If you want to express
>>the Hamiltonian in terms of quantum fields, that is fine too.
>>I can always go from fields to particle operators. I just want to avoid
>>any reference to classical Lagrangians and Hamiltonians. Otherwise
>>I am going to ask you many questions which are not relevant to the topic
>>of this discussion. If you agree that having a Hamiltonian operator is
>>sufficient for describing the time evolution in quantum mechanics or
>>QFT, then let us start with the Hamiltonian. I trust that you can
>>derive fine quantum Hamiltonians using your quantization procedures.
>>Let us just skip this step. Please write the quantum Hamiltonian, and
>>we'll start talking.
>
>
> Unfortunately, quantum Hamiltonians don't grow on trees and they have to
> satisfy the correspondence principle. They must reproduce the intended
> classical system in the \hbar -> limit. However, by the law of
> conservation of suffering, before we can construct a quantum system that
> reproduces a given classical one, we must know what the solutions of the
> classical system look like. Quantum systems are more general than
> classical ones, but one cannot expect to do the general case before
> solving the simpler one. So far you have not asked any such questions,
> but I would consider them on topic and try to answer to the best of my
> ability.
>
>
>>If you are saying that
>>calculations of time evolution in QM (or QFT) requires something in
>>addition to the Hamiltonian, then I disagree with you strongly.
>
>
> No, we only need the Hamiltonian, but before the right Hamiltonian is
> written down, one must solve the classical equations of motion. Just
> like before constructing a second quantized theory, one must describe
> the states of the first quantized one. The standard form of the QED
> Hamiltonian in the momentum representation is useless here since
> momentum states are no longer free eigenstates, due to the external
> potential.
>
>
>>Are you saying that we cannot solve quantum dynamical problem without
>>first doing classical calculation? This sounds very suspicious. I always
>>thought that classical theory is just an approximation to the quantum
>>one. So, let us start with the exact approach from the beginning and
>>skip the unnecessary classical steps.
>
>
> The classical steps are quite necessary. As I've already said, without
> them you don't even know which Hamiltonian to work with. Besides, the
> Green functions of the classical equations of motion will give you the
> propagators that you'll use in evaluating Feynman diagrams.

This thread is not about teaching me how to do canonical quantization.
If you insist on doing that, then we both wasting our time.
I see that you do not want to write down an explicit expression for the
quantum Hamiltonian of your theory. This just confirms my suspicion
that your theory (just as QED) does not have a well-defined Hamiltonian
H that can be simply inserted in formula \exp(iHt) and directly used for
calculations of the time evolution.

If I understand your correcly, there is a lot of extra stuff required in
your approach to the time evolution: external potential, classical
equations of motion, Green functions, propagators, to name a few.
We haven't started to talk about regulators, counterterms and other fun
things yet. I conclude that all these smoke and mirrors are just hiding
the simple fact that you are not doing quantum mechanics. You are doing
something else. For me, the rules of quantum mechanics are rigid and
simple: there should be a well-defined Hamilton operator in the Hilbert
space, and the time evolution of any state |\Psi> should be expressed
by formula \exp(iHt)|\Psi>. If your theory does not have that, then
it is not quantum mechanics.

Eugene Stefanovich.

Arnold Neumaier

Apr9-05, 02:56 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> For me, the rules of quantum mechanics are rigid and\n> simple: there should be a well-defined Hamilton operator in the Hilbert\n> space, and the time evolution of any state |Psi> should be expressed\n> by formula exp(iHt)|Psi>.\n\n\nThis rigidity prevents you from learning.\n\nYou cannot expect to get help in understanding tradition\nif you insist on ignoring it.\n\nPhysics is also a social enterprise. If you are content with having\nthe key to relativistic quantum mechanics in your private world,\nyou may ignore our advice. But if you want the recognition\nof phycisists you need to speak their language, and learn it if\nyou don\'t.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> For me, the rules of quantum mechanics are rigid and
> simple: there should be a well-defined Hamilton operator in the Hilbert
> space, and the time evolution of any state |\Psi> should be expressed
> by formula \exp(iHt)|\Psi>.

This rigidity prevents you from learning.

You cannot expect to get help in understanding tradition
if you insist on ignoring it.

Physics is also a social enterprise. If you are content with having
the key to relativistic quantum mechanics in your private world,
you may ignore our advice. But if you want the recognition
of phycisists you need to speak their language, and learn it if
you don't.

Arnold Neumaier

Igor Khavkine

Apr9-05, 02:56 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-08, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> This thread is not about teaching me how to do canonical quantization.\n> If you insist on doing that, then we both wasting our time.\n> I see that you do not want to write down an explicit expression for the\n> quantum Hamiltonian of your theory. This just confirms my suspicion\n> that your theory (just as QED) does not have a well-defined Hamiltonian\n> H that can be simply inserted in formula exp(iHt) and directly used for\n> calculations of the time evolution.\n\nUnfortunately, your suspicion can only be alleviated by rolling up your\nsleeves and doing some of the calculations you are trying to avoid.\nIn any case, I am sorry for wasting your time and am glad I need no\nlonger be wasting mine. Now, I hope this thread is finally put to rest.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-08, Eugene Stefanovich <eugenev@synopsys.com> wrote:

> This thread is not about teaching me how to do canonical quantization.
> If you insist on doing that, then we both wasting our time.
> I see that you do not want to write down an explicit expression for the
> quantum Hamiltonian of your theory. This just confirms my suspicion
> that your theory (just as QED) does not have a well-defined Hamiltonian
> H that can be simply inserted in formula \exp(iHt) and directly used for
> calculations of the time evolution.

Unfortunately, your suspicion can only be alleviated by rolling up your
sleeves and doing some of the calculations you are trying to avoid.
In any case, I am sorry for wasting your time and am glad I need no
longer be wasting mine. Now, I hope this thread is finally put to rest.

Igor

Eugene Stefanovich

Apr11-05, 09:14 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>For me, the rules of quantum mechanics are rigid and\n>>simple: there should be a well-defined Hamilton operator in the Hilbert\n>>space, and the time evolution of any state |Psi> should be expressed\n>>by formula exp(iHt)|Psi>.\n>\n>\n>\n> This rigidity prevents you from learning.\n\nThere are some physical principles which I am not willing to\nabandon unless I am forced to do that by overwhelming contradicting\nevidence. The principles\nof quantum mechanics are among those fundamental principles.\nThe description of time evolution by the unitary operator exp(iHt)\nis a central principle of quantum mechanics. That\'s where my rigidity\ncomes from.\n\n>\n> You cannot expect to get help in understanding tradition\n> if you insist on ignoring it.\n\nIf you are saying that QFT is beyond laws of quantum mechanics, then\nyour position is much more revolutionary than mine. The whole point\nof my book was to demonstrate that QFT and the laws of quantum mechanics\n(such as the existence of the Hamiltonian and the time evolution\noperator in the Hilbert space) can be happily reconciled. So, I am\nstaying closer to the tradition established by fathers of quantum\nmechanics.\n\nEugene Stefanovich\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>For me, the rules of quantum mechanics are rigid and
>>simple: there should be a well-defined Hamilton operator in the Hilbert
>>space, and the time evolution of any state |\Psi> should be expressed
>>by formula \exp(iHt)|\Psi>.
>
>
>
> This rigidity prevents you from learning.

There are some physical principles which I am not willing to
abandon unless I am forced to do that by overwhelming contradicting
evidence. The principles
of quantum mechanics are among those fundamental principles.
The description of time evolution by the unitary operator \exp(iHt)
is a central principle of quantum mechanics. That's where my rigidity
comes from.

>
> You cannot expect to get help in understanding tradition
> if you insist on ignoring it.

If you are saying that QFT is beyond laws of quantum mechanics, then
your position is much more revolutionary than mine. The whole point
of my book was to demonstrate that QFT and the laws of quantum mechanics
(such as the existence of the Hamiltonian and the time evolution
operator in the Hilbert space) can be happily reconciled. So, I am
staying closer to the tradition established by fathers of quantum
mechanics.

Eugene Stefanovich

Eugene Stefanovich

Apr12-05, 12:14 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote:\n> On 2005-04-08, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>\n>>This thread is not about teaching me how to do canonical quantization.\n>>If you insist on doing that, then we both wasting our time.\n>>I see that you do not want to write down an explicit expression for the\n>>quantum Hamiltonian of your theory. This just confirms my suspicion\n>>that your theory (just as QED) does not have a well-defined Hamiltonian\n>>H that can be simply inserted in formula exp(iHt) and directly used for\n>>calculations of the time evolution.\n>\n>\n> Unfortunately, your suspicion can only be alleviated by rolling up your\n> sleeves and doing some of the calculations you are trying to avoid.\n> In any case, I am sorry for wasting your time and am glad I need no\n> longer be wasting mine. Now, I hope this thread is finally put to rest.\n\nI was trying to avoid the calculations you suggested because, in my\nopinion, they have no relationship to the topic of this thread.\nI know pretty well how the Hamiltonian of QED is derived. This is\ndescribed very clearly in sections 8.1 - 8.4 of Weinberg\'s "The quantum\ntheory of fields" vol. 1. The resulting Hamiltonian is written\nexplicitly\nin eq. (8.4.22) - (8.4.25) of the book. I think that nobody doubts\nthat this is the *correct* Hamiltonian of QED. My suggestion was\nto start discussion directly from this Hamiltonian (or equivalent\nHamiltonian of your scalar particle theory). In my opinion, deriving\nthis Hamiltonian by repeating calculations in sections 8.1 - 8.4 on this\nnewsgroup would be a waste of time. I hope that you agree that the\nresult of this derivation would be not different from (8.4.22) - (8.4.25).\n\nMy point was that this Hamiltonian and even its version with\ncounterterms H^c cannot be considered as generator of time evolution,\nbecause it creates particles out of vacuum and one-particle states.\nI would agree with you that this problem can be solved by redefinition\nof particles. You haven\'t specified how exactly you are going to perform\nsuch transition to physical particles. However, no matter how you do\nthat, your theory must satisfy at least 3 axioms:\n\n1. relativistic invariance,\n2. trivial time evolution of the "physical" vacuum and 1-particle\nstates,\n3. the S-matrix in you theory must be exactly the same as that\ncalculated with the Hamiltonian H^c.\n\nIf you agree that these axioms are satisfied in your theory, then\ncongratulations! Your Hamiltonian H^d of physical particles belongs to\nthe class of admissible Hamiltonians described in section 12.1 of my\nbook. Then we are the on the same page, and we can start discussing\nthe physical implications of H^d.\n\nIf you do not accept the above 3 axioms, then you theory disagrees\nwith experiment, and has no physical relevance.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On 2005-04-08, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>
>>This thread is not about teaching me how to do canonical quantization.
>>If you insist on doing that, then we both wasting our time.
>>I see that you do not want to write down an explicit expression for the
>>quantum Hamiltonian of your theory. This just confirms my suspicion
>>that your theory (just as QED) does not have a well-defined Hamiltonian
>>H that can be simply inserted in formula \exp(iHt) and directly used for
>>calculations of the time evolution.
>
>
> Unfortunately, your suspicion can only be alleviated by rolling up your
> sleeves and doing some of the calculations you are trying to avoid.
> In any case, I am sorry for wasting your time and am glad I need no
> longer be wasting mine. Now, I hope this thread is finally put to rest.

I was trying to avoid the calculations you suggested because, in my
opinion, they have no relationship to the topic of this thread.
I know pretty well how the Hamiltonian of QED is derived. This is
described very clearly in sections 8.1 - 8.4 of Weinberg's "The quantum
theory of fields" vol. 1. The resulting Hamiltonian is written
explicitly
in eq. (8.4.22) - (8.4.25) of the book. I think that nobody doubts
that this is the *correct* Hamiltonian of QED. My suggestion was
to start discussion directly from this Hamiltonian (or equivalent
Hamiltonian of your scalar particle theory). In my opinion, deriving
this Hamiltonian by repeating calculations in sections 8.1 - 8.4 on this
newsgroup would be a waste of time. I hope that you agree that the
result of this derivation would be not different from (8.4.22) - (8.4.25).

My point was that this Hamiltonian and even its version with
counterterms H^c cannot be considered as generator of time evolution,
because it creates particles out of vacuum and one-particle states.
I would agree with you that this problem can be solved by redefinition
of particles. You haven't specified how exactly you are going to perform
such transition to physical particles. However, no matter how you do
that, your theory must satisfy at least 3 axioms:

1. relativistic invariance,
2. trivial time evolution of the "physical" vacuum and 1-particle
states,
3. the S-matrix in you theory must be exactly the same as that
calculated with the Hamiltonian H^c.

If you agree that these axioms are satisfied in your theory, then
congratulations! Your Hamiltonian H^d of physical particles belongs to
the class of admissible Hamiltonians described in section 12.1 of my
book. Then we are the on the same page, and we can start discussing
the physical implications of H^d.

If you do not accept the above 3 axioms, then you theory disagrees
with experiment, and has no physical relevance.

Eugene Stefanovich.

Arnold Neumaier

Apr12-05, 12:15 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Arnold Neumaier wrote:\n>\n>> Eugene Stefanovich wrote:\n>>\n>>> For me, the rules of quantum mechanics are rigid and\n>>> simple: there should be a well-defined Hamilton operator in the Hilbert\n>>> space, and the time evolution of any state |Psi> should be expressed\n>>> by formula exp(iHt)|Psi>.\n>>\n>> This rigidity prevents you from learning.\n>\n> There are some physical principles which I am not willing to\n> abandon unless I am forced to do that by overwhelming contradicting\n> evidence.\n\nI know that you aren\'t willing. But you need to will in order to learn.\n\nYou are like a young man who wants to learn mathematics but\ndoesn\'t want to accept any axioms because he only wants to start\nwith things that are established for sure. He\'ll never get started!\n\nThe rule for those new in a field is that they accept for some time\nthe wisdom of their supervisors and learn as much as possible that way.\nThere is enough time afterwards (or even in parallel) to explore\none\'s own thoughts about a subject. But learning under supervision\n- even things about which one has secret (or public) second thoughts\n- is part of any successful career.\n\n\n> The principles\n> of quantum mechanics are among those fundamental principles.\n> The description of time evolution by the unitary operator exp(iHt)\n> is a central principle of quantum mechanics. That\'s where my rigidity\n> comes from.\n\nThis is not violated in relativistic qunatum field theory;\nso there is no need to be rigid.\n\nI suggested that you learn about the 2D case from Glimm and\nJaffe (I gave references earlier). There you can see everything\nin full rigor, complete with Hamiltonian, renormalization and\neverything a mathematician could desire.\n\nThere are also some such results in 3D, though more technical.\nThe challenge is to do this in 4D. Even a 1 million dollar prize\nis waiting for the prince who turns the thorns into roses.\n\nBut doing it with success and without ideosyncratic speculations\nrequires that one first understood the 2D case.\nAnd this cannot be done in a week or so. That\'s why I had suggested\nthat you learn a year along the lines suggested by tradition.\n\nIf you don\'t learn in that year what I promised you to get out\nof it you can still go back to your ways that no one else\nis willing to follow.\n\nBut I am convinced that it will change your way of\nthinking about QFT in a way that you can\'t see at the moment.\nAnd it will make your insight and energy available to the\nphysics community, while as you work now, it is all lost to\na fight against windmills.\n\n\n>> You cannot expect to get help in understanding tradition\n>> if you insist on ignoring it.\n>\n> If you are saying that QFT is beyond laws of quantum mechanics,\n\nI never even hinted at that. QFT _is_ the quantum mechanics of fields.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:
>
>> Eugene Stefanovich wrote:
>>
>>> For me, the rules of quantum mechanics are rigid and
>>> simple: there should be a well-defined Hamilton operator in the Hilbert
>>> space, and the time evolution of any state |\Psi> should be expressed
>>> by formula \exp(iHt)|\Psi>.
>>
>> This rigidity prevents you from learning.
>
> There are some physical principles which I am not willing to
> abandon unless I am forced to do that by overwhelming contradicting
> evidence.

I know that you aren't willing. But you need to will in order to learn.

You are like a young man who wants to learn mathematics but
doesn't want to accept any axioms because he only wants to start
with things that are established for sure. He'll never get started!

The rule for those new in a field is that they accept for some time
the wisdom of their supervisors and learn as much as possible that way.
There is enough time afterwards (or even in parallel) to explore
one's own thoughts about a subject. But learning under supervision
- even things about which one has secret (or public) second thoughts
- is part of any successful career.

> The principles
> of quantum mechanics are among those fundamental principles.
> The description of time evolution by the unitary operator \exp(iHt)
> is a central principle of quantum mechanics. That's where my rigidity
> comes from.

This is not violated in relativistic qunatum field theory;
so there is no need to be rigid.

I suggested that you learn about the 2D case from Glimm and
Jaffe (I gave references earlier). There you can see everything
in full rigor, complete with Hamiltonian, renormalization and
everything a mathematician could desire.

There are also some such results in 3D, though more technical.
The challenge is to do this in 4D. Even a 1 million dollar prize
is waiting for the prince who turns the thorns into roses.

But doing it with success and without ideosyncratic speculations
requires that one first understood the 2D case.
And this cannot be done in a week or so. That's why I had suggested
that you learn a year along the lines suggested by tradition.

If you don't learn in that year what I promised you to get out
of it you can still go back to your ways that no one else
is willing to follow.

But I am convinced that it will change your way of
thinking about QFT in a way that you can't see at the moment.
And it will make your insight and energy available to the
physics community, while as you work now, it is all lost to
a fight against windmills.

>> You cannot expect to get help in understanding tradition
>> if you insist on ignoring it.
>
> If you are saying that QFT is beyond laws of quantum mechanics,

I never even hinted at that. QFT _is_ the quantum mechanics of fields.

Arnold Neumaier

Eugene Stefanovich

Apr12-05, 02:30 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n\n>>The principles\n>>of quantum mechanics are among those fundamental principles.\n>>The description of time evolution by the unitary operator exp(iHt)\n>>is a central principle of quantum mechanics. That\'s where my rigidity\n>>comes from.\n>\n> This is not violated in relativistic qunatum field theory;\n> so there is no need to be rigid.\n\nI desperately wish to believe you (and Igor), however I remain\nsuspicious until this time. Each time I ask you to write down the\nHamiltonian of QFT and start discussion of the time evolution you\nfind some excuse not to do that.\n\nI have learned from you how uneducated and arrogant I am,\nbut I haven\'t heard any good arguments why interactions in QFT must be\nretarded.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:

>>The principles
>>of quantum mechanics are among those fundamental principles.
>>The description of time evolution by the unitary operator \exp(iHt)
>>is a central principle of quantum mechanics. That's where my rigidity
>>comes from.
>
> This is not violated in relativistic qunatum field theory;
> so there is no need to be rigid.

I desperately wish to believe you (and Igor), however I remain
suspicious until this time. Each time I ask you to write down the
Hamiltonian of QFT and start discussion of the time evolution you
find some excuse not to do that.

I have learned from you how uneducated and arrogant I am,
but I haven't heard any good arguments why interactions in QFT must be
retarded.

Eugene Stefanovich.

Igor Khavkine

Apr13-05, 12:22 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-12, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> I was trying to avoid the calculations you suggested because, in my\n> opinion, they have no relationship to the topic of this thread.\n> I know pretty well how the Hamiltonian of QED is derived. This is\n> described very clearly in sections 8.1 - 8.4 of Weinberg\'s "The quantum\n> theory of fields" vol. 1. The resulting Hamiltonian is written\n> explicitly\n> in eq. (8.4.22) - (8.4.25) of the book. I think that nobody doubts\n> that this is the *correct* Hamiltonian of QED. My suggestion was\n> to start discussion directly from this Hamiltonian (or equivalent\n> Hamiltonian of your scalar particle theory). In my opinion, deriving\n> this Hamiltonian by repeating calculations in sections 8.1 - 8.4 on this\n> newsgroup would be a waste of time. I hope that you agree that the\n> result of this derivation would be not different from (8.4.22) - (8.4.25).\n\nAllow me to disagree. The Hamiltonian will be different. Different\npotential, different Hamiltonian. That\'s exactly why I wanted to derive\nit and at the same time do some work that will turn out to be useful\nlater (namely propagators aka classical Green functions). And without\nthe external localizing potential, it is very hard to prepare\nlocalized *multiparticle* states at finite times. At least I don\'t know\nhow.\n\n> My point was that this Hamiltonian and even its version with\n> counterterms H^c cannot be considered as generator of time evolution,\n> because it creates particles out of vacuum and one-particle states.\n> I would agree with you that this problem can be solved by redefinition\n> of particles. You haven\'t specified how exactly you are going to perform\n> such transition to physical particles.\n\nA free state at t=-oo gets evolved into an eigenstate state of the full\nHamiltonian at t=0. That is how the transition will be performed. What\nwe haven\'t gotten to is how this statement is translated into\nmathematical language. This is the part where you have to write down and\nsolve the equations of motion for operators and states in the\ninteraction picture.\n\n> However, no matter how you do\n> that, your theory must satisfy at least 3 axioms:\n>\n> 1. relativistic invariance,\n\nWhat is required is relativistic covariance. However, there is now a\npreferred frame due to the external potential. Calculations or\nexperiments in different frames need not be the same, there simply needs\nto be a way to translate between them. The motion of the walls of the\nexternal potential will tell you which frame you are in and how to\ntranslate your results into the frame where it is stationary. Consider\nthe walls of the confining potential to be the walls of a space ship.\nThe scalar field is like a ball bouncing inside. If in your frame the\nspace ship is moving, while in mine it is stationary, we don\'t see the\nsame motion of the ball. But we can agree on our observations\nnonetheless by doing an explicit Lorentz transformation or by\ncalculating invariant quantities such as the ball\'s proper time between\nbounces.\n\n> 2. trivial time evolution of the "physical" vacuum and 1-particle\n> states,\n\nIn other words, 1-particle states are stationary of the full\nHamiltonian. Sure they are, we just won\'t calculate what they are. Of\ncourse, such states exist only if the Hamiltonian itself is invariant\nunder time translations. However, we\'ll be dealing with a time-dependent\nperturbation potential. Hence the Hamiltonian will have no stationary\nstates except the physical vacuum, which we also won\'t calculate\nexplicitly.\n\n> 3. the S-matrix in you theory must be exactly the same as that\n> calculated with the Hamiltonian H^c.\n\nNo. Different potential, different Hamiltonian. Different Hamiltonian,\ndifferent S-matrix. We won\'t calculate the S-matrix.\n\n> If you do not accept the above 3 axioms, then you theory disagrees\n> with experiment, and has no physical relevance.\n\nOn the contrary. The model I proposed is based on a simple and\njustifiable physical assumption. The experimental apparatus is not\ndescribed by the theory. Without similar assumptions, most of the\nexamples and problems from any text on quantum mechanics have to be\nthrown out the window as well. If you wish to model the apparatus\ndynamically within your theory, good luck. At least I don\'t know how to\ndo it. However, I\'m in good company with all the people working on the\nquantum measurement problem.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-12, Eugene Stefanovich <eugenev@synopsys.com> wrote:

> I was trying to avoid the calculations you suggested because, in my
> opinion, they have no relationship to the topic of this thread.
> I know pretty well how the Hamiltonian of QED is derived. This is
> described very clearly in sections 8.1 - 8.4 of Weinberg's "The quantum
> theory of fields" vol. 1. The resulting Hamiltonian is written
> explicitly
> in eq. (8.4.22) - (8.4.25) of the book. I think that nobody doubts
> that this is the *correct* Hamiltonian of QED. My suggestion was
> to start discussion directly from this Hamiltonian (or equivalent
> Hamiltonian of your scalar particle theory). In my opinion, deriving
> this Hamiltonian by repeating calculations in sections 8.1 - 8.4 on this
> newsgroup would be a waste of time. I hope that you agree that the
> result of this derivation would be not different from (8.4.22) - (8.4.25).

Allow me to disagree. The Hamiltonian will be different. Different
potential, different Hamiltonian. That's exactly why I wanted to derive
it and at the same time do some work that will turn out to be useful
later (namely propagators aka classical Green functions). And without
the external localizing potential, it is very hard to prepare
localized *multiparticle* states at finite times. At least I don't know
how.

> My point was that this Hamiltonian and even its version with
> counterterms H^c cannot be considered as generator of time evolution,
> because it creates particles out of vacuum and one-particle states.
> I would agree with you that this problem can be solved by redefinition
> of particles. You haven't specified how exactly you are going to perform
> such transition to physical particles.

A free state at t=-oo gets evolved into an eigenstate state of the full
Hamiltonian at t=0. That is how the transition will be performed. What
we haven't gotten to is how this statement is translated into
mathematical language. This is the part where you have to write down and
solve the equations of motion for operators and states in the
interaction picture.

> However, no matter how you do
> that, your theory must satisfy at least 3 axioms:
>
> 1. relativistic invariance,

What is required is relativistic covariance. However, there is now a
preferred frame due to the external potential. Calculations or
experiments in different frames need not be the same, there simply needs
to be a way to translate between them. The motion of the walls of the
external potential will tell you which frame you are in and how to
translate your results into the frame where it is stationary. Consider
the walls of the confining potential to be the walls of a space ship.
The scalar field is like a ball bouncing inside. If in your frame the
space ship is moving, while in mine it is stationary, we don't see the
same motion of the ball. But we can agree on our observations
nonetheless by doing an explicit Lorentz transformation or by
calculating invariant quantities such as the ball's proper time between
bounces.

> 2. trivial time evolution of the "physical" vacuum and 1-particle
> states,

In other words, 1-particle states are stationary of the full
Hamiltonian. Sure they are, we just won't calculate what they are. Of
course, such states exist only if the Hamiltonian itself is invariant
under time translations. However, we'll be dealing with a time-dependent
perturbation potential. Hence the Hamiltonian will have no stationary
states except the physical vacuum, which we also won't calculate
explicitly.

> 3. the S-matrix in you theory must be exactly the same as that
> calculated with the Hamiltonian H^c.

No. Different potential, different Hamiltonian. Different Hamiltonian,
different S-matrix. We won't calculate the S-matrix.

> If you do not accept the above 3 axioms, then you theory disagrees
> with experiment, and has no physical relevance.

On the contrary. The model I proposed is based on a simple and
justifiable physical assumption. The experimental apparatus is not
described by the theory. Without similar assumptions, most of the
examples and problems from any text on quantum mechanics have to be
thrown out the window as well. If you wish to model the apparatus
dynamically within your theory, good luck. At least I don't know how to
do it. However, I'm in good company with all the people working on the
quantum measurement problem.

Igor

Arnold Neumaier

Apr13-05, 12:23 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> I know pretty well how the Hamiltonian of QED is derived. This is\n> described very clearly in sections 8.1 - 8.4 of Weinberg\'s "The quantum\n> theory of fields" vol. 1. The resulting Hamiltonian is written\n> explicitly\n> in eq. (8.4.22) - (8.4.25) of the book. I think that nobody doubts\n> that this is the *correct* Hamiltonian of QED.\n\nNo. The matter part is missing. See (8.4.1-3).\n\nMoreover, it is the correct Hamiltonian of _classical_ electrodynamics\nonly, since counterterms and renormalization are missing.\n\n\n> My point was that this Hamiltonian and even its version with\n> counterterms H^c cannot be considered as generator of time evolution,\n> because it creates particles out of vacuum and one-particle states.\n\n(8.4.1) is a valid _classical_ Hamiltonian, since classically\nnothing is created or annihilated. Creation and annihilation\noperators are irrelevant on the classical level.\n\nOn the quantum level, (8.4.1) is only a meaningless formal\nexpression. It is definitely _not_ the physical QED Hamiltonian!\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> I know pretty well how the Hamiltonian of QED is derived. This is
> described very clearly in sections 8.1 - 8.4 of Weinberg's "The quantum
> theory of fields" vol. 1. The resulting Hamiltonian is written
> explicitly
> in eq. (8.4.22) - (8.4.25) of the book. I think that nobody doubts
> that this is the *correct* Hamiltonian of QED.

No. The matter part is missing. See (8.4.1-3).

Moreover, it is the correct Hamiltonian of _classical_ electrodynamics
only, since counterterms and renormalization are missing.

> My point was that this Hamiltonian and even its version with
> counterterms H^c cannot be considered as generator of time evolution,
> because it creates particles out of vacuum and one-particle states.

(8.4.1) is a valid _classical_ Hamiltonian, since classically
nothing is created or annihilated. Creation and annihilation
operators are irrelevant on the classical level.

On the quantum level, (8.4.1) is only a meaningless formal
expression. It is definitely _not_ the physical QED Hamiltonian!

Arnold Neumaier

Arnold Neumaier

Apr13-05, 12:23 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>>Eugene Stefanovich wrote:\n>>>\n>>>>For me, the rules of quantum mechanics are rigid and\n>>>>simple: there should be a well-defined Hamilton operator in the Hilbert\n>>>>space, and the time evolution of any state |Psi> should be expressed\n>>>>by formula exp(iHt)|Psi>.\n>>>\n>>>This rigidity prevents you from learning.\n>>\n>>There are some physical principles which I am not willing to\n>>abandon unless I am forced to do that by overwhelming contradicting\n>>evidence.\n>\n> I know that you aren\'t willing. But you need to will in order to learn.\n\n\nMore precisely, your rigidity consists in requiring that H must\nnecessarily be given by an explicit expression in terms of creation\nand annihilation operators on a Hilbert space fixed in advance.\nIt isn\'t.\n\nIn any good physical theory involving space and time, there should\nnot only be a well-defined Hamiltonian H but also well-defined\nmomenta, angular momenta and boosts, forming either a projective\nunitary representation of the Galilei group or one of the Poincare\ngroup. Once one has this, one has a well-defined spacetime setting,\nincluding generators for translations in space and time.\nH is simply one of these generators. Therefore, whenever one has\nsuch a unitary representation, one has what you require:\n\'\'a well-defined Hamilton operator in the Hilbert space,\nand the time evolution of any state |Psi> should be\nexpressed by formula exp(iHt)|Psi>.\'\'\nThis holds no matter in which form the representation is defined.\nYour rigidity insists on the instant form, but is is obvious that\nthe point form and the front form also work, and they are frequently\nmore useful than the instant form.\n\nNow QFT is simply the construction of projective unitary\nrepresentations of the Galilei or Poincare group satisfying the\ncluster decomposition principle. See Weinberg\'s Volume 1, Chapter 4.\nOnce one has the representation, one has all that is needed for a\nphysical theory.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:

> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>>Eugene Stefanovich wrote:
>>>
>>>>For me, the rules of quantum mechanics are rigid and
>>>>simple: there should be a well-defined Hamilton operator in the Hilbert
>>>>space, and the time evolution of any state |\Psi> should be expressed
>>>>by formula \exp(iHt)|\Psi>.
>>>
>>>This rigidity prevents you from learning.
>>
>>There are some physical principles which I am not willing to
>>abandon unless I am forced to do that by overwhelming contradicting
>>evidence.
>
> I know that you aren't willing. But you need to will in order to learn.

More precisely, your rigidity consists in requiring that H must
necessarily be given by an explicit expression in terms of creation
and annihilation operators on a Hilbert space fixed in advance.
It isn't.

In any good physical theory involving space and time, there should
not only be a well-defined Hamiltonian H but also well-defined
momenta, angular momenta and boosts, forming either a projective
unitary representation of the Galilei group or one of the Poincare
group. Once one has this, one has a well-defined spacetime setting,
including generators for translations in space and time.
H is simply one of these generators. Therefore, whenever one has
such a unitary representation, one has what you require:
''a well-defined Hamilton operator in the Hilbert space,
and the time evolution of any state |\Psi> should be
expressed by formula \exp(iHt)|\Psi>.''
This holds no matter in which form the representation is defined.
Your rigidity insists on the instant form, but is is obvious that
the point form and the front form also work, and they are frequently
more useful than the instant form.

Now QFT is simply the construction of projective unitary
representations of the Galilei or Poincare group satisfying the
cluster decomposition principle. See Weinberg's Volume 1, Chapter 4.
Once one has the representation, one has all that is needed for a
physical theory.

Arnold Neumaier

Eugene Stefanovich

Apr13-05, 12:23 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n\n>\n>>>You cannot expect to get help in understanding tradition\n>>>if you insist on ignoring it.\n>>\n>>If you are saying that QFT is beyond laws of quantum mechanics,\n>\n>\n> I never even hinted at that. QFT _is_ the quantum mechanics of fields.\n\n\nQuantum mechanics is about particles and their interactions.\nParticles are observed in experiments, not fields.\nEven when experimentalists say that they measured the\nstrength of the electromagnetic field, they actually\nmeasured a force acting on test charges from other (charged)\nparticles. Fields are just convenient mathematical\nconstructions, no more than that.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:

>
>>>You cannot expect to get help in understanding tradition
>>>if you insist on ignoring it.
>>
>>If you are saying that QFT is beyond laws of quantum mechanics,
>
>
> I never even hinted at that. QFT _is_ the quantum mechanics of fields.

Quantum mechanics is about particles and their interactions.
Particles are observed in experiments, not fields.
Even when experimentalists say that they measured the
strength of the electromagnetic field, they actually
measured a force acting on test charges from other (charged)
particles. Fields are just convenient mathematical
constructions, no more than that.

Eugene Stefanovich.

Eugene Stefanovich

Apr13-05, 12:24 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>>\n>>>\n>>>>For me, the rules of quantum mechanics are rigid and\n>>>>simple: there should be a well-defined Hamilton operator in the Hilbert\n>>>>space, and the time evolution of any state |Psi> should be expressed\n>>>>by formula exp(iHt)|Psi>.\n>>>\n>>>This rigidity prevents you from learning.\n>>\n>>There are some physical principles which I am not willing to\n>>abandon unless I am forced to do that by overwhelming contradicting\n>>evidence.\n>\n>\n> I know that you aren\'t willing. But you need to will in order to learn.\n>\n> You are like a young man who wants to learn mathematics but\n> doesn\'t want to accept any axioms because he only wants to start\n> with things that are established for sure. He\'ll never get started!\n>\n> The rule for those new in a field is that they accept for some time\n> the wisdom of their supervisors and learn as much as possible that way.\n> There is enough time afterwards (or even in parallel) to explore\n> one\'s own thoughts about a subject. But learning under supervision\n> - even things about which one has secret (or public) second thoughts\n> - is part of any successful career.\n[...]\n>\n> I suggested that you learn about the 2D case from Glimm and\n> Jaffe (I gave references earlier). There you can see everything\n> in full rigor, complete with Hamiltonian, renormalization and\n> everything a mathematician could desire.\n>\n> There are also some such results in 3D, though more technical.\n> The challenge is to do this in 4D. Even a 1 million dollar prize\n> is waiting for the prince who turns the thorns into roses.\n>\n> But doing it with success and without ideosyncratic speculations\n> requires that one first understood the 2D case.\n> And this cannot be done in a week or so. That\'s why I had suggested\n> that you learn a year along the lines suggested by tradition.\n>\n> If you don\'t learn in that year what I promised you to get out\n> of it you can still go back to your ways that no one else\n> is willing to follow.\n>\n> But I am convinced that it will change your way of\n> thinking about QFT in a way that you can\'t see at the moment.\n> And it will make your insight and energy available to the\n> physics community, while as you work now, it is all lost to\n> a fight against windmills.\n\nCome on! You can do better than patronizing me and evade direct\ndiscussion. I presented to you a complete theory starting\nfrom postulates and ending with new physical predictions.\nI would appreciate if you can demonstrate that my postulates\nare wrong, or that my logic is "ideosyncratic", or that my predictions\ndisagree with experiment. The fact that my approach disagrees\nwith theoretical "tradition" or may insult physics community\ndoes not bother me much. Let\'s talk physics.\n\nIf you insist that I must study Glimm and Jaffe before talking\nwith you, I can accept that.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>>
>>>
>>>>For me, the rules of quantum mechanics are rigid and
>>>>simple: there should be a well-defined Hamilton operator in the Hilbert
>>>>space, and the time evolution of any state |\Psi> should be expressed
>>>>by formula \exp(iHt)|\Psi>.
>>>
>>>This rigidity prevents you from learning.
>>
>>There are some physical principles which I am not willing to
>>abandon unless I am forced to do that by overwhelming contradicting
>>evidence.
>
>
> I know that you aren't willing. But you need to will in order to learn.
>
> You are like a young man who wants to learn mathematics but
> doesn't want to accept any axioms because he only wants to start
> with things that are established for sure. He'll never get started!
>
> The rule for those new in a field is that they accept for some time
> the wisdom of their supervisors and learn as much as possible that way.
> There is enough time afterwards (or even in parallel) to explore
> one's own thoughts about a subject. But learning under supervision
> - even things about which one has secret (or public) second thoughts
> - is part of any successful career.
[...]
>
> I suggested that you learn about the 2D case from Glimm and
> Jaffe (I gave references earlier). There you can see everything
> in full rigor, complete with Hamiltonian, renormalization and
> everything a mathematician could desire.
>
> There are also some such results in 3D, though more technical.
> The challenge is to do this in 4D. Even a 1 million dollar prize
> is waiting for the prince who turns the thorns into roses.
>
> But doing it with success and without ideosyncratic speculations
> requires that one first understood the 2D case.
> And this cannot be done in a week or so. That's why I had suggested
> that you learn a year along the lines suggested by tradition.
>
> If you don't learn in that year what I promised you to get out
> of it you can still go back to your ways that no one else
> is willing to follow.
>
> But I am convinced that it will change your way of
> thinking about QFT in a way that you can't see at the moment.
> And it will make your insight and energy available to the
> physics community, while as you work now, it is all lost to
> a fight against windmills.

Come on! You can do better than patronizing me and evade direct
discussion. I presented to you a complete theory starting
from postulates and ending with new physical predictions.
I would appreciate if you can demonstrate that my postulates
are wrong, or that my logic is "ideosyncratic", or that my predictions
disagree with experiment. The fact that my approach disagrees
with theoretical "tradition" or may insult physics community
does not bother me much. Let's talk physics.

If you insist that I must study Glimm and Jaffe before talking
with you, I can accept that.

Eugene Stefanovich.

Arnold Neumaier

Apr13-05, 09:55 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> > does not bother me much. Let\'s talk physics.\n\nI _do_ talk physics but you don\'t listen but require me to talk\nyour theory. But your alleged complete theory doesn\'t bother\nme much. I spent enough time to understand what you did, and it\nis nothing fundamentally new though it has some interesting\naspects. These would become more understandable if put into\nthe traditional physics context rather than in your personal\nsetting. Therefore I stick to tradition, and do not need to\nargue against your ideosyncratic way of putting things.\n\n\n> If you insist that I must study Glimm and Jaffe before talking\n> with you, I can accept that.\n\nYes. It is essential. In particular Chapter 6 where he constructs the\nHamiltonian not as explicit as you may want it, but fully adequate.\nThe fact that it is not written in the explicit form you produced\ndoesn\'t mean it doesn\'t exist.\n\nIn the other posting you complained that\n\'\'Each time I ask you to write down the Hamiltonian of QFT\nand start discussion of the time evolution you find some excuse\nnot to do that.\'\'\n\nBut I repeatedly did by refering to the literature - you just found\nit not worthwhile to follow it up. Read in detail Chapter 6 of\nGlimm/Jaffe. Or understand that any representation of the Poincare\ngroup gives a Hamiltonian for free. And then review existing QFT\nin this light.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> > does not bother me much. Let's talk physics.

I _do_ talk physics but you don't listen but require me to talk
your theory. But your alleged complete theory doesn't bother
me much. I spent enough time to understand what you did, and it
is nothing fundamentally new though it has some interesting
aspects. These would become more understandable if put into
the traditional physics context rather than in your personal
setting. Therefore I stick to tradition, and do not need to
argue against your ideosyncratic way of putting things.

> If you insist that I must study Glimm and Jaffe before talking
> with you, I can accept that.

Yes. It is essential. In particular Chapter 6 where he constructs the
Hamiltonian not as explicit as you may want it, but fully adequate.
The fact that it is not written in the explicit form you produced
doesn't mean it doesn't exist.

In the other posting you complained that
''Each time I ask you to write down the Hamiltonian of QFT
and start discussion of the time evolution you find some excuse
not to do that.''

But I repeatedly did by refering to the literature - you just found
it not worthwhile to follow it up. Read in detail Chapter 6 of
Glimm/Jaffe. Or understand that any representation of the Poincare
group gives a Hamiltonian for free. And then review existing QFT
in this light.

Arnold Neumaier

Arnold Neumaier

Apr13-05, 09:55 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> Arnold Neumaier wrote:\n>>\n>>I never even hinted at that. QFT _is_ the quantum mechanics of fields.\n>\n> Quantum mechanics is about particles and their interactions.\n\nNot only. It is about quantities defined by operators and states\ndefined by wave functions or density matrices.\nParticles are just a particluar case of the formalism;\nfields are another. Second-quantized nonrelativisitc QM is\nstill quantum mechanics but a quantum mechanics of fields.\nAnd relativistic QFT just substitutes one symmetry group for\nanother, nothing else.\n\n> Particles are observed in experiments, not fields.\n\nWith the same support by the facts one could say that fields\nare observed in experiments, not particles.\nParticles are inferred from measured fields.\n\n\n\n> Even when experimentalists say that they measured the\n> strength of the electromagnetic field, they actually\n> measured a force acting on test charges from other (charged)\n> particles.\n\nAnd a force is just the force field at some point.\nOne cannot avoid fields. Mechanics is continuum mechanics,\nand only in an idealization few-particle mechanics.\n\n\n> Fields are just convenient mathematical\n> constructions, no more than that.\n\nMore likely, particles are just convenient mathematical\nconstructions, no more than that.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> Arnold Neumaier wrote:
>>
>>I never even hinted at that. QFT _is_ the quantum mechanics of fields.
>
> Quantum mechanics is about particles and their interactions.

Not only. It is about quantities defined by operators and states
defined by wave functions or density matrices.
Particles are just a particluar case of the formalism;
fields are another. Second-quantized nonrelativisitc QM is
still quantum mechanics but a quantum mechanics of fields.
And relativistic QFT just substitutes one symmetry group for
another, nothing else.

> Particles are observed in experiments, not fields.

With the same support by the facts one could say that fields
are observed in experiments, not particles.
Particles are inferred from measured fields.

> Even when experimentalists say that they measured the
> strength of the electromagnetic field, they actually
> measured a force acting on test charges from other (charged)
> particles.

And a force is just the force field at some point.
One cannot avoid fields. Mechanics is continuum mechanics,
and only in an idealization few-particle mechanics.

> Fields are just convenient mathematical
> constructions, no more than that.

More likely, particles are just convenient mathematical
constructions, no more than that.

Arnold Neumaier

Eugene Stefanovich

Apr13-05, 12:45 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n\n>\n> More precisely, your rigidity consists in requiring that H must\n> necessarily be given by an explicit expression in terms of creation\n> and annihilation operators on a Hilbert space fixed in advance.\n\nThat\'s exactly right.\n\n> It isn\'t.\n\n??\n\n>\n> In any good physical theory involving space and time, there should\n> not only be a well-defined Hamiltonian H but also well-defined\n> momenta, angular momenta and boosts, forming either a projective\n> unitary representation of the Galilei group or one of the Poincare\n> group. Once one has this, one has a well-defined spacetime setting,\n> including generators for translations in space and time.\n> H is simply one of these generators. Therefore, whenever one has\n> such a unitary representation, one has what you require:\n> \'\'a well-defined Hamilton operator in the Hilbert space,\n> and the time evolution of any state |Psi> should be\n> expressed by formula exp(iHt)|Psi>.\'\'\n> This holds no matter in which form the representation is defined.\n\nSo far I agree with you 100%\n\n> Your rigidity insists on the instant form, but is is obvious that\n> the point form and the front form also work, and they are frequently\n> more useful than the instant form.\n\nHere we start to disagree. It is true that, e.g., in the point form one\ncan write explicitly well-defined operators H, P, J, K. where H and P\nhave interaction terms (dynamical) while J and K keep their free\nparticle form (kinematical). However, I disagree that\nthis is just a matter of convenience which generators are chosen\ndynamical\nand which are kinematical. The difference can be detected in\nexperiments. So far, everything we know about interacting particles\npoints to the instant form as to the most natural choice.\n\nThe idea about equivalence of different forms of dynamics comes from\nscattering theory where, indeed, the same S-matrix can be reproduced in\ndifferent forms. There one can choose any form based on mathematical\nconvenience. Sometimes, point form or front form may be more useful.\nHowever, when you go beyond scattering theory, e.g., consider\nthe explicit time evolution, the non-equivalence of different forms\nbecomes apparent. Using point form or front form you\'ll get into\ncontradiction with experiment. So far nobody has observed any dynamical\neffect of space translations.\n\n\n\n>\n> Now QFT is simply the construction of projective unitary\n> representations of the Galilei or Poincare group satisfying the\n> cluster decomposition principle. See Weinberg\'s Volume 1, Chapter 4.\n> Once one has the representation, one has all that is needed for a\n> physical theory.\n\nThis is exactly right! The only point where we disagree (do we?) is your\nstatement that different forms of dynamics suit equally well for\ndescribing physics. The representation Weinberg is talking about\nis not just ANY representation.\nThere is a unique (among infinite set of possibilities) projective\nunitary representation of the Poincare group in the Fock space that\ndescribes physical reality. This representation has 10 unique generators\nH, P, J, K. These generators can be explicitly expressed as functions of\nparticle creation and annihilation operators. As soon as you fixed\nthe basis of one-, two-, ... n-particle states in the Fock space,\nthe expressions for generators are fixed. There is no free choice at\nall. Any other set of generators will lead to a different theory that\ncan be, in principle, distinguished experimentally from the unique\ncorrect theory.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:

>
> More precisely, your rigidity consists in requiring that H must
> necessarily be given by an explicit expression in terms of creation
> and annihilation operators on a Hilbert space fixed in advance.

That's exactly right.

> It isn't.

??

>
> In any good physical theory involving space and time, there should
> not only be a well-defined Hamiltonian H but also well-defined
> momenta, angular momenta and boosts, forming either a projective
> unitary representation of the Galilei group or one of the Poincare
> group. Once one has this, one has a well-defined spacetime setting,
> including generators for translations in space and time.
> H is simply one of these generators. Therefore, whenever one has
> such a unitary representation, one has what you require:
> ''a well-defined Hamilton operator in the Hilbert space,
> and the time evolution of any state |\Psi> should be
> expressed by formula \exp(iHt)|\Psi>.''
> This holds no matter in which form the representation is defined.

So far I agree with you 100%

> Your rigidity insists on the instant form, but is is obvious that
> the point form and the front form also work, and they are frequently
> more useful than the instant form.

Here we start to disagree. It is true that, e.g., in the point form one
can write explicitly well-defined operators H, P, J, K. where H and P
have interaction terms (dynamical) while J and K keep their free
particle form (kinematical). However, I disagree that
this is just a matter of convenience which generators are chosen
dynamical
and which are kinematical. The difference can be detected in
experiments. So far, everything we know about interacting particles
points to the instant form as to the most natural choice.

The idea about equivalence of different forms of dynamics comes from
scattering theory where, indeed, the same S-matrix can be reproduced in
different forms. There one can choose any form based on mathematical
convenience. Sometimes, point form or front form may be more useful.
However, when you go beyond scattering theory, e.g., consider
the explicit time evolution, the non-equivalence of different forms
becomes apparent. Using point form or front form you'll get into
contradiction with experiment. So far nobody has observed any dynamical
effect of space translations.

>
> Now QFT is simply the construction of projective unitary
> representations of the Galilei or Poincare group satisfying the
> cluster decomposition principle. See Weinberg's Volume 1, Chapter 4.
> Once one has the representation, one has all that is needed for a
> physical theory.

This is exactly right! The only point where we disagree (do we?) is your
statement that different forms of dynamics suit equally well for
describing physics. The representation Weinberg is talking about
is not just ANY representation.
There is a unique (among infinite set of possibilities) projective
unitary representation of the Poincare group in the Fock space that
describes physical reality. This representation has 10 unique generators
H, P, J, K. These generators can be explicitly expressed as functions of
particle creation and annihilation operators. As soon as you fixed
the basis of one-, two-, ... n-particle states in the Fock space,
the expressions for generators are fixed. There is no free choice at
all. Any other set of generators will lead to a different theory that
can be, in principle, distinguished experimentally from the unique
correct theory.

Eugene Stefanovich.

Eugene Stefanovich

Apr13-05, 03:33 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> I spent enough time to understand what you did, and it\n> is nothing fundamentally new though it has some interesting\n> aspects. These would become more understandable if put into\n> the traditional physics context rather than in your personal\n> setting. Therefore I stick to tradition, and do not need to\n> argue against your ideosyncratic way of putting things.\n\nOK, let\'s assume that I haven\'t done anything fundamentally\nnew, and everything I did can be put "into the traditional\nphysics context", as you say. As long as we are concerned with\npurely theoretical\nthings, like Hilbert spaces, Hamiltonians, wavefunctions, etc.\nevrything can be formulated in a variety of equivalent ways.\nI can buy that.\n\nHowever, there is only one correct way to talk about experimental\npredictions. Nature doesn\'t give us any freedom in terms of\nexperimental outcomes. One prediction of my approach is that\nelectromagnetic interactions propagate instantaneously. An experiment\ncan be performed (though it hasn\'t been done yet) that can either\nconfirm or reject this prediction. As I understand, you maintain\nthat "traditional physics" predicts retarded electromagnetic\ninteractions. So, we have a contradiction here. This suggests\none of the following possibilities.\n\n1. My interpretation of the RQD approach is wrong, and correct analysis\nwould result in retarded interactions just as in the standard\napproach.\n\n2. Your interpretation of the standard approach is wrong, and correct\nanalysis would result in instantaneous interactions just as in the\nRQD approach.\n\n3. RQD and the standard approach are fundamentally different, and only\none of them has the chance to withstand the experimental testing.\n\nPick your favorite option, and we\'ll discuss that.\n\n>\n>\n>> If you insist that I must study Glimm and Jaffe before talking\n>> with you, I can accept that.\n>\n>\n> Yes. It is essential. In particular Chapter 6 where he constructs the\n> Hamiltonian not as explicit as you may want it, but fully adequate.\n> The fact that it is not written in the explicit form you produced\n> doesn\'t mean it doesn\'t exist.\n>\n> In the other posting you complained that\n> \'\'Each time I ask you to write down the Hamiltonian of QFT\n> and start discussion of the time evolution you find some excuse\n> not to do that.\'\'\n>\n> But I repeatedly did by refering to the literature - you just found\n> it not worthwhile to follow it up. Read in detail Chapter 6 of\n> Glimm/Jaffe.\n\nOK, I\'ll do that. Though it\'ll take some time because I do not have\naccess to the library any time I want. Based on what I heard from you\nabout this chapter, I am not sure how worthwile this reading would be.\nI am not thrilled to learn how things can be done in 2D, I think\nI understand pretty well the realistic (3+1)D case. I also do not\nknow how useful are Hamiltonians in "implicit" form. I am pretty\nhappy with the explicit expression of the dressed particle Hamiltonian\nin particle operators. I don\'t want to criticize the book before I\nread it, but based on your descriptions, my expectations are rather low.\n\n\n> Or understand that any representation of the Poincare\n> group gives a Hamiltonian for free. And then review existing QFT\n> in this light.\n\nThat\'s the central point of my approach to QFT. Once you constructed\na representation of the Poincare group you get the generators\nH, P, J, K "for free". Inversely, once you constructed 10 operators\nH, P, J, K satisfying Poincare commutation relations, you get\nthe representation "for free". In both cases, you need to pay rather\nhigh price for the initial construction, though it can be done, e.g.,\nin QED.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> I spent enough time to understand what you did, and it
> is nothing fundamentally new though it has some interesting
> aspects. These would become more understandable if put into
> the traditional physics context rather than in your personal
> setting. Therefore I stick to tradition, and do not need to
> argue against your ideosyncratic way of putting things.

OK, let's assume that I haven't done anything fundamentally
new, and everything I did can be put "into the traditional
physics context", as you say. As long as we are concerned with
purely theoretical
things, like Hilbert spaces, Hamiltonians, wavefunctions, etc.
evrything can be formulated in a variety of equivalent ways.
I can buy that.

However, there is only one correct way to talk about experimental
predictions. Nature doesn't give us any freedom in terms of
experimental outcomes. One prediction of my approach is that
electromagnetic interactions propagate instantaneously. An experiment
can be performed (though it hasn't been done yet) that can either
confirm or reject this prediction. As I understand, you maintain
that "traditional physics" predicts retarded electromagnetic
interactions. So, we have a contradiction here. This suggests
one of the following possibilities.

1. My interpretation of the RQD approach is wrong, and correct analysis
would result in retarded interactions just as in the standard
approach.

2. Your interpretation of the standard approach is wrong, and correct
analysis would result in instantaneous interactions just as in the
RQD approach.

3. RQD and the standard approach are fundamentally different, and only
one of them has the chance to withstand the experimental testing.

Pick your favorite option, and we'll discuss that.

>
>
>> If you insist that I must study Glimm and Jaffe before talking
>> with you, I can accept that.
>
>
> Yes. It is essential. In particular Chapter 6 where he constructs the
> Hamiltonian not as explicit as you may want it, but fully adequate.
> The fact that it is not written in the explicit form you produced
> doesn't mean it doesn't exist.
>
> In the other posting you complained that
> ''Each time I ask you to write down the Hamiltonian of QFT
> and start discussion of the time evolution you find some excuse
> not to do that.''
>
> But I repeatedly did by refering to the literature - you just found
> it not worthwhile to follow it up. Read in detail Chapter 6 of
> Glimm/Jaffe.

OK, I'll do that. Though it'll take some time because I do not have
access to the library any time I want. Based on what I heard from you
about this chapter, I am not sure how worthwile this reading would be.
I am not thrilled to learn how things can be done in 2D, I think
I understand pretty well the realistic (3+1)D case. I also do not
know how useful are Hamiltonians in "implicit" form. I am pretty
happy with the explicit expression of the dressed particle Hamiltonian
in particle operators. I don't want to criticize the book before I
read it, but based on your descriptions, my expectations are rather low.

> Or understand that any representation of the Poincare
> group gives a Hamiltonian for free. And then review existing QFT
> in this light.

That's the central point of my approach to QFT. Once you constructed
a representation of the Poincare group you get the generators
H, P, J, K "for free". Inversely, once you constructed 10 operators
H, P, J, K satisfying Poincare commutation relations, you get
the representation "for free". In both cases, you need to pay rather
high price for the initial construction, though it can be done, e.g.,
in QED.

Eugene Stefanovich.

Eugene Stefanovich

Apr14-05, 01:13 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote:\n> On 2005-04-12, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>\n>>I was trying to avoid the calculations you suggested because, in my\n>>opinion, they have no relationship to the topic of this thread.\n>>I know pretty well how the Hamiltonian of QED is derived. This is\n>>described very clearly in sections 8.1 - 8.4 of Weinberg\'s "The quantum\n>>theory of fields" vol. 1. The resulting Hamiltonian is written\n>>explicitly\n>>in eq. (8.4.22) - (8.4.25) of the book. I think that nobody doubts\n>>that this is the *correct* Hamiltonian of QED. My suggestion was\n>>to start discussion directly from this Hamiltonian (or equivalent\n>>Hamiltonian of your scalar particle theory). In my opinion, deriving\n>>this Hamiltonian by repeating calculations in sections 8.1 - 8.4 on this\n>>newsgroup would be a waste of time. I hope that you agree that the\n>>result of this derivation would be not different from (8.4.22) - (8.4.25).\n>\n>\n> Allow me to disagree. The Hamiltonian will be different. Different\n> potential, different Hamiltonian. That\'s exactly why I wanted to derive\n> it and at the same time do some work that will turn out to be useful\n> later (namely propagators aka classical Green functions). And without\n> the external localizing potential, it is very hard to prepare\n> localized *multiparticle* states at finite times. At least I don\'t know\n> how.\n\nI am interested in interactions of free particles. In fact, I am even\nmore interested in the simplest time evolution of one free particle, so\nno localizing potential is needed. I understand that you\'ll have a\ntrouble preparing a single particle state at finite time. That\'s one\nreason why the RQD approach is better. In this approach, a single localized\nparticle at time t=0 is described by the state a^*(x) |0>, where\nx is the vector of position. a^*(x) is a simple Fourier transform of\nusual momentum-space creation operators a^*(p).\n\n>\n>\n>>My point was that this Hamiltonian and even its version with\n>>counterterms H^c cannot be considered as generator of time evolution,\n>>because it creates particles out of vacuum and one-particle states.\n>>I would agree with you that this problem can be solved by redefinition\n>>of particles. You haven\'t specified how exactly you are going to perform\n>>such transition to physical particles.\n>\n>\n> A free state at t=-oo gets evolved into an eigenstate state of the full\n> Hamiltonian at t=0. That is how the transition will be performed.\n\nAre you saying that the state\n\nexp[iH(0 - -oo)] a^*(p) |0>\n\nis the eigenstate of the Hamiltonian H? Why?\nIs it also an eigenstate of H at t/= 0?\n\n> What\n> we haven\'t gotten to is how this statement is translated into\n> mathematical language. This is the part where you have to write down and\n> solve the equations of motion for operators and states in the\n> interaction picture.\n>\n>\n>>However, no matter how you do\n>>that, your theory must satisfy at least 3 axioms:\n>>\n>>1. relativistic invariance,\n>\n>\n> What is required is relativistic covariance.\n\nDefine what is "relativistic covariance". By "relativistic invariance"\nI mean that 10 operators H, P, J, K satisfy Poincare commutation\nrelations.\n\n> However, there is now a\n> preferred frame due to the external potential. Calculations or\n> experiments in different frames need not be the same, there simply needs\n> to be a way to translate between them. The motion of the walls of the\n> external potential will tell you which frame you are in and how to\n> translate your results into the frame where it is stationary. Consider\n> the walls of the confining potential to be the walls of a space ship.\n> The scalar field is like a ball bouncing inside. If in your frame the\n> space ship is moving, while in mine it is stationary, we don\'t see the\n> same motion of the ball. But we can agree on our observations\n> nonetheless by doing an explicit Lorentz transformation or by\n> calculating invariant quantities such as the ball\'s proper time between\n> bounces.\n\nBy introducing external potential you destroy translational invariance\nwhich is a part of the relativistic invariance. Different positions in\nspace (those inside the walls and those outside the walls) are clearly\nnon-equivalent. It seems that we have very different ideas about the\nmeaning of the principle of relativity.\n\n\n\n>>2. trivial time evolution of the "physical" vacuum and 1-particle\n>>states,\n>\n>\n> In other words, 1-particle states are stationary of the full\n> Hamiltonian. Sure they are, we just won\'t calculate what they are. Of\n> course, such states exist only if the Hamiltonian itself is invariant\n> under time translations. However, we\'ll be dealing with a time-dependent\n> perturbation potential. Hence the Hamiltonian will have no stationary\n> states except the physical vacuum, which we also won\'t calculate\n> explicitly.\n\nNo, I want an explicit calculation for a one-physical-particle state.\nI would like to see, for example, the spreading of the particle\'s\nwave packet. You won\'t see it if you place the particle in the\npotential well.\n\n>\n>\n>>3. the S-matrix in you theory must be exactly the same as that\n>> calculated with the Hamiltonian H^c.\n>\n>\n> No. Different potential, different Hamiltonian. Different Hamiltonian,\n> different S-matrix. We won\'t calculate the S-matrix.\n\nI totally disagree. There should be only one Hamiltonian describing\ninteraction of particles in the Fock space. The same Hamiltonian\nshould be used for calculations of the time evolution, bound states,\nS-matrix, and everything else. Otherwise you don\'t have a theory,\nyou have a bunch of cooking book recipes.\n\n>\n>\n>>If you do not accept the above 3 axioms, then you theory disagrees\n>>with experiment, and has no physical relevance.\n>\n>\n> On the contrary. The model I proposed is based on a simple and\n> justifiable physical assumption. The experimental apparatus is not\n> described by the theory. Without similar assumptions, most of the\n> examples and problems from any text on quantum mechanics have to be\n> thrown out the window as well. If you wish to model the apparatus\n> dynamically within your theory, good luck. At least I don\'t know how to\n> do it. However, I\'m in good company with all the people working on the\n> quantum measurement problem.\n\nThen explain me why non-relativistic quantum mechanics doesn\'t have\nsimilar problems. In non-relativistic QM one can easily describe the\ntime evolution of\na single particle (the spreading of the wave packet) without introducing\nany localizing potential and without theoretical modeling of the\nmeasuring apparatus. All that is needed is the Hamiltonian H and\nformula |Psi(t)> = exp(iHt) |Psi(0)>. The many-particle case is not\ndifferent. I say that the relativistic case is not different either.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On 2005-04-12, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>
>>I was trying to avoid the calculations you suggested because, in my
>>opinion, they have no relationship to the topic of this thread.
>>I know pretty well how the Hamiltonian of QED is derived. This is
>>described very clearly in sections 8.1 - 8.4 of Weinberg's "The quantum
>>theory of fields" vol. 1. The resulting Hamiltonian is written
>>explicitly
>>in eq. (8.4.22) - (8.4.25) of the book. I think that nobody doubts
>>that this is the *correct* Hamiltonian of QED. My suggestion was
>>to start discussion directly from this Hamiltonian (or equivalent
>>Hamiltonian of your scalar particle theory). In my opinion, deriving
>>this Hamiltonian by repeating calculations in sections 8.1 - 8.4 on this
>>newsgroup would be a waste of time. I hope that you agree that the
>>result of this derivation would be not different from (8.4.22) - (8.4.25).
>
>
> Allow me to disagree. The Hamiltonian will be different. Different
> potential, different Hamiltonian. That's exactly why I wanted to derive
> it and at the same time do some work that will turn out to be useful
> later (namely propagators aka classical Green functions). And without
> the external localizing potential, it is very hard to prepare
> localized *multiparticle* states at finite times. At least I don't know
> how.

I am interested in interactions of free particles. In fact, I am even
more interested in the simplest time evolution of one free particle, so
no localizing potential is needed. I understand that you'll have a
trouble preparing a single particle state at finite time. That's one
reason why the RQD approach is better. In this approach, a single localized
particle at time t=0 is described by the state a^*(x) |0>, where
x is the vector of position. a^*(x) is a simple Fourier transform of
usual momentum-space creation operators a^*(p).

>
>
>>My point was that this Hamiltonian and even its version with
>>counterterms H^c cannot be considered as generator of time evolution,
>>because it creates particles out of vacuum and one-particle states.
>>I would agree with you that this problem can be solved by redefinition
>>of particles. You haven't specified how exactly you are going to perform
>>such transition to physical particles.
>
>
> A free state at t=-oo gets evolved into an eigenstate state of the full
> Hamiltonian at t=0. That is how the transition will be performed.

Are you saying that the state

\exp[iH(0 - -oo)] a^*(p) |0>

is the eigenstate of the Hamiltonian H? Why?
Is it also an eigenstate of H at t/= ?

> What
> we haven't gotten to is how this statement is translated into
> mathematical language. This is the part where you have to write down and
> solve the equations of motion for operators and states in the
> interaction picture.
>
>
>>However, no matter how you do
>>that, your theory must satisfy at least 3 axioms:
>>
>>1. relativistic invariance,
>
>
> What is required is relativistic covariance.

Define what is "relativistic covariance". By "relativistic invariance"
I mean that 10 operators H, P, J, K satisfy Poincare commutation
relations.

> However, there is now a
> preferred frame due to the external potential. Calculations or
> experiments in different frames need not be the same, there simply needs
> to be a way to translate between them. The motion of the walls of the
> external potential will tell you which frame you are in and how to
> translate your results into the frame where it is stationary. Consider
> the walls of the confining potential to be the walls of a space ship.
> The scalar field is like a ball bouncing inside. If in your frame the
> space ship is moving, while in mine it is stationary, we don't see the
> same motion of the ball. But we can agree on our observations
> nonetheless by doing an explicit Lorentz transformation or by
> calculating invariant quantities such as the ball's proper time between
> bounces.

By introducing external potential you destroy translational invariance
which is a part of the relativistic invariance. Different positions in
space (those inside the walls and those outside the walls) are clearly
non-equivalent. It seems that we have very different ideas about the
meaning of the principle of relativity.

>>2. trivial time evolution of the "physical" vacuum and 1-particle
>>states,
>
>
> In other words, 1-particle states are stationary of the full
> Hamiltonian. Sure they are, we just won't calculate what they are. Of
> course, such states exist only if the Hamiltonian itself is invariant
> under time translations. However, we'll be dealing with a time-dependent
> perturbation potential. Hence the Hamiltonian will have no stationary
> states except the physical vacuum, which we also won't calculate
> explicitly.

No, I want an explicit calculation for a one-physical-particle state.
I would like to see, for example, the spreading of the particle's
wave packet. You won't see it if you place the particle in the
potential well.

>
>
>>3. the S-matrix in you theory must be exactly the same as that
>> calculated with the Hamiltonian H^c.
>
>
> No. Different potential, different Hamiltonian. Different Hamiltonian,
> different S-matrix. We won't calculate the S-matrix.

I totally disagree. There should be only one Hamiltonian describing
interaction of particles in the Fock space. The same Hamiltonian
should be used for calculations of the time evolution, bound states,
S-matrix, and everything else. Otherwise you don't have a theory,
you have a bunch of cooking book recipes.

>
>
>>If you do not accept the above 3 axioms, then you theory disagrees
>>with experiment, and has no physical relevance.
>
>
> On the contrary. The model I proposed is based on a simple and
> justifiable physical assumption. The experimental apparatus is not
> described by the theory. Without similar assumptions, most of the
> examples and problems from any text on quantum mechanics have to be
> thrown out the window as well. If you wish to model the apparatus
> dynamically within your theory, good luck. At least I don't know how to
> do it. However, I'm in good company with all the people working on the
> quantum measurement problem.

Then explain me why non-relativistic quantum mechanics doesn't have
similar problems. In non-relativistic QM one can easily describe the
time evolution of
a single particle (the spreading of the wave packet) without introducing
any localizing potential and without theoretical modeling of the
measuring apparatus. All that is needed is the Hamiltonian H and
formula |\Psi(t)> = \exp(iHt) |\Psi(0)>. The many-particle case is not
different. I say that the relativistic case is not different either.

Eugene Stefanovich.

Arnold Neumaier

Apr14-05, 01:13 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> but I haven\'t heard any good arguments why interactions in QFT must be\n> retarded.\n\nHow can you get the well-established retarded classical limit if you\ndon\'t have it on the quantum level???\n\nShow us how to do it starting from _your_ Hamiltonian by deriving\nthe relativistic hydrodynamic equations, say.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> but I haven't heard any good arguments why interactions in QFT must be
> retarded.

How can you get the well-established retarded classical limit if you
don't have it on the quantum level???

Show us how to do it starting from _your_ Hamiltonian by deriving
the relativistic hydrodynamic equations, say.

Arnold Neumaier

Eugene Stefanovich

Apr14-05, 01:13 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>> Arnold Neumaier wrote:\n>>\n>>>\n>>> I never even hinted at that. QFT _is_ the quantum mechanics of fields.\n>>\n>>\n>> Quantum mechanics is about particles and their interactions.\n>\n>\n> Not only. It is about quantities defined by operators and states\n> defined by wave functions or density matrices.\n> Particles are just a particluar case of the formalism;\n> fields are another. Second-quantized nonrelativisitc QM is\n> still quantum mechanics but a quantum mechanics of fields.\n> And relativistic QFT just substitutes one symmetry group for\n> another, nothing else.\n>\n>> Particles are observed in experiments, not fields.\n>\n>\n> With the same support by the facts one could say that fields\n> are observed in experiments, not particles.\n> Particles are inferred from measured fields.\n>\n>\n>\n>> Even when experimentalists say that they measured the\n>> strength of the electromagnetic field, they actually\n>> measured a force acting on test charges from other (charged)\n>> particles.\n>\n>\n> And a force is just the force field at some point.\n> One cannot avoid fields. Mechanics is continuum mechanics,\n> and only in an idealization few-particle mechanics.\n>\n>\n>> Fields are just convenient mathematical\n>> constructions, no more than that.\n>\n>\n> More likely, particles are just convenient mathematical\n> constructions, no more than that.\n\nOK, I see your point. Our philosophical positions are diametrically\nopposite. I may say that clicks of the Geiger counter or tracks in the\nWilson camera are much easier to comprehend from the particle point\nof view, but I don\'t think this argument will have much impact on\nyou. Let\'s see which standpoint is more suitable for mathematical\ndescription of reality.\n\nI say that the particle and field standpoints\nare almost equally suitable for the description of scattering. However,\nthere are great advantages in using particles rather than fields when\nthe time evolution is studied. In particular, the dressed particle\nHamiltonian H (which is the only way to do time evolution known to me)\ncan be naturally expressed in terms of particle creation and\nannihilation operators. Each term has direct and clear physical meaning.\n\nOf course, you can say that one can express particle operators through\nquantum fields. However, the expression for H in terms of fields\nwill be very cumbersome and will lack any physical transparency.\nThere is nothing wrong in describing the Solar system in terms of\nPtolemy\'s epicycles. This description is just not very suitable for\nphysical analysis, that\'s all. That\'s why it was abandoned.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>> Arnold Neumaier wrote:
>>
>>>
>>> I never even hinted at that. QFT _is_ the quantum mechanics of fields.
>>
>>
>> Quantum mechanics is about particles and their interactions.
>
>
> Not only. It is about quantities defined by operators and states
> defined by wave functions or density matrices.
> Particles are just a particluar case of the formalism;
> fields are another. Second-quantized nonrelativisitc QM is
> still quantum mechanics but a quantum mechanics of fields.
> And relativistic QFT just substitutes one symmetry group for
> another, nothing else.
>
>> Particles are observed in experiments, not fields.
>
>
> With the same support by the facts one could say that fields
> are observed in experiments, not particles.
> Particles are inferred from measured fields.
>
>
>
>> Even when experimentalists say that they measured the
>> strength of the electromagnetic field, they actually
>> measured a force acting on test charges from other (charged)
>> particles.
>
>
> And a force is just the force field at some point.
> One cannot avoid fields. Mechanics is continuum mechanics,
> and only in an idealization few-particle mechanics.
>
>
>> Fields are just convenient mathematical
>> constructions, no more than that.
>
>
> More likely, particles are just convenient mathematical
> constructions, no more than that.

OK, I see your point. Our philosophical positions are diametrically
opposite. I may say that clicks of the Geiger counter or tracks in the
Wilson camera are much easier to comprehend from the particle point
of view, but I don't think this argument will have much impact on
you. Let's see which standpoint is more suitable for mathematical
description of reality.

I say that the particle and field standpoints
are almost equally suitable for the description of scattering. However,
there are great advantages in using particles rather than fields when
the time evolution is studied. In particular, the dressed particle
Hamiltonian H (which is the only way to do time evolution known to me)
can be naturally expressed in terms of particle creation and
annihilation operators. Each term has direct and clear physical meaning.

Of course, you can say that one can express particle operators through
quantum fields. However, the expression for H in terms of fields
will be very cumbersome and will lack any physical transparency.
There is nothing wrong in describing the Solar system in terms of
Ptolemy's epicycles. This description is just not very suitable for
physical analysis, that's all. That's why it was abandoned.

Eugene Stefanovich.

Eugene Stefanovich

Apr14-05, 01:44 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>but I haven\'t heard any good arguments why interactions in QFT must be\n>>retarded.\n>\n>\n> How can you get the well-established retarded classical limit if you\n> don\'t have it on the quantum level???\n\nI would delete the word "well-established" from the above sentence.\n\n>\n> Show us how to do it starting from _your_ Hamiltonian by deriving\n> the relativistic hydrodynamic equations, say.\n\nWhat the heck is "relativistic hydrodynamics"? We are struglling to\nunderstand the dynamics of simple systems of 0, 1, or 2 particles,\nand you are asking me to describe the dynamics of liquids?\nThat\'s too much for me.\n\nThe structure of interaction in RQD Hamiltonian clearly shows that\nthe force acting on any given particle is determined by instantaneous\npositions and velocities of other particles in the system. This\nmeans action-at-a-distance at both quantum and classical levels.\n\nEugene Stefanovich\n\nP.S. By the way, I had a chance to browse the Glimm-Jaffe book.\nIf you excuse me, I would like to stay away from the discussion\nof this book. It has 99% math and 1% physics. Too much math for\nmy modest mental abilities.\n\nI think I know why Glimm-Jaffe and axiomatic QFT, in general,\nhave a trouble to go beyond 2D models. They place too much emphasis on\nfields. The underlying assumption of all their postulates is that\neverything\nin the world is made of fields, and the language of local Lorentz\ninvariant quantum fields is the preferred language of Nature.\n\nWeinberg is right that so far all available examples of\ncluster separable unitary representations of the Poincare group\nin the Fock space were given by the field approach.\nI agree that this approach is a powerful mathematical tool\nwhich allows us to build satisfactory models relatively easily.\nI also agree that, historically, the introduction of fields\nwas rather logical (formal similarity to Maxwell\'s theory, etc.)\nHowever, I disagree that the idea of fields should be cast into\npostulate, and any deviations from this idea called heresy.\nI don\'t see any compelling physical reason to limit ourselves to\nlocal quantum fields. If we can construct a cluster separable\nunitary representation of the Poincare group in the Fock space\nwithout using fields, this theory would be just as acceptable.\nIn my opinion, that\'s the way to go beyond 2D models to physically\nrelevant theories.\nRQD is an example of such a theory. It\'s Hamiltonian has no simple\nexpression in terms of quantum fields and their derivatives.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>but I haven't heard any good arguments why interactions in QFT must be
>>retarded.
>
>
> How can you get the well-established retarded classical limit if you
> don't have it on the quantum level???

I would delete the word "well-established" from the above sentence.

>
> Show us how to do it starting from _your_ Hamiltonian by deriving
> the relativistic hydrodynamic equations, say.

What the heck is "relativistic hydrodynamics"? We are struglling to
understand the dynamics of simple systems of 0, 1, or 2 particles,
and you are asking me to describe the dynamics of liquids?
That's too much for me.

The structure of interaction in RQD Hamiltonian clearly shows that
the force acting on any given particle is determined by instantaneous
positions and velocities of other particles in the system. This
means action-at-a-distance at both quantum and classical levels.

Eugene Stefanovich

P.S. By the way, I had a chance to browse the Glimm-Jaffe book.
If you excuse me, I would like to stay away from the discussion
of this book. It has 99% math and 1% physics. Too much math for
my modest mental abilities.

I think I know why Glimm-Jaffe and axiomatic QFT, in general,
have a trouble to go beyond 2D models. They place too much emphasis on
fields. The underlying assumption of all their postulates is that
everything
in the world is made of fields, and the language of local Lorentz
invariant quantum fields is the preferred language of Nature.

Weinberg is right that so far all available examples of
cluster separable unitary representations of the Poincare group
in the Fock space were given by the field approach.
I agree that this approach is a powerful mathematical tool
which allows us to build satisfactory models relatively easily.
I also agree that, historically, the introduction of fields
was rather logical (formal similarity to Maxwell's theory, etc.)
However, I disagree that the idea of fields should be cast into
postulate, and any deviations from this idea called heresy.
I don't see any compelling physical reason to limit ourselves to
local quantum fields. If we can construct a cluster separable
unitary representation of the Poincare group in the Fock space
without using fields, this theory would be just as acceptable.
In my opinion, that's the way to go beyond 2D models to physically
relevant theories.
RQD is an example of such a theory. It's Hamiltonian has no simple
expression in terms of quantum fields and their derivatives.

Igor Khavkine

Apr14-05, 06:42 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-14, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n>> On 2005-04-12, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n\n> I am interested in interactions of free particles. In fact, I am even\n> more interested in the simplest time evolution of one free particle, so\n> no localizing potential is needed. I understand that you\'ll have a\n> trouble preparing a single particle state at finite time. That\'s one\n> reason why the RQD approach is better. In this approach, a single localized\n> particle at time t=0 is described by the state a^*(x) |0>, where\n> x is the vector of position. a^*(x) is a simple Fourier transform of\n> usual momentum-space creation operators a^*(p).\n\nI started this thread to show you how to calculate the speed of\npropagation of interaction in scalar QED (you claimed to have never seen\non before). For this purpose, calculating explicit expressions for one\nparticle states is not necessary. Once you learn the general technique,\nyou can apply it to whatever other problems interest you as well.\n\n>> A free state at t=-oo gets evolved into an eigenstate state of the full\n>> Hamiltonian at t=0. That is how the transition will be performed.\n>\n> Are you saying that the state\n>\n> exp[iH(0 - -oo)] a^*(p) |0>\n>\n> is the eigenstate of the Hamiltonian H? Why?\n> Is it also an eigenstate of H at t/= 0?\n\nYes. Replace 0 by any finite t and you\'ll get the same answer.\nLook up the Gellman-Low theorem. I\'ve already given references.\nNote that (0 - -oo) is illdefined. You have to take limits to make it\nwell defined.\n\n>>>If you do not accept the above 3 axioms, then you theory disagrees\n>>>with experiment, and has no physical relevance.\n>>\n>>\n>> On the contrary. The model I proposed is based on a simple and\n>> justifiable physical assumption. The experimental apparatus is not\n>> described by the theory. Without similar assumptions, most of the\n>> examples and problems from any text on quantum mechanics have to be\n>> thrown out the window as well. If you wish to model the apparatus\n>> dynamically within your theory, good luck. At least I don\'t know how to\n>> do it. However, I\'m in good company with all the people working on the\n>> quantum measurement problem.\n>\n> Then explain me why non-relativistic quantum mechanics doesn\'t have\n> similar problems. In non-relativistic QM one can easily describe the\n> time evolution of\n> a single particle (the spreading of the wave packet) without introducing\n> any localizing potential and without theoretical modeling of the\n> measuring apparatus. All that is needed is the Hamiltonian H and\n> formula |Psi(t)> = exp(iHt) |Psi(0)>. The many-particle case is not\n> different. I say that the relativistic case is not different either.\n\nYou want me to explain why apples are not oranges. That\'s because they\nare different fruit. Usual quantum mechanics corresponds to finitely\nmany classical degrees of freedom. QFT corresponds to infinitely many\nclassical degrees of freedom. Relativistic invariance has nothing to do\nwith it.\n\nIf you want to criticise the model I propose please phrase it as a\ncritique of the ball inside the space ship model. Is the ball\nHamiltonian translation invariant? Is there or not a preferred phrame\nfor the ball? Can different observers still agree on what\'s going to\nhappen to the ball? How?\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-14, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:
>> On 2005-04-12, Eugene Stefanovich <eugenev@synopsys.com> wrote:

> I am interested in interactions of free particles. In fact, I am even
> more interested in the simplest time evolution of one free particle, so
> no localizing potential is needed. I understand that you'll have a
> trouble preparing a single particle state at finite time. That's one
> reason why the RQD approach is better. In this approach, a single localized
> particle at time t=0 is described by the state a^*(x) |0>, where
> x is the vector of position. a^*(x) is a simple Fourier transform of
> usual momentum-space creation operators a^*(p).

I started this thread to show you how to calculate the speed of
propagation of interaction in scalar QED (you claimed to have never seen
on before). For this purpose, calculating explicit expressions for one
particle states is not necessary. Once you learn the general technique,
you can apply it to whatever other problems interest you as well.

>> A free state at t=-oo gets evolved into an eigenstate state of the full
>> Hamiltonian at t=0. That is how the transition will be performed.
>
> Are you saying that the state
>
> \exp[iH(0 - -oo)] a^*(p) |0>
>
> is the eigenstate of the Hamiltonian H? Why?
> Is it also an eigenstate of H at t/= ?

Yes. Replace by any finite t and you'll get the same answer.
Look up the Gellman-Low theorem. I've already given references.
Note that (0 - -oo) is illdefined. You have to take limits to make it
well defined.

>>>If you do not accept the above 3 axioms, then you theory disagrees
>>>with experiment, and has no physical relevance.
>>
>>
>> On the contrary. The model I proposed is based on a simple and
>> justifiable physical assumption. The experimental apparatus is not
>> described by the theory. Without similar assumptions, most of the
>> examples and problems from any text on quantum mechanics have to be
>> thrown out the window as well. If you wish to model the apparatus
>> dynamically within your theory, good luck. At least I don't know how to
>> do it. However, I'm in good company with all the people working on the
>> quantum measurement problem.
>
> Then explain me why non-relativistic quantum mechanics doesn't have
> similar problems. In non-relativistic QM one can easily describe the
> time evolution of
> a single particle (the spreading of the wave packet) without introducing
> any localizing potential and without theoretical modeling of the
> measuring apparatus. All that is needed is the Hamiltonian H and
> formula |\Psi(t)> = \exp(iHt) |\Psi(0)>. The many-particle case is not
> different. I say that the relativistic case is not different either.

You want me to explain why apples are not oranges. That's because they
are different fruit. Usual quantum mechanics corresponds to finitely
many classical degrees of freedom. QFT corresponds to infinitely many
classical degrees of freedom. Relativistic invariance has nothing to do
with it.

If you want to criticise the model I propose please phrase it as a
critique of the ball inside the space ship model. Is the ball
Hamiltonian translation invariant? Is there or not a preferred phrame
for the ball? Can different observers still agree on what's going to
happen to the ball? How?

Igor

Eugene Stefanovich

Apr15-05, 08:42 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n>\n>\n>>>A free state at t=-oo gets evolved into an eigenstate state of the full\n>>>Hamiltonian at t=0. That is how the transition will be performed.\n>>\n>>Are you saying that the state\n>>\n>>exp[iH(0 - -oo)] a^*(p) |0>\n>>\n>>is the eigenstate of the Hamiltonian H? Why?\n>>Is it also an eigenstate of H at t/= 0?\n>\n>\n> Yes. Replace 0 by any finite t and you\'ll get the same answer.\n> Look up the Gellman-Low theorem. I\'ve already given references.\n> Note that (0 - -oo) is illdefined. You have to take limits to make it\n> well defined.\n\n\nThe Gell-Mann-Low theorem says that if interaction is adiabatically\nturned on, and |Phi> is eigenvector of H_0, then\n\n|Psi> = exp[iH(0 - -oo)] |Phi>\n\nis eigenvector of H with the same\neigenvalue (let us assume for simplicity that the spectra of H_0 and\nH are the same). This works for vacuum and one-particle states.\nSo, it seems that you achieved the desired dressing. Not so fast.\nIt appears that two-, three-, etc. particle states transformed by\nexp[iH(0 - -oo)] also become eigenstates of H. E.g.,\n\nexp[iH(0 - -oo)] a^*(p) a^*(q) |0>\n\nis an eigenstate of H.\nSo, you found a representation that diagonalizes H.\nThis means that there is no scattering anymore.\nThe scattering operator is identically unity S=1.\n\nWhere is the problem? You satisfied the requirement that vacuum and\none-particle states are eigenvalues of H (this is condition (A) in\nsubsection 12.1.2 of my book), but you got nonsense, because\nthere is (at least) one more requirement to be satisfied in a\ncorrect dressed\nparticle theory: the dressing transformation should not disturb the\nscattering operator (condition (C)). I could reproduce your "dressing"\ntransformation in my approach by setting all factors \\zeta_i in\n(12.29) to be zero off the energy shell. However condition (C)\n(and condition (B)) does not allow me to do that. The factors \\zeta_i\nmust be smooth, so they must gradually tend to zero off the energy\nshell.\n\n\n\n>\n>\n>>>>If you do not accept the above 3 axioms, then you theory disagrees\n>>>>with experiment, and has no physical relevance.\n>>>\n>>>\n>>>On the contrary. The model I proposed is based on a simple and\n>>>justifiable physical assumption. The experimental apparatus is not\n>>>described by the theory. Without similar assumptions, most of the\n>>>examples and problems from any text on quantum mechanics have to be\n>>>thrown out the window as well. If you wish to model the apparatus\n>>>dynamically within your theory, good luck. At least I don\'t know how to\n>>>do it. However, I\'m in good company with all the people working on the\n>>>quantum measurement problem.\n>>\n>>Then explain me why non-relativistic quantum mechanics doesn\'t have\n>>similar problems. In non-relativistic QM one can easily describe the\n>> time evolution of\n>>a single particle (the spreading of the wave packet) without introducing\n>>any localizing potential and without theoretical modeling of the\n>>measuring apparatus. All that is needed is the Hamiltonian H and\n>>formula |Psi(t)> = exp(iHt) |Psi(0)>. The many-particle case is not\n>>different. I say that the relativistic case is not different either.\n>\n>\n> You want me to explain why apples are not oranges. That\'s because they\n> are different fruit. Usual quantum mechanics corresponds to finitely\n> many classical degrees of freedom. QFT corresponds to infinitely many\n> classical degrees of freedom. Relativistic invariance has nothing to do\n> with it.\n\nThat\'s unfortunate that QFT requires infinitely meny degrees of freedom\nto describe a single particle. Let me propose another view on the\ndifference between quantum mechanics and QFT. Both theories can be\nformulated in the same Fock space with the number of particles\nvarying from 0 to infinity. In quantum mechanics the number of particles\nis preserved, interactions may have only form a^*a^*aa, a^*a^*a^*aaa,\netc., so all sectors with fixed numbers of particles are\ninvariant wrt to the time evolution and other inertial transformations.\nSo, one is allowed to consider each sector separately. In QFT,\ninteractions may change the number of particles, e.g., interactions like\na^*a^*c^*aa are allowed, so one needs to consider entire Fock space at\nonce.\n\nAll this talk about the infinite number of degrees of freedom\nis the tribute to one particular (field) method to define interactions\nin the Fock space, and to the (wrong) idea that fields are physical\nentities having their own degrees of freedom. In RQD, fields are\njust convenient mathematical constructs, and only real particles have\ndegrees of freedom that count. A system with a finite (and generally\nvariable) number\nof particles has a finite (variable) number of degrees of freedom.\n\n>\n> If you want to criticise the model I propose please phrase it as a\n> critique of the ball inside the space ship model. Is the ball\n> Hamiltonian translation invariant?\n\nIf you consider ball as your physical system, then its Hamiltonian\nis not translationally invariant. If you consider ball + spaceship\nas your physical system, then its Hamiltonian is translationally\ninvariant.\n\n> Is there or not a preferred phrame\n> for the ball?\n\nThe equivalence of two observers (or reference frames) means that\nif you prepare the same system in both frames and perform the same\nmeasurement you get the same result. If your system is the ball,\nthen certainly there is a preferred reference frame. If observer\ninside the spaceship throws the ball it will bounce off the wall.\nIf observer outside the spaceship throws the ball, it will fly\nindefinitely far away. On the other hand, if two observers in different\nparts of space "prepare" the system "spaceship + ball", their\nmeasurements would be identical. So, the principle of relativity\nshould be applied only to isolated systems (like spaceship + ball).\nThe ball in a potential well is not an isolated system.\n\n\n> Can different observers still agree on what\'s going to\n> happen to the ball? How?\n\nOK, we agreed (I hope) to consider the "spaceship + ball" as\nour physical system. The question is how different observers would\ndescribe this system. In particular, how they would see the position of\nthe ball inside the spaceship. Let\'s take one (instantaneous) observer O\nand assume that he finds the ball at point r with velocity v.\n\nConsider 4 other\nobservers: xO (shifted by distance x wrt O), aO (rotated by angle a wrt\nO), tO (translated in time wrt O), vO (moving with velocity w wrt O).\nIt is easy to find the results of measurements by xO and aO,\ne.g. xO will find the ball with observables r-x and v. This means that\nspace translations and rotations are kinematical transformations\nindependent of what is the interaction between the ball and the\nspaceship.\n\nPrediction of the results of measurements by tO is not\nthat simple. We need to know the interaction between the ball and the\nspaceship in order to solve the dynamical equations and find the\nposition of\nthe ball at nonzero t. Time translations are dynamical transformations.\nThe generator of time translations (the Hamiltonian) has additional\nterms depending on the interaction in the system.\n\nThe same is true for boosts. Since there is interaction in the system\n"spaceship + ball", the generator of boosts contains interaction terms,\nand prediction of measurements by observer vO requires solution of\nnon-trivial "boost dynamics" equations. Contrary to Einstein\'s special\nrelativity, this solution would not result in simple linear Lorentz\ntransformations. We will obtain Lorentz transformations only in the\nlimit when the ball-spaceship interaction is weak.\n\nThis is the major point of RQD. If I am asked to describe the essence of\nmy approach in one sentence, I would say "Boosts are dynamical".\nThe rest just follows from this.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
>
>
>>>A free state at t=-oo gets evolved into an eigenstate state of the full
>>>Hamiltonian at t=0. That is how the transition will be performed.
>>
>>Are you saying that the state
>>
>>\exp[iH(0 - -oo)] a^*(p) |0>
>>
>>is the eigenstate of the Hamiltonian H? Why?
>>Is it also an eigenstate of H at t/= ?
>
>
> Yes. Replace by any finite t and you'll get the same answer.
> Look up the Gellman-Low theorem. I've already given references.
> Note that (0 - -oo) is illdefined. You have to take limits to make it
> well defined.

The Gell-Mann-Low theorem says that if interaction is adiabatically
turned on, and |\Phi> is eigenvector of H_0, then

|\Psi> = \exp[iH(0 - -oo)] |\Phi>

is eigenvector of H with the same
eigenvalue (let us assume for simplicity that the spectra of H_0 and
H are the same). This works for vacuum and one-particle states.
So, it seems that you achieved the desired dressing. Not so fast.
It appears that two-, three-, etc. particle states transformed by
\exp[iH(0 - -oo)] also become eigenstates of H. E.g.,

\exp[iH(0 - -oo)] a^*(p) a^*(q) |0>

is an eigenstate of H.
So, you found a representation that diagonalizes H.
This means that there is no scattering anymore.
The scattering operator is identically unity S=1.

Where is the problem? You satisfied the requirement that vacuum and
one-particle states are eigenvalues of H (this is condition (A) in
subsection 12.1.2 of my book), but you got nonsense, because
there is (at least) one more requirement to be satisfied in a
correct dressed
particle theory: the dressing transformation should not disturb the
scattering operator (condition (C)). I could reproduce your "dressing"
transformation in my approach by setting all factors \zeta_i in
(12.29) to be zero off the energy shell. However condition (C)
(and condition (B)) does not allow me to do that. The factors \zeta_i
must be smooth, so they must gradually tend to zero off the energy
shell.

>
>
>>>>If you do not accept the above 3 axioms, then you theory disagrees
>>>>with experiment, and has no physical relevance.
>>>
>>>
>>>On the contrary. The model I proposed is based on a simple and
>>>justifiable physical assumption. The experimental apparatus is not
>>>described by the theory. Without similar assumptions, most of the
>>>examples and problems from any text on quantum mechanics have to be
>>>thrown out the window as well. If you wish to model the apparatus
>>>dynamically within your theory, good luck. At least I don't know how to
>>>do it. However, I'm in good company with all the people working on the
>>>quantum measurement problem.
>>
>>Then explain me why non-relativistic quantum mechanics doesn't have
>>similar problems. In non-relativistic QM one can easily describe the
>> time evolution of
>>a single particle (the spreading of the wave packet) without introducing
>>any localizing potential and without theoretical modeling of the
>>measuring apparatus. All that is needed is the Hamiltonian H and
>>formula |\Psi(t)> = \exp(iHt) |\Psi(0)>. The many-particle case is not
>>different. I say that the relativistic case is not different either.
>
>
> You want me to explain why apples are not oranges. That's because they
> are different fruit. Usual quantum mechanics corresponds to finitely
> many classical degrees of freedom. QFT corresponds to infinitely many
> classical degrees of freedom. Relativistic invariance has nothing to do
> with it.

That's unfortunate that QFT requires infinitely meny degrees of freedom
to describe a single particle. Let me propose another view on the
difference between quantum mechanics and QFT. Both theories can be
formulated in the same Fock space with the number of particles
varying from to infinity. In quantum mechanics the number of particles
is preserved, interactions may have only form a^*a^*aa, a^*a^*a^*aaa,
etc., so all sectors with fixed numbers of particles are
invariant wrt to the time evolution and other inertial transformations.
So, one is allowed to consider each sector separately. In QFT,
interactions may change the number of particles, e.g., interactions like
a^*a^*c^*aa are allowed, so one needs to consider entire Fock space at
once.

All this talk about the infinite number of degrees of freedom
is the tribute to one particular (field) method to define interactions
in the Fock space, and to the (wrong) idea that fields are physical
entities having their own degrees of freedom. In RQD, fields are
just convenient mathematical constructs, and only real particles have
degrees of freedom that count. A system with a finite (and generally
variable) number
of particles has a finite (variable) number of degrees of freedom.

>
> If you want to criticise the model I propose please phrase it as a
> critique of the ball inside the space ship model. Is the ball
> Hamiltonian translation invariant?

If you consider ball as your physical system, then its Hamiltonian
is not translationally invariant. If you consider ball + spaceship
as your physical system, then its Hamiltonian is translationally
invariant.

> Is there or not a preferred phrame
> for the ball?

The equivalence of two observers (or reference frames) means that
if you prepare the same system in both frames and perform the same
measurement you get the same result. If your system is the ball,
then certainly there is a preferred reference frame. If observer
inside the spaceship throws the ball it will bounce off the wall.
If observer outside the spaceship throws the ball, it will fly
indefinitely far away. On the other hand, if two observers in different
parts of space "prepare" the system "spaceship + ball", their
measurements would be identical. So, the principle of relativity
should be applied only to isolated systems (like spaceship + ball).
The ball in a potential well is not an isolated system.

> Can different observers still agree on what's going to
> happen to the ball? How?

OK, we agreed (I hope) to consider the "spaceship + ball" as
our physical system. The question is how different observers would
describe this system. In particular, how they would see the position of
the ball inside the spaceship. Let's take one (instantaneous) observer O
and assume that he finds the ball at point r with velocity v.

Consider 4 other
observers: xO (shifted by distance x wrt O), aO (rotated by angle a wrt
O), tO (translated in time wrt O), vO (moving with velocity w wrt O).
It is easy to find the results of measurements by xO and aO,
e.g. xO will find the ball with observables r-x and v. This means that
space translations and rotations are kinematical transformations
independent of what is the interaction between the ball and the
spaceship.

Prediction of the results of measurements by tO is not
that simple. We need to know the interaction between the ball and the
spaceship in order to solve the dynamical equations and find the
position of
the ball at nonzero t. Time translations are dynamical transformations.
The generator of time translations (the Hamiltonian) has additional
terms depending on the interaction in the system.

The same is true for boosts. Since there is interaction in the system
"spaceship + ball", the generator of boosts contains interaction terms,
and prediction of measurements by observer vO requires solution of
non-trivial "boost dynamics" equations. Contrary to Einstein's special
relativity, this solution would not result in simple linear Lorentz
transformations. We will obtain Lorentz transformations only in the
limit when the ball-spaceship interaction is weak.

This is the major point of RQD. If I am asked to describe the essence of
my approach in one sentence, I would say "Boosts are dynamical".
The rest just follows from this.

Eugene Stefanovich.

Arnold Neumaier

Apr15-05, 08:42 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Arnold Neumaier wrote:\n>>\n>> More precisely, your rigidity consists in requiring that H must\n>> necessarily be given by an explicit expression in terms of creation\n>> and annihilation operators on a Hilbert space fixed in advance.\n>\n> That\'s exactly right.\n>\n>> It isn\'t.\n>\n> ??\n\nThere are implicit ways to define operators and spaces.\nYou are like saying pi isn\'t well-defined by mathematicians because\nthere is no explicit decimal expansion for it. But mathematicians\ndefine it implicitly in terms of zeros of sin(x).\n\nSimularly, they have many options to define operators, not only\nthe single one you chose to accept as physical.\n\n\n>> In any good physical theory involving space and time, there should\n>> not only be a well-defined Hamiltonian H but also well-defined\n>> momenta, angular momenta and boosts, forming either a projective\n>> unitary representation of the Galilei group or one of the Poincare\n>> group. Once one has this, one has a well-defined spacetime setting,\n>> including generators for translations in space and time.\n>> H is simply one of these generators. Therefore, whenever one has\n>> such a unitary representation, one has what you require:\n>> \'\'a well-defined Hamilton operator in the Hilbert space,\n>> and the time evolution of any state |Psi> should be\n>> expressed by formula exp(iHt)|Psi>.\'\'\n>> This holds no matter in which form the representation is defined.\n>\n> So far I agree with you 100%\n>\n>> Your rigidity insists on the instant form, but is is obvious that\n>> the point form and the front form also work, and they are frequently\n>> more useful than the instant form.\n>\n>\n> Here we start to disagree. It is true that, e.g., in the point form one\n> can write explicitly well-defined operators H, P, J, K. where H and P\n> have interaction terms (dynamical) while J and K keep their free\n> particle form (kinematical). However, I disagree that\n> this is just a matter of convenience which generators are chosen\n> dynamical\n> and which are kinematical.\n\nIf you agreed with me that any Poincare representation defines a\nphysical Hamiltonian on a Hilbert space, you must also agree that the\nmathematical form cannot matter. Representation is representation.\n\n\n\n> So far, everything we know about interacting particles\n> points to the instant form as to the most natural choice.\n\nI know many more things than you; and they point to the fact\nthat the form doesn\'t matter. If you insist on your view,\nthere is nothing more to discuss.\n\n\n>> Now QFT is simply the construction of projective unitary\n>> representations of the Galilei or Poincare group satisfying the\n>> cluster decomposition principle. See Weinberg\'s Volume 1, Chapter 4.\n>> Once one has the representation, one has all that is needed for a\n>> physical theory.\n>\n> This is exactly right! The only point where we disagree (do we?) is your\n> statement that different forms of dynamics suit equally well for\n> describing physics. The representation Weinberg is talking about\n> is not just ANY representation.\n> There is a unique (among infinite set of possibilities) projective\n> unitary representation of the Poincare group in the Fock space that\n> describes physical reality. This representation has 10 unique generators\n> H, P, J, K.\n\nAnd this representation can be written in many different, though\nmathematically equivalent forms. This is what you can\'t see,\nbut it is the consensus of the working physicists.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:
>>
>> More precisely, your rigidity consists in requiring that H must
>> necessarily be given by an explicit expression in terms of creation
>> and annihilation operators on a Hilbert space fixed in advance.
>
> That's exactly right.
>
>> It isn't.
>
> ??

There are implicit ways to define operators and spaces.
You are like saying \pi isn't well-defined by mathematicians because
there is no explicit decimal expansion for it. But mathematicians
define it implicitly in terms of zeros of sin(x).

Simularly, they have many options to define operators, not only
the single one you chose to accept as physical.

>> In any good physical theory involving space and time, there should
>> not only be a well-defined Hamiltonian H but also well-defined
>> momenta, angular momenta and boosts, forming either a projective
>> unitary representation of the Galilei group or one of the Poincare
>> group. Once one has this, one has a well-defined spacetime setting,
>> including generators for translations in space and time.
>> H is simply one of these generators. Therefore, whenever one has
>> such a unitary representation, one has what you require:
>> ''a well-defined Hamilton operator in the Hilbert space,
>> and the time evolution of any state |\Psi> should be
>> expressed by formula \exp(iHt)|\Psi>.''
>> This holds no matter in which form the representation is defined.
>
> So far I agree with you 100%
>
>> Your rigidity insists on the instant form, but is is obvious that
>> the point form and the front form also work, and they are frequently
>> more useful than the instant form.
>
>
> Here we start to disagree. It is true that, e.g., in the point form one
> can write explicitly well-defined operators H, P, J, K. where H and P
> have interaction terms (dynamical) while J and K keep their free
> particle form (kinematical). However, I disagree that
> this is just a matter of convenience which generators are chosen
> dynamical
> and which are kinematical.

If you agreed with me that any Poincare representation defines a
physical Hamiltonian on a Hilbert space, you must also agree that the
mathematical form cannot matter. Representation is representation.

> So far, everything we know about interacting particles
> points to the instant form as to the most natural choice.

I know many more things than you; and they point to the fact
that the form doesn't matter. If you insist on your view,
there is nothing more to discuss.

>> Now QFT is simply the construction of projective unitary
>> representations of the Galilei or Poincare group satisfying the
>> cluster decomposition principle. See Weinberg's Volume 1, Chapter 4.
>> Once one has the representation, one has all that is needed for a
>> physical theory.
>
> This is exactly right! The only point where we disagree (do we?) is your
> statement that different forms of dynamics suit equally well for
> describing physics. The representation Weinberg is talking about
> is not just ANY representation.
> There is a unique (among infinite set of possibilities) projective
> unitary representation of the Poincare group in the Fock space that
> describes physical reality. This representation has 10 unique generators
> H, P, J, K.

And this representation can be written in many different, though
mathematically equivalent forms. This is what you can't see,
but it is the consensus of the working physicists.

Arnold Neumaier

Arnold Neumaier

Apr15-05, 08:42 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Arnold Neumaier wrote:\n>\n>> Show us how to do it starting from _your_ Hamiltonian by deriving\n>> the relativistic hydrodynamic equations, say.\n>\n> What the heck is "relativistic hydrodynamics"? We are struglling to\n> understand the dynamics of simple systems of 0, 1, or 2 particles,\n> and you are asking me to describe the dynamics of liquids?\n> That\'s too much for me.\n\nBut not too much for established QED.\n\nIt is a theory of many-particle systems and has a reasonable\nclassical continuum limit which gives relativistic hydrodynamics.\n\nYou have always been claiming that QED has no good dynamics,\nbut that Your theory has.\n\nSo I display your ignorance by confronting you with all the\ndynamical stuff that QED can do and ask you to reproduce it.\nYou are allowed to employ all the kinds of approximations QED\npeople use in their derivations, but you must show that\nyour way of doing it is superior. Otherwise no one will\ncare about your claims.\n\n\n> P.S. By the way, I had a chance to browse the Glimm-Jaffe book.\n> If you excuse me, I would like to stay away from the discussion\n> of this book. It has 99% math and 1% physics. Too much math for\n> my modest mental abilities.\n\nThere is no royal road to the understanding of QFT.\n\nGlimm/Jaffe is worthwhile mathematics, necessary for a\ndeeper understanding of the physics.\n\nFor our present purposes, you can start directly in Chapter 6\nand refer for the math as needed to the appendix to Part 1.\nThe hard proofs of many of the statements are delegated to later\nchapters, and you can skip these, trusting that they know\ntheir math. But Chapter 6 is fairly elementary, and a must read\nfor anyone trying to be revolutionary.\n\nAnd you\'ll see that their book has much more\nphysics than you now imagine.\n\n\n> I think I know why Glimm-Jaffe and axiomatic QFT, in general,\n> have a trouble to go beyond 2D models. They place too much emphasis on\n> fields. The underlying assumption of all their postulates is that\n> everything\n> in the world is made of fields, and the language of local Lorentz\n> invariant quantum fields is the preferred language of Nature.\n\nThis is the established credo of particle physics since around 1975.\nSo it is natural that they assume it. Your negation of this assumption\nis neither wise nor well-founded, and you\'ll find hardly anyone\nfollowing your lead.\n\n\n\n>\n> Weinberg is right that so far all available examples of\n> cluster separable unitary representations of the Poincare group\n> in the Fock space were given by the field approach.\n\nAnd any such representaion is sufficient to give a Hamiltonian\nand hence a valid dynamical picture.\n\n\n> I agree that this approach is a powerful mathematical tool\n> which allows us to build satisfactory models relatively easily.\n> I also agree that, historically, the introduction of fields\n> was rather logical (formal similarity to Maxwell\'s theory, etc.)\n> However, I disagree that the idea of fields should be cast into\n> postulate, and any deviations from this idea called heresy.\n\nI call it not heresy but lack of insight.\n\n\n> I don\'t see any compelling physical reason to limit ourselves to\n> local quantum fields.\n\nThe reasons are given by Weinberg in Chapter 4, and you need this as a\nstarting point since you have no independent way to show cluster\nseparability of your theory. Thus you cannot avoid local quantum\nfields!\n\n\nArnold Neumaier\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:
>
>> Show us how to do it starting from _your_ Hamiltonian by deriving
>> the relativistic hydrodynamic equations, say.
>
> What the heck is "relativistic hydrodynamics"? We are struglling to
> understand the dynamics of simple systems of 0, 1, or 2 particles,
> and you are asking me to describe the dynamics of liquids?
> That's too much for me.

But not too much for established QED.

It is a theory of many-particle systems and has a reasonable
classical continuum limit which gives relativistic hydrodynamics.

You have always been claiming that QED has no good dynamics,
but that Your theory has.

So I display your ignorance by confronting you with all the
dynamical stuff that QED can do and ask you to reproduce it.
You are allowed to employ all the kinds of approximations QED
people use in their derivations, but you must show that
your way of doing it is superior. Otherwise no one will
care about your claims.

> P.S. By the way, I had a chance to browse the Glimm-Jaffe book.
> If you excuse me, I would like to stay away from the discussion
> of this book. It has 99% math and 1% physics. Too much math for
> my modest mental abilities.

There is no royal road to the understanding of QFT.

Glimm/Jaffe is worthwhile mathematics, necessary for a
deeper understanding of the physics.

For our present purposes, you can start directly in Chapter 6
and refer for the math as needed to the appendix to Part 1.
The hard proofs of many of the statements are delegated to later
chapters, and you can skip these, trusting that they know
their math. But Chapter 6 is fairly elementary, and a must read
for anyone trying to be revolutionary.

And you'll see that their book has much more
physics than you now imagine.

> I think I know why Glimm-Jaffe and axiomatic QFT, in general,
> have a trouble to go beyond 2D models. They place too much emphasis on
> fields. The underlying assumption of all their postulates is that
> everything
> in the world is made of fields, and the language of local Lorentz
> invariant quantum fields is the preferred language of Nature.

This is the established credo of particle physics since around 1975.
So it is natural that they assume it. Your negation of this assumption
is neither wise nor well-founded, and you'll find hardly anyone
following your lead.

>
> Weinberg is right that so far all available examples of
> cluster separable unitary representations of the Poincare group
> in the Fock space were given by the field approach.

And any such representaion is sufficient to give a Hamiltonian
and hence a valid dynamical picture.

> I agree that this approach is a powerful mathematical tool
> which allows us to build satisfactory models relatively easily.
> I also agree that, historically, the introduction of fields
> was rather logical (formal similarity to Maxwell's theory, etc.)
> However, I disagree that the idea of fields should be cast into
> postulate, and any deviations from this idea called heresy.

I call it not heresy but lack of insight.

> I don't see any compelling physical reason to limit ourselves to
> local quantum fields.

The reasons are given by Weinberg in Chapter 4, and you need this as a
starting point since you have no independent way to show cluster
separability of your theory. Thus you cannot avoid local quantum
fields!

Arnold Neumaier

Arnold Neumaier

Apr15-05, 08:46 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Arnold Neumaier wrote:\n>\n>> Eugene Stefanovich wrote:\n>>\n>>> I know pretty well how the Hamiltonian of QED is derived. [...]\n\n>> (8.4.1) is a valid _classical_ Hamiltonian, since classically\n>> nothing is created or annihilated. Creation and annihilation\n>> operators are irrelevant on the classical level.\n>>\n>> On the quantum level, (8.4.1) is only a meaningless formal\n>> expression. It is definitely _not_ the physical QED Hamiltonian!\n>\n> Disagreed. Look, for example at (8.4.22). It is expressed through\n> creation and annihilation operators. a^u and j_u can be also\n> expressed through creation and annihilation operators (or through free\n> quantum fields, if you like). That\'s what people do when they write down\n> the quantum Hamiltonian of QED in the Coulomb gauge.\n\nThis doesn\'t matter. What people write downd at first are only\napproximate expressions with the unspoken implication that these\nmust be regularized and afterwards a limit with regulaization\ndependent coupling constants and counterterms added and everything\n(including the wave function) renormalized. Only the resulting limit\ndefines the physical QED Hamiltonian!\n\nQuantization was\n> introduced in section 8.3, and formulas (8.4.22) - (8.4.25) are in\n> the later section 8.4. The quantum Hamiltonian (8.4.22) - (8.4.25)\n> (plus my\n> formula (1) plus necessary counterterms) in spite of trilinear\n> interactions and infinite coefficients can be used nicely\n> to calculate the S-matrix. All infinities cancel out.\n\nAfter renormalization only. This already involves all the\nregularization stuff. You are _not_ allowed to conclude that\nthe ill-defined formal Hamiltonian is the physical one.\n\n> That\'s what\n> Weinberg does later in chapter 8 and in chapter 11. However,\n> I have no idea how somebody can use this Hamiltonian to calculate\n> the time evolution via exp(iHt).\n\nIt cannot, because it isn\'t the right H.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:
>
>> Eugene Stefanovich wrote:
>>
>>> I know pretty well how the Hamiltonian of QED is derived. [...]

>> (8.4.1) is a valid _classical_ Hamiltonian, since classically
>> nothing is created or annihilated. Creation and annihilation
>> operators are irrelevant on the classical level.
>>
>> On the quantum level, (8.4.1) is only a meaningless formal
>> expression. It is definitely _not_ the physical QED Hamiltonian!
>
> Disagreed. Look, for example at (8.4.22). It is expressed through
> creation and annihilation operators. a^u and j_u can be also
> expressed through creation and annihilation operators (or through free
> quantum fields, if you like). That's what people do when they write down
> the quantum Hamiltonian of QED in the Coulomb gauge.

This doesn't matter. What people write downd at first are only
approximate expressions with the unspoken implication that these
must be regularized and afterwards a limit with regulaization
dependent coupling constants and counterterms added and everything
(including the wave function) renormalized. Only the resulting limit
defines the physical QED Hamiltonian!

Quantization was
> introduced in section 8.3, and formulas (8.4.22) - (8.4.25) are in
> the later section 8.4. The quantum Hamiltonian (8.4.22) - (8.4.25)
> (plus my
> formula (1) plus necessary counterterms) in spite of trilinear
> interactions and infinite coefficients can be used nicely
> to calculate the S-matrix. All infinities cancel out.

After renormalization only. This already involves all the
regularization stuff. You are _not_ allowed to conclude that
the ill-defined formal Hamiltonian is the physical one.

> That's what
> Weinberg does later in chapter 8 and in chapter 11. However,
> I have no idea how somebody can use this Hamiltonian to calculate
> the time evolution via \exp(iHt).

It cannot, because it isn't the right H.

Arnold Neumaier

Eugene Stefanovich

Apr15-05, 05:03 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>\n>> Arnold Neumaier wrote:\n>>\n>>> Eugene Stefanovich wrote:\n>>>\n>>>> I know pretty well how the Hamiltonian of QED is derived. [...]\n>>>\n>\n>>> (8.4.1) is a valid _classical_ Hamiltonian, since classically\n>>> nothing is created or annihilated. Creation and annihilation\n>>> operators are irrelevant on the classical level.\n>>>\n>>> On the quantum level, (8.4.1) is only a meaningless formal\n>>> expression. It is definitely _not_ the physical QED Hamiltonian!\n>>\n>>\n>> Disagreed. Look, for example at (8.4.22). It is expressed through\n>> creation and annihilation operators. a^u and j_u can be also\n>> expressed through creation and annihilation operators (or through free\n>> quantum fields, if you like). That\'s what people do when they write down\n>> the quantum Hamiltonian of QED in the Coulomb gauge.\n>\n>\n> This doesn\'t matter. What people write downd at first are only\n> approximate expressions with the unspoken implication that these\n> must be regularized and afterwards a limit with regulaization\n> dependent coupling constants and counterterms added and everything\n> (including the wave function) renormalized. Only the resulting limit\n> defines the physical QED Hamiltonian!\n>\n> Quantization was\n>\n>> introduced in section 8.3, and formulas (8.4.22) - (8.4.25) are in\n>> the later section 8.4. The quantum Hamiltonian (8.4.22) - (8.4.25)\n>> (plus my\n>> formula (1) plus necessary counterterms) in spite of trilinear\n>> interactions and infinite coefficients can be used nicely\n>> to calculate the S-matrix. All infinities cancel out.\n>\n>\n> After renormalization only.\n\nRenormalization is nothing more than adding counterterms to the\nHamiltonian (8.4.22) - (8.4.25). These counterterms have the same\noperator structure as the original terms (they are corrections to\nmasses and charges) so the dangerous trilinear terms remain in\nthe QED Hamiltonian after renormalization.\n\n> This already involves all the\n> regularization stuff.\n\nRegularization is nothing more than a temporary detour\nto the imaginary world with cut-off interactions or\nnon-integer dimension. This just simplifies intermediate\nsteps of integral calculations (makes them temporary finite\nand allows to make all necessary cancellations with\nfinite expressions). At the end, the cutoff should be taken to\ninfinity or space-time dimension returned to 4. All terms\nthat cancelled each other in the regularized world, become\n(individually) infinite in this limit. But we (correctly)\nassume that, nevertheless, the cancellations we did in\nthe regularized theory still hold when the cutoff goes to infinity.\n\nRegularization and renormalization allow one to obtain finite\nand accurate S-matrix in the limit. However, in the same limit\nthe counterterms in the Hamiltonian of QED acquire infinite\ncoefficients (bare masses and charges are infinite). Even if we\nclose our eyes at these infinities, the Hamiltonian still contains\ndangerous trilinear terms which make it useless for calculations of the\ntime evolution.\n\n> You are _not_ allowed to conclude that\n> the ill-defined formal Hamiltonian is the physical one.\n[...]\n>> I have no idea how somebody can use this Hamiltonian to calculate\n>> the time evolution via exp(iHt).\n>\n>\n> It cannot, because it isn\'t the right H.\n\nYou are absolutely right. The idea of RQD is to construct the right\nHamiltonian. I call it the dressed particle Hamiltonian and obtain it\nvia unitary transformation of H (not the original H in\n(8.4.22) - (8.4.25), but the Hamiltonian with counterterms).\n\nYou may be right (though I doubt it) that the same is achieved in\nthe traditional approach, but somehow hidden in the jungles of\nregularization and wave function renormalization. Then RQD simply\nmakes explicit this non-trivial fact and gives everybody\na clean and physically transparent Hamiltonian that can be used\nfrom now on for all kinds of physical calculations without\nthe need for regularization and renormalization.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>
>> Arnold Neumaier wrote:
>>
>>> Eugene Stefanovich wrote:
>>>
>>>> I know pretty well how the Hamiltonian of QED is derived. [...]
>>>
>
>>> (8.4.1) is a valid _classical_ Hamiltonian, since classically
>>> nothing is created or annihilated. Creation and annihilation
>>> operators are irrelevant on the classical level.
>>>
>>> On the quantum level, (8.4.1) is only a meaningless formal
>>> expression. It is definitely _not_ the physical QED Hamiltonian!
>>
>>
>> Disagreed. Look, for example at (8.4.22). It is expressed through
>> creation and annihilation operators. a^u and j_u can be also
>> expressed through creation and annihilation operators (or through free
>> quantum fields, if you like). That's what people do when they write down
>> the quantum Hamiltonian of QED in the Coulomb gauge.
>
>
> This doesn't matter. What people write downd at first are only
> approximate expressions with the unspoken implication that these
> must be regularized and afterwards a limit with regulaization
> dependent coupling constants and counterterms added and everything
> (including the wave function) renormalized. Only the resulting limit
> defines the physical QED Hamiltonian!
>
> Quantization was
>
>> introduced in section 8.3, and formulas (8.4.22) - (8.4.25) are in
>> the later section 8.4. The quantum Hamiltonian (8.4.22) - (8.4.25)
>> (plus my
>> formula (1) plus necessary counterterms) in spite of trilinear
>> interactions and infinite coefficients can be used nicely
>> to calculate the S-matrix. All infinities cancel out.
>
>
> After renormalization only.

Renormalization is nothing more than adding counterterms to the
Hamiltonian (8.4.22) - (8.4.25). These counterterms have the same
operator structure as the original terms (they are corrections to
masses and charges) so the dangerous trilinear terms remain in
the QED Hamiltonian after renormalization.

> This already involves all the
> regularization stuff.

Regularization is nothing more than a temporary detour
to the imaginary world with cut-off interactions or
non-integer dimension. This just simplifies intermediate
steps of integral calculations (makes them temporary finite
and allows to make all necessary cancellations with
finite expressions). At the end, the cutoff should be taken to
infinity or space-time dimension returned to 4. All terms
that cancelled each other in the regularized world, become
(individually) infinite in this limit. But we (correctly)
assume that, nevertheless, the cancellations we did in
the regularized theory still hold when the cutoff goes to infinity.

Regularization and renormalization allow one to obtain finite
and accurate S-matrix in the limit. However, in the same limit
the counterterms in the Hamiltonian of QED acquire infinite
coefficients (bare masses and charges are infinite). Even if we
close our eyes at these infinities, the Hamiltonian still contains
dangerous trilinear terms which make it useless for calculations of the
time evolution.

> You are _not_ allowed to conclude that
> the ill-defined formal Hamiltonian is the physical one.
[...]
>> I have no idea how somebody can use this Hamiltonian to calculate
>> the time evolution via \exp(iHt).
>
>
> It cannot, because it isn't the right H.

You are absolutely right. The idea of RQD is to construct the right
Hamiltonian. I call it the dressed particle Hamiltonian and obtain it
via unitary transformation of H (not the original H in
(8.4.22) - (8.4.25), but the Hamiltonian with counterterms).

You may be right (though I doubt it) that the same is achieved in
the traditional approach, but somehow hidden in the jungles of
regularization and wave function renormalization. Then RQD simply
makes explicit this non-trivial fact and gives everybody
a clean and physically transparent Hamiltonian that can be used
from now on for all kinds of physical calculations without
the need for regularization and renormalization.

Eugene Stefanovich.

Arnold Neumaier

Apr16-05, 07:54 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Arnold Neumaier wrote:\n\n>> spent enough time to understand what you did, and it\n>> is nothing fundamentally new though it has some interesting\n>> aspects. These would become more understandable if put into\n>> the traditional physics context rather than in your personal\n>> setting. Therefore I stick to tradition, and do not need to\n>> argue against your ideosyncratic way of putting things.\n>\n> OK, let\'s assume that I haven\'t done anything fundamentally\n> new, and everything I did can be put "into the traditional\n> physics context", as you say. As long as we are concerned with\n> purely theoretical\n> things, like Hilbert spaces, Hamiltonians, wavefunctions, etc.\n> evrything can be formulated in a variety of equivalent ways.\n> I can buy that.\n>\n> However, there is only one correct way to talk about experimental\n> predictions. Nature doesn\'t give us any freedom in terms of\n> experimental outcomes. One prediction of my approach is that\n> electromagnetic interactions propagate instantaneously.\n\nI am not interested in your weird predictions that stem from a\nmisinterpretation of the underlying physics.\n\n\n>> In the other posting you complained that\n>> \'\'Each time I ask you to write down the Hamiltonian of QFT\n>> and start discussion of the time evolution you find some excuse\n>> not to do that.\'\'\n>>\n>> But I repeatedly did by refering to the literature - you just found\n>> it not worthwhile to follow it up. Read in detail Chapter 6 of\n>> Glimm/Jaffe.\n>\n> OK, I\'ll do that. Though it\'ll take some time because I do not have\n> access to the library any time I want. Based on what I heard from you\n> about this chapter, I am not sure how worthwile this reading would be.\n> I am not thrilled to learn how things can be done in 2D, I think\n> I understand pretty well the realistic (3+1)D case.\n\nNo. You completely misunderstand tradition.\n\n> I also do not know how useful are Hamiltonians in "implicit" form.\n\nUseful or not, it shows that the physical Hamiltonian exists\nand is well-defined in standard field theory, and that no\nalterations to tradition is needed to have a good time evolution.\nThis contradicts one of your consistent claims.\n\n\n> I am pretty\n> happy with the explicit expression of the dressed particle Hamiltonian\n> in particle operators. I don\'t want to criticize the book before I\n> read it, but based on your descriptions, my expectations are rather low.\n\nOne can see from the descripteion that the form of the representation\nis completely immaterial. This contradicts another one of your\nconsistent claims.\n\n>> Or understand that any representation of the Poincare\n>> group gives a Hamiltonian for free. And then review existing QFT\n>> in this light.\n>\n> That\'s the central point of my approach to QFT.\n\nThat\'s the central point of _any_ approach to QFT. You are doing\nnothing new here.\n\n> Once you constructed\n> a representation of the Poincare group you get the generators\n> H, P, J, K "for free". Inversely, once you constructed 10 operators\n> H, P, J, K satisfying Poincare commutation relations, you get\n> the representation "for free".\n\nAnd if you construct the representation in the point form or the front\nform, exactly the same applies. And all these forms are equivalent\nin the sense that they can be transformed into each other by unitary\ntransformations. Thus their choice is purely a matter of convenience.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:

>> spent enough time to understand what you did, and it
>> is nothing fundamentally new though it has some interesting
>> aspects. These would become more understandable if put into
>> the traditional physics context rather than in your personal
>> setting. Therefore I stick to tradition, and do not need to
>> argue against your ideosyncratic way of putting things.
>
> OK, let's assume that I haven't done anything fundamentally
> new, and everything I did can be put "into the traditional
> physics context", as you say. As long as we are concerned with
> purely theoretical
> things, like Hilbert spaces, Hamiltonians, wavefunctions, etc.
> evrything can be formulated in a variety of equivalent ways.
> I can buy that.
>
> However, there is only one correct way to talk about experimental
> predictions. Nature doesn't give us any freedom in terms of
> experimental outcomes. One prediction of my approach is that
> electromagnetic interactions propagate instantaneously.

I am not interested in your weird predictions that stem from a
misinterpretation of the underlying physics.

>> In the other posting you complained that
>> ''Each time I ask you to write down the Hamiltonian of QFT
>> and start discussion of the time evolution you find some excuse
>> not to do that.''
>>
>> But I repeatedly did by refering to the literature - you just found
>> it not worthwhile to follow it up. Read in detail Chapter 6 of
>> Glimm/Jaffe.
>
> OK, I'll do that. Though it'll take some time because I do not have
> access to the library any time I want. Based on what I heard from you
> about this chapter, I am not sure how worthwile this reading would be.
> I am not thrilled to learn how things can be done in 2D, I think
> I understand pretty well the realistic (3+1)D case.

No. You completely misunderstand tradition.

> I also do not know how useful are Hamiltonians in "implicit" form.

Useful or not, it shows that the physical Hamiltonian exists
and is well-defined in standard field theory, and that no
alterations to tradition is needed to have a good time evolution.
This contradicts one of your consistent claims.

> I am pretty
> happy with the explicit expression of the dressed particle Hamiltonian
> in particle operators. I don't want to criticize the book before I
> read it, but based on your descriptions, my expectations are rather low.

One can see from the descripteion that the form of the representation
is completely immaterial. This contradicts another one of your
consistent claims.

>> Or understand that any representation of the Poincare
>> group gives a Hamiltonian for free. And then review existing QFT
>> in this light.
>
> That's the central point of my approach to QFT.

That's the central point of _any_ approach to QFT. You are doing
nothing new here.

> Once you constructed
> a representation of the Poincare group you get the generators
> H, P, J, K "for free". Inversely, once you constructed 10 operators
> H, P, J, K satisfying Poincare commutation relations, you get
> the representation "for free".

And if you construct the representation in the point form or the front
form, exactly the same applies. And all these forms are equivalent
in the sense that they can be transformed into each other by unitary
transformations. Thus their choice is purely a matter of convenience.

Arnold Neumaier

Arnold Neumaier

Apr16-05, 07:54 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> Of course, you can say that one can express particle operators through\n> quantum fields. However, the expression for H in terms of fields\n> will be very cumbersome and will lack any physical transparency.\n> There is nothing wrong in describing the Solar system in terms of\n> Ptolemy\'s epicycles.\n\nEpicycles are to ellipses what power series are to implicit equations.\n\nYour expansion is like the epicycles of Ptolemy.\nYou need infinitely many terms to define a quantity that\ncan be defined much easier implicitly.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> Of course, you can say that one can express particle operators through
> quantum fields. However, the expression for H in terms of fields
> will be very cumbersome and will lack any physical transparency.
> There is nothing wrong in describing the Solar system in terms of
> Ptolemy's epicycles.

Epicycles are to ellipses what power series are to implicit equations.

Your expansion is like the epicycles of Ptolemy.
You need infinitely many terms to define a quantity that
can be defined much easier implicitly.

Arnold Neumaier

Igor Khavkine

Apr16-05, 07:55 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-15, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n> The Gell-Mann-Low theorem says that if interaction is adiabatically\n> turned on, and |Phi> is eigenvector of H_0, then\n>\n>|Psi> = exp[iH(0 - -oo)] |Phi>\n>\n> is eigenvector of H with the same\n> eigenvalue (let us assume for simplicity that the spectra of H_0 and\n> H are the same). This works for vacuum and one-particle states.\n> So, it seems that you achieved the desired dressing. Not so fast.\n> It appears that two-, three-, etc. particle states transformed by\n> exp[iH(0 - -oo)] also become eigenstates of H. E.g.,\n>\n> exp[iH(0 - -oo)] a^*(p) a^*(q) |0>\n>\n> is an eigenstate of H.\n> So, you found a representation that diagonalizes H.\n> This means that there is no scattering anymore.\n> The scattering operator is identically unity S=1.\n\nYou misunderstand the theorem. First, you cannot use exp(iHt) as the\ntime evolution operator since H is time dependent (it contains the\nadiabatic switching function). Second, |Phi> and |Psi> will not have the\nsame energy, they\'ll be different because of the interaction. Third, you\ncan\'t make any conclusions about what kind of state |Psi> is just from\nknowing what kind of state |Phi> is. The bare ground state need not\nevolve into the true ground state (superconductivity or electroweak\ntheory). An n-particle state need not evolve into an n-particle state\n(QED, only charge is conserved, not particle number in fact, the\nparticle number will be indeterminate). You have no way of controlling\nlocalization properties of particle states at finite time.\n\nSome states can still be identified by their quantum numbers. For\ninstance, in my proposed model, if at t=-oo a particle of charge +1 is\nplace in each well, then at t=0 we\'ll have an eigenstate of the\ninteracting Hamiltonian with charge +1 in each well, but we can\'t say\nmuch more. However, this is sufficient for the purposes of the proposed\nexperiment.\n\n> That\'s unfortunate that QFT requires infinitely meny degrees of freedom\n> to describe a single particle. Let me propose another view on the\n> difference between quantum mechanics and QFT. Both theories can be\n> formulated in the same Fock space with the number of particles\n> varying from 0 to infinity. In quantum mechanics the number of particles\n> is preserved, interactions may have only form a^*a^*aa, a^*a^*a^*aaa,\n> etc., so all sectors with fixed numbers of particles are\n> invariant wrt to the time evolution and other inertial transformations.\n> So, one is allowed to consider each sector separately. In QFT,\n> interactions may change the number of particles, e.g., interactions like\n> a^*a^*c^*aa are allowed, so one needs to consider entire Fock space at\n> once.\n>\n> All this talk about the infinite number of degrees of freedom\n> is the tribute to one particular (field) method to define interactions\n> in the Fock space, and to the (wrong) idea that fields are physical\n> entities having their own degrees of freedom. In RQD, fields are\n> just convenient mathematical constructs, and only real particles have\n> degrees of freedom that count. A system with a finite (and generally\n> variable) number\n> of particles has a finite (variable) number of degrees of freedom.\n\nQuantum mechanics with indefinite number of identical particles == quantum\nmechanics of a field. Whether or not you like fields, they offer a\nmathematically equivalent formulation of the same theory that you\naccept. Lets leave philosophy to the philosophers and reserve physical\ninterpretation to numbers obtained from measurements and not to\nscribbles on paper.\n\n>> If you want to criticise the model I propose please phrase it as a\n>> critique of the ball inside the space ship model. Is the ball\n>> Hamiltonian translation invariant?\n>\n> If you consider ball as your physical system, then its Hamiltonian\n> is not translationally invariant. If you consider ball + spaceship\n> as your physical system, then its Hamiltonian is translationally\n> invariant.\n\nBingo.\n\n>> Is there or not a preferred phrame\n>> for the ball?\n>\n> The equivalence of two observers (or reference frames) means that\n> if you prepare the same system in both frames and perform the same\n> measurement you get the same result. If your system is the ball,\n> then certainly there is a preferred reference frame. If observer\n> inside the spaceship throws the ball it will bounce off the wall.\n> If observer outside the spaceship throws the ball, it will fly\n> indefinitely far away. On the other hand, if two observers in different\n> parts of space "prepare" the system "spaceship + ball", their\n> measurements would be identical. So, the principle of relativity\n> should be applied only to isolated systems (like spaceship + ball).\n> The ball in a potential well is not an isolated system.\n\nCorrect, except that an observer does not throw the ball, only records\nits position. Moreover, if the analogy with an infinite potential well\nis taken literally, the ball is completely constrained to be inside the\nship, never outside.\n\nThe ball inside the ship is not an isolated system, but that in no way\nprevents the calculation of the ball\'s trajectory.\n\n>> Can different observers still agree on what\'s going to\n>> happen to the ball? How?\n>\n> OK, we agreed (I hope) to consider the "spaceship + ball" as\n> our physical system. [...]\n\nLeave preparation to the preparers and observation to the observers.\nThere is only one spaceship, but some observers see it move and others\ndon\'t. They can all still compare notes and agree on what trajectory the\nball took if the relative motion of the ship is taken into account as\nwell as how they measured time intervals and distances. In short, for\nsimplicity and without loss of generality, we can choose to do\ncalculations only in a frame where the ship is stationary.\n\nDo you need any more justification for the use of an external potential?\nIf not, my offer is still open and you have the following steps to do:\n\n1. Find the stationary states of the free classical system.\n2. Promote the mode coefficients (and their conjugate momenta)\nto operators and obtain a diagonal expression for the Hamiltonian.\n3. Write down the equations of motion for states and operators in the\ninteraction picture. Write down the formal solution. (This does not\nrequire an explicit expression for the Hamiltonian).\n\nMore later, but at least some of the above has to be done first.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-15, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

> The Gell-Mann-Low theorem says that if interaction is adiabatically
> turned on, and |\Phi> is eigenvector of H_0, then
>
>|\Psi> = \exp[iH(0 - -oo)] |\Phi>
>
> is eigenvector of H with the same
> eigenvalue (let us assume for simplicity that the spectra of H_0 and
> H are the same). This works for vacuum and one-particle states.
> So, it seems that you achieved the desired dressing. Not so fast.
> It appears that two-, three-, etc. particle states transformed by
> \exp[iH(0 - -oo)] also become eigenstates of H. E.g.,
>
> \exp[iH(0 - -oo)] a^*(p) a^*(q) |0>
>
> is an eigenstate of H.
> So, you found a representation that diagonalizes H.
> This means that there is no scattering anymore.
> The scattering operator is identically unity S=1.

You misunderstand the theorem. First, you cannot use \exp(iHt) as the
time evolution operator since H is time dependent (it contains the
adiabatic switching function). Second, |\Phi> and |\Psi> will not have the
same energy, they'll be different because of the interaction. Third, you
can't make any conclusions about what kind of state |\Psi> is just from
knowing what kind of state |\Phi> is. The bare ground state need not
evolve into the true ground state (superconductivity or electroweak
theory). An n-particle state need not evolve into an n-particle state
(QED, only charge is conserved, not particle number in fact, the
particle number will be indeterminate). You have no way of controlling
localization properties of particle states at finite time.

Some states can still be identified by their quantum numbers. For
instance, in my proposed model, if at t=-oo a particle of charge +1 is
place in each well, then at t=0 we'll have an eigenstate of the
interacting Hamiltonian with charge +1 in each well, but we can't say
much more. However, this is sufficient for the purposes of the proposed
experiment.

> That's unfortunate that QFT requires infinitely meny degrees of freedom
> to describe a single particle. Let me propose another view on the
> difference between quantum mechanics and QFT. Both theories can be
> formulated in the same Fock space with the number of particles
> varying from to infinity. In quantum mechanics the number of particles
> is preserved, interactions may have only form a^*a^*aa, a^*a^*a^*aaa,
> etc., so all sectors with fixed numbers of particles are
> invariant wrt to the time evolution and other inertial transformations.
> So, one is allowed to consider each sector separately. In QFT,
> interactions may change the number of particles, e.g., interactions like
> a^*a^*c^*aa are allowed, so one needs to consider entire Fock space at
> once.
>
> All this talk about the infinite number of degrees of freedom
> is the tribute to one particular (field) method to define interactions
> in the Fock space, and to the (wrong) idea that fields are physical
> entities having their own degrees of freedom. In RQD, fields are
> just convenient mathematical constructs, and only real particles have
> degrees of freedom that count. A system with a finite (and generally
> variable) number
> of particles has a finite (variable) number of degrees of freedom.

Quantum mechanics with indefinite number of identical particles == quantum
mechanics of a field. Whether or not you like fields, they offer a
mathematically equivalent formulation of the same theory that you
accept. Lets leave philosophy to the philosophers and reserve physical
interpretation to numbers obtained from measurements and not to
scribbles on paper.

>> If you want to criticise the model I propose please phrase it as a
>> critique of the ball inside the space ship model. Is the ball
>> Hamiltonian translation invariant?
>
> If you consider ball as your physical system, then its Hamiltonian
> is not translationally invariant. If you consider ball + spaceship
> as your physical system, then its Hamiltonian is translationally
> invariant.

Bingo.

>> Is there or not a preferred phrame
>> for the ball?
>
> The equivalence of two observers (or reference frames) means that
> if you prepare the same system in both frames and perform the same
> measurement you get the same result. If your system is the ball,
> then certainly there is a preferred reference frame. If observer
> inside the spaceship throws the ball it will bounce off the wall.
> If observer outside the spaceship throws the ball, it will fly
> indefinitely far away. On the other hand, if two observers in different
> parts of space "prepare" the system "spaceship + ball", their
> measurements would be identical. So, the principle of relativity
> should be applied only to isolated systems (like spaceship + ball).
> The ball in a potential well is not an isolated system.

Correct, except that an observer does not throw the ball, only records
its position. Moreover, if the analogy with an infinite potential well
is taken literally, the ball is completely constrained to be inside the
ship, never outside.

The ball inside the ship is not an isolated system, but that in no way
prevents the calculation of the ball's trajectory.

>> Can different observers still agree on what's going to
>> happen to the ball? How?
>
> OK, we agreed (I hope) to consider the "spaceship + ball" as
> our physical system. [...]

Leave preparation to the preparers and observation to the observers.
There is only one spaceship, but some observers see it move and others
don't. They can all still compare notes and agree on what trajectory the
ball took if the relative motion of the ship is taken into account as
well as how they measured time intervals and distances. In short, for
simplicity and without loss of generality, we can choose to do
calculations only in a frame where the ship is stationary.

Do you need any more justification for the use of an external potential?
If not, my offer is still open and you have the following steps to do:

1. Find the stationary states of the free classical system.
2. Promote the mode coefficients (and their conjugate momenta)
to operators and obtain a diagonal expression for the Hamiltonian.
3. Write down the equations of motion for states and operators in the
interaction picture. Write down the formal solution. (This does not
require an explicit expression for the Hamiltonian).

More later, but at least some of the above has to be done first.

Igor

Eugene Stefanovich

Apr17-05, 07:54 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>Of course, you can say that one can express particle operators through\n>>quantum fields. However, the expression for H in terms of fields\n>>will be very cumbersome and will lack any physical transparency.\n>>There is nothing wrong in describing the Solar system in terms of\n>>Ptolemy\'s epicycles.\n>\n>\n> Epicycles are to ellipses what power series are to implicit equations.\n>\n> Your expansion is like the epicycles of Ptolemy.\n> You need infinitely many terms to define a quantity that\n> can be defined much easier implicitly.\n\nI still haven\'t seen how you are going to describe the time evolution\n(e.g., spreading of the wave packet) of a single particle with your\n"implicit" Hamiltonian.\n\nEugene Stefanovich.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>Of course, you can say that one can express particle operators through
>>quantum fields. However, the expression for H in terms of fields
>>will be very cumbersome and will lack any physical transparency.
>>There is nothing wrong in describing the Solar system in terms of
>>Ptolemy's epicycles.
>
>
> Epicycles are to ellipses what power series are to implicit equations.
>
> Your expansion is like the epicycles of Ptolemy.
> You need infinitely many terms to define a quantity that
> can be defined much easier implicitly.

I still haven't seen how you are going to describe the time evolution
(e.g., spreading of the wave packet) of a single particle with your
"implicit" Hamiltonian.

Eugene Stefanovich.

Eugene Stefanovich

Apr17-05, 07:54 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n> On 2005-04-15, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>>Igor Khavkine wrote:\n>\n>\n>>The Gell-Mann-Low theorem says that if interaction is adiabatically\n>>turned on, and |Phi> is eigenvector of H_0, then\n>>\n>>|Psi> = exp[iH(0 - -oo)] |Phi>\n>>\n>>is eigenvector of H with the same\n>>eigenvalue (let us assume for simplicity that the spectra of H_0 and\n>>H are the same). This works for vacuum and one-particle states.\n>>So, it seems that you achieved the desired dressing. Not so fast.\n>>It appears that two-, three-, etc. particle states transformed by\n>>exp[iH(0 - -oo)] also become eigenstates of H. E.g.,\n>>\n>>exp[iH(0 - -oo)] a^*(p) a^*(q) |0>\n>>\n>>is an eigenstate of H.\n>>So, you found a representation that diagonalizes H.\n>>This means that there is no scattering anymore.\n>>The scattering operator is identically unity S=1.\n>\n>\n> You misunderstand the theorem. First, you cannot use exp(iHt) as the\n> time evolution operator since H is time dependent (it contains the\n> adiabatic switching function). Second, |Phi> and |Psi> will not have the\n> same energy, they\'ll be different because of the interaction. Third, you\n> can\'t make any conclusions about what kind of state |Psi> is just from\n> knowing what kind of state |Phi> is. The bare ground state need not\n> evolve into the true ground state (superconductivity or electroweak\n> theory). An n-particle state need not evolve into an n-particle state\n> (QED, only charge is conserved, not particle number in fact, the\n> particle number will be indeterminate). You have no way of controlling\n> localization properties of particle states at finite time.\n>\n> Some states can still be identified by their quantum numbers. For\n> instance, in my proposed model, if at t=-oo a particle of charge +1 is\n> place in each well, then at t=0 we\'ll have an eigenstate of the\n> interacting Hamiltonian with charge +1 in each well, but we can\'t say\n> much more. However, this is sufficient for the purposes of the proposed\n> experiment.\n\nI basically agree with everything you are saying. However, you haven\'t\naddressed my point. I said that if you form the state of two physical\nparticles at t=0 by formula (adiabatic switching of interaction is\nassumed)\n\nexp[iH(0 - -oo)] a^*(p) a^*(q) |0>\n\nthen according to the Gell-Mann-Low theorem this state is an eigenstate\nof the full Hamiltonian H. Then the time evolution of this state is\njust a multiplication by a trivial phase factor, which is clearly wrong.\n\n>>>Can different observers still agree on what\'s going to\n>>>happen to the ball? How?\n>>\n>>OK, we agreed (I hope) to consider the "spaceship + ball" as\n>>our physical system. [...]\n>\n>\n> Leave preparation to the preparers and observation to the observers.\n> There is only one spaceship, but some observers see it move and others\n> don\'t. They can all still compare notes and agree on what trajectory the\n> ball took if the relative motion of the ship is taken into account as\n> well as how they measured time intervals and distances. In short, for\n> simplicity and without loss of generality, we can choose to do\n> calculations only in a frame where the ship is stationary.\n>\n> Do you need any more justification for the use of an external potential?\n> If not, my offer is still open and you have the following steps to do:\n>\n> 1. Find the stationary states of the free classical system.\n> 2. Promote the mode coefficients (and their conjugate momenta)\n> to operators and obtain a diagonal expression for the Hamiltonian.\n> 3. Write down the equations of motion for states and operators in the\n> interaction picture. Write down the formal solution. (This does not\n> require an explicit expression for the Hamiltonian).\n>\n> More later, but at least some of the above has to be done first.\n\nI am sorry, I am starting to lose the thread of this discussion.\nI don\'t see what\'s the connection between balls, spaceships, external\npotentials, etc. and the simple question I posed: describe the time\nevolution of a single particle in QED.\n\nLet me try to pose the question in another way. Do you agree or not\nagree that the Hamiltonian (8.4.1) derived in Weinberg\'s section 8.4\n(plus counterterms that can be obtained from formula (11.1.9)) is the\ncorrect Hamiltonian of QED?\nYour choices are:\n1) You agree, and we start discussion right away based on this\nHamiltonian\n2) You disagree, then please explain which of the following you think is\ntrue:\na) The above Hamiltonian should be modified (please explain how)\nb) The above Hamiltonian is useless for time evolution calculations,\nsuch calculations should be done not by formula exp(iHt), but in\nentirely different way.\n\nMy feeling is that your choice is 2)b). Am I right?\n\nEugene Stefanovich.\n\n\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On 2005-04-15, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>>Igor Khavkine wrote:
>
>
>>The Gell-Mann-Low theorem says that if interaction is adiabatically
>>turned on, and |\Phi> is eigenvector of H_0, then
>>
>>|\Psi> = \exp[iH(0 - -oo)] |\Phi>
>>
>>is eigenvector of H with the same
>>eigenvalue (let us assume for simplicity that the spectra of H_0 and
>>H are the same). This works for vacuum and one-particle states.
>>So, it seems that you achieved the desired dressing. Not so fast.
>>It appears that two-, three-, etc. particle states transformed by
>>\exp[iH(0 - -oo)] also become eigenstates of H. E.g.,
>>
>>\exp[iH(0 - -oo)] a^*(p) a^*(q) |0>
>>
>>is an eigenstate of H.
>>So, you found a representation that diagonalizes H.
>>This means that there is no scattering anymore.
>>The scattering operator is identically unity S=1.
>
>
> You misunderstand the theorem. First, you cannot use \exp(iHt) as the
> time evolution operator since H is time dependent (it contains the
> adiabatic switching function). Second, |\Phi> and |\Psi> will not have the
> same energy, they'll be different because of the interaction. Third, you
> can't make any conclusions about what kind of state |\Psi> is just from
> knowing what kind of state |\Phi> is. The bare ground state need not
> evolve into the true ground state (superconductivity or electroweak
> theory). An n-particle state need not evolve into an n-particle state
> (QED, only charge is conserved, not particle number in fact, the
> particle number will be indeterminate). You have no way of controlling
> localization properties of particle states at finite time.
>
> Some states can still be identified by their quantum numbers. For
> instance, in my proposed model, if at t=-oo a particle of charge +1 is
> place in each well, then at t=0 we'll have an eigenstate of the
> interacting Hamiltonian with charge +1 in each well, but we can't say
> much more. However, this is sufficient for the purposes of the proposed
> experiment.

I basically agree with everything you are saying. However, you haven't
addressed my point. I said that if you form the state of two physical
particles at t=0 by formula (adiabatic switching of interaction is
assumed)

\exp[iH(0 - -oo)] a^*(p) a^*(q) |0>

then according to the Gell-Mann-Low theorem this state is an eigenstate
of the full Hamiltonian H. Then the time evolution of this state is
just a multiplication by a trivial phase factor, which is clearly wrong.

>>>Can different observers still agree on what's going to
>>>happen to the ball? How?
>>
>>OK, we agreed (I hope) to consider the "spaceship + ball" as
>>our physical system. [...]
>
>
> Leave preparation to the preparers and observation to the observers.
> There is only one spaceship, but some observers see it move and others
> don't. They can all still compare notes and agree on what trajectory the
> ball took if the relative motion of the ship is taken into account as
> well as how they measured time intervals and distances. In short, for
> simplicity and without loss of generality, we can choose to do
> calculations only in a frame where the ship is stationary.
>
> Do you need any more justification for the use of an external potential?
> If not, my offer is still open and you have the following steps to do:
>
> 1. Find the stationary states of the free classical system.
> 2. Promote the mode coefficients (and their conjugate momenta)
> to operators and obtain a diagonal expression for the Hamiltonian.
> 3. Write down the equations of motion for states and operators in the
> interaction picture. Write down the formal solution. (This does not
> require an explicit expression for the Hamiltonian).
>
> More later, but at least some of the above has to be done first.

I am sorry, I am starting to lose the thread of this discussion.
I don't see what's the connection between balls, spaceships, external
potentials, etc. and the simple question I posed: describe the time
evolution of a single particle in QED.

Let me try to pose the question in another way. Do you agree or not
agree that the Hamiltonian (8.4.1) derived in Weinberg's section 8.4
(plus counterterms that can be obtained from formula (11.1.9)) is the
correct Hamiltonian of QED?
Your choices are:
1) You agree, and we start discussion right away based on this
Hamiltonian
2) You disagree, then please explain which of the following you think is
true:
a) The above Hamiltonian should be modified (please explain how)
b) The above Hamiltonian is useless for time evolution calculations,
such calculations should be done not by formula \exp(iHt), but in
entirely different way.

My feeling is that your choice is 2)b). Am I right?

Eugene Stefanovich.

Eugene Stefanovich

Apr18-05, 04:23 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n\n\n>>Once you constructed\n>>a representation of the Poincare group you get the generators\n>>H, P, J, K "for free". Inversely, once you constructed 10 operators\n>>H, P, J, K satisfying Poincare commutation relations, you get\n>>the representation "for free".\n>\n>\n> And if you construct the representation in the point form or the front\n> form, exactly the same applies. And all these forms are equivalent\n> in the sense that they can be transformed into each other by unitary\n> transformations. Thus their choice is purely a matter of convenience.\n\nUnitary transformations of generators yield different physics.\nLet\'s take a simple example of the Hamiltonian of the hydrogen atom\n\nH = p^2/2m + 1/r (1)\n\nLet\'s form another Hamiltonian by a unitary transformation\n\nH\' = U H U^{-1} (2)\n\nAre you saying that Hamiltonians (1) and (2) describe the same physics?\nI disagree.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:

>>Once you constructed
>>a representation of the Poincare group you get the generators
>>H, P, J, K "for free". Inversely, once you constructed 10 operators
>>H, P, J, K satisfying Poincare commutation relations, you get
>>the representation "for free".
>
>
> And if you construct the representation in the point form or the front
> form, exactly the same applies. And all these forms are equivalent
> in the sense that they can be transformed into each other by unitary
> transformations. Thus their choice is purely a matter of convenience.

Unitary transformations of generators yield different physics.
Let's take a simple example of the Hamiltonian of the hydrogen atom

H = p^2/2m + 1/r (1)

Let's form another Hamiltonian by a unitary transformation

H' = U H U^{-1} (2)

Are you saying that Hamiltonians (1) and (2) describe the same physics?
I disagree.

Eugene Stefanovich.

Arnold Neumaier

Apr18-05, 04:23 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Arnold Neumaier wrote:\n>\n>> Eugene Stefanovich wrote:\n>>\n>>>\n>>> Arnold Neumaier wrote:\n>>>\n>>>> Eugene Stefanovich wrote:\n>>>>\n>>>>> I know pretty well how the Hamiltonian of QED is derived. [...]\n>>\n>>>> (8.4.1) is a valid _classical_ Hamiltonian, since classically\n>>>> nothing is created or annihilated. Creation and annihilation\n>>>> operators are irrelevant on the classical level.\n>>>>\n>>>> On the quantum level, (8.4.1) is only a meaningless formal\n>>>> expression. It is definitely _not_ the physical QED Hamiltonian!\n>>>\n>>> Disagreed.\n\nBut below you agreed. It is impossible to argue with such a\nschizophrenic mind.\n\n\n>> This doesn\'t matter. What people write downd at first are only\n>> approximate expressions with the unspoken implication that these\n>> must be regularized and afterwards a limit with regulaization\n>> dependent coupling constants and counterterms added and everything\n>> (including the wave function) renormalized. Only the resulting limit\n>> defines the physical QED Hamiltonian!\n\n[...]\n>\n> Renormalization is nothing more than adding counterterms to the\n> Hamiltonian (8.4.22) - (8.4.25). These counterterms have the same\n> operator structure as the original terms (they are corrections to\n> masses and charges) so the dangerous trilinear terms remain in\n> the QED Hamiltonian after renormalization.\n>\n>> This already involves all the regularization stuff.\n>\n> Regularization is nothing more than a temporary detour\n> to the imaginary world with cut-off interactions or\n> non-integer dimension.\n\nNo. Renormalization is the process that defines the meaning\nof QFT. It is _not_ a detour but the heart that makes\neverything work.\n\n\n>> You are _not_ allowed to conclude that\n>> the ill-defined formal Hamiltonian is the physical one.\n>\n> [...]\n>\n>>> I have no idea how somebody can use this Hamiltonian to calculate\n>>> the time evolution via exp(iHt).\n>>\n>> It cannot, because it isn\'t the right H.\n>\n> You are absolutely right.\n\nBut in your previous mail you had claimed that it was, and\nbased on this you gave some argument that should refute me.\nOf course, from a wrong hypothesis one can conclude anything.\nSo the conclusion is irrelevant.\n\n\n> You may be right (though I doubt it) that the same is achieved in\n> the traditional approach, but somehow hidden in the jungles of\n> regularization and wave function renormalization.\n\nIt is not hidden for those who care about understanding the\nrelevant literature and not just the typical textbook setting\nwhich aims at the quickest road to relevant calculations rather\nthan at an in depth understanding.\n\nBut as long as you ignore the literature (and in this case,\nthe math physics literature since only the mathematical physicists\nwere interested in getting this right), you don\'t need to be\nsurprised that you scratch only the surface, and heance reach\nweird conclusions.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:
>
>> Eugene Stefanovich wrote:
>>
>>>
>>> Arnold Neumaier wrote:
>>>
>>>> Eugene Stefanovich wrote:
>>>>
>>>>> I know pretty well how the Hamiltonian of QED is derived. [...]
>>
>>>> (8.4.1) is a valid _classical_ Hamiltonian, since classically
>>>> nothing is created or annihilated. Creation and annihilation
>>>> operators are irrelevant on the classical level.
>>>>
>>>> On the quantum level, (8.4.1) is only a meaningless formal
>>>> expression. It is definitely _not_ the physical QED Hamiltonian!
>>>
>>> Disagreed.

But below you agreed. It is impossible to argue with such a
schizophrenic mind.

>> This doesn't matter. What people write downd at first are only
>> approximate expressions with the unspoken implication that these
>> must be regularized and afterwards a limit with regulaization
>> dependent coupling constants and counterterms added and everything
>> (including the wave function) renormalized. Only the resulting limit
>> defines the physical QED Hamiltonian!

[...]
>
> Renormalization is nothing more than adding counterterms to the
> Hamiltonian (8.4.22) - (8.4.25). These counterterms have the same
> operator structure as the original terms (they are corrections to
> masses and charges) so the dangerous trilinear terms remain in
> the QED Hamiltonian after renormalization.
>
>> This already involves all the regularization stuff.
>
> Regularization is nothing more than a temporary detour
> to the imaginary world with cut-off interactions or
> non-integer dimension.

No. Renormalization is the process that defines the meaning
of QFT. It is _not_ a detour but the heart that makes
everything work.

>> You are _not_ allowed to conclude that
>> the ill-defined formal Hamiltonian is the physical one.
>
> [...]
>
>>> I have no idea how somebody can use this Hamiltonian to calculate
>>> the time evolution via \exp(iHt).
>>
>> It cannot, because it isn't the right H.
>
> You are absolutely right.

But in your previous mail you had claimed that it was, and
based on this you gave some argument that should refute me.
Of course, from a wrong hypothesis one can conclude anything.
So the conclusion is irrelevant.

> You may be right (though I doubt it) that the same is achieved in
> the traditional approach, but somehow hidden in the jungles of
> regularization and wave function renormalization.

It is not hidden for those who care about understanding the
relevant literature and not just the typical textbook setting
which aims at the quickest road to relevant calculations rather
than at an in depth understanding.

But as long as you ignore the literature (and in this case,
the math physics literature since only the mathematical physicists
were interested in getting this right), you don't need to be
surprised that you scratch only the surface, and heance reach
weird conclusions.

Arnold Neumaier

Igor Khavkine

Apr18-05, 04:23 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-17, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n> I basically agree with everything you are saying. However, you haven\'t\n> addressed my point. I said that if you form the state of two physical\n> particles at t=0 by formula (adiabatic switching of interaction is\n> assumed)\n>\n> exp[iH(0 - -oo)] a^*(p) a^*(q) |0>\n\nYou are still using the wrong formula, H depends on time.\n\n> then according to the Gell-Mann-Low theorem this state is an eigenstate\n> of the full Hamiltonian H. Then the time evolution of this state is\n> just a multiplication by a trivial phase factor, which is clearly wrong.\n\nWhy is that wrong? What did you expect this state to be? Did you expect\nto have a state with two particles that will scatter? True that\'s not an\neigenstate, but there is no reason for you to expect such a state.\nI don\'t know what that state at finite times is, but it\'s likely a bound\nstate.\n\nBuilding a state at finite time that is not an eigen state and has a\nwell defined physical interpretation of having two localized particles\nthat can scatter is hard. In fact, I don\'t know of any way to do it\nexcept to confine the particles by an external potential.\n\n> I am sorry, I am starting to lose the thread of this discussion.\n> I don\'t see what\'s the connection between balls, spaceships, external\n> potentials, etc. and the simple question I posed: describe the time\n> evolution of a single particle in QED.\n\nTo spell it out: space ship == external infinite potential well,\nball == scalar charged particle. This analogy is very close, but as a\nprecaution, I must say that it must not be taken literally.\n\n> Let me try to pose the question in another way. Do you agree or not\n> agree that the Hamiltonian (8.4.1) derived in Weinberg\'s section 8.4\n> (plus counterterms that can be obtained from formula (11.1.9)) is the\n> correct Hamiltonian of QED?\n> Your choices are:\n> 1) You agree, and we start discussion right away based on this\n> Hamiltonian\n> 2) You disagree, then please explain which of the following you think is\n> true:\n> a) The above Hamiltonian should be modified (please explain how)\n> b) The above Hamiltonian is useless for time evolution calculations,\n\nI prefer not to make claims of uselessness against things that have\nindeed been very useful for many decades.\n\n> such calculations should be done not by formula exp(iHt), but in\n> entirely different way.\n\nThe way is different by not entirely. The formula exp(iHt) is wrong\nbecause it only applies to time independent Hamiltonians. Neither is the\ncase when you deal with adiabatic interaction switching or the\ninteraction picture.\n\nAs to how the Hamiltonian should be modified, instructions are below:\n\n>> 1. Find the stationary states of the free classical system.\n>> 2. Promote the mode coefficients (and their conjugate momenta)\n>> to operators and obtain a diagonal expression for the Hamiltonian.\n>> 3. Write down the equations of motion for states and operators in the\n>> interaction picture. Write down the formal solution. (This does not\n>> require an explicit expression for the Hamiltonian).\n\nTime to get out paper and pencil.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-17, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

> I basically agree with everything you are saying. However, you haven't
> addressed my point. I said that if you form the state of two physical
> particles at t=0 by formula (adiabatic switching of interaction is
> assumed)
>
> \exp[iH(0 - -oo)] a^*(p) a^*(q) |0>

You are still using the wrong formula, H depends on time.

> then according to the Gell-Mann-Low theorem this state is an eigenstate
> of the full Hamiltonian H. Then the time evolution of this state is
> just a multiplication by a trivial phase factor, which is clearly wrong.

Why is that wrong? What did you expect this state to be? Did you expect
to have a state with two particles that will scatter? True that's not an
eigenstate, but there is no reason for you to expect such a state.
I don't know what that state at finite times is, but it's likely a bound
state.

Building a state at finite time that is not an eigen state and has a
well defined physical interpretation of having two localized particles
that can scatter is hard. In fact, I don't know of any way to do it
except to confine the particles by an external potential.

> I am sorry, I am starting to lose the thread of this discussion.
> I don't see what's the connection between balls, spaceships, external
> potentials, etc. and the simple question I posed: describe the time
> evolution of a single particle in QED.

To spell it out: space ship == external infinite potential well,
ball == scalar charged particle. This analogy is very close, but as a
precaution, I must say that it must not be taken literally.

> Let me try to pose the question in another way. Do you agree or not
> agree that the Hamiltonian (8.4.1) derived in Weinberg's section 8.4
> (plus counterterms that can be obtained from formula (11.1.9)) is the
> correct Hamiltonian of QED?
> Your choices are:
> 1) You agree, and we start discussion right away based on this
> Hamiltonian
> 2) You disagree, then please explain which of the following you think is
> true:
> a) The above Hamiltonian should be modified (please explain how)
> b) The above Hamiltonian is useless for time evolution calculations,

I prefer not to make claims of uselessness against things that have
indeed been very useful for many decades.

> such calculations should be done not by formula \exp(iHt), but in
> entirely different way.

The way is different by not entirely. The formula \exp(iHt) is wrong
because it only applies to time independent Hamiltonians. Neither is the
case when you deal with adiabatic interaction switching or the
interaction picture.

As to how the Hamiltonian should be modified, instructions are below:

>> 1. Find the stationary states of the free classical system.
>> 2. Promote the mode coefficients (and their conjugate momenta)
>> to operators and obtain a diagonal expression for the Hamiltonian.
>> 3. Write down the equations of motion for states and operators in the
>> interaction picture. Write down the formal solution. (This does not
>> require an explicit expression for the Hamiltonian).

Time to get out paper and pencil.

Igor

Arnold Neumaier

Apr18-05, 04:26 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Arnold Neumaier wrote:\n>\n>> Eugene Stefanovich wrote:\n>>\n>>> Of course, you can say that one can express particle operators through\n>>> quantum fields. However, the expression for H in terms of fields\n>>> will be very cumbersome and will lack any physical transparency.\n>>> There is nothing wrong in describing the Solar system in terms of\n>>> Ptolemy\'s epicycles.\n>>\n>> Epicycles are to ellipses what power series are to implicit equations.\n>>\n>> Your expansion is like the epicycles of Ptolemy.\n>> You need infinitely many terms to define a quantity that\n>> can be defined much easier implicitly.\n>\n> I still haven\'t seen how you are going to describe the time evolution\n> (e.g., spreading of the wave packet) of a single particle with your\n> "implicit" Hamiltonian.\n\nIf the Hamiltonian is H (defined implicitly but uniquely) then\npsi(t)=exp(-i/hbar H) psi(0),\nas always in quantum mechanics. This makes sense for every selfadjoint\noperator H, whether it has the form preferred by you or whether it is\nconstructed in any other way. See any book on functional analysis,\ne.g. the treatise by Reed and Simon.\n\nIf you find this too hard, you are not qualified for criticising\ntradition.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:
>
>> Eugene Stefanovich wrote:
>>
>>> Of course, you can say that one can express particle operators through
>>> quantum fields. However, the expression for H in terms of fields
>>> will be very cumbersome and will lack any physical transparency.
>>> There is nothing wrong in describing the Solar system in terms of
>>> Ptolemy's epicycles.
>>
>> Epicycles are to ellipses what power series are to implicit equations.
>>
>> Your expansion is like the epicycles of Ptolemy.
>> You need infinitely many terms to define a quantity that
>> can be defined much easier implicitly.
>
> I still haven't seen how you are going to describe the time evolution
> (e.g., spreading of the wave packet) of a single particle with your
> "implicit" Hamiltonian.

If the Hamiltonian is H (defined implicitly but uniquely) then
\psi(t)=\exp(-i/\hbar H) \psi(0),
as always in quantum mechanics. This makes sense for every selfadjoint
operator H, whether it has the form preferred by you or whether it is
constructed in any other way. See any book on functional analysis,
e.g. the treatise by Reed and Simon.

If you find this too hard, you are not qualified for criticising
tradition.

Arnold Neumaier

jsolomon@mail.com

Apr18-05, 09:20 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nEugene Stefanovich wrote:\n>\n> Unitary transformations of generators yield different physics.\n> Let\'s take a simple example of the Hamiltonian of the hydrogen atom\n>\n> H = p^2/2m + 1/r (1)\n>\n> Let\'s form another Hamiltonian by a unitary transformation\n>\n> H\' = U H U^{-1} (2)\n>\n> Are you saying that Hamiltonians (1) and (2) describe the same\nphysics?\n> I disagree.\n>\n> Eugene Stefanovich.\n\nThe reason a unitary transformation does not change the physics is\nbecause a unitary transformation in the Hamiltonian will be compensated\nfor by a unitary transformation in the state vector, so that the\nexpectation value is unchanged. For example, the energy of the state\n|s> is <s|H|s>. Apply the unitary transformation so that the new\nHamiltonian is\n\nH\' = U H U^{-1}\n\nApply the same transformation to the state vectors to obtain,\n\n|s\'>=U|s>\n\nTherefore <s|H|s>=<s\'|H\'|s\'> and the energy expectation values do not\nchange. A similar situation holds for all other expectatoin values\n(i.e. current, momentum, charge, etc.).\n\nA Unitary transformation changes your "point of view", not the physics.\nIt crudely analogous to changing your reference frame. This should\nnot be considered unimportant. Working in the correct reference frame\ngreatly simplifies many problems and is essential to obtaining physical\ninsight.\n\nDan Solomon\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Unitary transformations of generators yield different physics.
> Let's take a simple example of the Hamiltonian of the hydrogen atom
>
> H = p^2/2m + 1/r (1)
>
> Let's form another Hamiltonian by a unitary transformation
>
> H' = U H U^{-1} (2)
>
> Are you saying that Hamiltonians (1) and (2) describe the same
physics?
> I disagree.
>
> Eugene Stefanovich.

The reason a unitary transformation does not change the physics is
because a unitary transformation in the Hamiltonian will be compensated
for by a unitary transformation in the state vector, so that the
expectation value is unchanged. For example, the energy of the state
|s> is <s|H|s>. Apply the unitary transformation so that the new
Hamiltonian is

H' = U H U^{-1}

Apply the same transformation to the state vectors to obtain,

|s'>=U|s>

Therefore <s|H|s>=<s'|H'|s'> and the energy expectation values do not
change. A similar situation holds for all other expectatoin values
(i.e. current, momentum, charge, etc.).

A Unitary transformation changes your "point of view", not the physics.
It crudely analogous to changing your reference frame. This should
not be considered unimportant. Working in the correct reference frame
greatly simplifies many problems and is essential to obtaining physical
insight.

Dan Solomon

Eugene Stefanovich

Apr18-05, 09:20 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n\n>>I basically agree with everything you are saying. However, you haven\'t\n>>addressed my point. I said that if you form the state of two physical\n>>particles at t=0 by formula (adiabatic switching of interaction is\n>>assumed)\n>>\n>>exp[iH(0 - -oo)] a^*(p) a^*(q) |0>\n>\n>\n> You are still using the wrong formula, H depends on time.\n>\n\nI wrote in parentheses "adiabatic switching of interaction is\nassumed", though haven\'t indicated this explicitly. Is there any\nother time dependence in H?\n\n>\n>>then according to the Gell-Mann-Low theorem this state is an eigenstate\n>>of the full Hamiltonian H. Then the time evolution of this state is\n>>just a multiplication by a trivial phase factor, which is clearly wrong.\n>\n>\n> Why is that wrong? What did you expect this state to be? Did you expect\n> to have a state with two particles that will scatter? True that\'s not an\n> eigenstate, but there is no reason for you to expect such a state.\n> I don\'t know what that state at finite times is, but it\'s likely a bound\n> state.\n>\n> Building a state at finite time that is not an eigen state and has a\n> well defined physical interpretation of having two localized particles\n> that can scatter is hard. In fact, I don\'t know of any way to do it\n> except to confine the particles by an external potential.\n\nSuch scattering states are obtained naturally in the RQD approach\nwithout the need for an artificial external potential.\n\n>\n>\n>>I am sorry, I am starting to lose the thread of this discussion.\n>>I don\'t see what\'s the connection between balls, spaceships, external\n>>potentials, etc. and the simple question I posed: describe the time\n>>evolution of a single particle in QED.\n>\n>\n> To spell it out: space ship == external infinite potential well,\n> ball == scalar charged particle. This analogy is very close, but as a\n> precaution, I must say that it must not be taken literally.\n\nI understand this analogy, but I maintain that a rigorous theory\nshould describe the time evolution, in general, and scattering,\nin particular, using a single well-defined Hamiltonian without\nany tricks with external potentials. You don\'t need to introduce\nexternal potentials when you calculate the S-matrix in QED,\nthen why should you use them in calculations of the time evolution?\n\n>\n>\n>>Let me try to pose the question in another way. Do you agree or not\n>>agree that the Hamiltonian (8.4.1) derived in Weinberg\'s section 8.4\n>>(plus counterterms that can be obtained from formula (11.1.9)) is the\n>>correct Hamiltonian of QED?\n>>Your choices are:\n>>1) You agree, and we start discussion right away based on this\n>> Hamiltonian\n>>2) You disagree, then please explain which of the following you think is\n>> true:\n>> a) The above Hamiltonian should be modified (please explain how)\n>> b) The above Hamiltonian is useless for time evolution calculations,\n>\n>\n> I prefer not to make claims of uselessness against things that have\n> indeed been very useful for many decades.\n>\n>\n>> such calculations should be done not by formula exp(iHt), but in\n>> entirely different way.\n>\n>\n> The way is different by not entirely. The formula exp(iHt) is wrong\n> because it only applies to time independent Hamiltonians. Neither is the\n> case when you deal with adiabatic interaction switching or the\n> interaction picture.\n\nThe Hamiltonians describing isolated systems (that\'s the only kind of\nsystems I am willing to consider) are time-independent. There is no\nadiabatic switching of interactions in nature. In theory, the adiabatic\nswitching is just a mathematical trick that allows us to avoid\ndiscussion of wave packets, and thereby simplifies calculations.\nIn principle, everything can be done without the adiabatic switching,\nwith time-independent Hamiltonians. The amount of calculational work\nwould increase, but this has no fundamental significance.\n\n>\n> As to how the Hamiltonian should be modified, instructions are below:\n>\n>\n>>>1. Find the stationary states of the free classical system.\n>>>2. Promote the mode coefficients (and their conjugate momenta)\n>>> to operators and obtain a diagonal expression for the Hamiltonian.\n>>>3. Write down the equations of motion for states and operators in the\n>>> interaction picture. Write down the formal solution. (This does not\n>>> require an explicit expression for the Hamiltonian).\n>>\n>\n> Time to get out paper and pencil.\n\nNot so fast. I cannot follow your derivation without understanding\nphysical and/or mathematical premises of proposed steps. Even if I\nagree that your steps lead to some kind of "time evolution" you must\nconvince me that this evolution is unitary (probabilities are preserved)\nand relativistically invariant (satisfies Poincare group relationships\nwith space translations, rotations, and boosts). I am wondering, how\nyou are going to do that without explicit expressions for the\nHamiltonian and other Poincare generators.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:

>>I basically agree with everything you are saying. However, you haven't
>>addressed my point. I said that if you form the state of two physical
>>particles at t=0 by formula (adiabatic switching of interaction is
>>assumed)
>>
>>\exp[iH(0 - -oo)] a^*(p) a^*(q) |0>
>
>
> You are still using the wrong formula, H depends on time.
>

I wrote in parentheses "adiabatic switching of interaction is
assumed", though haven't indicated this explicitly. Is there any
other time dependence in H?

>
>>then according to the Gell-Mann-Low theorem this state is an eigenstate
>>of the full Hamiltonian H. Then the time evolution of this state is
>>just a multiplication by a trivial phase factor, which is clearly wrong.
>
>
> Why is that wrong? What did you expect this state to be? Did you expect
> to have a state with two particles that will scatter? True that's not an
> eigenstate, but there is no reason for you to expect such a state.
> I don't know what that state at finite times is, but it's likely a bound
> state.
>
> Building a state at finite time that is not an eigen state and has a
> well defined physical interpretation of having two localized particles
> that can scatter is hard. In fact, I don't know of any way to do it
> except to confine the particles by an external potential.

Such scattering states are obtained naturally in the RQD approach
without the need for an artificial external potential.

>
>
>>I am sorry, I am starting to lose the thread of this discussion.
>>I don't see what's the connection between balls, spaceships, external
>>potentials, etc. and the simple question I posed: describe the time
>>evolution of a single particle in QED.
>
>
> To spell it out: space ship == external infinite potential well,
> ball == scalar charged particle. This analogy is very close, but as a
> precaution, I must say that it must not be taken literally.

I understand this analogy, but I maintain that a rigorous theory
should describe the time evolution, in general, and scattering,
in particular, using a single well-defined Hamiltonian without
any tricks with external potentials. You don't need to introduce
external potentials when you calculate the S-matrix in QED,
then why should you use them in calculations of the time evolution?

>
>
>>Let me try to pose the question in another way. Do you agree or not
>>agree that the Hamiltonian (8.4.1) derived in Weinberg's section 8.4
>>(plus counterterms that can be obtained from formula (11.1.9)) is the
>>correct Hamiltonian of QED?
>>Your choices are:
>>1) You agree, and we start discussion right away based on this
>> Hamiltonian
>>2) You disagree, then please explain which of the following you think is
>> true:
>> a) The above Hamiltonian should be modified (please explain how)
>> b) The above Hamiltonian is useless for time evolution calculations,
>
>
> I prefer not to make claims of uselessness against things that have
> indeed been very useful for many decades.
>
>
>> such calculations should be done not by formula \exp(iHt), but in
>> entirely different way.
>
>
> The way is different by not entirely. The formula \exp(iHt) is wrong
> because it only applies to time independent Hamiltonians. Neither is the
> case when you deal with adiabatic interaction switching or the
> interaction picture.

The Hamiltonians describing isolated systems (that's the only kind of
systems I am willing to consider) are time-independent. There is no
adiabatic switching of interactions in nature. In theory, the adiabatic
switching is just a mathematical trick that allows us to avoid
discussion of wave packets, and thereby simplifies calculations.
In principle, everything can be done without the adiabatic switching,
with time-independent Hamiltonians. The amount of calculational work
would increase, but this has no fundamental significance.

>
> As to how the Hamiltonian should be modified, instructions are below:
>
>
>>>1. Find the stationary states of the free classical system.
>>>2. Promote the mode coefficients (and their conjugate momenta)
>>> to operators and obtain a diagonal expression for the Hamiltonian.
>>>3. Write down the equations of motion for states and operators in the
>>> interaction picture. Write down the formal solution. (This does not
>>> require an explicit expression for the Hamiltonian).
>>
>
> Time to get out paper and pencil.

Not so fast. I cannot follow your derivation without understanding
physical and/or mathematical premises of proposed steps. Even if I
agree that your steps lead to some kind of "time evolution" you must
convince me that this evolution is unitary (probabilities are preserved)
and relativistically invariant (satisfies Poincare group relationships
with space translations, rotations, and boosts). I am wondering, how
you are going to do that without explicit expressions for the
Hamiltonian and other Poincare generators.

Eugene Stefanovich.

Eugene Stefanovich

Apr19-05, 02:48 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n\n>>>>>\n>>>>>On the quantum level, (8.4.1) is only a meaningless formal\n>>>>>expression. It is definitely _not_ the physical QED Hamiltonian!\n>>>>\n>>>>Disagreed.\n>>>\n>\n> But below you agreed. It is impossible to argue with such a\n> schizophrenic mind.\n\nI have explained that a few times already. The Hamiltonian H (8.4.1)\nplus counterterms in (11.1.9) is the true Hamiltonian of QED.\nIf you insert this Hamiltonian in the standard expression for the\nS-operator, you obtain finite and accurate results for scattering\namplitudes, energies of bound states etc. without any artificial\nadjustments.\n\nAlthough H is the true Hamiltonian\nof QED, it is not the true Hamiltonian\nof interacting charged particles and photons. Because this Hamiltonian\ncannot be used for time evolution calculations without\nencountering unphysical instability of vacuum and one-particle states.\nThis schizophrenic situation is easily explained. QED was never\ninterested in time evolution. All properties calculated in QED\nrequired integration of certain t-dependent terms from minus infinity to\nplus infinity. This results in cancellation of all bad terms in the\nS-operator. Only good properties of H are at work in QED, its bad\nproperties are suppressed.\n\nIf we want to go beyond the S-matrix and energies of bound states,\nwe cannot use the Hamiltonian H. The way to get rid of the bad\nproperties of H is to perform a unitary dressing transformation\n\nH\' = U H U^{-1}\n\nThen H\' can be used for both dynamical and S-matrix calculations.\n\n>>\n>>Regularization is nothing more than a temporary detour\n>>to the imaginary world with cut-off interactions or\n>>non-integer dimension.\n>\n>\n> No. Renormalization is the process that defines the meaning\n> of QFT. It is _not_ a detour but the heart that makes\n> everything work.\n\nI was talking about regularization. In my vocabulary\n"regularization" is modifying the theory so that all integrals\nbecome temporarily finite, and one can do necessary cancellations.\nAt the end the regulators should be removed, and the final results\ndo not bear any signs of what kind of regularization was performed.\n\n"Renormalization" is adding (generally infinite) counterterms to the\noriginal Hamiltonian (with physical masses and charges).\nRenormalization makes the Hamiltonian infinite,\nhowever it allows to obtain finite and accurate scattering amplitudes\nand energies of bound states.\n\nSo, regularization is a "detour" and I am right.\nRenormalization is "the heart that makes everything work",\nand you are right too.\n\n\n>>>>I have no idea how somebody can use this Hamiltonian to calculate\n>>>>the time evolution via exp(iHt).\n>>>\n>>>It cannot, because it isn\'t the right H.\n>>\n>>You are absolutely right.\n>\n>\n> But in your previous mail you had claimed that it was, and\n> based on this you gave some argument that should refute me.\n> Of course, from a wrong hypothesis one can conclude anything.\n> So the conclusion is irrelevant.\n\nI have explained above in what sense the Hamiltonian H of QED\nis right, and in what sense it is wrong.\n\n\n>>You may be right (though I doubt it) that the same is achieved in\n>>the traditional approach, but somehow hidden in the jungles of\n>>regularization and wave function renormalization.\n>\n>\n> It is not hidden for those who care about understanding the\n> relevant literature and not just the typical textbook setting\n> which aims at the quickest road to relevant calculations rather\n> than at an in depth understanding.\n>\n> But as long as you ignore the literature (and in this case,\n> the math physics literature since only the mathematical physicists\n> were interested in getting this right), you don\'t need to be\n> surprised that you scratch only the surface, and heance reach\n> weird conclusions.\n\nThen, possibly you can find an error in my derivations,\nand prove that my "weird conclusions" are not related to the physical\nworld? Can you?\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:

>>>>>
>>>>>On the quantum level, (8.4.1) is only a meaningless formal
>>>>>expression. It is definitely _not_ the physical QED Hamiltonian!
>>>>
>>>>Disagreed.
>>>
>
> But below you agreed. It is impossible to argue with such a
> schizophrenic mind.

I have explained that a few times already. The Hamiltonian H (8.4.1)
plus counterterms in (11.1.9) is the true Hamiltonian of QED.
If you insert this Hamiltonian in the standard expression for the
S-operator, you obtain finite and accurate results for scattering
amplitudes, energies of bound states etc. without any artificial
adjustments.

Although H is the true Hamiltonian
of QED, it is not the true Hamiltonian
of interacting charged particles and photons. Because this Hamiltonian
cannot be used for time evolution calculations without
encountering unphysical instability of vacuum and one-particle states.
This schizophrenic situation is easily explained. QED was never
interested in time evolution. All properties calculated in QED
required integration of certain t-dependent terms from minus infinity to
plus infinity. This results in cancellation of all bad terms in the
S-operator. Only good properties of H are at work in QED, its bad
properties are suppressed.

If we want to go beyond the S-matrix and energies of bound states,
we cannot use the Hamiltonian H. The way to get rid of the bad
properties of H is to perform a unitary dressing transformation

H' = U H U^{-1}

Then H' can be used for both dynamical and S-matrix calculations.

>>
>>Regularization is nothing more than a temporary detour
>>to the imaginary world with cut-off interactions or
>>non-integer dimension.
>
>
> No. Renormalization is the process that defines the meaning
> of QFT. It is _not_ a detour but the heart that makes
> everything work.

I was talking about regularization. In my vocabulary
"regularization" is modifying the theory so that all integrals
become temporarily finite, and one can do necessary cancellations.
At the end the regulators should be removed, and the final results
do not bear any signs of what kind of regularization was performed.

"Renormalization" is adding (generally infinite) counterterms to the
original Hamiltonian (with physical masses and charges).
Renormalization makes the Hamiltonian infinite,
however it allows to obtain finite and accurate scattering amplitudes
and energies of bound states.

So, regularization is a "detour" and I am right.
Renormalization is "the heart that makes everything work",
and you are right too.

>>>>I have no idea how somebody can use this Hamiltonian to calculate
>>>>the time evolution via \exp(iHt).
>>>
>>>It cannot, because it isn't the right H.
>>
>>You are absolutely right.
>
>
> But in your previous mail you had claimed that it was, and
> based on this you gave some argument that should refute me.
> Of course, from a wrong hypothesis one can conclude anything.
> So the conclusion is irrelevant.

I have explained above in what sense the Hamiltonian H of QED
is right, and in what sense it is wrong.

>>You may be right (though I doubt it) that the same is achieved in
>>the traditional approach, but somehow hidden in the jungles of
>>regularization and wave function renormalization.
>
>
> It is not hidden for those who care about understanding the
> relevant literature and not just the typical textbook setting
> which aims at the quickest road to relevant calculations rather
> than at an in depth understanding.
>
> But as long as you ignore the literature (and in this case,
> the math physics literature since only the mathematical physicists
> were interested in getting this right), you don't need to be
> surprised that you scratch only the surface, and heance reach
> weird conclusions.

Then, possibly you can find an error in my derivations,
and prove that my "weird conclusions" are not related to the physical
world? Can you?

Eugene Stefanovich.

Eugene Stefanovich

Apr19-05, 02:48 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Arnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>Eugene Stefanovich wrote:\n>>>\n>>>\n>>>>Of course, you can say that one can express particle operators through\n>>>>quantum fields. However, the expression for H in terms of fields\n>>>>will be very cumbersome and will lack any physical transparency.\n>>>>There is nothing wrong in describing the Solar system in terms of\n>>>>Ptolemy\'s epicycles.\n>>>\n>>>Epicycles are to ellipses what power series are to implicit equations.\n>>>\n>>>Your expansion is like the epicycles of Ptolemy.\n>>>You need infinitely many terms to define a quantity that\n>>>can be defined much easier implicitly.\n>>\n>>I still haven\'t seen how you are going to describe the time evolution\n>>(e.g., spreading of the wave packet) of a single particle with your\n>>"implicit" Hamiltonian.\n>\n>\n> If the Hamiltonian is H (defined implicitly but uniquely) then\n> psi(t)=exp(-i/hbar H) psi(0),\n> as always in quantum mechanics. This makes sense for every selfadjoint\n> operator H, whether it has the form preferred by you or whether it is\n> constructed in any other way. See any book on functional analysis,\n> e.g. the treatise by Reed and Simon.\n>\n> If you find this too hard, you are not qualified for criticising\n> tradition.\n\nI fully agree with your formula\n\npsi(t)=exp(-i/hbar H t) psi(0)\n\n(let\'s not argue about i or -i in the exponent, that\'s not relevant\nhere).\nThe tiny little problem we have is what should be used in this\nformula in place of H. We both agree that Weinberg\'s H in (8.4.1)\nplus counterterms cannot be used there. I suggest to use the RQD\nHamiltonian H\' = U H U^{-1}, which satisfies all physical conditions\nknown to me. I still haven\'t heard from you any alternative suggestion,\napart from some vague references to "implicit" definition\nand "wave function renormalization".\n\nEugene Stefanovich\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>>Arnold Neumaier wrote:
>>
>>
>>>Eugene Stefanovich wrote:
>>>
>>>
>>>>Of course, you can say that one can express particle operators through
>>>>quantum fields. However, the expression for H in terms of fields
>>>>will be very cumbersome and will lack any physical transparency.
>>>>There is nothing wrong in describing the Solar system in terms of
>>>>Ptolemy's epicycles.
>>>
>>>Epicycles are to ellipses what power series are to implicit equations.
>>>
>>>Your expansion is like the epicycles of Ptolemy.
>>>You need infinitely many terms to define a quantity that
>>>can be defined much easier implicitly.
>>
>>I still haven't seen how you are going to describe the time evolution
>>(e.g., spreading of the wave packet) of a single particle with your
>>"implicit" Hamiltonian.
>
>
> If the Hamiltonian is H (defined implicitly but uniquely) then
> \psi(t)=\exp(-i/\hbar H) \psi(0),
> as always in quantum mechanics. This makes sense for every selfadjoint
> operator H, whether it has the form preferred by you or whether it is
> constructed in any other way. See any book on functional analysis,
> e.g. the treatise by Reed and Simon.
>
> If you find this too hard, you are not qualified for criticising
> tradition.

I fully agree with your formula

\psi(t)=\exp(-i/\hbar H t) \psi(0)

(let's not argue about i or -i in the exponent, that's not relevant
here).
The tiny little problem we have is what should be used in this
formula in place of H. We both agree that Weinberg's H in (8.4.1)
plus counterterms cannot be used there. I suggest to use the RQD
Hamiltonian H' = U H U^{-1}, which satisfies all physical conditions
known to me. I still haven't heard from you any alternative suggestion,
apart from some vague references to "implicit" definition
and "wave function renormalization".

Eugene Stefanovich

Eugene Stefanovich

Apr19-05, 02:49 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>jsolomon@mail.com wrote:\n> Eugene Stefanovich wrote:\n>\n>>Unitary transformations of generators yield different physics.\n>>Let\'s take a simple example of the Hamiltonian of the hydrogen atom\n>>\n>>H = p^2/2m + 1/r (1)\n>>\n>>Let\'s form another Hamiltonian by a unitary transformation\n>>\n>>H\' = U H U^{-1} (2)\n>>\n>>Are you saying that Hamiltonians (1) and (2) describe the same\n>\n> physics?\n>\n>>I disagree.\n>>\n>>Eugene Stefanovich.\n>\n>\n> The reason a unitary transformation does not change the physics is\n> because a unitary transformation in the Hamiltonian will be compensated\n> for by a unitary transformation in the state vector, so that the\n> expectation value is unchanged. For example, the energy of the state\n> |s> is <s|H|s>. Apply the unitary transformation so that the new\n> Hamiltonian is\n>\n> H\' = U H U^{-1}\n>\n> Apply the same transformation to the state vectors to obtain,\n>\n> |s\'>=U|s>\n>\n> Therefore <s|H|s>=<s\'|H\'|s\'> and the energy expectation values do not\n> change. A similar situation holds for all other expectatoin values\n> (i.e. current, momentum, charge, etc.).\n>\n> A Unitary transformation changes your "point of view", not the physics.\n> It crudely analogous to changing your reference frame. This should\n> not be considered unimportant. Working in the correct reference frame\n> greatly simplifies many problems and is essential to obtaining physical\n> insight.\n\nYou are absolutely right that if both Hamiltonian and state vectors\nare changed by a unitary transformation, then physical results do not\nchange. However, I was referring to the situation when only the\nHamiltonian H (and 9 other generators of the Poincare group P, J, and\nK) is transformed. Then the physics is different.\n\nThis question arose in the context of our discussion with Arnold\nNeumaier about (non)equivalence of different relativistic forms of\ndynamics. For example, in a 2-particle system described by 1-particle\nobservables\n\np_1, j_2, k_1, h_1\np_2, j_2, k_2, h_2\n\none can define two kinds of representations of the Poincare group:\none is the "instant form dynamics"\n\nP = p_1 + p_2\nJ = j_1 + j_2\nK = k_1 + k_2 + Z\nH = h_1 + h_2 + V\n\nwhere Z and V are interaction operators depending on 1-particle\nobservables. Another is the "point form dynamics"\n\nP\' = p_1 + p_2 + Y\nJ\' = j_1 + j_2\nK\' = k_1 + k_2\nH\' = h_1 + h_2 + W\n\nwhere Y and W are interaction operators. A theorem proven\nby Sokolov and Shatnii states that one can find a unitary\ntransformation U such that\n\nU {P, J, K, H} U^{-1} = {P\', J\', K\', H\'}\n\n(the 1-particle observables are NOT transformed) and that S-matrices\ncomputed with H and H\' are the same. I say that two sets of generators\n{P, J, K, H} and {P\', J\', K\', H\'} describe different physics.\nTo understand that, it is important to realize that the states and\n1-particle operators should no be transformed by U. Otherwise, if\nyou apply U to 1-particle observables, the new set of\ngenerators\n\nP\' = U P U^{-1} = U (p_1 + p_2) U^{-1} = p\'_1 + p\'_2\nJ\' = j\'_1 + j\'_2\nK\' = k\'_1 + k\'_2 + Z\'\nH\' = h\'_1 + h\'_2 + V\'\n\nis still in the instant form, because the definition of the form\nof interaction is determined by the relationship between the generators\nof the Poincare group and 1-particle observables. Such a relationship\ndoes not change if the transformation is applied to both of them.\n\nEugene Stefanovich\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>jsolomon@mail.com wrote:
> Eugene Stefanovich wrote:
>
>>Unitary transformations of generators yield different physics.
>>Let's take a simple example of the Hamiltonian of the hydrogen atom
>>
>>H = p^2/2m + 1/r (1)
>>
>>Let's form another Hamiltonian by a unitary transformation
>>
>>H' = U H U^{-1} (2)
>>
>>Are you saying that Hamiltonians (1) and (2) describe the same
>
> physics?
>
>>I disagree.
>>
>>Eugene Stefanovich.
>
>
> The reason a unitary transformation does not change the physics is
> because a unitary transformation in the Hamiltonian will be compensated
> for by a unitary transformation in the state vector, so that the
> expectation value is unchanged. For example, the energy of the state
> |s> is <s|H|s>. Apply the unitary transformation so that the new
> Hamiltonian is
>
> H' = U H U^{-1}
>
> Apply the same transformation to the state vectors to obtain,
>
> |s'>=U|s>
>
> Therefore <s|H|s>=<s'|H'|s'> and the energy expectation values do not
> change. A similar situation holds for all other expectatoin values
> (i.e. current, momentum, charge, etc.).
>
> A Unitary transformation changes your "point of view", not the physics.
> It crudely analogous to changing your reference frame. This should
> not be considered unimportant. Working in the correct reference frame
> greatly simplifies many problems and is essential to obtaining physical
> insight.

You are absolutely right that if both Hamiltonian and state vectors
are changed by a unitary transformation, then physical results do not
change. However, I was referring to the situation when only the
Hamiltonian H (and 9 other generators of the Poincare group P, J, and
K) is transformed. Then the physics is different.

This question arose in the context of our discussion with Arnold
Neumaier about (non)equivalence of different relativistic forms of
dynamics. For example, in a 2-particle system described by 1-particle
observables

p_1, j_2, k_1, h_1p_2, j_2, k_2, h_2

one can define two kinds of representations of the Poincare group:
one is the "instant form dynamics"

P = p_1 + p_2J = j_1 + j_2K = k_1 + k_2 + ZH = h_1 + h_2 + V

where Z and V are interaction operators depending on 1-particle
observables. Another is the "point form dynamics"

P' = p_1 + p_2 + YJ' = j_1 + j_2K' = k_1 + k_2H' = h_1 + h_2 + W

where Y and W are interaction operators. A theorem proven
by Sokolov and Shatnii states that one can find a unitary
transformation U such that

U {P, J, K, H} U^{-1} = {P',[/itex] J', K', H'}

(the 1-particle observables are NOT transformed) and that S-matrices
computed with H and H' are the same. I say that two sets of generators
{P, J, K, H} and {P', J', K', H'} describe different physics.
To understand that, it is important to realize that the states and
1-particle operators should no be transformed by U. Otherwise, if
you apply U to 1-particle observables, the new set of
generators

[itex]P' = U P U^{-1} = U (p_1 + p_2) U^{-1} = p'_1 + p'_2J' = j'_1 + j'_2K' = k'_1 + k'_2 + Z'H' = h'_1 + h'_2 + V'

is still in the instant form, because the definition of the form
of interaction is determined by the relationship between the generators
of the Poincare group and 1-particle observables. Such a relationship
does not change if the transformation is applied to both of them.

Eugene Stefanovich

Igor Khavkine

Apr19-05, 02:49 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-19, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n>> Building a state at finite time that is not an eigen state and has a\n>> well defined physical interpretation of having two localized particles\n>> that can scatter is hard. In fact, I don\'t know of any way to do it\n>> except to confine the particles by an external potential.\n>\n> Such scattering states are obtained naturally in the RQD approach\n> without the need for an artificial external potential.\n\nAh, but what if a natural external potential is used? Can your approach\nincorporate one at all? If you think all external potentials are\nartificial, I\'m sure the engineers who designed the transistors in the\nCPU of your computer will be surprised to learn that they\'ve been using\nnonsense for the past few decades.\n\nBTW, how do you know that these scattering states are what you think\nthey are? Can you tell me what is the expectation value and variance for\nthe position of each particle?\n\n>> To spell it out: space ship == external infinite potential well,\n>> ball == scalar charged particle. This analogy is very close, but as a\n>> precaution, I must say that it must not be taken literally.\n>\n> I understand this analogy, but I maintain that a rigorous theory\n> should describe the time evolution, in general, and scattering,\n> in particular, using a single well-defined Hamiltonian without\n> any tricks with external potentials. You don\'t need to introduce\n> external potentials when you calculate the S-matrix in QED,\n> then why should you use them in calculations of the time evolution?\n\nThere is nothing unrigorous about using an external potential. If you\nare really opposed to non-isolated systems, then I wonder how you would\ncalculate the trajectory of a falling stone. Would you solve Einstein\'s\nequations for the entire solar system, nay, galaxy, nay, universe,\nwith all the details down to the stone itself?\n\n> The Hamiltonians describing isolated systems (that\'s the only kind of\n> systems I am willing to consider) are time-independent. There is no\n> adiabatic switching of interactions in nature. In theory, the adiabatic\n> switching is just a mathematical trick that allows us to avoid\n> discussion of wave packets, and thereby simplifies calculations.\n> In principle, everything can be done without the adiabatic switching,\n> with time-independent Hamiltonians. The amount of calculational work\n> would increase, but this has no fundamental significance.\n\nIf you are comfortable with wavepackets and adiabatic switching is\nequivalent to their use, then you should be comfortable with adiabatic\nswitching as well.\n\n>> Time to get out paper and pencil.\n>\n> Not so fast. I cannot follow your derivation without understanding\n> physical and/or mathematical premises of proposed steps. Even if I\n> agree that your steps lead to some kind of "time evolution" you must\n> convince me that this evolution is unitary (probabilities are preserved)\n> and relativistically invariant (satisfies Poincare group relationships\n> with space translations, rotations, and boosts). I am wondering, how\n> you are going to do that without explicit expressions for the\n> Hamiltonian and other Poincare generators.\n\nOn the contrary, I need not convince you of anything. The only person\nwho can convince you of anything is yourself. What I can do is present\nyou with facts, it is up to you to absorb them. Moreover, you *can*\nfollow the outlined steps, I hope I can put at least that much\nconfidence in your technical ability, however you do not seem inclined\nto. That is unfortunate. You seek understanding but avoid calculations,\nthat is a path doomed to failure.\n\nThere are a thousand things you might want to check about the standard\nclaculational framework in QFT. However, it is not my intention to\nexpand on each individual one *before* an actual calculation has been\nperformed. All the things you are asking for are completely irrelevant\nfor the proposed calculation. However, once completed it will be a\ntemplate for any calculation you wish to do whose result can be\nexpressed in terms of expectation values of products of field operators,\nwhich includes all operators of physical interest. You are welcome to\nattempt any of these on your onwn time once you grasp the basic\nformalism.\n\nThe result of the particular proposed calculation will be the time\ndependence of the expectation value of the position operator for one of\nthe particles after the other was somehow perturbed. That is all I claim\nto deliver. Anything that is not necessary for the calculation will not\nmake an appearance. This includes an explicit expression for physical\nstates, an explicit expression for the matrix elements of the\nHamiltonian between them, or anything that has to do with translation or\nboost generators.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-19, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

>> Building a state at finite time that is not an eigen state and has a
>> well defined physical interpretation of having two localized particles
>> that can scatter is hard. In fact, I don't know of any way to do it
>> except to confine the particles by an external potential.
>
> Such scattering states are obtained naturally in the RQD approach
> without the need for an artificial external potential.

Ah, but what if a natural external potential is used? Can your approach
incorporate one at all? If you think all external potentials are
artificial, I'm sure the engineers who designed the transistors in the
CPU of your computer will be surprised to learn that they've been using
nonsense for the past few decades.

BTW, how do you know that these scattering states are what you think
they are? Can you tell me what is the expectation value and variance for
the position of each particle?

>> To spell it out: space ship == external infinite potential well,
>> ball == scalar charged particle. This analogy is very close, but as a
>> precaution, I must say that it must not be taken literally.
>
> I understand this analogy, but I maintain that a rigorous theory
> should describe the time evolution, in general, and scattering,
> in particular, using a single well-defined Hamiltonian without
> any tricks with external potentials. You don't need to introduce
> external potentials when you calculate the S-matrix in QED,
> then why should you use them in calculations of the time evolution?

There is nothing unrigorous about using an external potential. If you
are really opposed to non-isolated systems, then I wonder how you would
calculate the trajectory of a falling stone. Would you solve Einstein's
equations for the entire solar system, nay, galaxy, nay, universe,
with all the details down to the stone itself?

> The Hamiltonians describing isolated systems (that's the only kind of
> systems I am willing to consider) are time-independent. There is no
> adiabatic switching of interactions in nature. In theory, the adiabatic
> switching is just a mathematical trick that allows us to avoid
> discussion of wave packets, and thereby simplifies calculations.
> In principle, everything can be done without the adiabatic switching,
> with time-independent Hamiltonians. The amount of calculational work
> would increase, but this has no fundamental significance.

If you are comfortable with wavepackets and adiabatic switching is
equivalent to their use, then you should be comfortable with adiabatic
switching as well.

>> Time to get out paper and pencil.
>
> Not so fast. I cannot follow your derivation without understanding
> physical and/or mathematical premises of proposed steps. Even if I
> agree that your steps lead to some kind of "time evolution" you must
> convince me that this evolution is unitary (probabilities are preserved)
> and relativistically invariant (satisfies Poincare group relationships
> with space translations, rotations, and boosts). I am wondering, how
> you are going to do that without explicit expressions for the
> Hamiltonian and other Poincare generators.

On the contrary, I need not convince you of anything. The only person
who can convince you of anything is yourself. What I can do is present
you with facts, it is up to you to absorb them. Moreover, you *can*
follow the outlined steps, I hope I can put at least that much
confidence in your technical ability, however you do not seem inclined
to. That is unfortunate. You seek understanding but avoid calculations,
that is a path doomed to failure.

There are a thousand things you might want to check about the standard
claculational framework in QFT. However, it is not my intention to
expand on each individual one *before* an actual calculation has been
performed. All the things you are asking for are completely irrelevant
for the proposed calculation. However, once completed it will be a
template for any calculation you wish to do whose result can be
expressed in terms of expectation values of products of field operators,
which includes all operators of physical interest. You are welcome to
attempt any of these on your onwn time once you grasp the basic
formalism.

The result of the particular proposed calculation will be the time
dependence of the expectation value of the position operator for one of
the particles after the other was somehow perturbed. That is all I claim
to deliver. Anything that is not necessary for the calculation will not
make an appearance. This includes an explicit expression for physical
states, an explicit expression for the matrix elements of the
Hamiltonian between them, or anything that has to do with translation or
boost generators.

Igor

Arnold Neumaier

Apr19-05, 07:51 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n> Arnold Neumaier wrote:\n>\n>>And if you construct the representation in the point form or the front\n>>form, exactly the same applies. And all these forms are equivalent\n>>in the sense that they can be transformed into each other by unitary\n>>transformations. Thus their choice is purely a matter of convenience.\n>\n> Let\'s take a simple example of the Hamiltonian of the hydrogen atom\n>\n> H = p^2/2m + 1/r (1)\n>\n> Let\'s form another Hamiltonian by a unitary transformation\n>\n> H\' = U H U^{-1} (2)\n>\n> Are you saying that Hamiltonians (1) and (2) describe the same physics?\n\nYes, if the wave function is transformed, too, according to\npsi\'= U psi.\nThis is standard practice.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>And if you construct the representation in the point form or the front
>>form, exactly the same applies. And all these forms are equivalent
>>in the sense that they can be transformed into each other by unitary
>>transformations. Thus their choice is purely a matter of convenience.
>
> Let's take a simple example of the Hamiltonian of the hydrogen atom
>
> H = p^2/2m + 1/r (1)
>
> Let's form another Hamiltonian by a unitary transformation
>
> H' = U H U^{-1} (2)
>
> Are you saying that Hamiltonians (1) and (2) describe the same physics?

Yes, if the wave function is transformed, too, according to
\psi'= U \psi.
This is standard practice.

Arnold Neumaier

Eugene Stefanovich

Apr19-05, 02:56 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n> On 2005-04-19, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>>Igor Khavkine wrote:\n>\n>\n>>>Building a state at finite time that is not an eigen state and has a\n>>>well defined physical interpretation of having two localized particles\n>>>that can scatter is hard. In fact, I don\'t know of any way to do it\n>>>except to confine the particles by an external potential.\n>>\n>>Such scattering states are obtained naturally in the RQD approach\n>>without the need for an artificial external potential.\n>\n>\n> Ah, but what if a natural external potential is used? Can your approach\n> incorporate one at all? If you think all external potentials are\n> artificial, I\'m sure the engineers who designed the transistors in the\n> CPU of your computer will be surprised to learn that they\'ve been using\n> nonsense for the past few decades.\n\nI think we were discussing the ways how an exact and rigorous theory\ncan be formulated. I think you would agree with me that exact and\nrigorous formulation is possible only for isolated systems. In practical\nproblems, such as transistor calculations, this exact and rigorous\nformulation becomes a burden, and one needs to introduce approximations\nand simplifications,\nsuch as external potentials. An approximation can be trusted if it is\nderived from the ab initio approach via series of controlled and\njustified steps. Otherwise, if you use a heuristic approach without deep\nroots in exact and rigorous formalism, it is very easy to make a\nmistake. Your approach may work well in certain areas, but there is a\ngood chance that it will fail somewhere.\n\n>\n> BTW, how do you know that these scattering states are what you think\n> they are? Can you tell me what is the expectation value and variance for\n> the position of each particle?\n\nYes, the expectation values of position are given by matrix elements of\nthe Newton-Wigner position operator. For example, I can prepare a state\nof two particles one of which is localized at point x, and another is\nlocalized at point y:\n\na^*(x) a^*(y) |0>\n\nThe creation operators in the position representation are Fourier images\nof the usual momentum-space creation operators.\n\n>\n>\n>>>To spell it out: space ship == external infinite potential well,\n>>>ball == scalar charged particle. This analogy is very close, but as a\n>>>precaution, I must say that it must not be taken literally.\n>>\n>>I understand this analogy, but I maintain that a rigorous theory\n>>should describe the time evolution, in general, and scattering,\n>>in particular, using a single well-defined Hamiltonian without\n>>any tricks with external potentials. You don\'t need to introduce\n>>external potentials when you calculate the S-matrix in QED,\n>>then why should you use them in calculations of the time evolution?\n>\n>\n> There is nothing unrigorous about using an external potential. If you\n> are really opposed to non-isolated systems,\n\nAs I said, I do not oppose the use of external potentials in complex\nsystems where such an approximation is justified. We were discussing\nthe simplest systems of all: one particle, or, perhaps, two particles.\nI hope that at least in such cases we can do without any approximation\nat all.\n\nYour idea of placing particles in some boxes does not look like\napproximation to me. These boxes change the physics of the problem\ncompletely. I would like to discuss the time evolution of a single\nparticle which does not interact with anything else. No boxes, please.\n\n\n\n>>>Time to get out paper and pencil.\n>>\n>>Not so fast. I cannot follow your derivation without understanding\n>>physical and/or mathematical premises of proposed steps. Even if I\n>>agree that your steps lead to some kind of "time evolution" you must\n>>convince me that this evolution is unitary (probabilities are preserved)\n>>and relativistically invariant (satisfies Poincare group relationships\n>>with space translations, rotations, and boosts). I am wondering, how\n>>you are going to do that without explicit expressions for the\n>>Hamiltonian and other Poincare generators.\n>\n>\n> On the contrary, I need not convince you of anything. The only person\n> who can convince you of anything is yourself. What I can do is present\n> you with facts, it is up to you to absorb them. Moreover, you *can*\n> follow the outlined steps, I hope I can put at least that much\n> confidence in your technical ability, however you do not seem inclined\n> to. That is unfortunate. You seek understanding but avoid calculations,\n> that is a path doomed to failure.\n\nYes, I think I *can* follow mathematical steps, but only if I understand\nthe physical reason behind them. I do not understand the physical reason\nbehind the steps you are proposing, so I am afraid to be involved in\nwitchcraft.\n\n>\n> There are a thousand things you might want to check about the standard\n> claculational framework in QFT. However, it is not my intention to\n> expand on each individual one *before* an actual calculation has been\n> performed. All the things you are asking for are completely irrelevant\n> for the proposed calculation. However, once completed it will be a\n> template for any calculation you wish to do whose result can be\n> expressed in terms of expectation values of products of field operators,\n> which includes all operators of physical interest. You are welcome to\n> attempt any of these on your onwn time once you grasp the basic\n> formalism.\n>\n> The result of the particular proposed calculation will be the time\n> dependence of the expectation value of the position operator for one of\n> the particles after the other was somehow perturbed. That is all I claim\n> to deliver. Anything that is not necessary for the calculation will not\n> make an appearance. This includes an explicit expression for physical\n> states, an explicit expression for the matrix elements of the\n> Hamiltonian between them, or anything that has to do with translation or\n> boost generators.\n\nHow can I trust you if the first step proposed by you is to put\nparticles in "boxes"? By doing this you completely modify the nature of\nthe problem.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On 2005-04-19, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>>Igor Khavkine wrote:
>
>
>>>Building a state at finite time that is not an eigen state and has a
>>>well defined physical interpretation of having two localized particles
>>>that can scatter is hard. In fact, I don't know of any way to do it
>>>except to confine the particles by an external potential.
>>
>>Such scattering states are obtained naturally in the RQD approach
>>without the need for an artificial external potential.
>
>
> Ah, but what if a natural external potential is used? Can your approach
> incorporate one at all? If you think all external potentials are
> artificial, I'm sure the engineers who designed the transistors in the
> CPU of your computer will be surprised to learn that they've been using
> nonsense for the past few decades.

I think we were discussing the ways how an exact and rigorous theory
can be formulated. I think you would agree with me that exact and
rigorous formulation is possible only for isolated systems. In practical
problems, such as transistor calculations, this exact and rigorous
formulation becomes a burden, and one needs to introduce approximations
and simplifications,
such as external potentials. An approximation can be trusted if it is
derived from the ab initio approach via series of controlled and
justified steps. Otherwise, if you use a heuristic approach without deep
roots in exact and rigorous formalism, it is very easy to make a
mistake. Your approach may work well in certain areas, but there is a
good chance that it will fail somewhere.

>
> BTW, how do you know that these scattering states are what you think
> they are? Can you tell me what is the expectation value and variance for
> the position of each particle?

Yes, the expectation values of position are given by matrix elements of
the Newton-Wigner position operator. For example, I can prepare a state
of two particles one of which is localized at point x, and another is
localized at point y:

a^*(x) a^*(y) |0>

The creation operators in the position representation are Fourier images
of the usual momentum-space creation operators.

>
>
>>>To spell it out: space ship == external infinite potential well,
>>>ball == scalar charged particle. This analogy is very close, but as a
>>>precaution, I must say that it must not be taken literally.
>>
>>I understand this analogy, but I maintain that a rigorous theory
>>should describe the time evolution, in general, and scattering,
>>in particular, using a single well-defined Hamiltonian without
>>any tricks with external potentials. You don't need to introduce
>>external potentials when you calculate the S-matrix in QED,
>>then why should you use them in calculations of the time evolution?
>
>
> There is nothing unrigorous about using an external potential. If you
> are really opposed to non-isolated systems,

As I said, I do not oppose the use of external potentials in complex
systems where such an approximation is justified. We were discussing
the simplest systems of all: one particle, or, perhaps, two particles.
I hope that at least in such cases we can do without any approximation
at all.

Your idea of placing particles in some boxes does not look like
approximation to me. These boxes change the physics of the problem
completely. I would like to discuss the time evolution of a single
particle which does not interact with anything else. No boxes, please.

>>>Time to get out paper and pencil.
>>
>>Not so fast. I cannot follow your derivation without understanding
>>physical and/or mathematical premises of proposed steps. Even if I
>>agree that your steps lead to some kind of "time evolution" you must
>>convince me that this evolution is unitary (probabilities are preserved)
>>and relativistically invariant (satisfies Poincare group relationships
>>with space translations, rotations, and boosts). I am wondering, how
>>you are going to do that without explicit expressions for the
>>Hamiltonian and other Poincare generators.
>
>
> On the contrary, I need not convince you of anything. The only person
> who can convince you of anything is yourself. What I can do is present
> you with facts, it is up to you to absorb them. Moreover, you *can*
> follow the outlined steps, I hope I can put at least that much
> confidence in your technical ability, however you do not seem inclined
> to. That is unfortunate. You seek understanding but avoid calculations,
> that is a path doomed to failure.

Yes, I think I *can* follow mathematical steps, but only if I understand
the physical reason behind them. I do not understand the physical reason
behind the steps you are proposing, so I am afraid to be involved in
witchcraft.

>
> There are a thousand things you might want to check about the standard
> claculational framework in QFT. However, it is not my intention to
> expand on each individual one *before* an actual calculation has been
> performed. All the things you are asking for are completely irrelevant
> for the proposed calculation. However, once completed it will be a
> template for any calculation you wish to do whose result can be
> expressed in terms of expectation values of products of field operators,
> which includes all operators of physical interest. You are welcome to
> attempt any of these on your onwn time once you grasp the basic
> formalism.
>
> The result of the particular proposed calculation will be the time
> dependence of the expectation value of the position operator for one of
> the particles after the other was somehow perturbed. That is all I claim
> to deliver. Anything that is not necessary for the calculation will not
> make an appearance. This includes an explicit expression for physical
> states, an explicit expression for the matrix elements of the
> Hamiltonian between them, or anything that has to do with translation or
> boost generators.

How can I trust you if the first step proposed by you is to put
particles in "boxes"? By doing this you completely modify the nature of
the problem.

Eugene Stefanovich.

Arnold Neumaier

Apr20-05, 12:30 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> Arnold Neumaier wrote:\n>\n>>>>>>On the quantum level, (8.4.1) is only a meaningless formal\n>>>>>>expression. It is definitely _not_ the physical QED Hamiltonian!\n>>>>>\n>>>>>Disagreed.\n>>>>\n>>But below you agreed. It is impossible to argue with such a\n>>schizophrenic mind.\n>\n> Although H is the true Hamiltonian\n> of QED, it is not the true Hamiltonian\n> of interacting charged particles and photons.\n\nThere is only one true Hamiltonian of QED, and it describes\ninteracting charged particles and photons.\n\n\n> This schizophrenic situation is easily explained. QED was never\n> interested in time evolution.\n\nYou don\'t know what you are talking about. QED is not only\nscattering calculations from standard rtextbooks, but a lot more!\n\nYou explicitly confessed your ignorance about much I told you,\nwhere QED is applied to time-dependent processes (Dirac-Fock,\nBoltzmann, relativistic hydrodynamics), and I could name others,\ne.g. search http://scholar.google.com/ for\n"molecular quantum electrodynamics".\n\n\n>>>Regularization is nothing more than a temporary detour\n>>>to the imaginary world with cut-off interactions or\n>>>non-integer dimension.\n>>\n>>No. Renormalization is the process that defines the meaning\n>>of QFT. It is _not_ a detour but the heart that makes\n>>everything work.\n>\n>\n> I was talking about regularization. In my vocabulary\n> "regularization" is modifying the theory so that all integrals\n> become temporarily finite, and one can do necessary cancellations.\n> At the end the regulators should be removed, and the final results\n> do not bear any signs of what kind of regularization was performed.\n>\n> "Renormalization" is adding (generally infinite) counterterms to the\n> original Hamiltonian (with physical masses and charges).\n> Renormalization makes the Hamiltonian infinite,\n> however it allows to obtain finite and accurate scattering amplitudes\n> and energies of bound states.\n>\n> So, regularization is a "detour" and I am right.\n> Renormalization is "the heart that makes everything work",\n> and you are right too.\n\nRenormalization involves regularization, since the counterterms make\nsense only in the regularized version. Hence regularization is part\nof the heart.\n\n\n> Then, possibly you can find an error in my derivations,\n\nI am not interested in finding errors in your derivations.\nIt is clear from your conclusions that you are not consistent\nwith tradition, and I see no reason to distrust the latter in this\nrespect.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> Arnold Neumaier wrote:
>
>>>>>>On the quantum level, (8.4.1) is only a meaningless formal
>>>>>>expression. It is definitely _not_ the physical QED Hamiltonian!
>>>>>
>>>>>Disagreed.
>>>>
>>But below you agreed. It is impossible to argue with such a
>>schizophrenic mind.
>
> Although H is the true Hamiltonian
> of QED, it is not the true Hamiltonian
> of interacting charged particles and photons.

There is only one true Hamiltonian of QED, and it describes
interacting charged particles and photons.

> This schizophrenic situation is easily explained. QED was never
> interested in time evolution.

You don't know what you are talking about. QED is not only
scattering calculations from standard rtextbooks, but a lot more!

You explicitly confessed your ignorance about much I told you,
where QED is applied to time-dependent processes (Dirac-Fock,
Boltzmann, relativistic hydrodynamics), and I could name others,
e.g. search http://scholar.google.com/ for
"molecular quantum electrodynamics".

>>>Regularization is nothing more than a temporary detour
>>>to the imaginary world with cut-off interactions or
>>>non-integer dimension.
>>
>>No. Renormalization is the process that defines the meaning
>>of QFT. It is _not_ a detour but the heart that makes
>>everything work.
>
>
> I was talking about regularization. In my vocabulary
> "regularization" is modifying the theory so that all integrals
> become temporarily finite, and one can do necessary cancellations.
> At the end the regulators should be removed, and the final results
> do not bear any signs of what kind of regularization was performed.
>
> "Renormalization" is adding (generally infinite) counterterms to the
> original Hamiltonian (with physical masses and charges).
> Renormalization makes the Hamiltonian infinite,
> however it allows to obtain finite and accurate scattering amplitudes
> and energies of bound states.
>
> So, regularization is a "detour" and I am right.
> Renormalization is "the heart that makes everything work",
> and you are right too.

Renormalization involves regularization, since the counterterms make
sense only in the regularized version. Hence regularization is part
of the heart.

> Then, possibly you can find an error in my derivations,

I am not interested in finding errors in your derivations.
It is clear from your conclusions that you are not consistent
with tradition, and I see no reason to distrust the latter in this
respect.

Arnold Neumaier

Arnold Neumaier

Apr20-05, 12:31 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n\n> Arnold Neumaier wrote:\n\n>>If the Hamiltonian is H (defined implicitly but uniquely) then\n>> psi(t)=exp(-i/hbar H) psi(0),\n>>as always in quantum mechanics. This makes sense for every selfadjoint\n>>operator H, whether it has the form preferred by you or whether it is\n>>constructed in any other way. See any book on functional analysis,\n>>e.g. the treatise by Reed and Simon.\n>>\n>>If you find this too hard, you are not qualified for criticising\n>>tradition.\n>\n>\n> I fully agree with your formula\n>\n> psi(t)=exp(-i/hbar H t) psi(0)\n>\n> (let\'s not argue about i or -i in the exponent, that\'s not relevant\n> here).\n> The tiny little problem we have is what should be used in this\n> formula in place of H.\n\nThe implicitly defined H, e.g. from Glimm and Jaffe.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:

> Arnold Neumaier wrote:

>>If the Hamiltonian is H (defined implicitly but uniquely) then
>> \psi(t)=\exp(-i/\hbar H) \psi(0),
>>as always in quantum mechanics. This makes sense for every selfadjoint
>>operator H, whether it has the form preferred by you or whether it is
>>constructed in any other way. See any book on functional analysis,
>>e.g. the treatise by Reed and Simon.
>>
>>If you find this too hard, you are not qualified for criticising
>>tradition.
>
>
> I fully agree with your formula
>
> \psi(t)=\exp(-i/\hbar H t) \psi(0)
>
> (let's not argue about i or -i in the exponent, that's not relevant
> here).
> The tiny little problem we have is what should be used in this
> formula in place of H.

The implicitly defined H, e.g. from Glimm and Jaffe.

Igor Khavkine

Apr20-05, 01:00 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-19, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n\n> I think we were discussing the ways how an exact and rigorous theory\n> can be formulated. I think you would agree with me that exact and\n> rigorous formulation is possible only for isolated systems. In practical\n> problems, such as transistor calculations, this exact and rigorous\n> formulation becomes a burden, and one needs to introduce approximations\n> and simplifications,\n> such as external potentials. An approximation can be trusted if it is\n> derived from the ab initio approach via series of controlled and\n> justified steps. Otherwise, if you use a heuristic approach without deep\n> roots in exact and rigorous formalism, it is very easy to make a\n> mistake. Your approach may work well in certain areas, but there is a\n> good chance that it will fail somewhere.\n\nPhysics is about practical problems and their solutions. At least that\'s\nwhat I was discussing. Physics is also the science of approximations.\nThis includes applying them where appropriate and also not applying them\nwhere not appropriate. To a working physicist, a physically justified\napproximation is just as valid and sometimes more desirable than purely\nmathematically derived one.\n\n> Your idea of placing particles in some boxes does not look like\n> approximation to me. These boxes change the physics of the problem\n> completely. I would like to discuss the time evolution of a single\n> particle which does not interact with anything else. No boxes, please.\n\nI\'ve already said what I intend to discuss, and single particles is not\nit. The boxes may be removed at any finite time, the only change is once\nagain a time dependent potential. However, their removal introduces\nextra labor into the calculation. I do not intend to consider it.\n\n> How can I trust you if the first step proposed by you is to put\n> particles in "boxes"? By doing this you completely modify the nature of\n> the problem.\n\nThe relevant features of the model are preserved:\n\n* relativistic equations for the electromagnetic field\n* same form of interaction as in QED\n* short distance behavior of either the EM or scalar field are preserved\n\nGaining your trust is not something I am particularly intersted in. I\nhave answered your questions about the physical justification of the\nmodel. The general formalism admits external and even time dependent\npotentials, hence there it is mathematically consistent as well. I have\nalso told you what you should expect to see and not to see in the course\nof the calculation. The remaining objections to the use of this model\nare in the realm of philosophy, that is two floors up from the Physics\ndepartment.\n\nThe choice is yours, either you start following the outlined calculation\nor this thread has no reason to continue.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-19, Eugene Stefanovich <eugenev@synopsys.com> wrote:

> I think we were discussing the ways how an exact and rigorous theory
> can be formulated. I think you would agree with me that exact and
> rigorous formulation is possible only for isolated systems. In practical
> problems, such as transistor calculations, this exact and rigorous
> formulation becomes a burden, and one needs to introduce approximations
> and simplifications,
> such as external potentials. An approximation can be trusted if it is
> derived from the ab initio approach via series of controlled and
> justified steps. Otherwise, if you use a heuristic approach without deep
> roots in exact and rigorous formalism, it is very easy to make a
> mistake. Your approach may work well in certain areas, but there is a
> good chance that it will fail somewhere.

Physics is about practical problems and their solutions. At least that's
what I was discussing. Physics is also the science of approximations.
This includes applying them where appropriate and also not applying them
where not appropriate. To a working physicist, a physically justified
approximation is just as valid and sometimes more desirable than purely
mathematically derived one.

> Your idea of placing particles in some boxes does not look like
> approximation to me. These boxes change the physics of the problem
> completely. I would like to discuss the time evolution of a single
> particle which does not interact with anything else. No boxes, please.

I've already said what I intend to discuss, and single particles is not
it. The boxes may be removed at any finite time, the only change is once
again a time dependent potential. However, their removal introduces
extra labor into the calculation. I do not intend to consider it.

> How can I trust you if the first step proposed by you is to put
> particles in "boxes"? By doing this you completely modify the nature of
> the problem.

The relevant features of the model are preserved:

* relativistic equations for the electromagnetic field
* same form of interaction as in QED
* short distance behavior of either the EM or scalar field are preserved

Gaining your trust is not something I am particularly intersted in. I
have answered your questions about the physical justification of the
model. The general formalism admits external and even time dependent
potentials, hence there it is mathematically consistent as well. I have
also told you what you should expect to see and not to see in the course
of the calculation. The remaining objections to the use of this model
are in the realm of philosophy, that is two floors up from the Physics
department.

The choice is yours, either you start following the outlined calculation
or this thread has no reason to continue.

Igor

Eugene Stefanovich

Apr20-05, 01:57 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>Arnold Neumaier wrote:\n>>\n>>\n>>>>>>>On the quantum level, (8.4.1) is only a meaningless formal\n>>>>>>>expression. It is definitely _not_ the physical QED Hamiltonian!\n>>>>>>\n>>>>>>Disagreed.\n>>>>>\n>>> But below you agreed. It is impossible to argue with such a\n>>>schizophrenic mind.\n>>\n>>Although H is the true Hamiltonian\n>>of QED, it is not the true Hamiltonian\n>>of interacting charged particles and photons.\n>\n>\n> There is only one true Hamiltonian of QED, and it describes\n> interacting charged particles and photons.\n\nThat\'s right. And this Hamiltonian cannot describe even the time\nevolution of a lousy single particle.\n\n>\n>\n>\n>>This schizophrenic situation is easily explained. QED was never\n>>interested in time evolution.\n>\n>\n> You don\'t know what you are talking about. QED is not only\n> scattering calculations from standard rtextbooks, but a lot more!\n>\n> You explicitly confessed your ignorance about much I told you,\n> where QED is applied to time-dependent processes (Dirac-Fock,\n> Boltzmann, relativistic hydrodynamics), and I could name others,\n> e.g. search http://scholar.google.com/ for\n> "molecular quantum electrodynamics".\n\nWhat about the time evolution of a single particle?\nStandard QED Hamiltonian cannot do that.\n\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>Arnold Neumaier wrote:
>>
>>
>>>>>>>On the quantum level, (8.4.1) is only a meaningless formal
>>>>>>>expression. It is definitely _not_ the physical QED Hamiltonian!
>>>>>>
>>>>>>Disagreed.
>>>>>
>>>But below you agreed. It is impossible to argue with such a
>>>schizophrenic mind.
>>
>>Although H is the true Hamiltonian
>>of QED, it is not the true Hamiltonian
>>of interacting charged particles and photons.
>
>
> There is only one true Hamiltonian of QED, and it describes
> interacting charged particles and photons.

That's right. And this Hamiltonian cannot describe even the time
evolution of a lousy single particle.

>
>
>
>>This schizophrenic situation is easily explained. QED was never
>>interested in time evolution.
>
>
> You don't know what you are talking about. QED is not only
> scattering calculations from standard rtextbooks, but a lot more!
>
> You explicitly confessed your ignorance about much I told you,
> where QED is applied to time-dependent processes (Dirac-Fock,
> Boltzmann, relativistic hydrodynamics), and I could name others,
> e.g. search http://scholar.google.com/ for
> "molecular quantum electrodynamics".

What about the time evolution of a single particle?
Standard QED Hamiltonian cannot do that.

Eugene Stefanovich.

Eugene Stefanovich

Apr20-05, 01:57 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n> Eugene Stefanovich wrote:\n>\n>\n>>Arnold Neumaier wrote:\n>\n>\n>>>If the Hamiltonian is H (defined implicitly but uniquely) then\n>>> psi(t)=exp(-i/hbar H) psi(0),\n>>>as always in quantum mechanics. This makes sense for every selfadjoint\n>>>operator H, whether it has the form preferred by you or whether it is\n>>>constructed in any other way. See any book on functional analysis,\n>>>e.g. the treatise by Reed and Simon.\n>>>\n>>>If you find this too hard, you are not qualified for criticising\n>>>tradition.\n>>\n>>\n>>I fully agree with your formula\n>>\n>>psi(t)=exp(-i/hbar H t) psi(0)\n>>\n>>(let\'s not argue about i or -i in the exponent, that\'s not relevant\n>>here).\n>>The tiny little problem we have is what should be used in this\n>>formula in place of H.\n>\n>\n> The implicitly defined H, e.g. from Glimm and Jaffe.\n>\n\nI don\'t know what\'s the point of this implicit definition if it\ncannot be used to evaluate the above formula even in the simplest case\nwhen psi is a state of one free particle.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:
> Eugene Stefanovich wrote:
>
>
>>Arnold Neumaier wrote:
>
>
>>>If the Hamiltonian is H (defined implicitly but uniquely) then
>>> \psi(t)=\exp(-i/\hbar H) \psi(0),
>>>as always in quantum mechanics. This makes sense for every selfadjoint
>>>operator H, whether it has the form preferred by you or whether it is
>>>constructed in any other way. See any book on functional analysis,
>>>e.g. the treatise by Reed and Simon.
>>>
>>>If you find this too hard, you are not qualified for criticising
>>>tradition.
>>
>>
>>I fully agree with your formula
>>
>>\psi(t)=\exp(-i/\hbar H t) \psi(0)
>>
>>(let's not argue about i or -i in the exponent, that's not relevant
>>here).
>>The tiny little problem we have is what should be used in this
>>formula in place of H.
>
>
> The implicitly defined H, e.g. from Glimm and Jaffe.
>

I don't know what's the point of this implicit definition if it
cannot be used to evaluate the above formula even in the simplest case
when \psi is a state of one free particle.

Eugene Stefanovich.

Eugene Stefanovich

Apr20-05, 01:57 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nIgor Khavkine wrote:\n> On 2005-04-19, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n>\n>\n>>I think we were discussing the ways how an exact and rigorous theory\n>>can be formulated. I think you would agree with me that exact and\n>>rigorous formulation is possible only for isolated systems. In practical\n>> problems, such as transistor calculations, this exact and rigorous\n>>formulation becomes a burden, and one needs to introduce approximations\n>>and simplifications,\n>>such as external potentials. An approximation can be trusted if it is\n>>derived from the ab initio approach via series of controlled and\n>>justified steps. Otherwise, if you use a heuristic approach without deep\n>>roots in exact and rigorous formalism, it is very easy to make a\n>>mistake. Your approach may work well in certain areas, but there is a\n>>good chance that it will fail somewhere.\n>\n>\n> Physics is about practical problems and their solutions. At least that\'s\n> what I was discussing. Physics is also the science of approximations.\n> This includes applying them where appropriate and also not applying them\n> where not appropriate. To a working physicist, a physically justified\n> approximation is just as valid and sometimes more desirable than purely\n> mathematically derived one.\n\nGenerally, I agree with you, but there should be some physical\nproblems (like the time evolution of a single free particle) that\nshould be solvable without any approximation. Otherwise the sanity of\nthe theory becomes questionable.\n\n>\n>\n>>Your idea of placing particles in some boxes does not look like\n>>approximation to me. These boxes change the physics of the problem\n>>completely. I would like to discuss the time evolution of a single\n>>particle which does not interact with anything else. No boxes, please.\n>\n>\n> I\'ve already said what I intend to discuss, and single particles is not\n> it.\n\nI am not going to trust your 2-particle calculation until you\nshow that you can reproduce the time evolution of one isolated particle.\n\n> The boxes may be removed at any finite time, the only change is once\n> again a time dependent potential. However, their removal introduces\n> extra labor into the calculation. I do not intend to consider it.\n>\n>\n>>How can I trust you if the first step proposed by you is to put\n>>particles in "boxes"? By doing this you completely modify the nature of\n>>the problem.\n>\n>\n> The relevant features of the model are preserved:\n>\n> * relativistic equations for the electromagnetic field\n> * same form of interaction as in QED\n> * short distance behavior of either the EM or scalar field are preserved\n>\n> Gaining your trust is not something I am particularly intersted in. I\n> have answered your questions about the physical justification of the\n> model. The general formalism admits external and even time dependent\n> potentials, hence there it is mathematically consistent as well. I have\n> also told you what you should expect to see and not to see in the course\n> of the calculation. The remaining objections to the use of this model\n> are in the realm of philosophy, that is two floors up from the Physics\n> department.\n>\n> The choice is yours, either you start following the outlined calculation\n> or this thread has no reason to continue.\n\nIt is clear to me that using the Hamiltonian of QED (rigorously derived\nby Weinberg\nin eq. (8.4.1) using all bells and whistles of the canonical quantization\ntheory) one cannot give any sensible description of the time evolution\nof a single free particle. I don\'t see how your attempts are going to\nsave the situation.\n\nEugene Stefanovich.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On 2005-04-19, Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>
>>I think we were discussing the ways how an exact and rigorous theory
>>can be formulated. I think you would agree with me that exact and
>>rigorous formulation is possible only for isolated systems. In practical
>> problems, such as transistor calculations, this exact and rigorous
>>formulation becomes a burden, and one needs to introduce approximations
>>and simplifications,
>>such as external potentials. An approximation can be trusted if it is
>>derived from the ab initio approach via series of controlled and
>>justified steps. Otherwise, if you use a heuristic approach without deep
>>roots in exact and rigorous formalism, it is very easy to make a
>>mistake. Your approach may work well in certain areas, but there is a
>>good chance that it will fail somewhere.
>
>
> Physics is about practical problems and their solutions. At least that's
> what I was discussing. Physics is also the science of approximations.
> This includes applying them where appropriate and also not applying them
> where not appropriate. To a working physicist, a physically justified
> approximation is just as valid and sometimes more desirable than purely
> mathematically derived one.

Generally, I agree with you, but there should be some physical
problems (like the time evolution of a single free particle) that
should be solvable without any approximation. Otherwise the sanity of
the theory becomes questionable.

>
>
>>Your idea of placing particles in some boxes does not look like
>>approximation to me. These boxes change the physics of the problem
>>completely. I would like to discuss the time evolution of a single
>>particle which does not interact with anything else. No boxes, please.
>
>
> I've already said what I intend to discuss, and single particles is not
> it.

I am not going to trust your 2-particle calculation until you
show that you can reproduce the time evolution of one isolated particle.

> The boxes may be removed at any finite time, the only change is once
> again a time dependent potential. However, their removal introduces
> extra labor into the calculation. I do not intend to consider it.
>
>
>>How can I trust you if the first step proposed by you is to put
>>particles in "boxes"? By doing this you completely modify the nature of
>>the problem.
>
>
> The relevant features of the model are preserved:
>
> * relativistic equations for the electromagnetic field
> * same form of interaction as in QED
> * short distance behavior of either the EM or scalar field are preserved
>
> Gaining your trust is not something I am particularly intersted in. I
> have answered your questions about the physical justification of the
> model. The general formalism admits external and even time dependent
> potentials, hence there it is mathematically consistent as well. I have
> also told you what you should expect to see and not to see in the course
> of the calculation. The remaining objections to the use of this model
> are in the realm of philosophy, that is two floors up from the Physics
> department.
>
> The choice is yours, either you start following the outlined calculation
> or this thread has no reason to continue.

It is clear to me that using the Hamiltonian of QED (rigorously derived
by Weinberg
in eq. (8.4.1) using all bells and whistles of the canonical quantization
theory) one cannot give any sensible description of the time evolution
of a single free particle. I don't see how your attempts are going to
save the situation.

Eugene Stefanovich.

Igor Khavkine

Apr20-05, 04:08 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On 2005-04-20, Eugene Stefanovich <eugenev@synopsys.com> wrote:\n> Igor Khavkine wrote:\n\n>> I\'ve already said what I intend to discuss, and single particles is not\n>> it.\n>\n> I am not going to trust your 2-particle calculation until you\n> show that you can reproduce the time evolution of one isolated particle.\n\nYour trust is yours to give and not mine to covet.\n\n>> The choice is yours, either you start following the outlined calculation\n>> or this thread has no reason to continue.\n>\n> It is clear to me that using the Hamiltonian of QED (rigorously derived\n> by Weinberg\n> in eq. (8.4.1) using all bells and whistles of the canonical quantization\n> theory) one cannot give any sensible description of the time evolution\n> of a single free particle. I don\'t see how your attempts are going to\n> save the situation.\n\nSince you don\'t see it after this extensive discussion and are not\nwilling to just give it a try, perhaps you never will. That is\nunfortunate, but no longer a concern of mine. I am retiring from this\nconversation.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On 2005-04-20, Eugene Stefanovich <eugenev@synopsys.com> wrote:
> Igor Khavkine wrote:

>> I've already said what I intend to discuss, and single particles is not
>> it.
>
> I am not going to trust your 2-particle calculation until you
> show that you can reproduce the time evolution of one isolated particle.

Your trust is yours to give and not mine to covet.

>> The choice is yours, either you start following the outlined calculation
>> or this thread has no reason to continue.
>
> It is clear to me that using the Hamiltonian of QED (rigorously derived
> by Weinberg
> in eq. (8.4.1) using all bells and whistles of the canonical quantization
> theory) one cannot give any sensible description of the time evolution
> of a single free particle. I don't see how your attempts are going to
> save the situation.

Since you don't see it after this extensive discussion and are not
willing to just give it a try, perhaps you never will. That is
unfortunate, but no longer a concern of mine. I am retiring from this
conversation.

Igor

Eugene Stefanovich

Apr21-05, 04:27 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Igor Khavkine wrote:\n\n>>>The choice is yours, either you start following the outlined calculation\n>>>or this thread has no reason to continue.\n>>\n>>It is clear to me that using the Hamiltonian of QED (rigorously derived\n>>by Weinberg\n>>in eq. (8.4.1) using all bells and whistles of the canonical quantization\n>>theory) one cannot give any sensible description of the time evolution\n>>of a single free particle. I don\'t see how your attempts are going to\n>>save the situation.\n>\n>\n> Since you don\'t see it after this extensive discussion and are not\n> willing to just give it a try, perhaps you never will. That is\n> unfortunate, but no longer a concern of mine. I am retiring from this\n> conversation.\n\nI am not trying your approach because I do not understand how are you\ngoing to describe the time evolution of a *free* particle but putting it\nin a *box*. Explain the physics of what you are doing and we\'ll have a\nnice conversation.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:

>>>The choice is yours, either you start following the outlined calculation
>>>or this thread has no reason to continue.
>>
>>It is clear to me that using the Hamiltonian of QED (rigorously derived
>>by Weinberg
>>in eq. (8.4.1) using all bells and whistles of the canonical quantization
>>theory) one cannot give any sensible description of the time evolution
>>of a single free particle. I don't see how your attempts are going to
>>save the situation.
>
>
> Since you don't see it after this extensive discussion and are not
> willing to just give it a try, perhaps you never will. That is
> unfortunate, but no longer a concern of mine. I am retiring from this
> conversation.

I am not trying your approach because I do not understand how are you
going to describe the time evolution of a *free* particle but putting it
in a *box*. Explain the physics of what you are doing and we'll have a
nice conversation.

Eugene.

Arnold Neumaier

Apr21-05, 04:28 PM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich wrote:\n>\n> Arnold Neumaier wrote:\n>\n>> Eugene Stefanovich wrote:\n>>\n>>> Arnold Neumaier wrote:\n>>>\n>>>>>>>> On the quantum level, (8.4.1) is only a meaningless formal\n>>>>>>>> expression. It is definitely _not_ the physical QED Hamiltonian!\n>>>>>>>\n>>>>>>> Disagreed.\n>>>>>>\n>>>> But below you agreed. It is impossible to argue with such a\n>>>> schizophrenic mind.\n>>>\n>>> Although H is the true Hamiltonian\n>>> of QED, it is not the true Hamiltonian\n>>> of interacting charged particles and photons.\n>>\n>> There is only one true Hamiltonian of QED, and it describes\n>> interacting charged particles and photons.\n>\n> That\'s right. And this Hamiltonian cannot describe even the time\n> evolution of a lousy single particle.\n\nOf course it does. Your remark only shows how little you understand.\n\nBut the discussion is again degenerating, and I\'ll stop contributing\nto this topic until you showed significant signs of a readiness to\nlearn the state of the art.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:
>
>> Eugene Stefanovich wrote:
>>
>>> Arnold Neumaier wrote:
>>>
>>>>>>>> On the quantum level, (8.4.1) is only a meaningless formal
>>>>>>>> expression. It is definitely _not_ the physical QED Hamiltonian!
>>>>>>>
>>>>>>> Disagreed.
>>>>>>
>>>> But below you agreed. It is impossible to argue with such a
>>>> schizophrenic mind.
>>>
>>> Although H is the true Hamiltonian
>>> of QED, it is not the true Hamiltonian
>>> of interacting charged particles and photons.
>>
>> There is only one true Hamiltonian of QED, and it describes
>> interacting charged particles and photons.
>
> That's right. And this Hamiltonian cannot describe even the time
> evolution of a lousy single particle.

Of course it does. Your remark only shows how little you understand.

But the discussion is again degenerating, and I'll stop contributing
to this topic until you showed significant signs of a readiness to
learn the state of the art.

Arnold Neumaier

Eugene Stefanovich

Apr22-05, 12:02 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nArnold Neumaier wrote:\n\n>>>>\n>>>>Although H is the true Hamiltonian\n>>>>of QED, it is not the true Hamiltonian\n>>>>of interacting charged particles and photons.\n>>>\n>>>There is only one true Hamiltonian of QED, and it describes\n>>>interacting charged particles and photons.\n>>\n>>That\'s right. And this Hamiltonian cannot describe even the time\n>>evolution of a lousy single particle.\n>\n>\n> Of course it does. Your remark only shows how little you understand.\n>\n> But the discussion is again degenerating, and I\'ll stop contributing\n> to this topic until you showed significant signs of a readiness to\n> learn the state of the art.\n\n\nI will do the same until you showed me a Hamiltonian of QED and\nhow one can calculate the time evolution of a single free particle\nwith this Hamiltonian.\n\nEugene Stefanovich.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Arnold Neumaier wrote:

>>>>
>>>>Although H is the true Hamiltonian
>>>>of QED, it is not the true Hamiltonian
>>>>of interacting charged particles and photons.
>>>
>>>There is only one true Hamiltonian of QED, and it describes
>>>interacting charged particles and photons.
>>
>>That's right. And this Hamiltonian cannot describe even the time
>>evolution of a lousy single particle.
>
>
> Of course it does. Your remark only shows how little you understand.
>
> But the discussion is again degenerating, and I'll stop contributing
> to this topic until you showed significant signs of a readiness to
> learn the state of the art.

I will do the same until you showed me a Hamiltonian of QED and
how one can calculate the time evolution of a single free particle
with this Hamiltonian.

Eugene Stefanovich.

Arnold Neumaier

Oct11-06, 02:39 PM

Eugene Stefanovich wrote:
> Arnold Neumaier wrote:
>
>>And if you construct the representation in the point form or the front
>>form, exactly the same applies. And all these forms are equivalent
>>in the sense that they can be transformed into each other by unitary
>>transformations. Thus their choice is purely a matter of convenience.
>
> Let's take a simple example of the Hamiltonian of the hydrogen atom
>
> H = p^2/2m + 1/r (1)
>
> Let's form another Hamiltonian by a unitary transformation
>
> H' = U H U^{-1} (2)
>
> Are you saying that Hamiltonians (1) and (2) describe the same physics?

Yes, if the wave function is transformed, too, according to
psi'= U psi.
This is standard practice.

Arnold Neumaier

Arnold Neumaier

Oct11-06, 02:40 PM

Eugene Stefanovich wrote:

> Arnold Neumaier wrote:
>
>>>>>>On the quantum level, (8.4.1) is only a meaningless formal
>>>>>>expression. It is definitely _not_ the physical QED Hamiltonian!
>>>>>
>>>>>Disagreed.
>>>>
>>But below you agreed. It is impossible to argue with such a
>>schizophrenic mind.
>
> Although H is the true Hamiltonian
> of QED, it is not the true Hamiltonian
> of interacting charged particles and photons.

There is only one true Hamiltonian of QED, and it describes
interacting charged particles and photons.

> This schizophrenic situation is easily explained. QED was never
> interested in time evolution.

You don't know what you are talking about. QED is not only
scattering calculations from standard rtextbooks, but a lot more!

You explicitly confessed your ignorance about much I told you,
where QED is applied to time-dependent processes (Dirac-Fock,
Boltzmann, relativistic hydrodynamics), and I could name others,
e.g. search http://scholar.google.com/ for
"molecular quantum electrodynamics".

>>>Regularization is nothing more than a temporary detour
>>>to the imaginary world with cut-off interactions or
>>>non-integer dimension.
>>
>>No. Renormalization is the process that defines the meaning
>>of QFT. It is _not_ a detour but the heart that makes
>>everything work.
>
>
> I was talking about regularization. In my vocabulary
> "regularization" is modifying the theory so that all integrals
> become temporarily finite, and one can do necessary cancellations.
> At the end the regulators should be removed, and the final results
> do not bear any signs of what kind of regularization was performed.
>
> "Renormalization" is adding (generally infinite) counterterms to the
> original Hamiltonian (with physical masses and charges).
> Renormalization makes the Hamiltonian infinite,
> however it allows to obtain finite and accurate scattering amplitudes
> and energies of bound states.
>
> So, regularization is a "detour" and I am right.
> Renormalization is "the heart that makes everything work",
> and you are right too.

Renormalization involves regularization, since the counterterms make
sense only in the regularized version. Hence regularization is part
of the heart.

> Then, possibly you can find an error in my derivations,

I am not interested in finding errors in your derivations.
It is clear from your conclusions that you are not consistent
with tradition, and I see no reason to distrust the latter in this
respect.

Arnold Neumaier

Arnold Neumaier

Oct11-06, 02:40 PM

Eugene Stefanovich wrote:

> Arnold Neumaier wrote:

>>If the Hamiltonian is H (defined implicitly but uniquely) then
>> psi(t)=exp(-i/hbar H) psi(0),
>>as always in quantum mechanics. This makes sense for every selfadjoint
>>operator H, whether it has the form preferred by you or whether it is
>>constructed in any other way. See any book on functional analysis,
>>e.g. the treatise by Reed and Simon.
>>
>>If you find this too hard, you are not qualified for criticising
>>tradition.
>
>
> I fully agree with your formula
>
> psi(t)=exp(-i/hbar H t) psi(0)
>
> (let's not argue about i or -i in the exponent, that's not relevant
> here).
> The tiny little problem we have is what should be used in this
> formula in place of H.

The implicitly defined H, e.g. from Glimm and Jaffe.

Arnold Neumaier

Oct11-06, 02:40 PM

Eugene Stefanovich wrote:
>
> Arnold Neumaier wrote:
>
>> Eugene Stefanovich wrote:
>>
>>> Arnold Neumaier wrote:
>>>
>>>>>>>> On the quantum level, (8.4.1) is only a meaningless formal
>>>>>>>> expression. It is definitely _not_ the physical QED Hamiltonian!
>>>>>>>
>>>>>>> Disagreed.
>>>>>>
>>>> But below you agreed. It is impossible to argue with such a
>>>> schizophrenic mind.
>>>
>>> Although H is the true Hamiltonian
>>> of QED, it is not the true Hamiltonian
>>> of interacting charged particles and photons.
>>
>> There is only one true Hamiltonian of QED, and it describes
>> interacting charged particles and photons.
>
> That's right. And this Hamiltonian cannot describe even the time
> evolution of a lousy single particle.

Of course it does. Your remark only shows how little you understand.

But the discussion is again degenerating, and I'll stop contributing
to this topic until you showed significant signs of a readiness to
learn the state of the art.

Arnold Neumaier

Eugene Stefanovich

Oct11-06, 02:40 PM

Arnold Neumaier wrote:

>>>>
>>>>Although H is the true Hamiltonian
>>>>of QED, it is not the true Hamiltonian
>>>>of interacting charged particles and photons.
>>>
>>>There is only one true Hamiltonian of QED, and it describes
>>>interacting charged particles and photons.
>>
>>That's right. And this Hamiltonian cannot describe even the time
>>evolution of a lousy single particle.
>
>
> Of course it does. Your remark only shows how little you understand.
>
> But the discussion is again degenerating, and I'll stop contributing
> to this topic until you showed significant signs of a readiness to
> learn the state of the art.

I will do the same until you showed me a Hamiltonian of QED and
how one can calculate the time evolution of a single free particle
with this Hamiltonian.

Eugene Stefanovich.