What is the velocity of a relativistic electron?by Juan R. GonzálezÁlvarez Tags: electron, relativistic, velocity 
#1
Feb1408, 05:00 AM

P: n/a

I find different answers:
((Classical electrodynamics)) v < c ((Relativistic Quantum Mechanics)) v = c using Dirac equation for PSI>. ((Spacetime approach to QED)) v = c because the spacetime kernel K_+ is derived from the Dirac equation for PSI>. Feynman even writes in his wellknown textbook on QED [1] that {BLOCKQUOTE This result is sometimes made pausible by the argument that a precise determination of velocity implies precise determinations of position at two times. Then by the uncertainty principle, the momentum is completely uncertain and all values are equally likely, this is seen to imply that velocities near the speed of light are more probable, so that in the limit the expected value of the velocity is the speed of light. } Then in a footnote, Feynman recognizes that argument is invalid since v commutes with p. Thus, an incorrect value is justified using an incorrect argument! ((Field approach to QED)) I have computed the value of v using also the field approach to QED. I.e the value for the v corresponding to a Dirac field PHI(x,t) is again c. Conventional thinking says that position is not an observable in field QED (just a parameter) and thus v cannot be indentified with the velocity of a particle. This argument is difficult to accept because is if x and v are not observables at some fundamental level of theory, then, i) why there exist observables x and v at level of nonrelativistic quantum mechanics? ii) How does one obtain classical observables x and v? Already Dirac warned about QED but his quote is almost ignored (only his related quote on renormalization is usually cited). The quote on incompatibility of QED with previous theories is {BLOCKQUOTE Most physicists are very satisfied with this situation. They argue that if one has rules for doing calculations and the results agree with observation, that is all that one requires. But it is not all that one requires. One requires a single comprehensive theory applying to all physical phenomena. Not one theory for dealing with non relativistic effects and a separate disjoint theory for dealing with certain relativistic effects. [...] For these reasons I find the present quantum electrodynamics quite unsatisfactory. } 


#2
Feb1608, 05:00 AM

P: n/a

On 20080213, Juan R. GonzàlezÀlvarez <Juan@canonicalscience.com> wrote:
> I find different answers: > > ((Classical electrodynamics)) > > v < c > > ((Relativistic Quantum Mechanics)) > > v = c using Dirac equation for PSI>. Eh, no. You are confused by an archaic paradox. The paradox consists of identifying the operator c*alpha (in the usual 3+1 Dirac equation notation) with the physical velocity, which quite obviously doesn't work. For a modern discussion of the Dirac electron's velocity operator, see [1] where the author discusses the FoldyWouthuysen transformation and how to use it to get the correct operator obeservable. See also [2] for some more discussion and references. The velocity operator thus obtained, satisfies the Heisenberg equations of motion v = dx/dt and the well known relation v = p/m in the nonrelativistic limit. Note, here x is not the spatial argument of the usual Dirac wave function. It is also obtained through the FoldyWouthuysen transformation, or from the formula given by Newton and Wigner (again, see [2]). Once a particular time axis has been chosen, this operator x is unique (up to translations of the origin of coordinates). [...] > i) why there exist observables x and v at level of nonrelativistic > quantum mechanics? They exist alright, as evidenced by any elementary treatment of QM. What you are perhaps wondering is how are the x and v observables generalized to relativistic single particle QM and to QFT. The answer to the first part of this question is contained in the references I gave above. The generalization from relativistic singleparticle QM to QFT follows standard methods of second quantization [3]. > ii) How does one obtain classical observables x and v? Given that x and v exist and QM operators, the corresponding classical observables are obtained as expectation values of these operators with respect to coherent sates, which embed the classical phase space in the QM Hilbert space of states. It is important to note that the cascade QFT > rel QM > nonrel QM > CM is only a limiting procedure. There are quantum effects that become nonnegligible when a particle is no longer described by a coherent state. There are relativistic effects that become nonnegligible when the particles speed aproaches the speed of light. There are multiparticle effects that become nonnegligible when the singleparticle sector cannot be sufficiently decoupled from the rest of the QFT Fock space. > Already Dirac warned about QED but his quote is almost ignored (only his > related quote on renormalization is usually cited). The quote on > incompatibility of QED with previous theories is > > {BLOCKQUOTE > Most physicists are very satisfied with this situation. They argue that if > one has rules for doing calculations and the results agree with > observation, that is all that one requires. But it is not all that one > requires. One requires a single comprehensive theory applying to all > physical phenomena. Not one theory for dealing with non relativistic > effects and a separate disjoint theory for dealing with certain > relativistic effects. [...] For these reasons I find the present quantum > electrodynamics quite unsatisfactory. } I think Dirac would have been happy today, since his dream has been achieved. [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998). Chapter 7. [2] J.P. Costellay and B.H.J. McKellarz The FoldyWouthuysen transformation arXiv:hepph/9503416 [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_. Igor 


#3
Feb1708, 05:00 AM

P: n/a

Igor Khavkine wrote on Fri, 15 Feb 2008 20:47:23 +0000:
> On 20080213, Juan R. Gonzlezlvarez <Juan@canonicalscience.com> wrote: >> I find different answers: >> >> ((Classical electrodynamics)) >> >> v < c >> >> ((Relativistic Quantum Mechanics)) >> >> v = c using Dirac equation for PSI>. > > Eh, no. > > You are confused by an archaic paradox. The paradox consists of > identifying the operator c*alpha (in the usual 3+1 Dirac equation > notation) with the physical velocity, which quite obviously doesn't > work. For a modern discussion of the Dirac electron's velocity operator, > see [1] where the author discusses the FoldyWouthuysen transformation > and how to use it to get the correct operator obeservable. See also [2] > for some more discussion and references. > > The velocity operator thus obtained, satisfies the Heisenberg equations > of motion v = dx/dt and the well known relation v = p/m in the > nonrelativistic limit. Note, here x is not the spatial argument of the > usual Dirac wave function. It is also obtained through the > FoldyWouthuysen transformation, or from the formula given by Newton and > Wigner (again, see [2]). Once a particular time axis has been chosen, > this operator x is unique (up to translations of the origin of > coordinates). Actually the author of reference [1] does not maintain your own distinction between "correct" and "incorrect" velocity operators. He is saying something *different* and, in fact, he is supporting my point. By velocity operator i did mean its standard definition, obviously. >From [1] one reads: {BLOCKQUOTE [emphasis mine] We have seen that the *standard* velocity operator gives the electron speed as equal to the velocity of the light. } Next he discusses a *different* velocity operator (his eq. 7.41) and adds {BLOCKQUOTE [emphasis mine] These two can be *reconciled* if we identify (7.41) with the *average* velocity of the particle. } Then the *full* motion of the electron can be divided into two parts {BLOCKQUOTE [emphasis mine] Firstly there is the *average* velocity (7.41) and secondly there is very rapid oscillatory motion which ensures that if an instantaneous measurement of the *velocity* of the electron could be done it would give c. } It seems that the author of [1] think one cannot measure velocities when says "if an instantaneous measurement [...] could be done". This argument usually goes over discussion of sequential measurements, thus position at two times and then taking the limit when time interval goes to zero. This is also the point on [1]. This argument however is not acceptable, which is also the remark done by Feynman in the footnote on the page i cited (see sci.physics.foundations, where my post is not truncated, for the reference). Feynman (and myself) point is that v commutes with p, therefore, instantaneous measurement of v are possible (at least in theory). Now we would analyze the author [1] is really showing. He splits the "velocity of the electron" into two parts, say v = <v> + Dv he identifies <v>, the *average* velocity, with {BLOCKQUOTE the average motion of the particle, i.e. that given by classical relativistic formulae } Thus <v> does not describe the full velocity of the *quantum* particle. And the author gives a value c for the latter. A simple transformation cannot change that fact. >> i) why there exist observables x and v at level of nonrelativistic >> quantum mechanics? > > They exist alright, as evidenced by any elementary treatment of QM. What > you are perhaps wondering is how are the x and v observables generalized > to relativistic single particle QM and to QFT. > > The answer to the first part of this question is contained in the > references I gave above. The generalization from relativistic > singleparticle QM to QFT follows standard methods of second > quantization [3]. I was just asking an inverse question. We know x is a classical parameter in relativistic quantum field theory. Thus the (unsolved) question is: How a nonobservable transform into an observable in the non relativistic limit? This is one of inconsistencies troubled Dirac. >> ii) How does one obtain classical observables x and v? > > Given that x and v exist and QM operators, the corresponding classical > observables are obtained as expectation values of these operators with > respect to coherent sates, which embed the classical phase space in the > QM Hilbert space of states. > > It is important to note that the cascade QFT > rel QM > nonrel QM > > CM is only a limiting procedure. There are quantum effects that become > nonnegligible when a particle is no longer described by a coherent > state. There are relativistic effects that become nonnegligible when > the particles speed aproaches the speed of light. There are > multiparticle effects that become nonnegligible when the > singleparticle sector cannot be sufficiently decoupled from the rest of > the QFT Fock space. You missed my point, again. If relativistic quantum field theory says that position is not an observable. How it is supposed that something cannot be observed at some supposed fundamental level turns into a observable in the classical limit? >> Already Dirac warned about QED but his quote is almost ignored (only >> his related quote on renormalization is usually cited). The quote on >> incompatibility of QED with previous theories is >> >> {BLOCKQUOTE >> Most physicists are very satisfied with this situation. They argue that >> if one has rules for doing calculations and the results agree with >> observation, that is all that one requires. But it is not all that one >> requires. One requires a single comprehensive theory applying to all >> physical phenomena. Not one theory for dealing with non relativistic >> effects and a separate disjoint theory for dealing with certain >> relativistic effects. [...] For these reasons I find the present >> quantum electrodynamics quite unsatisfactory. } > > I think Dirac would have been happy today, since his dream has been > achieved. Difficult to believe since Dirac issues remain unsolved at present date. In [hepth0501222], It is stated that Dirac maintained his criticism on QED in a talk on 1983. Moreover the quote is rather large and finalizes with Dirac hope for the development of a full and consistent relativistic quantum mechanics. Several groups in all the world are now searching one. > [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998). > Chapter 7. > > [2] J.P. Costellay and B.H.J. McKellarz > The FoldyWouthuysen transformation > arXiv:hepph/9503416 > > [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_. > > Igor  I follow http://canonicalscience.org/en/misce...guidelines.txt 


#4
Feb1708, 05:00 AM

P: n/a

What is the velocity of a relativistic electron?
On 20080216, Juan R. <juanrgonzaleza@canonicalscience.com> wrote:
> Igor Khavkine wrote on Fri, 15 Feb 2008 20:47:23 +0000: > >> On 20080213, Juan R. Gonzlezlvarez <Juan@canonicalscience.com> wrote: >>> I find different answers: >>> >>> ((Classical electrodynamics)) >>> >>> v < c >>> >>> ((Relativistic Quantum Mechanics)) >>> >>> v = c using Dirac equation for PSI>. >> >> Eh, no. >> >> You are confused by an archaic paradox. The paradox consists of >> identifying the operator c*alpha (in the usual 3+1 Dirac equation >> notation) with the physical velocity, which quite obviously doesn't >> work. For a modern discussion of the Dirac electron's velocity operator, >> see [1] where the author discusses the FoldyWouthuysen transformation >> and how to use it to get the correct operator obeservable. See also [2] >> for some more discussion and references. >> >> The velocity operator thus obtained, satisfies the Heisenberg equations >> of motion v = dx/dt and the well known relation v = p/m in the >> nonrelativistic limit. Note, here x is not the spatial argument of the >> usual Dirac wave function. It is also obtained through the >> FoldyWouthuysen transformation, or from the formula given by Newton and >> Wigner (again, see [2]). Once a particular time axis has been chosen, >> this operator x is unique (up to translations of the origin of >> coordinates). > > Actually the author of reference [1] does not maintain your own > distinction between "correct" and "incorrect" velocity operators. He > is saying something *different* and, in fact, he is supporting my > point. > > By velocity operator i did mean its standard definition, obviously. >>From [1] one reads: > > {BLOCKQUOTE [emphasis mine] > We have seen that the *standard* velocity operator gives the electron > speed as equal to the velocity of the light. > } Velocity is defined by experiment. If the "standard" velocity operator does not reproduce experimental measurements, then it is still not the *correct* one. It is only unfortunate that it has somehow become the "standard" one. In any case, reference [1] describes how to obtain an operator that does reproduce experimental observations of electron velocities. If you were wondering if such an operator exists, then that's your answer. If you were wondering why this is not the "standard" velocity operator, then the answer has more to do with history than physics. [...] > Now we would analyze the author [1] is really showing. He splits the > "velocity of the electron" into two parts, say > > v = <v> + Dv > > he identifies <v>, the *average* velocity, with > > {BLOCKQUOTE > the average motion of the particle, i.e. that given by classical > relativistic formulae > } > > Thus <v> does not describe the full velocity of the *quantum* > particle. > > And the author gives a value c for the latter. A simple transformation > cannot change that fact. I am eagerly awaiting your publications on the measurement of the "full velocity of the quantum particle". Until then, we only have the usual velocity measurement experiments, which are already well reproduced by the velocity operator obtained with the help of the FoldyWouthuysen transformation. >>> i) why there exist observables x and v at level of nonrelativistic >>> quantum mechanics? >> >> They exist alright, as evidenced by any elementary treatment of QM. What >> you are perhaps wondering is how are the x and v observables generalized >> to relativistic single particle QM and to QFT. >> >> The answer to the first part of this question is contained in the >> references I gave above. The generalization from relativistic >> singleparticle QM to QFT follows standard methods of second >> quantization [3]. > > I was just asking an inverse question. We know x is a classical > parameter in relativistic quantum field theory. Thus the (unsolved) > question is: > > How a nonobservable transform into an observable in the non > relativistic limit? Let x  position coordinates (parameter) X  position observable (operator) x>  an eigenstate of the X operator chi(x)  positionspace wave function Psi(x)  field operator chi>  singleparticle state 0>  Fock vacuum 1. Project onto position eigenstates: chi(x) = <xchi>. 2. Embed state in Fock space: chi> = int dx chi(x) Psi(x)^* 0>. 3. Promote X to operator on Fock space: X = int dx x Psi(x)^* Psi(x). On the LHS of step 3, we have an operator X (an observable), while on the RHS of the same equality we have integration over a coordinate x (a field theoretic parameter). Step 3 is the answer to your question. This is a generic method for promoting singleparticle operators to Fock operators. In any particular situation, there will be differences. Again, the FoldyWouthuysen transformation helps you pick out the right position operator. See sections 5965 of [3], for the general prescription. >> I think Dirac would have been happy today, since his dream has been >> achieved. > > Difficult to believe since Dirac issues remain unsolved at present > date. > > In [hepth0501222], It is stated that Dirac maintained his > criticism on QED in a talk on 1983. Yes, it is unfortunate that in 1983 theoretical physics lost one of its great thinkers. It is also unfortunate that Dirac can no longer join us in a discussion of how his dream of a consistent treatment of relativistic quantum mechanics has been achieved. I guess all we can do now is rely on existing evidence and our own judgement, rather than the opinions of someone whose mind can no longer be altered. > Moreover the quote is rather large and finalizes with Dirac hope for > the development of a full and consistent relativistic quantum > mechanics. > > Several groups in all the world are now searching one. I wish them the best of luck. >> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998). >> Chapter 7. >> >> [2] J.P. Costellay and B.H.J. McKellarz >> The FoldyWouthuysen transformation >> arXiv:hepph/9503416 >> >> [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_. Igor 


#5
Feb2708, 05:00 AM

P: n/a

On Feb 17, 9:01 am, Igor Khavkine <igor...@gmail.com> wrote:
> On 20080216, Juan R. <juanrgonzal...@canonicalscience.com> wrote: > > By velocity operator i did mean its standard definition, obviously. > >>From [1] one reads: > > > {BLOCKQUOTE [emphasis mine] > > We have seen that the *standard* velocity operator gives the electron > > speed as equal to the velocity of the light. > > } > > Velocity is defined by experiment. Velocity is defined by theory. Experimental outcomes are always interpreted *inside* the framework of some theory. > If the "standard" velocity operator > does not reproduce experimental measurements, then it is still not the > *correct* one. It is only unfortunate that it has somehow become the > "standard" one. There is nothing "unfortunate" except your insistence on cataloging operators into "correct" and "incorrect". The terminology used in [1] is more accurate:  velocity operator for v  *average* velocity operator for <v>. As explained in [1] the *average* velocity operator corresponds to measurements of the average velocity. Author on [1] does also clear that the average measurement correspond to {BLOCKQUOTE [1] the average motion of the particle } Moreover, your claim that the standard velocity operator disagrees with measurements is plain false. As explained in my previous message (and in quotations from [1]) the velocity operator can be divided into two components (see also relevant equations on [1]) {BLOCKQUOTE [emphasis mine] Firstly there is the *average* velocity (7.41) and secondly there is very rapid oscillatory motion } v = <v> + Dv If you are measuring *only* the *average* motion of the particle, then the Dv term may vanish and v = <v> > In any case, reference [1] describes how to obtain an operator that does > reproduce experimental observations of electron velocities. Yes, it describes how to define an *average* operator for observations of *average* velocities. I already explained this before also. Just search in my previous comments when i wrote about the equation (7.41) on reference [1] But how the title of this thread reminder us, my question was "what is the velocity of an relativistic electron?" This is also answered on reference [1]: {BLOCKQUOTE We have seen that the standard velocity operator gives the electron speed as equal to the velocity of the light. } [1] also remarks that velocity operator corresponds to actual velocity of the electron, whereas the average operator only corresponds to measurements of average velocities. > I am eagerly awaiting your publications on the measurement of the "full > velocity of the quantum particle". Until then, we only have the usual > velocity measurement experiments, which are already well reproduced by > the velocity operator obtained with the help of the FoldyWouthuysen > transformation. As explained in [1] those "usual velocity measurement" correspond to {BLOCKQUOTE [emphasis mine] the *average* motion of the particle, i.e. that given by classical relativistic formulae } I was asking for the actual velocity of the electron. And i already said this before (but you ignored twice) Feynman (and myself also) shows that twotimes arguments are not acceptable because of the commutation relations between v and p. > > How a nonobservable transform into an observable in the non > > relativistic limit? > > Let x  position coordinates (parameter) > X  position observable (operator) > x>  an eigenstate of the X operator > chi(x)  positionspace wave function > Psi(x)  field operator > chi>  singleparticle state > 0>  Fock vacuum > > 1. Project onto position eigenstates: chi(x) = <xchi>. > 2. Embed state in Fock space: chi> = int dx chi(x) Psi(x)^* 0>. > 3. Promote X to operator on Fock space: X = int dx x Psi(x)^* Psi(x). > > On the LHS of step 3, we have an operator X (an observable), while on > the RHS of the same equality we have integration over a coordinate x (a > field theoretic parameter). Step 3 is the answer to your question. > > This is a generic method for promoting singleparticle operators to Fock > operators. In any particular situation, there will be differences. > Again, the FoldyWouthuysen transformation helps you pick out the right > position operator. See sections 5965 of [3], for the general > prescription. Unfortunately my complaints already start with your step 1. It seems you do not understand difference between classical parameters and quantum observables. It seems also you lack knowledge of basic relativistic localization issues. But let us continue. I am not sure if my question was understood. First I will repeat it by third time. Relativistic quantum field theory (supposedly a fundamental theory) treat position x as a classical parameter (i.e. *not* an observable) whereas nonrelativistic quantum mechanics treat position x as an *observable*. The difficulty arises in a natural way. How does something (x) is not observable (not even being quantum!) transform into something is quantum and observable by making approximations? This makes no sense for me. That position is not a quantum observable in relativistic quantum field theory is noticed in several places. E.g. [I] {BLOCKQUOTE [emphasis in the original] The symmetry between space and time is restored by regarding position, as well, as time as a /classical/ parameter, so that the classical fourvector (x, t) is the spacetime point at or around which we make our /field/ measurement. } That position observables are mathematically incompatible with relativistic requirements is also known. See, for example, section "1.6 The Position operator" on [II]. The section finalizes with {BLOCKQUOTE The outcome, then, is that a position operator is inconsistent with relativity. [...] In field theory, we model localization by making the observables dependent on position in spacetime. } But that spacetime is a classical construct. This introduces foundational issues [I] {BLOCKQUOTE [emphasis in the original] Fieldtheorists do not solve the nonlocality of the oneparticle [quantum mechanical] theory. Rather, they avoid the problem [...] } Thus Dirac complaints still hold. {BLOCKQUOTE [emphasis mine] Most physicists are very satisfied with this situation. They argue that if one has rules for doing calculations and the results agree with observation, that is all that one requires. But it is not all that one requires. One requires *a single comprehensive theory* applying to *all* physical phenomena. Not *one theory for* dealing with *non relativistic effects* and *a separate disjoint theory for* dealing with *certain relativistic effects*. [...] For these reasons I find the present quantum electrodynamics quite unsatisfactory. } The solution? Dirac already advanced which one would be acceptable. We may develop a full and consistent relativistic quantum mechanics. Relativistic quantum field theory would be a *special* case derived from a really fundamental relativistic quantum mechanics theory. The scheme wold be Relativistic quantum mechanics > Relativistic quantum field theory   v 'Wave' relativistic quantum mechanics   v Nonrelativistic quantum mechanics There are several recent relativistic quantum mechanics theory proposals on literature. I have illustrated the basis for some in a recent posting in this newsgroup. To be remarked all of them address questions are *avoided* by relativistic quantum field theory. In my opinion the current search for a unification of General Relativity and quantum theory is a waste of time (and it has been indeed). One wold unify special relativity and the quantum first. > Yes, it is unfortunate that in 1983 theoretical physics lost one of its > great thinkers. It is also unfortunate that Dirac can no longer join us > in a discussion of how his dream of a consistent treatment of > relativistic quantum mechanics has been achieved. First, QED has not changed in any fundamental way since 1983. And, second, no single comprehensive theory applying to both relativistic and nonrelativistic phenomena" is available today (QED continues to work only for infinite time scattering processes between free fields over a classical spacetime). I believe that Dirac very much would agree with me today. > >> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998). > >> Chapter 7. > > >> [2] J.P. Costellay and B.H.J. McKellarz > >> The FoldyWouthuysen transformation > >> arXiv:hepph/9503416 > > >> [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_. > > Igor [I] Why axiomatic field Theory? Ray F. Streater. In Philosophical foundations of Quantum field Theory. 1990. Oxford University Press, USA. Harvey R. Brown; Rom Harre (Editors). p.143 [II] Quantum Field Theory for Mathematicians. 1999. Cambridge University Press. Robin Ticciati  I follow http://canonicalscience.org/en/misce...guidelines.txt 


#6
Feb2808, 05:00 AM

P: n/a

Juan R. wrote;
> (QED continues to > work only for infinite time scattering processes between free fields > over a classical spacetime). This is not true. One can compute  nonrigorously, in renormalized perturbation theory  many timedependent things, namely via the SchwingerKeldysh (or closed time path = CPT) formalism; see, e.g., http://theory.gsi.de/~vanhees/publ/green.pdf See also the entry S9c. What about relativistic QFT at finite times? in my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physicsfaq.txt Arnold Neumaier 


#7
Mar308, 05:00 AM

P: n/a

On 20080226, Juan R. <juanrgonzaleza@canonicalscience.com> wrote:
> On Feb 17, 9:01 am, Igor Khavkine <igor...@gmail.com> wrote: >> On 20080216, Juan R. <juanrgonzal...@canonicalscience.com> wrote: >> Velocity is defined by experiment. > > Velocity is defined by theory. Experimental outcomes are always > interpreted *inside* the framework of some theory. Well, then I guess that settles the question. I'm sure that people who do electron transport measurements will be happy to hear that they need to interpret their data so that electron velocities always come out equal to the speed of light (to be consistent with the theoretical definition given in your original post), instead of the centimeters per second that their instruments have been telling them all along. >> I am eagerly awaiting your publications on the measurement of the "full >> velocity of the quantum particle". Until then, we only have the usual >> velocity measurement experiments, which are already well reproduced by >> the velocity operator obtained with the help of the FoldyWouthuysen >> transformation. > > As explained in [1] those "usual velocity measurement" correspond to > > {BLOCKQUOTE [emphasis mine] > the *average* motion of the particle, i.e. that given by classical > relativistic formulae > } > > I was asking for the actual velocity of the electron. I see, so you want to talk about a quantity that *has not* been measured and then compare it to a quantity that *has* been measured? A rather strange endeavour. > And i already said this before (but you ignored twice) Feynman (and > myself also) shows that twotimes arguments are not acceptable because > of the commutation relations between v and p. Juan R. originally wrote: > Feynman even writes in his wellknown textbook on QED [1] that > > {BLOCKQUOTE > This result is sometimes made pausible by the argument that a precise > determination of velocity implies precise determinations of position at > two times. Then by the uncertainty principle, the momentum is completely > uncertain and all values are equally likely, this is seen to imply that > velocities near the speed of light are more probable, so that in the > limit > the expected value of the velocity is the speed of light. } > > Then in a footnote, Feynman recognizes that argument is invalid since v > commutes with p. Thus, an incorrect value is justified using an > incorrect > argument! Let me get this straight. You are saying that, for an electron, v is not equal to c. Also, both you and Feynman are saying that the argument presented above does not actually show that v = c. I completely agree with both these statements, as long as v is the same velocity that measures at centimeters per second in typical copper wire. >> > How a nonobservable transform into an observable in the non >> > relativistic limit? >> >> Let x  position coordinates (parameter) >> X  position observable (operator) >> x>  an eigenstate of the X operator >> chi(x)  positionspace wave function >> Psi(x)  field operator >> chi>  singleparticle state >> 0>  Fock vacuum >> >> 1. Project onto position eigenstates: chi(x) = <xchi>. >> 2. Embed state in Fock space: chi> = int dx chi(x) Psi(x)^* 0>. >> 3. Promote X to operator on Fock space: X = int dx x Psi(x)^* Psi(x). >> >> On the LHS of step 3, we have an operator X (an observable), while on >> the RHS of the same equality we have integration over a coordinate x (a >> field theoretic parameter). Step 3 is the answer to your question. >> >> This is a generic method for promoting singleparticle operators to Fock >> operators. In any particular situation, there will be differences. >> Again, the FoldyWouthuysen transformation helps you pick out the right >> position operator. See sections 5965 of [3], for the general >> prescription. > > Unfortunately my complaints already start with your step 1. > > It seems you do not understand difference between classical parameters > and quantum observables. It seems also you lack knowledge of basic > relativistic localization issues. But let us continue. I will let my record attest to my understanding of the matter (or lack thereof). As to your own record, you've yet to say much concrete on the subject at all. While I may not be an expert in relativistic localization, I understand the topic well enough to point out that the question below has nothing in particular to do with relativity. > Relativistic quantum field theory (supposedly a fundamental theory) > treat position x as a classical parameter (i.e. *not* an observable) > whereas nonrelativistic quantum mechanics treat position x as an > *observable*. > > The difficulty arises in a natural way. How does something (x) is not > observable (not even being quantum!) transform into something is > quantum and observable by making approximations? The answer, as I've already stated, is in the formula of step 3 above. There is no approximation, and this method works for nonrelativistic as well as relativistic theories. It describes the transition from particle theory to field theory. The relativistic > nonrelativistic approximation may be made at will at either end of the transition. > Thus Dirac complaints still hold. Unfortunately, as I've pointed out, Dirac is not here to be convinced otherwise. Neither are any of the other authors whose objections you have also quoted. I've outlined a number of steps that, through a sequence of well defined transformations and approximations, convert operators defined on the relativistic fieldtheoretic Fock space to the usual position and velocity operators on the nonrelativistic particle Hilbert space, as well as the reverse steps. In my understanding of your question, there should be enough information here to give you an answer. While you do not seem to be satisfied with this answer, I've yet to see you raise concrete objections to any of these steps. >> >> [1] Paul Strange, _Relativistic Quantum Mechanics_, Cambridge (1998). >> >> Chapter 7. >> >> [3] Paul A. M. Dirac, _Principles of Quantum Mechanics_. Igor 


#8
Mar708, 05:00 AM

P: n/a

Arnold Neumaier wrote on Wed, 27 Feb 2008 12:33:20 +0000:
> Juan R. wrote; > >> (QED continues to >> work only for infinite time scattering processes between free fields >> over a classical spacetime). > > This is not true. One can compute  nonrigorously, in renormalized > perturbation theory  many timedependent things, namely via the > SchwingerKeldysh (or closed time path = CPT) formalism; see, e.g., > http://theory.gsi.de/~vanhees/publ/green.pdf {BLOCKQUOTE [1] The closed time path formalism proposed by Schwinger and refined by many others, notable Keldysh, has been frequently used to study nonequilibrium situations. But a close inspection of the formalism reveals that it is more of a steady state theory than a truly timedependent theory. } "steady state theory" = "timeindependent theory" See also the prose before and after (eq 13) and after (eq 36) in the pfd you are citing. Moreover, the SK formalism is based in inconsistent mixture of elements from contradictory theories, e.g. Feynman diagrams for Smatrix (only defined for infinite times) are mixed with a LvN equation for finite times. See also for extra criticism on SK [1]. > > See also the entry > S9c. What about relativistic QFT at finite times? > in my theoretical physics FAQ at > http://www.mat.univie.ac.at/~neum/physicsfaq.txt I agree when says: {BLOCKQUOTE Current 4D QFT is based on perturbation theory for free (i.e., asymptotic in and out) states; therefore it gives only predictions that relate the in and outstates. [...] In nonrelativistic QM, one has a welldefined dynamics at finite times, given by the Schroedinger equation. This dynamics can be recast in terms of Feynman path integrals. Unfortunately, this does not extend to the relativistic case. [...] But I haven't seen a single article that gives meaning (i.e., infrared and ultraviolet finite, renormalization scheme independent properties) to, say, quantum electrodynamics states at finite t and their propagation in time. People don't even know what an initial state should be in a relativistic QFT (i.e., from which space to take the states at finite t); so how can they know how to propagate it... Thus the standard theory gives an Smatrix (or rather an asymptotic series for it) but not a dynamics at finite times. } and disagree with other claims. I can see you do not cite any of the classical results over which relativistic QFT was developed. For instance, the PeirlsLandau inequality is the reason S = U(infinity, +infinity) and that QFT lacks dynamical description at finite times. It is interesting you cite the recent work of my colleague Eugene V. Stefanovich. He agrees with me on the nonexistence of finite time relativistic QFT. E.g. see section 9.1.2 on [2]. You also write in the FAQ {BLOCKQUOTE The missing consistent dynamical theory in 4D relativistic QFT may also have consequences for the foundations of quantum mechanics. Clearly, measurements happen in finite time, hence cannot be described at present in a fundamental way (i.e., beyond the nonrelativistic QM approximation). Thus foundational studies based on nonrelativistic QM are naturally incomplete. This implies that it is quite possible that a solution of the unresolved issues in relativistic QFT are related to the unresolved issues in quantum measurement theory. } One of main points of Eugene (and others, including me) affirms there is *not* quantum dynamical relativistic theory based in Minkowski spacetime unification: "The EinsteinMinkowski 4dimensional spacetime is an approximate concept as well." Do not forget that relativistic QFT breaks with quantum mechanics and treats spacetime as a *classical* object. See references and quotations in my previous message. [1] A unified formalism of thermal quantum field theory. 1994. Int. J. of Mod. Phys A 9(14), 2363. Chu, H; Umezawa, H. [2] http://arxiv.org/abs/physics/0504062  I apply http://canonicalscience.org/en/misce...guidelines.txt 


#9
Mar708, 05:00 AM

P: n/a

On 20080307, Juan R. GonzÃ¡lezÃlvarez <juan@canonicalscience.com> wrote: > Arnold Neumaier wrote on Wed, 27 Feb 2008 12:33:20 +0000: >> Juan R. wrote; >> >>> (QED continues to >>> work only for infinite time scattering processes between free fields >>> over a classical spacetime). >> >> This is not true. One can compute  nonrigorously, in renormalized >> perturbation theory  many timedependent things, namely via the >> SchwingerKeldysh (or closed time path = CPT) formalism; see, e.g., >> http://theory.gsi.de/~vanhees/publ/green.pdf > > {BLOCKQUOTE [1] > The closed time path formalism proposed by Schwinger and refined by many > others, notable Keldysh, has been frequently used to study nonequilibrium > situations. But a close inspection of the formalism reveals that it is > more of a steady state theory than a truly timedependent theory. > } > > "steady state theory" = "timeindependent theory" > [1] A unified formalism of thermal quantum field theory. 1994. Int. J. > of Mod. Phys A 9(14), 2363. Chu, H; Umezawa, H. This reference is from the authors of a formalism called "thermo field dynamics" (many papers and at least one textbook has been written on the subject). From your quote it appears that the authors perceive the SchwingerKeldysh formalism as a direct competitor and alternative, distinct from theirs. The point of both, BTW, is to capture time dependent phenomena in field theory. Unfortunately, the authors of [1] are mistaken. A very clear side by side comparison of "thermo field dynamics" and the SchwingerKeldysh formalism reveals that the latter completely reproduce and is more general than the former. Lawrie, I. D. On the relationship between thermo field dynamics and quantum statistical mechanics J. Phys. A: Math. Gen. 27 (1994) 14351452 > See also the prose before and after (eq 13) and after (eq 36) in the pfd > you are citing. > > Moreover, the SK formalism is based in inconsistent mixture of elements > from contradictory theories, e.g. Feynman diagrams for Smatrix (only > defined for infinite times) are mixed with a LvN equation for finite > times. See also for extra criticism on SK [1]. Perhaps you'd like to share some specific criticism of (eq 13) and (eq 36) from van Hees's notes and why the SchwingerKeldysh formalism appear inconsistent to you. >> See also the entry >> S9c. What about relativistic QFT at finite times? >> in my theoretical physics FAQ at >> http://www.mat.univie.ac.at/~neum/physicsfaq.txt > I can see you do not cite any of the classical results over which > relativistic QFT was developed. For instance, the PeirlsLandau > inequality is the reason S = U(infinity, +infinity) and that QFT lacks > dynamical description at finite times. It's unclear what you mean here. Peierls and Landau only wrote two papers together, neither of which dealt with this topic. > It is interesting you cite the recent work of my colleague Eugene V. > Stefanovich. He agrees with me on the nonexistence of finite time > relativistic QFT. E.g. see section 9.1.2 on [2]. Eugene's misconceptions regarding the content of QED have been pointed out numerous times in this group. Igor 


#10
Mar808, 05:00 AM

P: n/a

Juan R. GonzÃ¡lezÃlvarez wrote
(in the thread: What is the velocity of a relativistic electron?): > Arnold Neumaier wrote on Wed, 27 Feb 2008 12:33:20 +0000: >> Juan R. wrote; >> >>> (QED continues to >>> work only for infinite time scattering processes between free fields >>> over a classical spacetime). >> This is not true. One can compute  nonrigorously, in renormalized >> perturbation theory  many timedependent things, namely via the >> SchwingerKeldysh (or closed time path = CPT) formalism; see, e.g., >> http://theory.gsi.de/~vanhees/publ/green.pdf > > {BLOCKQUOTE [1] > The closed time path formalism proposed by Schwinger and refined by many > others, notable Keldysh, has been frequently used to study nonequilibrium > situations. But a close inspection of the formalism reveals that it is > more of a steady state theory than a truly timedependent theory. > } This ''close inspection'' cannot have a deep basis. How would it then be possible that the paper E. Calzetta and B. L. Hu, Nonequilibrium quantum fields: Closedtimepath effective action, Wigner function, and Boltzmann equation, Phys. Rev. D 37 (1988), 28782900. derives finitetime Boltzmanntype kinetic equations from quantum field theory using the CTP formalism? > Moreover, the SK formalism is based in inconsistent mixture of elements > from contradictory theories, e.g. Feynman diagrams for Smatrix (only > defined for infinite times) are mixed with a LvN equation for finite > times. See also for extra criticism on SK [1]. Any current formalism for interacting quantum field theory is inconsistent and illdefined when viewed from a logical point of view. Nevertheless, people derive from it predictions, often very accurate ones. >> See also the entry >> S9c. What about relativistic QFT at finite times? >> in my theoretical physics FAQ at >> http://www.mat.univie.ac.at/~neum/physicsfaq.txt > > I agree when says: > > {BLOCKQUOTE > Current 4D QFT is based on perturbation theory for free > (i.e., asymptotic in and out) states; therefore it gives only > predictions that relate the in and outstates. This referred to the standard textbook approach. I improved the FAQ by providing clarifying words. > But I haven't seen a single article that gives meaning (i.e., > infrared and ultraviolet finite, renormalization scheme independent > properties) to, say, quantum electrodynamics states at finite t and > their propagation in time. Nevertheless, there are very useful approximations towards that goal, which I had mentioned in the part that follows, but which you didn't quote. > I can see you do not cite any of the classical results over which > relativistic QFT was developed. For instance, the PeirlsLandau > inequality is the reason S = U(infinity, +infinity) and that QFT lacks > dynamical description at finite times. There is little point in citing old results which no longer reflect the current state of the art. > It is interesting you cite the recent work of my colleague Eugene V. > Stefanovich. He agrees with me on the nonexistence of finite time > relativistic QFT. E.g. see section 9.1.2 on [2]. Quoting a paper does not mean that I consent to every statement in it. A few years ago we had extensive discussions here on s.p.r. where you can find out what I think of his work. Many of his claims at that time were immature, and what you write below about his current views does not indicate that things have changed. > You also write in the FAQ > > {BLOCKQUOTE > The missing consistent dynamical theory in 4D relativistic QFT > may also have consequences for the foundations of quantum mechanics. The point here is ''consistent'', which means logically impeccable. The theory of Stefanovich has the same logical flaws as all current 4D quantum field theory  no convergence theory which would prove that the objects to which perturbative or nonperturbative approximations are computed are logically welldefined. I had read the first version of his online book (your reference [2]), and it presents a view which completely ignores much of what has been established in the field. His arguments are provably wrong in 2D quantum field theory. So there is no reason to trust them in 4D. > Clearly, measurements happen in finite time, hence cannot > be described at present in a fundamental way (i.e., beyond the > nonrelativistic QM approximation). Thus foundational > studies based on nonrelativistic QM are naturally incomplete. > This implies that it is quite possible that a solution of the > unresolved issues in relativistic QFT are related to the unresolved > issues in quantum measurement theory. > } > > One of main points of Eugene (and others, including me) affirms there is > *not* quantum dynamical relativistic theory based in Minkowski spacetime > unification: "The EinsteinMinkowski 4dimensional spacetime is an > approximate concept as well." I find this view mistaken and not supported by the existing research literature. > Do not forget that relativistic QFT breaks with quantum mechanics No. It is fully based on quantum mechanics, as my FAQ amply demonstrates. [2] http://arxiv.org/abs/physics/0504062 Arnold Neumaier 


#11
Mar908, 05:00 AM

P: n/a

Igor Khavkine wrote on Sun, 02 Mar 2008 23:24:23 +0000:
> Well, then I guess that settles the question. I'm sure that people who > do electron transport measurements will be happy to hear that they need > to interpret their data so that electron velocities always come out > equal to the speed of light (to be consistent with the theoretical > definition given in your original post), instead of the centimeters per > second that their instruments have been telling them all along. More of the same. What velocities are measuring? Instantaneous velocities or averaged velocities? Each one has its own operator. This was discussed four or five times before. It is ironic, you also seem completely unaware on how atomic and molecular physicists, and quantum chemists work. Take the classical (Darwin) expression U_D = 1/2 (v v' + v r v' r) (e^2/r) 1/c^2 For a Dirac electron the velocity operator is v_op = [H,x] = c alpha substitution {v > v_op} gives U_B = 1/2 (alpha alpha' + alpha r alpha' r) (e^2/r) This is the Dirac Breit operator broadly used in atomic and molecular physics, and quantum chemistry. Another example is the Gaunt potential, a standard in relativistic molecular hamiltonians for bound states. Again its derivation involves v > c alpha. Ironically, this is also the way Feynman derived QED now considered the most precise theory of physics. Take a look to his historical paper [5]. Feynman starts from the classical interaction Lagrangian (he uses c = 1) { 1  (v v') delta_+(s^2) } and substitutes velocity by its quantum operator {1  (alpha alpha') delta_+(s^2) } This gives [5] "our fundamental equation for electrodynamics." In all those *standard* cases, the rule has been v > {v_op = c alpha} Is not ironic physicists and chemists work with an velocity operator "which quite obviously doesn't work" in your own words? > I see, so you want to talk about a quantity that *has not* been measured > and then compare it to a quantity that *has* been measured? A rather > strange endeavour. Everything there is due to your own confusion. I have differentiated between velocity and average velocity, including explicit formulae for both. And the title for this thread is: What is the velocity of a relativistic electron? Why do I need to clarify the same point again and again in each post? > Let me get this straight. You are saying that, for an electron, v is not > equal to c. More confusion and superfitial reading of my posts. What v is not equal? I wrote four velocities on my original post in their respective theoretical contexts. And a fifth v was introduced during discussion! > Also, both you and Feynman are saying that the argument presented above > does not actually show that v = c. You got all completely wrong once more. Feynman first computes the velocity for a Dirac electron. As stated before this *wellknown* result is v = c Over the basis of that result Feynman writes: {BLOCKQUOTE This result is sometimes made pausible by the argument that a precise determination of velocity implies precise determinations of position at two times. } And just present a standard argument done in literature regarding the non commutativity of x and p. Then in a footnote Feynman recognizes that twotimes argument is actually invalid because v and p commute. But of course he continues using the result v > {v_op = c alpha} In fact, Feynman *got* QED in that way [5]. > While I may not be an expert in relativistic localization, I understand > the topic well enough to point out that the question below has nothing > in particular to do with relativity. I provided an adequate quotation "1.6 The Position operator" in a previous message. You deleted but i reintroduce here: {BLOCKQUOTE The outcome, then, is that a position operator is inconsistent with relativity. } What part, do you think, has nothing in particular to do with relativity? Maybe, do you thikÂ¡nk, when he says "is inconsistent with relativity.." > The answer, as I've already stated, is in the formula of step 3 above. > There is no approximation, and this method works for nonrelativistic as > well as relativistic theories. It describes the transition from particle > theory to field theory. The relativistic > nonrelativistic > approximation may be made at will at either end of the transition. As noticed before your answer is incorrect. And your 'derivation' inconsistent already at step 1. Also You seems to be not disturbed becaue i was explcitely asking for the QM wellknown result x =/= parameter whereas your 'answer' began with assumption x = parameter My previous reply to your 'answer' points some of your confussions. In your new reply you are adding new misunderstandings. Thanks because this remember me why i 'invented' guidelines http://canonicalscience.org/en/misce...guidelines.txt >> Thus Dirac complaints still hold. > > Unfortunately, as I've pointed out, Dirac is not here to be convinced > otherwise. Neither are any of the other authors whose objections you > have also quoted. Yes, but pointing the obvious Dirac is not here would not hidde the important: that his complaints remain in print. [5] SpaceTime Approach to Quantum Electrodynamics. 1949. Phys. Rev. 76, 769. Feynman, R. P.  I apply http://canonicalscience.org/en/misce...guidelines.txt 


#12
Mar908, 05:00 AM

P: n/a

On Mar 8, 10:27 am, "Juan R. GonzálezÁlvarez"
<j...@canonicalscience.com> wrote: > More of the same. > > What velocities are measuring? Instantaneous velocities or averaged > velocities? The answer is simple, and is founded on the theory of invariants as expressed in the Dirac algebra. The alphas are spacetime bivectors  surface elements in spacetime, thus alpha_i = gamma_0 gamma_i Thus, covariants formed from a spinor and the alphas cannot possibly represent the velocity of a material body, which is vectorial, not bivectorial (e.g. a free electromagnetic wave). The Dirac current, and by implication the ponderable velocity of the Dirac particle itself, is rather formed directly from the gammas, J_mu = psibar gamma_mu psi If you want to concoct a ponderable velocity from this, then divide through by the (positive definite) fourth component to produce a relativistic unit vector, which is the fourvelocity hiding in this current. That is all. So what does psibar alpha psi represent? It's a group velocity resulting from interference of the positive and negative energy components of the spinor field. It is not vectorial, as a true velocity must be, rather, the timespace part of a bivector, like the electric field. That the expected value of it is always c is physically interesting, but not in any way connected to the ponderable velocity of the massive Dirac particle. I would advise a study of the Foldy transformation. drl 


#13
Mar1108, 05:00 AM

P: n/a

On 20080308, DRLunsford <antimatter33@yahoo.com> wrote:
> I would advise a study of the Foldy transformation. Incidentally, here's an interesting, short memoir written by Foldy, concerning the origin of his work with Wouthuysen. Physics at a Research University, Case Western Reserve 18301990 http://www.phys.cwru.edu/history/book%20page.htm Appendix G, Foldy's Essay: Origins of FoldyWouthuysen Transformation http://www.phys.cwru.edu/history/boo...p%20G%20FW.pdf Igor 


#14
Mar1108, 05:00 AM

P: n/a

On 20080308, Juan R. <juan@canonicalscience.com> wrote:
> Igor Khavkine wrote on Sun, 02 Mar 2008 23:24:23 +0000: > >> Well, then I guess that settles the question. I'm sure that people who >> do electron transport measurements will be happy to hear that they need >> to interpret their data so that electron velocities always come out >> equal to the speed of light (to be consistent with the theoretical >> definition given in your original post), instead of the centimeters per >> second that their instruments have been telling them all along. > > More of the same. > > What velocities are measuring? Instantaneous velocities or averaged > velocities? These measurements correspond to what has been called the "averaged velocity" observable. Since no other kind of velocity has ever been measured, I am completely justified in calling it "the velocity". > Each one has its own operator. This was discussed four or five times > before. I've challenged you to come up with an experiment that has measured what you call the "instantaneous velocity" (so far without response). Until one can be found, the "instantaneous velocity" is merely an operator with units of m/s and with little physical relevance. > It is ironic, you also seem completely unaware on how atomic and molecular > physicists, and quantum chemists work. > > Take the classical (Darwin) expression > > U_D = 1/2 (v v' + v r v' r) (e^2/r) 1/c^2 > > For a Dirac electron the velocity operator is > > v_op = [H,x] = c alpha > > substitution {v > v_op} gives > > U_B = 1/2 (alpha alpha' + alpha r alpha' r) (e^2/r) > > This is the Dirac Breit operator broadly used in atomic and molecular > physics, and quantum chemistry. > > Another example is the Gaunt potential, a standard in relativistic > molecular hamiltonians for bound states. Again its derivation involves > > v > c alpha. These potentials are derived directly from the Dirac equation and remain in the same 4component spinor representation. These can also be subjected to the FoldyWouthuysen transformation and put into 2component PauliSchroedingerlike form (see [6]). Since the FW transformation is unitary, spectral properties of the Hamiltonian remain unchanged. However, the Hamiltonian does get rewritten in terms of the x and p canonical variables (different from the ones in the Dirac representation), where v = dx/dt = p/E, as expected. > Ironically, this is also the way Feynman derived QED now considered the > most precise theory of physics. Take a look to his historical paper [5]. > > Feynman starts from the classical interaction Lagrangian (he uses c = 1) > > { 1  (v v') delta_+(s^2) } > > and substitutes velocity by its quantum operator > > {1  (alpha alpha') delta_+(s^2) } > > This gives [5] "our fundamental equation for electrodynamics." > > In all those *standard* cases, the rule has been > > v > {v_op = c alpha} > > Is not ironic physicists and chemists work with an velocity operator > "which quite obviously doesn't work" in your own words? There are many equivalent formulations of quantum mechanics (encompassing QED), including Feynmans. What is true of one formulation, by equivalence, must be true of all other ones. The use of the Dirac representation, including its alpha and x operators, is not in the least ironic. Most of the time, only spectral properties of the Hamiltonian are of interest. The expectation values of the x and alpha are not measured, and hence need to be interpreted. However, whenever high order relativistic corrections are desired (to be computed in the PauliSchroedinger representation, where position and momentum observables have the common interpretations), these corrections are invariably written in the FW representation (again, see [6]). >> While I may not be an expert in relativistic localization, I understand >> the topic well enough to point out that the question below has nothing >> in particular to do with relativity. > > I provided an adequate quotation "1.6 The Position operator" in a previous > message. You deleted but i reintroduce here: > > {BLOCKQUOTE > The outcome, then, is that a position operator is inconsistent with > relativity. > } I do not have access to Ticciatti's book at the moment, so I cannot judge his statement in context. However, whatever his position is, it doesn't change the fact that massive relativistic particles (whether formulated on their own or in the context of field theory) possess a well defined, frame dependent position operator (granted that the situation does become more complicated once multiparticle effects are taken into account). > What part, do you think, has nothing in particular to do with relativity? The part where a field theoretic parameter (x) gets converted into a single particle operator (X), or vice versa. I wrote: >> Let x  position coordinates (parameter) >> X  position observable (operator) >> x>  an eigenstate of the X operator < Xx> = xx> >> chi(x)  positionspace wave function >> Psi(x)  field operator >> chi>  singleparticle state >> 0>  Fock vacuum >> >> 1. Project onto position eigenstates: chi(x) = <xchi>. >> 2. Embed state in Fock space: chi> = int dx chi(x) Psi(x)^* 0>. >> 3. Promote X to operator on Fock space: X = int dx x Psi(x)^* Psi(x). >> >> The answer, as I've already stated, is in the formula of step 3 above. >> There is no approximation, and this method works for nonrelativistic as >> well as relativistic theories. It describes the transition from particle >> theory to field theory. The relativistic > nonrelativistic >> approximation may be made at will at either end of the transition. > > As noticed before your answer is incorrect. And your 'derivation' > inconsistent already at step 1. > > Also You seems to be not disturbed becaue i was explcitely asking for the > QM wellknown result > > x =/= parameter > > whereas your 'answer' began with assumption > > x = parameter Aha, perhaps you missed the second line of what I wrote above, where I introduce X (capital X) as a single particle position operator observable. Are you perchange denying the fact that a selfadjoint operator has a real spectrum or that any state vector can be written as a linear superposition of its eigenstates? You may want to read over what I wrote above somewhat more carefully. [6] W. A. Barker, F. N. Glover, Reduction of Relativistic TwoParticle Wave Equations to Approximate Forms. III (1955) Phys.Rev. v.99, p.317 Igor 


#15
Mar1208, 05:00 AM

P: n/a

Thus spake DRLunsford <antimatter33@yahoo.com>
>I would advise a study of the Foldy transformation. Agreed. This link to the mathematical detail may also be helpful http://www.physics.ucdavis.edu/~cheng/230A/RQM7.pdf Regards  Charles Francis moderator sci.physics.foundations. charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and braces) http://www.teleconnection.info/rqg/MainIndex 


#16
Mar1408, 05:00 AM

P: n/a

On Mar 8, 2:59 pm, DRLunsford <antimatte...@yahoo.com> wrote:
> On Mar 8, 10:27 am, "Juan R. GonzálezÁlvarez" > So what does psibar alpha psi represent? It's a group velocity > resulting from interference of the positive and negative energy > components of the spinor field. It is not vectorial, as a true > velocity must be, rather, the timespace part of a bivector, like the > electric field. That the expected value of it is always c is > physically interesting, but not in any way connected to the ponderable > velocity of the massive Dirac particle. Oh I forgot to add, c is interesting in this context because it amounts to an invariant expression of constraint upon the colocal presence of matter and antimatter. These are tied together by the conservation of charge, and also angular momentum. Hmm. drl 



#17
Mar1408, 06:59 AM

Sci Advisor
P: 1,136

Igor Khavkine wrote:
>> v = c using Dirac equation for PSI>.[/color] >Eh, no. > >The paradox consists of identifying the operator c*alpha (in the >usual 3+1 Dirac equation notation) with the physical velocity, Actually, this IS the right operator and it DOES produce v when applied on a plane wave solution of the Dirac equation in the Chiral representation. 1/(2E) ( phi* c*alpha phi ) = v http://www.physicsforums.com/showpos...50&postcount=6 Regards, Hans 


#18
Mar2308, 05:41 AM

P: n/a

[Moderator's note: Text reformatted. Try to both limit your unquoted
lines to 72 characters AND have paragraphs have lines with almost 72 characters (to avoid breaking words) and not some long lines and some really short ones mixed together. P.H.] DRLunsford wrote on Sat, 08 Mar 2008 18:59:13 +0000: >> What velocities are measuring? Instantaneous velocities or averaged >> velocities? > > The answer is simple, and is founded on the theory of invariants as > expressed in the Dirac algebra. > > The alphas are spacetime bivectors  surface elements in spacetime, thus > > alpha_i = gamma_0 gamma_i > > Thus, covariants formed from a spinor and the alphas cannot possibly > represent the velocity of a material body, which is vectorial, not > bivectorial (e.g. a free electromagnetic wave). The answer is not so simple. As a starting point, one would remember existence of *two* Dirac equations first: wavefunction Dirac (QM) and field Dirac (QFT). The wavefunction equation is not defined for classical spacetime since x is a quantum observable (Hilbert space) and time a classical parameter. In Dirac QM [1,3] velocity is defined like v_op = [H_op,x_op] and taking Dirac QM Hamiltonian this gives [1,3] v_op = [H_op,x_op] = {c alpha}_op. Thus {c alpha} represent velocity of the Dirac electron [1,3]. But in this case it is incorrect speak about "surface elements in spacetime" because there is not spacetime. Precisely the lack of spacetime concept in QM was the reason for its explicit introduction latter in QFT. The Dirac field equation of QFT is defined for fields over classical spacetime, because x was downgraded from quantum observable to classical parameter during the development of QFT. Since position x is not an operator in QFT (but a spacetime label), the above QM definition using the commutator does *not* apply. You can still compute some concept of velocity in QFT for parameters x and t using QFT hamiltonian and QFT momenta for a Dirac field. You get c alpha again (just read my original message). You can interpret alphas (becoming from Dirac field equation) on a spacetime basis. > I would advise a study of the Foldy transformation. In one sense, the F/W transformation accounts for the difference between quantum space used on QM and the classical space used on QFT. As explained in page 208 of [1] the FW transformation splits QM actual electronic motion into two components: average motion and the so called /Zitterbewegung/. Actual (quantum) position is splinted into two components: averaged position term more a quantum correction term. As explained in same page 208 the averaged position does *not* equals the exact position of the electron. This averaged position essentially corresponding to the *classical* space worked on QFT. Again, the concept of spacetime in QFT is classical. See references cited in previous messages. That is the reason velocity associated to changes on the average position (i.e. the average velocity [1]) correspond to {BLOCKQUOTE [1] the average motion of the particle, i.e. that given by classical relativistic formulae } [1] Paul Strange. Relativistic Quantum Mechanics. Cambridge (1998). Chapter 7. [3] Quantum Electrodynamics. 1998. Advanced Book Classics; Perseus Book Group. R. P. Feynman. Page 47.  http://canonicalscience.org/en/misce...guidelines.txt 


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