<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&gt; On 2005-06-03, Eugene Stefanovich &lt;eugenev@synopsys.com&gt; wrote:\n&gt;\n&gt;\n&gt;&gt;The representation of the Lorentz group used in the Dirac\'s construction\n&gt;&gt;is 4-dimensional, and therefore not unitary. Lorentz group is not\n&gt;&gt;compact, so any non-trivial unitary representation must be infinite-\n&gt;&gt;dimensional. Thus, there cannot be a unitary mapping between Dirac\'s\n&gt;&gt;and Wigner\'s wavefunctions. The condition of unitarity is essential,\n&gt;&gt;because it guarantees the preservation of probabilities in different\n&gt;&gt;inertial reference frames, and therefore only unitary representations\n&gt;&gt;satisfy the relativity principle. Only Wigner\'s construction\n&gt;&gt;satisfies this condition. Dirac\'s functions cannot be regarded as\n&gt;&gt;probability amplitudes.\n&gt;\n&gt;\n&gt; This argument shows a deep misunderstanding of how the Dirac wave\n&gt; function represents the Poincare group. The wave function *itself* is an\n&gt; element of the (infinite dimensional) unitary representation of the\n&gt; Poincare group. The transformations act as\n&gt;\n&gt; [(Lambda,a)psi](x) = psi(Lambda^{-1}x - a).\n\n&gt;\n&gt; These transformations are unitary, since the action of the Poincare\n&gt; group preserve the inner product\n&gt;\n&gt; (psi,phi) = int psi(x)* phi(x) dx.\n\nYou forgot the transformation of the field components\n\n[(Lambda,a)psi](x) = D(Lambda) psi(Lambda^{-1}x - a). (1)\n\nwhere D(Lambda) are 4-by-4 matrices representing the Lorentz subgroup\nof the Poincare group.\nThe action (1) is not unitary, because matrices D(Lambda) are not\nunitary.\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">&nbsp;&nbsp;View this Usenet post in original ASCII form </a></div><P></jabberwocky>Igor Khavkine wrote:
> On $2005-06-03,$ Eugene Stefanovich <eugenev@synopsys.com> wrote:
>
>
>>The representation of the Lorentz group used in the Dirac's construction
>>is 4-dimensional, and therefore not unitary. Lorentz group is not
>>compact, so any non-trivial unitary representation must be infinite-
>>dimensional. Thus, there cannot be a unitary mapping between Dirac's
>>and Wigner's wavefunctions. The condition of unitarity is essential,
>>because it guarantees the preservation of probabilities in different
>>inertial reference frames, and therefore only unitary representations
>>satisfy the relativity principle. Only Wigner's construction
>>satisfies this condition. Dirac's functions cannot be regarded as
>>probability amplitudes.

>
>
> This argument shows a deep misunderstanding of how the Dirac wave
> function represents the Poincare group. The wave function *itself* is an
> element of the (infinite dimensional) unitary representation of the
> Poincare group. The transformations act as
>
> $[(\Lambda,a)\psi](x) = \psi(\Lambda^{-1}x - a)$.

>
> These transformations are unitary, since the action of the Poincare
> group preserve the inner product
>
> $(\psi,\phi) = \int \psi(x)* \phi(x) dx$.

You forgot the transformation of the field components

$[(\Lambda,a)\psi](x) = D(\Lambda) \psi(\Lambda^{-1}x - a)$. (1)

where $D(\Lambda)$ are $4-by-4$ matrices representing the Lorentz subgroup
of the Poincare group.
The action (1) is not unitary, because matrices $D(\Lambda)$ are not
unitary.

Eugene.



On $2005-06-03,$ Eugene Stefanovich wrote: > Yes, you can obtain numerical functions satisfying the Dirac equation > using Weinberg's formulas (14.1.$4) - (14.1.5).$ However, these functions > have nothing to do with probability amplitudes describing states. > I learned in quantum mechanics that wave functions are expansion > coefficients of the state vectors in certain orthonormal basis. > I haven't seen the definition of a wave function as a matrix element > of quantum field, like in (14.1.$4) - (14.1.5).$ This looks like a > completely new approach to quantum mechanics. How would you write > the wave function of a 2-particle system in this notation? No, this is not how quantum mechanics is usually taught, but that shouldn't stop you from learning to look at it this way when you have the opportunity. This approach is not new at all. Have you been exposed to second quantization outside the treatment of QED? Schroedinger's wave function can be obtained in exactly the same way, but as matrix elements of the Schroedinger field operators. Have you read Dirac's book _Principles of Quantum Mechanics_? He deals with this kind of equivalence issue in the chapter on identical particles. The probabilistic interpretation only needs a positive definite inner product. In both the Schroedinger and Dirac wave pictures, one is readily available. It is easy to write multiplarticle states as well. They correspond to multivariable wave functions. For instance: $$\psi++(x,y) = <0|\PSI(x) \PSI(y)[/itex] |2-particle> $\psi-+(x,y) = <0|\PSI(x)* \PSI(y)$ |2-particle> $\psi+-(x,y) = <0|\PSI(x) \PSI(y)*|2-particle>$ \psi--(x,y) $= <0|\PSI(x)* \PSI(y)*|2-particle>$$ Will give you, respectively, the pos-pos, neg-pos, pos-neg, and neg-neg frequency components of a 2-particle antisymmetric wave function. > I am surprised that Weinberg took this route in his chapter 14. > There is a more consistent way to the stationary states of > the hydrogen atom and Lamb shifts. This is exactly the kind of statement that I criticised earlier. You have failed to point out any inconsistency in the equivalence between the Dirac wave function and Fock space state points of view. Yet you continue to make unsubstantiated claims about its consistency, despite an explicit demonstration of the equivalence. If you make this statement because you are confident of your opinion, this casts a shadow on the validity of other statements that you believe to be true. Igor  On [itex]2005-06-05,$ Eugene Stefanovich wrote: > Igor Khavkine wrote: > >> If you agree that fields are a good starting point for the description >> of quasiparticles in a metal (as you seem to in the above quote), you >> have to agree that fields are also a good starting point for the >> description of electrons in a vacuum. Because if you don't, you'll have >> proverbial egg on your face if the vacuum is discovered to have some >> "more fundamental" structure. > > I don't think that vacuum has some fundamental structure. > There is no experimental evidence to suggest that, and your favorite > Occam's razor cuts out any remaining doubts. It is true that we have not observed any non-trivial structure in the vacuum. In this case Occam's razor dismisses theories that do postulate some fundamental structure to the vacuum in order to describe observations. And I completely agree with that. For the purposes of explaining observations, the Standard Model is plenty good enough and it's vacuum has no structure. Occam's razor is irrelevant to the present discussion because we are talking about which mathematical starting point (local fields or Poincare invariant particle states) is advantageous in formulating field theory. In QED, or any Poincare invariant theory, the two formulations are equivalent. But this is no longer so when looking at other applications of QFT. The particle interpretation changes wildly from theory to theory, while the field interpretation provides a common ground for all of them (given examples were curved space times, accelerated frames of reference, emergent systems like metals and superconductors). Taking insight gained elsewhere and coming back to QED, fields provide a better formulation of the theory. In addition, fields provide their users with an additional layer of security agains getting proverbial egg on their faces *if* the vacuum of QED is discovered to have some structure. Igor



Igor Khavkine wrote: > Occam's razor is irrelevant to the present discussion because we are > talking about which mathematical starting point (local fields or > Poincare invariant particle states) is advantageous in formulating field > theory. In QED, or any Poincare invariant theory, the two formulations > are equivalent. But this is no longer so when looking at other > applications of QFT. The particle interpretation changes wildly from > theory to theory, while the field interpretation provides a common > ground for all of them (given examples were curved space times, > accelerated frames of reference, emergent systems like metals and > superconductors). I can't tell anything about curved space-times or accelerated frames. Though, I would like to mention that there is a significant difference between QFT applications to solids and QED. For example, the electron-phonon interaction operator in QFT treatment of solids formally resembles the electron-photon trilinear interaction in QED. The difference is that an electron in a solid *really* can spontaneously emit a phonon. Electrons in a solid are *really* dressed with a coat of virtual phonons, thus forming a polaron. The mass and charge of the polaron are *really* different from those of the bare particle. In this case, the renormalization describes $a *real*$ physical phenomenon of interaction between electron and physical medium (crystal lattice). This is not the case in QED. You cannot take the physical electron out of the medium (vacuum). Unlike in solid state, bare electrons in QED are just mathematical artifacts. The "dressed particle" formalism describes all experimental data (e.g., the S-matrix) without involving the concept of renormalization. My point is that you cannot invoke analogy with solid state applications to justify you statements about QED. There are similarities, but there are importanrt differences as well. Eugene.



On $2005-06-05,$ Eugene Stefanovich wrote: > Igor Khavkine wrote: >> This argument shows a deep misunderstanding of how the Dirac wave >> function represents the Poincare group. The wave function *itself* is an >> element of the (infinite dimensional) unitary representation of the >> Poincare group. The transformations act as >> >> $[(\Lambda,a)\psi](x) = \psi(\Lambda^{-1}x - a)$. > >> >> These transformations are unitary, since the action of the Poincare >> group preserve the inner product >> >> $(\psi,\phi) = \int \psi(x)* \phi(x) dx$. > > You forgot the transformation of the field components > > $[(\Lambda,a)\psi](x) = D(\Lambda) \psi(\Lambda^{-1}x - a)$. (1) > > where $D(\Lambda)$ are $4-by-4$ matrices representing the Lorentz subgroup > of the Poincare group. > The action (1) is not unitary, because matrices $D(\Lambda)$ are not > unitary. No I have not forgotten this. To be explicit: $(\psi,\phi) = sum_s \int d^{3x} \psi_s(x,t)* \phi_s(x,t)$. The integral is over the spacial slice at time t. Clearly, translations are represented unitarily. Since spacial rotations are represented unitarily over the spinor indices, the are also represented unitarily with respect to the above inner product. Time translations are represented unitarily as a consequence of the equation of motion (the Dirac equation). The only non-trivial case to check are the boost transformations. Boost transformations mix x and t coordinates in the argument of $\psi(x,t)$. But the integral in the inner product is only over the values of of $\psi(x,t)at a$ fixed time. However, the equation of motion comes to the rescue again and keeps the above inner product invariant. This can be checked explicitly by verifying that the infinitesimal boost generator (combining the parts acting on the spinor indices s and the space-time coordinates (x,t)) is hermitian with respect to the above inner product. This covers the full algebra of Poincare generators. The important lesson here is that the above inner product is defined on the set of solutions of the Dirac equation, not for arbitrary functions $\psi(x,t)$. Igor



Aaron Bergman wrote: >>Could we stay within QED, please? > > > No. If you want to talk about QFT, you have to talk about QFT. I want to talk only about QED. That's the subject I know and understand pretty well. Let me then say that the quantum theory of electromagnetic interactions can be formulated as a theory of particles rather than fields. This "dressed particle" approach has important advantages: existence of a finite Hamiltonian with clear physical meaning; renormalization is not needed; bound states and time evolution can be calculated directly by standard formulas of quantum mechanics, etc. Eugene



Igor Khavkine wrote: > On $2005-06-03,$ Eugene Stefanovich wrote: > > >>Yes, you can obtain numerical functions satisfying the Dirac equation >>using Weinberg's formulas (14.1.$4) - (14.1.5).$ However, these functions >>have nothing to do with probability amplitudes describing states. >>I learned in quantum mechanics that wave functions are expansion >>coefficients of the state vectors in certain orthonormal basis. >>I haven't seen the definition of a wave function as a matrix element >>of quantum field, like in (14.1.$4) - (14.1.5).$ This looks like a >>completely new approach to quantum mechanics. How would you write >>the wave function of a 2-particle system in this notation? > > > No, this is not how quantum mechanics is usually taught, but that > shouldn't stop you from learning to look at it this way when you have > the opportunity. This approach is not new at all. Have you been exposed > to second quantization outside the treatment of QED? Schroedinger's wave > function can be obtained in exactly the same way, but as matrix elements > of the Schroedinger field operators. Have you read Dirac's book > _Principles of Quantum Mechanics_? He deals with this kind of > equivalence issue in the chapter on identical particles. The > probabilistic interpretation only needs a positive definite inner > product. In both the Schroedinger and Dirac wave pictures, one is > readily available. I may agree with you if you can prove that the representation of the Poincare group built on Dirac's functions $U(\Lambda, a) \psi(x) = D(\Lambda^{-1}) \psi(\Lambda x + a)$ (1) is unitarily equivalent to the Wigner's representation. But this is not true. Representation (1) is not unitary. All irreducible unitary representation were built in Wigner's 1939 paper, and (1) is not one of them. I know exactly what is the meaning of 2 components of the Wigner's wave function. Their squares give the probabilities of measuring spin up and spin down, respectively. These are the probabilities that the particle will go up (or down) in the Stern-Gerlach apparatus. Could you explain me what is the physical meaning of 4 components of the Dirac's wave function? > > It is easy to write multiplarticle states as well. They correspond to > multivariable wave functions. For instance: > > $\psi++(x,y) = <0|\PSI(x) \PSI(y)$ |2-particle> > $\psi-+(x,y) = <0|\PSI(x)* \PSI(y)$ |2-particle> > $\psi+-(x,y) = <0|\PSI(x) \PSI(y)*|2-particle>$ > \psi--(x,y) $= <0|\PSI(x)* \PSI(y)*|2-particle>$ > > Will give you, respectively, the pos-pos, neg-pos, pos-neg, and neg-neg > frequency components of a 2-particle antisymmetric wave function. Again, the physical meaning of positive and negative frequency components is rather obscure. What about pos-pos, etc. components? Is there an experimental setup that can measure these components? > > >>I am surprised that Weinberg took this route in his chapter 14. >>There is a more consistent way to the stationary states of >>the hydrogen atom and Lamb shifts. > > > This is exactly the kind of statement that I criticised earlier. You > have failed to point out any inconsistency in the equivalence between > the Dirac wave function and Fock space state points of view. Yet you > continue to make unsubstantiated claims about its consistency, despite > an explicit demonstration of the equivalence. You haven't demonstrated the equivalence yet. I don't think there is an unitary equivalence. Eugene.



On $2005-06-06,$ Eugene Stefanovich wrote: > I can't tell anything about curved space-times or accelerated frames. > Though, I would like to mention that there is a significant > difference between QFT applications to solids and QED. Yes, there is. It's called Poincare invariance. All the rest is pretty much the same. > For example, the electron-phonon interaction operator in QFT treatment > of solids formally resembles the electron-photon trilinear interaction > in QED. The difference is that an electron in a solid *really* can > spontaneously emit a phonon. Electrons in a solid are *really* dressed > with a coat of virtual phonons, thus forming a polaron. The mass and > charge of the polaron are *really* different from those of the bare > particle. In this case, the renormalization describes $a *real*$ physical > phenomenon of interaction between electron and physical medium (crystal > lattice). You are right about one thing. Physical electron states in a crystal (those with definite crystal momentum) can spontaneously emit a phonon. But lets examine the reasons: * Finite temperature. There is a probability of either absorbing from or emitting into the phonon thermal bath. * Finite size. Electrons are scattered off boundaries of the sample. * Impurities. Electrons are scattered off impurities. Now, lets try to remove each of these obstacles one by one. First synthesize a perfectly pure crystal. No foreign substances, dislocations or the like. This is actually possible to a high degree of precision with modern experimental techniques. Even if the crystal is not 100% pure, it's safe to make this assumption formulating a theory. Assume that the crystal is infinite. Obviously, this is impossible in reality. But infinity is a relative term, at least for physicists. Atomic spacing in a crystal is of the order of Angstroms while macroscopic samples are of the order of centimeters. That's $a 10^8$ ratio of scales, for many purposes that's as good as infinity. Again, a safe theoretical assumption to make is that the unerlying crystal lattice is unbounded. Last but not least, for physicists, like infinity, zero is relative number. The relevant temperature scale in metals is the Fermi energy. Its typical values are in the $10^4's$ of Kelvin. Modern liquid Helium cryostats can easily get below the 1K mark. Again, an easy theoretical assumption to make is that we work at zero temperature. What you get as a result is a zero temperature theory (just like QED) with space and time translation invariance (just like QED). But without rotational or boost symmetry (unlike QED). Due to translational invariance, physical electron states (described in this approximation) are indeed stable and do not spontaneously emit phonons (just like QED). Calculations in this theory are performed in exactly the same way as any other QFT (including QED). Feynman diagrams, propagators, renormalization, and all the other field theory tools play as much of a role there as anywhere else. As I've claimed previously, mathematically these theories are if not identical at least very close to each other. Even more differences can be swept away when considering only low energy excitations above the Fermi surface. Since they only involve long wavelength modes, the underlying crystal structure could be neglected and replaced by a continuum. In this case, renormalization is indispensable because ultraviolet infinities can make an appearance as well. However, we always have a recourse to the underlying discrete theory that tells us to which finite values should diverging transitions amplitudes must be renormalized. This is not possible, because at the moment we have no underlying theory capable of making these predictions. But an appeal to underlying theory need not even be made since these transition amplitudes can be measured experimentally, and the resulting values could be used instead. Thus I'm quite justified in saying that QED and the theory of electrons in a metal are very similar. The fact that one is assumed to be fundamental while the other is not is irrelevant. As you can see, both can be formulated on their own terms without reference to any underlying theory, simply by appealing to experiment. Neither theory looks inside its own vacuum, thus its composition is quite irrelevant to calculations. > This is not the case in QED. You cannot take the physical electron > out of the medium (vacuum). Unlike in solid state, bare electrons in > QED are just mathematical artifacts. The "dressed particle" formalism > describes all experimental data (e.g., the S-matrix) without involving > the concept of renormalization. The distinction between the vacuua of QED and metals is a matter of physics, not mathematics. It has not once been claimed in this or previous conversations that bare particles are assigned physical meaning. Thus you are not contradicting anyone. Let me remind you that the point of contention was your claim that a mathematical formulation of the QED in terms of bare particles is impotent when it comes to various kinds of calculations. This claim is what I've been contradicting and falsifying with examples not only from QED but from other applications of QFT, including condensed matter systems. As to the issue of renormalization, you are once again lapsing into the habit of repeating statements that have been shown to be incorrect. No, you cannot do without renormalization. You start with the standard QED Hamiltonian and use renormalization to evaluate coefficients in your own Hamiltonian. Thus until you calculate your Hamiltonian to all orders in perturbation theory, you will still need to refer to the original Hamiltonian and apply renormalization. If you do not wish to refer to the original formulation of QED, you have to use experimental input. However this approach is not much different from writing down a phenomenological effective field theory, which can be done from scratch without any extra work. > My point is that you cannot invoke analogy with solid state > applications to justify you statements about QED. > There are similarities, but there are importanrt differences as well. And what statements would that be? I've said the following. QED: particle and field formulations are equivalent. Other QFT's, including those from condensed matter physics: sometimes the particle description breaks down, but the field one doesn't. Coming back to QED: insight from other QFT's tells us that fields are a better starting point for the formulation. Igor



In article <42A3F65C.8040401@synopsys.com>, Eugene Stefanovich wrote: > Aaron Bergman wrote: > > >>Could we stay within QED, please? > > > > > > No. If you want to talk about QFT, you have to talk about QFT. > > I want to talk only about QED. That's the subject I know and understand > pretty well. Let me then say that the quantum theory of electromagnetic > interactions can be formulated as a theory of particles rather than > fields. This "dressed particle" approach has important advantages: > existence of a finite Hamiltonian with clear physical meaning; > renormalization is not needed; bound states and time evolution can > be calculated directly by standard formulas of quantum mechanics, > etc. The disadvantages to QED are, of course, that it probably does not exist except as an effective field theory. I have not followed this thread (and don't have the energy to get too involved in it), but it sounds like you're exactly doing renormalization using different words. That you can do QED in flat space in a first quantized form isn't a surprise to anyone, but I tend to think that quarks are important, too. Aaron



Igor Khavkine wrote: > You are right about one thing. Physical electron states in a crystal > (those with definite crystal momentum) can spontaneously emit a phonon. > But lets examine the reasons: > > * Finite temperature. There is a probability of either absorbing from or > emitting into the phonon thermal bath. > * Finite size. Electrons are scattered off boundaries of the sample. > * Impurities. Electrons are scattered off impurities. > No, these are not the only reasons. Electrons are coupled to phonons even in the perfect infinite crystal at zero temperatures. Physically, the coupling occurs because the electron has charge and crystal lattice is polarizable (this applies mostly to dielectrics, and at much lesser extent to metals). Extra electron in the conduction band polarizes the lattice around itself. Atoms (or ions) in the lattice become displaced from their normal positions. In QFT language this is described as a virtual phonon "coat" that follows the electron wherever it goes. Another name for this object is "polaron". Polaron is the analogue of the "dressed electron" in QED. The difference is that 1) Polaron dressing does not not produce infinite corrections to the mass and charge (as in QED), due to the underlying granular structure of the lattice. 2) Polaron dressing is perfectly physical and even observable, unlike in QED. Eugene.



On $2005-06-07,$ Eugene Stefanovich wrote: > > > Igor Khavkine wrote: > >> You are right about one thing. Physical electron states in a crystal >> (those with definite crystal momentum) can spontaneously emit a phonon. >> But lets examine the reasons: >> $>> *$ Finite temperature. There is a probability of either absorbing from or >> emitting into the phonon thermal bath. $>> *$ Finite size. Electrons are scattered off boundaries of the sample. $>> *$ Impurities. Electrons are scattered off impurities. > > No, these are not the only reasons. Electrons are coupled to phonons > even in the perfect infinite crystal at zero temperatures. As I described, a single particle electron state in this situation will never emit a phonon, simply because it is an eigenstate of the Hamiltonian. > Physically, > the coupling occurs because the electron has charge and crystal > lattice is > polarizable (this applies mostly to dielectrics, and at much lesser > extent to metals). Extra electron in the conduction band polarizes > the lattice around itself. Atoms (or ions) in the lattice become > displaced from their normal positions. In QFT language this is described > as a virtual phonon "coat" that follows the electron wherever it goes. > Another name for this object is "polaron". Polaron is the analogue of > the "dressed electron" in QED. The difference is that > > 1) Polaron dressing does not not produce infinite corrections to the > mass and charge (as in QED), due to the underlying granular structure > of the lattice. > > 2) Polaron dressing is perfectly physical and > even observable, unlike in QED. Once you remove the lattice all together (take the continuum limit and restrict yourself to long wavelength measurements only), you run into the same ultraviolet difficulties as you would expect in a continuum theory. Then you get infinite renormalizing corrections. They are infinite compared to what would be the mass and charge of a bare electron. In this case a bare electron is not the same as an electron outside the crystal. But it doesn't matter, because at the assumed level of experimental precision you can't observe an electron outside the lattice anyway. Formulating the theory in terms of bare electrons is perfectly reasonable since they are not observed. This is the same situation as in QED. Our instruments tell us that we can't take an electron out of the vacuum and look at its properties without interaction. Hence the theory need not make any sensible statments about what, say, the mass and charge of a bare electron are. So a mathematical formulation of the theory where the bare mass and charge are infnite is acceptable (with all the fine print provided by renormalization). But we are getting farther and farther from the topic. This subthread is about the relative merits of the field and particle formulations. So on topic, let me throw someting else into the mix. The mathematical formulation of QFT can handle a large variety of theories, with or without symmetries. Take for instance a quantum field interacting with a time and space varying external field. There is neither space nor time translation symmetry in this theory. So particle states cannot be identified as eigenstates of any sort of symmetry generators. However, a satisfactory quantization can be constructed by solving partial differential equations for the wave functions obtained as matrix elements of field operators. And yes, such a model has numerous physical applications. Igor



Igor Khavkine wrote: > On $2005-06-05,$ Eugene Stefanovich wrote: > >>Igor Khavkine wrote: > > >>>This argument shows a deep misunderstanding of how the Dirac wave >>>function represents the Poincare group. The wave function *itself* is an >>>element of the (infinite dimensional) unitary representation of the >>>Poincare group. The transformations act as >>> >>> $[(\Lambda,a)\psi](x) = \psi(\Lambda^{-1}x - a)$. >> >>>These transformations are unitary, since the action of the Poincare >>>group preserve the inner product >>> >>> $(\psi,\phi) = \int \psi(x)* \phi(x) dx$. >> >>You forgot the transformation of the field components >> $>>[(\Lambda,a)\psi](x) = D(\Lambda) \psi(\Lambda^{-1}x - a).$ (1) >> >>where $D(\Lambda)$ are $4-by-4$ matrices representing the Lorentz subgroup >>of the Poincare group. >>The action (1) is not unitary, because matrices $D(\Lambda)$ are not >>unitary. > > > No I have not forgotten this. To be explicit: > > $(\psi,\phi) = sum_s \int d^{3x} \psi_s(x,t)* \phi_s(x,t)$. > > The integral is over the spacial slice at time t. Clearly, translations > are represented unitarily. Since spacial rotations are represented > unitarily over the spinor indices, the are also represented unitarily > with respect to the above inner product. Time translations are > represented unitarily as a consequence of the equation of motion (the > Dirac equation). > > The only non-trivial case to check are the boost transformations. Boost > transformations mix x and t coordinates in the argument of $\psi(x,t)$. But > the integral in the inner product is only over the values of of $\psi(x,t)$ > at a fixed time. However, the equation of motion comes to the rescue > again and keeps the above inner product invariant. This can be checked > explicitly by verifying that the infinitesimal boost generator > (combining the parts acting on the spinor indices s and the space-time > coordinates (x,t)) is hermitian with respect to the above inner product. > This covers the full algebra of Poincare generators. > > The important lesson here is that the above inner product is defined on > the set of solutions of the Dirac equation, not for arbitrary functions > $\psi(x,t)$. Let me first agree with you. Yes, you can project out 2 extra components, and restrict the space of functions using Dirac's equation, and show that resulting functions afford an unitary irreducible representation of the Poincare group, just as Wigner's functions do. You can also apply the Foldy-Wouthuysen transformation to Dirac's functions and arrive to the operators of velocity and position that make certain physical sense. Yes, you can do all that. My question is whether all these non-trivial manipulations have any physical meaning? It seems to me that we have started from something (solutions of Dirac's equation) that didn't have much physical sense and then applied a series of patches (like Foldy-Wouthuysen transformations) to put our creation in a better shape. Why should we go through all this hassle when we have a perfect solution for the 1-particle problem given by the Wigner's prescription? The physical foundations for this prescription are impeccable: 1. The Hilbert space of the system must contain an unitary (= preservation of probabilities) representation of the Poincare group (= principle of relativity). 2. Since the system is elementary, this should be the "smallest" (= irreducible) Hilbert space. From these two postulate one immediately obtains the classification of elementary particles by mass and spin, wave functions without extra unphysical components, the transformation laws (wrt all inertial transformations) preserving the inner product. Explicit expressions for all operators of observables (momentum, energy, velocity, spin, position, etc.) are obtained in a straightforward fashion without any adjustments. Unfortunately, most textbooks prefer to talk about Dirac's wave functions, and don't pay any attention to the brilliant Wigner's approach. I think, this is just for historical reasons. Simply, Dirac was first. I remember reading somewhere that Wigner's paper was not accepted in the major physical journal of that time (Phys. Rev.?) and he was forced to publish in a mathematical journal. I find it weird, because to me his article (E. Wigner, On Unitary Representations of the Inhomogeneous Lorentz Group, Ann. of Math., 40, 149, (1939)) represents one of the few finest works of theoretical physics in the 20th century. Please understand me right. I don't want to diminish the hugh contribution that Dirac's equation made to the development of theoretical physics. However, I think, in 2005, almost 80 years after its discovery, it is naive to think that Dirac's equation is a relativistic equivalent of the Schroedinger equation. Now we know that in relativistic physics the energy of a free particle with mass m is related to the momentum by formula $E = \sqrt(m^2 + p^2)$. Therefore, the Hamiltonian in the momentum representation is simply $H = \sqrt(M^2 + P^2),$ and not the 4x4 matrix written by Dirac. If it is not a good idea to describe 1-electron states by Dirac's "wavefunctions", then maybe these functions have no place in relativistic quantum physics? No, not at all. There is a place where you simply cannot do without Dirac's functions. As Weinberg showed in his Chapter 5, Dirac's functions (this time you need to consider operator functions, or quantum fields, built as linear combinations of creation and annihilation operators) are absolutely essential for construction of Poincare invariant cluster separable interactions in the Fock space. So, my suggestion is 1) use Wigner's approach for description of 1-particle states, 2) build n-particle Hilbert spaces as tensor product of Wigner's 1-particle spaces 3) Build the Fock space as a direct sum of n-particle spaces. 4) Use Dirac's (and other) quantum fields to construct non-trivial interparticle interactions in the Fock space. Eugene. Eugene.





Aaron Bergman wrote: > The disadvantages to QED are, of course, that it probably does not exist > except as an effective field theory. I often hear this statement, but I don't understand what it means exactly. What are your grievances about QED? 1. renormalization? 2. high-energy behavior (Landau pole)? 3. non-convergence of the perturbative series? 4. ??? I can't help you much with 2$. - 3$. (I am not convinced that these are real problems), but I am pretty sure that 1. is not a problem at all. With the help of the "dressed particle" approach, QED can be made a fully logical and self-consistent theory in the range of experimentally accessible energies, giving predictions agreeing with experiment within error bounds. What else do you expect from a successful physical theory? Eugene.



In article <42A60AD5.9090005@synopsys.com>, Eugene Stefanovich wrote: > Aaron Bergman wrote: > > > The disadvantages to QED are, of course, that it probably does not exist > > except as an effective field theory. > > I often hear this statement, but I don't understand what it means > exactly. What are your grievances about QED? > > 1. renormalization? Neither regularization nor renormalization is a mystery. > 2. high-energy behavior (Landau pole)? Bingo > 3. non-convergence of the perturbative series? Not an issue. Aaron



In article <42A5F921.5020408@synopsys.com>, Eugene Stefanovich wrote: > Renormalization is one important part of this story, but not the > whole story. There is another big part called "dressing" which > is less known. This part clarifies the description of physical > particles and clears up numerous paradoxes left by the renormalization > approach. The description of physical particles doesn't need clearing up. And, as you say, dressing isn't new (I thought it was in Weinberg, even, but I must have picked up the term elsewhere.) What is in Weinberg is sections 10.$2-10$.3 (among others) where he discussed the physical poles vs. the bare fields. Aaron



Igor Khavkine wrote: > > But we are getting farther and farther from the topic. This subthread is > about the relative merits of the field and particle formulations. So on > topic, let me throw someting else into the mix. The mathematical > formulation of QFT can handle a large variety of theories, with or > without symmetries. Take for instance a quantum field interacting with a > time and space varying external field. There is neither space nor time > translation symmetry in this theory. So particle states cannot be > identified as eigenstates of any sort of symmetry generators. However, a > satisfactory quantization can be constructed by solving partial > differential equations for the wave functions obtained as matrix > elements of field operators. And yes, such a model has numerous physical > applications. Again, I would prefer to limit our discussion to straight QED. If we remove philosophical noise, then the question boils down to this: is it better to represent operators in the Fock space as functions of quantum fields or as functions of creation/annihilation operators of particles? Strictly mathematically, these two representations are interchangeable: fields can be uniquely obtained from particle operators and vice versa. Then we can ask which way is more convenient and physically transparent? If we speak about traditional QED which is interested, almost exclusively, in the S-matrix, then fields are certainly more convenient. Scattering amplitudes can be obtained from Feynman rules that can be read directly from the Lagrangian. These formulas involve propagators that are expectation values of field products. For these calculations one even does not need an explicit expression for the Hamiltonian. Of course, the same results could be obtained by using interaction terms in the Hamiltonian expressed through creation and annihilation operators. But this involves much more labor and, understandably, is not popular. The story is different if we try to calculate the time evolution. For that, we need the Hamiltonian in the "dressed particle" representation. The "dressing transformation" requires sorting of different operators into three groups: renorm, phys, and unphys. Operators in these groups are most easily identified by their composition in terms of creation and annihilation operators. For example $a^*a$ is renorm $a^*ac$ is unphys $a^*a^*aa$ is phys. So, for time evolution and bound states calculations in RQD, the representation in terms of creation and annihilation operators is certainly more preferable. Eugene.