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Very simple QFT questions |
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| Mar17-07, 06:50 PM | #1 |
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Very simple QFT questions
I have some extremely basic questions in QFT.
First, P&S discuss causality in QFT in the first chapter of the book and, after showing that [itex] <0| \phi(x)\phi(y)|0> [/itex] does not vanish for spacelike intervals, they say "to really discuss causality, however, we should ask not whether particles can propagate over spacelike intervals but whether a measurement performed at one point....." What do they mean by this? I have my own interpretation but it's nontrivial and it may be wrong (they mention that so casually that it seems obvious to them). Of course, classically, it would make no sense to say that a a particle could propagate over spacelike intervals. In the quantum (and relativistic) world, the only way for me to make sense of what they are saying is that measuring the position of a particle precisely involves energies that necessarily lead to the creation of more particles than the ones already present in the system. Therefore, the very idea of checking whether a single particle that was at a point "x" at t=0 is now located at a point "y" at a time t is impossible in QFT. Second, how is a measurement actually defined in QFT? In NRQM, it's pretty clear. For a given hermitian operator, one finds it's eigenvalues and eigenstates, and so on. How is this defined in QFT? For example, consider the field [itex] \phi(x) [/itex] itself. Now, I always thought that this is not in itself an observable, so it does not make sense to talk about *measuring* phi. And yet, P&S talk about measurements of phi in the first chapter. Is that an abuse of language? One problem I find with QFT books is that there is no effort devoted to making the connection with NRQM. This is strange, it's the equivalent of teaching GR and never talking about how one may recover Newton's gravitation. One should be able to treat a simple sysytem (let's say the infinite square well!) and show in what way one may recover the NRQM result, within some limit! Does anyone know of a book that does that type of connection? Another question is about the classical limit of QED. People mention that a coherent state of photons correspond to the classical limit of classical E&M. But how does that work, exactly? I know that one can then replace the creation and annihilation photon operators by their expectation values, but the field obtained is still imaginary so it's not the classical field. There has to be much more work (with many subtleties involved, no doubt) before connecting with the classical treatment of E&M. |
| Mar19-07, 07:50 AM | #2 |
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I discuss this problem in a somewhat different context in Sec. VIII of http://arxiv.org/abs/quant-ph/0609163 Sec. IX is also relevant for that issue. I also propose a solution of this problem in http://arxiv.org/abs/quant-ph/0406173 |
| Mar19-07, 09:29 AM | #3 |
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I started studying QFT from P&S too, and got stuck with nearly the same things. It's starting to look that there aren't very good answers to these questions, although I'm waiting replies to this thread still hopefully.
Instead of answering, I'll throw another question about the basics of QFT (or about relativistic theory in general). Peskin & Shroeder explain in the beginning of their book, that a propagator [tex] K(x,y,T)=\int\frac{d^3p}{(2\pi)^3}e^{-i(E_{\boldsymbol{p}}T-\boldsymbol{p}\cdot(\boldsymbol{x}-\boldsymbol{y}))} [/tex] cannot be used, because it violates causality. Really so? If I assume a wavefunction (like in nonrelativistic theory) [tex]\phi(t,x)[/tex] to have a time evolution defined with [tex] \phi(t+T,x)=\int d^3y\; K(x,y,T)\phi(t,y) [/tex] a brief calculation then shows that the wavefunction satisfies the KG-equation, which is Lorentz invariant. Doesn't this mean, that this propagator gives Lorentz invariant time evolution? And how could Lorentz invariant time evolution violate causality? Besides, the integral in the propagator does not converge, but it merely behaves as a distribution when used correctly. Hence it doesn't look very smart to simply estimate if it looks zero or nonzero outside the lightcone. My point is, that not only is the explanation on how the causality is conserved, with the measurement interpretation, confusing, but so is also the explanation on why another propagator would instead violate causality. |
| Mar19-07, 01:38 PM | #4 |
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Very simple QFT questions |
| Mar19-07, 07:31 PM | #5 |
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I just couple of minutes ago happened to hit the url http://www.physics.ucsb.edu/~mark/qft.html on these forums, started reading it, and noticed that Srednicki explains neccecity of commutator vanishing outside the lightcone quite differently. I haven't understood it myself yet, but it certainly looks worth cheking out. On the page 46 of the pdf.
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| Mar19-07, 10:53 PM | #6 |
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Different QFT books treat this subject quite differently (Zee, Feynman, P&S) and with different results. A thorough treatment should handle this entirely analytically instead of using approximations as done in most textbook. checked with numerical simulations. I did extensive numerical simulations of Klein Gordon propagation (in many different spacial dimensions) and one never sees any propagation outside the light cone. Also analytically one doesn't see anything outside the light cone. The concise mathematical expression for the Green's function in 3+1 dimensions, for forward propagation is: [tex]\Theta(t) \left(\ \frac{1}{2\pi}\delta(s^2)\ + \frac{m}{4\pi s} \Theta(s^2)\ \mbox{\huge J}_1(ms)\ \right), \qquad \mbox{with:}\ \ \ s^2=t^2-x^2 [/tex] Where Theta is the Heaviside step function and J1 is the Bessel J function of the first order. The Theta at the left selects the forward propagating half while the other cuts off any propagation outside the light cone. Analytically, the Bessel function goes imaginary outside the lightcone and this is generally what becomes the part "outside the light cone" in the form of the Bessel I and Bessel K functions which become [tex] \mbox{\huge I}_1(mx)\ \rightarrow\ \frac{1}{\sqrt{2\pi mx}}\ \ e^{mx}, \ \ \ \ \mbox{\huge K}_1(mx)\ \rightarrow\ \sqrt{\frac{\pi}{2 mx}}\ \ e^{-mx}[/tex] for large x, typically K1 becomes exp(-mx) which is then usually given as the part outside the light cone. However, the concise derivation of the Green's function does produce the Heaviside step function which eliminates the propagation outside the lightcone. (Note that in the limit of m=0 the propagation outside the lightcone would become infinite!) One can find many variations of the analytical expression of the Klein Gordon propagator given above, sometimes with a negative sign for the Dirac delta function, which is wrong since this part becomes the photon propagator in the limit case where where m goes to zero, and should be positive. Sometimes one sees a different normalization factor. Also the Bessel function changes from text to text. Feynman, in 1948, with paper and pencil as the only tools to calculate (!) plus mathematical table books came to: [tex]-\frac{1}{4\pi}\delta(s^2)\ + \frac{m}{8\pi s} \mbox{\huge H}_1^{(2)}(ms)\ \right), \qquad \mbox{with:}\ \ \mbox{\huge H}_1^{(2)}(ms)= \mbox{\huge J}_1(ms)-i\mbox{\huge Y}_1(ms) [/tex] There's the sign, a factor 1/2 and the Hankel function obtained from the tables which is the Bessel equivalent of exp(ix)=cos(x)+isin(x). This is where the "propagation outside the lightcone" started historically: http://chip-architect.com/physics/KG...or_Feynman.jpg Another very popular (modern) textbook (Zee) handles it in I.3 formula (23). This is a also a hand waving approximation leading to the exp(-mr) light cone leaking. P&S then use a particular argument with particles and anti particles which would cancel out each others propagation outside the lightcone. to restore causality. (In chapter 2.4) This after they get the exp(-mr) term from a similar approximation. The simplest way in which you can convince yourself that there is no propagation outside the lightcone is by expanding like this: [tex]\frac{1}{p^2-m^2}\ =\ \frac{1}{p^2}+\frac{m^2}{p^4}+\frac{m^4}{p^6}+\frac{m^6}{p^8}+.....[/tex] The Fourier transform of this series leads to a series representing the Bessel J function. The first term is the massless propagator which is strictly on the light cone only. It's Fourier transform is the Dirac function in the space-time version of the propagator. The second term represent a massless propagator acting on a massless propagator, thus the propagation on the light cone becomes a source itself which is again propagated on the lightcone, and so on. Thus: None of the terms in the series has any propagation outside the light cone, and neither does the sum of the geometric series, The Klein Gordon propagator. Regards, Hans. PS: related stuff: http://functions.wolfram.com/PDF/BesselJ.pdf (also has the KG propagator) http://en.wikipedia.org/wiki/Bessel_function http://www.chip-architect.com/physic..._radiation.pdf With the latter paper and the series expansion you can derive the KG propagator in any dimension. |
| Mar20-07, 01:09 AM | #7 |
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| Mar20-07, 03:46 AM | #8 |
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"In order to associate any physical content with this mathematical result , we must assume that it makes sense to attach physical meaning to the measurement of a field strength at at point, a concept already criticized in earlier paragraphs". So I think we are in good company when we are confused ! (B and J then quote a paper by Bohr and Rosenfeld which I don't have access too, but it sounds as if it might be quite useful - Phys Rev 78 p794 (1950) ) |
| Mar20-07, 04:33 PM | #9 |
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| Mar24-07, 10:15 AM | #10 |
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I was confused about the same thing after reading Peskin and Schroeder. I found a good explanation (at least I found it helpful) on the bottom Page 198 of Weinberg's "The Quantum Theory of Fields, Volume 1". Although you need to following the arguements Weinberg had been making in the first four chapters, he basically says it is best to think of the causality condition as something which is needed for the S-matrix to be Lorentz invariant, rather than thinking of it terms of measurements of field values at different points.
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| Mar24-07, 12:33 PM | #11 |
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It is refreshing to see a QFT book that does not feel like a simple repeat of the same presentation again and again. It's clear that the author spent time thinking about presenting things from scratch and in a logical way. I especially dislike the conventional presentation which starts with the non sequitur that one must quantize classical fields (even if the fields on starts with have no classical correspondence at all, like the Dirac field!). It's only when I read Weinberg that I found that finally there was a textbook presenting QFT in a logical way, with the starting point that one must allow the number of particles to vary so one builds a Fock space and then it is the requirement of Lorentz invariance that leads to the need of introducing fields!! This book is closer in spirit to Weinberg but more transparent (who is very good but, let's admit it, quite heavy to follow sometimes) |
| Mar24-07, 09:59 PM | #12 |
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Can anyone tell me whether the published version has a decent index? (Amazon doesn't have any "look inside" images, so I can't find out that way.) |
| Jul9-07, 06:06 PM | #13 |
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Great thread! It is refreshing to see so many people having the same questions that I had. It is sad that QFT textbooks did such a poor job in addressing these questions.
Then what about superluminal propagation? It seems to be well-established that a wave function of a localized particle spreads out faster than the speed of light. And the usual wisdom says that this contradicts causality, because from the point of view of a moving observer the events of particle creation and absorption would change their time order. G. C. Hegerfeldt, "Instantaneous spreading and Einstein causality in quantum theory", Ann. Phys. (Leipzig), 7, (1998) 716. Below I will try to explain that the conclusion about violation of causality could be premature. Instantaneous spreading and causality do not necessarily contradict each other. There are two key points in my analysis. First is that "instantaneous spreading" refers to the wavefunction, and wavefunctions must be interpreted probabilistically. The second point is that particle localization is relative. If observer at rest sees the particle as localized, then the moving observer sees that particle's wave function is spread out over entire space. F. Strocchi, "Relativistic quantum mechanics and field theory", http://www.arxiv.org/hep-th/0401143 First take the point of view of an observer at rest O. This observer prepares particle localized at point A at t=0. At time instant t>0 he sees that the wave function has spread out superluminally. This means that the probability of finding the particle at a point B, whose distance from A is greater than ct, is non-zero. So far there is no contradiction. Now take the point of view of observer O', which moves with a high speed relative to O. As I said above, at time t=0 (by his own clock) observer O would see particle's wave function as spread out in space. There will be a maximum at point A. But there will be also a non-zero probability of finding the particle at point B. At later times the wave function will spread out even more. However, it is important that observer O' cannot definitely say that he sees the particle as propagating from B to A. He is seeing some diffuse probability distributions at all times, from which it is impossible to say exactly what is the speed and direction of particle's propagation. So, the situation in quantum mechanics is quite different from the classical mechanics, where propagation direction and speed always have a well-defined meaning, and causality is not compatible with superluminal propagation. |
| Jul10-07, 04:26 AM | #14 |
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| Jul10-07, 12:20 PM | #15 |
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T. D. Newton and E. P. Wigner, "Localized states for elementary systems", Rev. Mod. Phys., 21, (1949) 400. |
| Jul10-07, 01:43 PM | #16 |
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with a concise analytical treatment giving exact solutions in configuration space, nor with extensive numerical simulations. See my old post above. The simplest way to convince oneself may be this series development of the Klein Gordon propagator: [tex]\frac{1}{p^2-m^2}\ =\ \frac{1}{p^2}+\frac{m^2}{p^4}+\frac{m^4}{p^6}+\frac{m^6}{p^8}+.....[/tex] Which becomes the following operator in configuration space: [tex]\Box^{-1}\ \ -\ \ m^2\Box^{-2}\ \ +\ \ m^4\Box^{-3}\ \ -\ \ m^6\Box^{-4}\ \ +\ \ .... [/tex] Where [tex]\Box^{-1}[/tex] is the inverse d'Alembertian, which spreads the wave function out on the lightcone as if it was a massless field. The second term then retransmits it, opposing the original effect, again purely on the light cone. The third term is the second retransmission, et-cetera, ad-infinitum. All propagators in this series are on the lightcone. The wave function does spread within the light cone because of the retransmission, but it does never spread outside the light cone, with superluminal speed. Regards, Hans. |
| Jul10-07, 01:52 PM | #17 |
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