# Very simple QFT questions

by nrqed
Tags: simple
P: 1,192
 Quote by Micha Could somebody say then in one sentence, what is the exact physical meaning of the Feynman propagator?
The Feynman propagator has no direct physical meaning. It simply appears as a component in the calculation of infinite-time scattering amplitudes.

 Quote by Micha I'd say, [the Feynman propagator] is the amplitude to find a particle at spacetime point y, when you have found one at earlier spacetime point x.
I don't think this is correct; at least, I've never seen a calculation that shows it to be correct. (One has to be careful about the meaning of position in quantum field theory, so there are some subtleties. But it doesn't even have the right dimensions.)
P: 1,746
Hi Avodyne,

 Quote by Avodyne The Feynman propagator has no direct physical meaning. It simply appears as a component in the calculation of infinite-time scattering amplitudes.
I agree with you completely. Propagators are formal quantities used in calculations of the S-matrix amplitudes. Position-space propagators cannot be interpreted as propagation amplitudes (from point to point) or time-dependent wave functions. Such interpretation can be found in some QFT textbooks, but it 1) has zero experimental support; 2) leads to numerous theoretical contradictions.

Eugene.
 Sci Advisor P: 1,135 For those interested: from Pauli's famous 1940 paper, Spin and Statistics: Pauli's Spin and Statistics To be compared with Feynman's: Feynman's propagator in position space. Although Pauli's propagators are worse. (zero'th order Bessels rather than first order). Pauli, quote, "expressively postulates" commutation outside the light cone to overrule the Green's function. Peshkin & Schroeder's remarks about anti-particles canceling the non-causality stem from the second link. Chapter 18 of "Fundamental processes": Taking only one pole violates relativity, any physical process has diagrams with the other pole as well (anti-particle) to restore Lorentz invariance. Regards, Hans
 Sci Advisor P: 1,677 Hi Hans, what paper or book is that Feynman link too? Hellishly hard to find some of his old papers nowdays. Anyway that should settle the confusion as expected.
P: 1,135
 Quote by Haelfix Hi Hans, what paper or book is that Feynman link too? Hellishly hard to find some of his old papers nowdays. Anyway that should settle the confusion as expected.
It's in this nice book from his 1959-60 Caltech lectures:

The Theory of Fundamental Processes

Regards, Hans

PS: more copies here: amazon.com
P: 145
 Quote by Haelfix Anyway that should settle the confusion as expected.
Notice, that in the link Feyman does take serious the leaking out of the lightcone of the propagator named after him as a physical effect.

If the modern view is apperently different, ok.

EDIT: What I ask myself, is, how to we design an experiment to check this?
P: 1,746
 Quote by Micha EDIT: What I ask myself, is, how to we design an experiment to check this?
That's exactly the point. How can you measure propagators?

Eugene.
 Sci Advisor P: 1,192 Feynman clearly thought of the propagator as representing the amplitude for a particle to start at one spacetime point and end at another. When faced with the difficulty that this amplitude does not vanish outside the lightcone, he made up something about precise measurement of position leading to pair production, and this making it OK that his amplitude was nonzero outside the lightcone. None of this holds up to close scrutiny. Feynman was making up the formalism (for what we now call Feynman diagrams) as he went along; he had no deep justification. Only later did Dyson show how you could get Feynman's formalism from quantum field theory. But one of the things you lose when you do this is the notion that the propagator is an actual amplitude. We can calculate that amplitude in quantum field theory for a free massive particle. (There are extra issues in the massless case.) For simplicity of notation, I will work in one space dimension, and set hbar=c=1. The generalization to more dimensions is obvious. We know what the one-particle momentum eigenstates are: $$|k\rangle = a^\dagger(k)|0\rangle.$$ The only issue is normalization. Let us use the commutation relations $$[a(k),a^\dagger(k')]=f(k)\delta(k-k'),$$ where f(k) is a positive-definite function that is otherwise arbitrary. Common choices in the literature include f(k)=1 and f(k)=(2pi)32E(k), where E(k)=(k2+m2)1/2. But any positive-definite function is acceptable; this is simply a matter of convention. I will leave f(k) unspecified. The one-particle momentum eigenstates then have the normalization $$\langle k'|k\rangle = f(k)\delta(k'-k).$$ Correspondingly, the completeness relation is $$\int {dk\over f(k)}|k\rangle\langle k| = 1.$$ Now we need to decide what a position eigenstate is. Certainly two eigenstates at different positions should be orthogonal, so we have $$\langle x'|x\rangle = h(x)\delta(x'-x),$$ where h(x) is a postivie-definite normalization function analogous to f(k). The choice h(x)=constant is the only one consistent with translation invariance, so we will take h(x)=1: $$\langle x'|x\rangle = \delta(x'-x).$$ Next, we need to know the inner product of a position eigenstate and a momentum eigenstate. We will take $$\langle x|k\rangle =g(k)e^{ikx}.$$ The x dependence is again the only one consistent with translation invariance. Given a choice of f(k), we can determine g(k) as follows: $$\delta(x'-x)=\langle x'|x\rangle =\int {dk\over f(k)}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{ik(x'-x)}.$$ This only holds if |g(k)|2/f(k)=1/2pi, so we will make that choice. Note that a one-particle position eigenstate can be expressed as a linear combination of one-particle momentum eigenstates; there is no "pair production", because there are no interactions to produce any pairs. Now let's compute the propagation amplitude. This is given by $$\langle x'|e^{-iHt}|x\rangle =\int {dk\over f(k)}\langle x'|e^{-iHt}|k\rangle\langle k|x\rangle =\int {dk\over f(k)}e^{-iE(k)t}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x'-x)} =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x'-x)}.$$ This is not what you would get from the Feynman propagator, which would involve integrating over dk/E(k) instead of dk. Also, this does not vanish outside the lightcone. This has nothing whatsoever to do with pair production, because we have done the calculation entirely within the one-particle subspace. So, is it a problem? Only if you can measure it. Can you? It depends on what you mean by "measurement" in quantum field theory.
P: 1,746
 Quote by Avodyne Now let's compute the propagation amplitude. This is given by $$\langle x'|e^{-iHt}|x\rangle =\int {dk\over f(k)}\langle x'|e^{-iHt}|k\rangle\langle k|x\rangle =\int {dk\over f(k)}e^{-iE(k)t}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x'-x)} =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x'-x)}.$$ This is not what you would get from the Feynman propagator, which would involve integrating over dk/E(k) instead of dk. Also, this does not vanish outside the lightcone. This has nothing whatsoever to do with pair production, because we have done the calculation entirely within the one-particle subspace. So, is it a problem? Only if you can measure it. Can you? It depends on what you mean by "measurement" in quantum field theory.

I think your derivation is correct and what you got is exactly the "amplitude of finding the particle at point x' at time t if it was released from point x at time 0". Indeed, this amplitude does not vanish outside the lightcone. And I am ready to accept that this fact can be measured, in principle. But there is no contradiction with the principle of causality yet. You still need to prove that superluminal propagation of wavefunctions can be used for sending signals back to the past. Can you do that?

Eugene.
P: 304
 Quote by Avodyne Now let's compute the propagation amplitude. This is given by $$\langle x'|e^{-iHt}|x\rangle =\int {dk\over f(k)}\langle x'|e^{-iHt}|k\rangle\langle k|x\rangle =\int {dk\over f(k)}e^{-iE(k)t}\langle x'|k\rangle\langle k|x\rangle =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x'-x)} =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x'-x)}.$$ This is not what you would get from the Feynman propagator, which would involve integrating over dk/E(k) instead of dk. Also, this does not vanish outside the lightcone.
I think this is no surprise, since the Hamiltonian corresponding to the energy E(k)=(k2+m2)1/2 is known to be non-local. The acausal transition amplitude you calculate is just an expression of this fact. Wasn't the traditional escape from this nightmare to question, whether this is the right one-particle Hamiltonian ?
P: 1,746
 Quote by OOO I think this is no surprise, since the Hamiltonian corresponding to the energy E(k)=(k2+m2)1/2 is known to be non-local. The acausal transition amplitude you calculate is just an expression of this fact. Wasn't the traditional escape from this nightmare to question, whether this is the right one-particle Hamiltonian ?
There are many good reasons to believe that E(k)=(k2+m2)1/2 is the correct 1-particle Hamiltonian:

1. This form of the Hamiltonian follows from Wigner's theory of irreducible representations of the Poincare group;

2. This form is used throughout QFT with great success in calculations of scattering cross-sections, etc.

I think that the "escape from this nightmare" should be sought in another direction. Most importantly, there is no nightmare yet. The superluminal propagation is not a paradox by itself. The only real paradox is violation of causality, e.g., if one can build a machine that influences the past. I haven't seen a convincing proof that one can build such a machine by using superluminally propagating wave functions.

Eugene.
P: 304
 Quote by meopemuk There are many good reasons to believe that E(k)=(k2+m2)1/2 is the correct 1-particle Hamiltonian: 1. This form of the Hamiltonian follows from Wigner's theory of irreducible representations of the Poincare group; 2. This form is used throughout QFT with great success in calculations of scattering cross-sections, etc. I think that the "escape from this nightmare" should be sought in another direction. Most importantly, there is no nightmare yet. The superluminal propagation is not a paradox by itself. The only real paradox is violation of causality, e.g., if one can build a machine that influences the past. I haven't seen a convincing proof that one can build such a machine by using superluminally propagating wave functions. Eugene.
I for one find the prospect of describing the propagation of a single particle by
an equation like

$$i\partial_t \psi = \sqrt{-\partial_x^2+m^2} \psi$$

somewhat "itchy". Of course, my objection doesn't mean much. And maybe I'm just too narrow-minded.
P: 1,746
 Quote by OOO I for one find the prospect of describing the propagation of a single particle by an equation like $$i\partial_t \psi = \sqrt{-\partial_x^2+m^2} \psi$$ somewhat "itchy".
There is no way around it. The principle of relativity (the Poincare group) + quantum mechanics lead directly to this equation. All details of the proof can be found in first five chapters of http://www.arxiv.org/abs/physics/0504062

Eugene.
P: 304
 Quote by meopemuk There is no way around it. The principle of relativity (the Poincare group) + quantum mechanics lead directly to this equation. All details of the proof can be found in first five chapters of http://www.arxiv.org/abs/physics/0504062 Eugene.
P: 1,192
 Quote by meopemuk I think your derivation is correct and what you got is exactly the "amplitude of finding the particle at point x' at time t if it was released from point x at time 0". Indeed, this amplitude does not vanish outside the lightcone. And I am ready to accept that this fact can be measured, in principle.
Well, I'm not even sure that's possible. We really need a better model of what it means to measure something. The obvious thing to do is model particle detectors as external sources coupled to the field. I strongly suspect that this will render the effect unobservable.

 Quote by meopemuk But there is no contradiction with the principle of causality yet. You still need to prove that superluminal propagation of wavefunctions can be used for sending signals back to the past. Can you do that?
Certainly not! And I don't think it's possible, but this can only be answered in the context of a specific model of measurement.
 P: 145 Avodyne, I think your result breaks Lorentz invariance. The measure dk for the integral is only the spatial momentum component, and this will we different for different Lorentz frames. I suppose, this has to do with the fact, that you excluded pair production. Do you agree and if so, do you have any good reasons, why we should trust the formula anyway?
 Sci Advisor P: 1,192 Here is another calculation one could do. Suppose we have a free scalar field $$\varphi(x,t).$$ We could couple it to a time-dependent source $$J(x,t)$$ by adding a term $$\textstyle\int dx\,\varphi(x,t)J(x,t)$$ to the hamiltonian. Suppose J(x,t) is zero for t<0 at all x, and for |x|>L for all t, where L is some fixed finite length. (I'm still in one dimension.) Suppose also that the theory is in its ground state for t<0. At t=0, the source turns on, and the state changes. Now compute the time-dependent expectation value of the field. It will of course be zero for t<0. I conjecture that it will remain exactly zero for |x|>L+t; that is, outside the lightcone of the disturbance by the source. This calculation can be done exactly. I will do it when I get a chance.