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Very simple QFT questionsby nrqed
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#181
Oct2207, 06:44 PM

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#182
Oct2207, 07:00 PM

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Hi Avodyne,
Eugene. 


#183
Oct2207, 07:00 PM

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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 antiparticles canceling the noncausality 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 (antiparticle) to restore Lorentz invariance. Regards, Hans 


#184
Oct2307, 12:49 AM

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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. 


#185
Oct2307, 03:47 AM

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The Theory of Fundamental Processes Regards, Hans PS: more copies here: amazon.com 


#186
Oct2307, 10:04 AM

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If the modern view is apperently different, ok. EDIT: What I ask myself, is, how to we design an experiment to check this? 


#187
Oct2307, 10:31 AM

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Eugene. 


#188
Oct2307, 01:25 PM

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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 oneparticle momentum eigenstates are: [tex]k\rangle = a^\dagger(k)0\rangle.[/tex] The only issue is normalization. Let us use the commutation relations [tex][a(k),a^\dagger(k')]=f(k)\delta(kk'),[/tex] where f(k) is a positivedefinite function that is otherwise arbitrary. Common choices in the literature include f(k)=1 and f(k)=(2pi)^{3}2E(k), where E(k)=(k^{2}+m^{2})^{1/2}. But any positivedefinite function is acceptable; this is simply a matter of convention. I will leave f(k) unspecified. The oneparticle momentum eigenstates then have the normalization [tex]\langle k'k\rangle = f(k)\delta(k'k).[/tex] Correspondingly, the completeness relation is [tex]\int {dk\over f(k)}k\rangle\langle k = 1.[/tex] Now we need to decide what a position eigenstate is. Certainly two eigenstates at different positions should be orthogonal, so we have [tex]\langle x'x\rangle = h(x)\delta(x'x),[/tex] where h(x) is a postiviedefinite 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: [tex]\langle x'x\rangle = \delta(x'x).[/tex] Next, we need to know the inner product of a position eigenstate and a momentum eigenstate. We will take [tex]\langle xk\rangle =g(k)e^{ikx}.[/tex] 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: [tex]\delta(x'x)=\langle x'x\rangle =\int {dk\over f(k)}\langle x'k\rangle\langle kx\rangle =\int {dk\over f(k)}g(k)^2 e^{ik(x'x)}.[/tex] This only holds if g(k)^{2}/f(k)=1/2pi, so we will make that choice. Note that a oneparticle position eigenstate can be expressed as a linear combination of oneparticle 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 [tex]\langle x'e^{iHt}x\rangle =\int {dk\over f(k)}\langle x'e^{iHt}k\rangle\langle kx\rangle =\int {dk\over f(k)}e^{iE(k)t}\langle x'k\rangle\langle kx\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)}.[/tex] 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 oneparticle 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. 


#189
Oct2307, 02:21 PM

P: 1,746

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. 


#190
Oct2307, 03:03 PM

P: 304




#191
Oct2307, 03:31 PM

P: 1,746

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 crosssections, 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. 


#192
Oct2307, 03:46 PM

P: 304

an equation like [tex]i\partial_t \psi = \sqrt{\partial_x^2+m^2} \psi[/tex] somewhat "itchy". Of course, my objection doesn't mean much. And maybe I'm just too narrowminded. 


#193
Oct2307, 03:54 PM

P: 1,746

Eugene. 


#194
Oct2307, 04:08 PM

P: 304




#195
Oct2307, 04:17 PM

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#196
Oct2307, 05:09 PM

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? 


#197
Oct2307, 05:45 PM

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On the other hand, the calculation is entirely within the quantum field theory of a free scalar field, which is manifestly Lorentz invariant. 


#198
Oct2307, 06:10 PM

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P: 1,190

Here is another calculation one could do. Suppose we have a free scalar field [tex]\varphi(x,t).[/tex] We could couple it to a timedependent source [tex]J(x,t)[/tex] by adding a term [tex]\textstyle\int dx\,\varphi(x,t)J(x,t)[/tex] 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 timedependent 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. 


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