How can the Cauchy integral and Fourier integral produce the same result?

[Micha]

Maybe I just didn't understand what expression you had in mind.

In 1+1 dimensions it would be:

harm(k0,k1) = delta(k0-p0)*delta(p0^2-*k1^2-m^2)

for some arbitrary p0

 

[OOO]

Yes, I can subscribe to this statement. But this raises a question.
If you can not get all boundary conditions by moving the singularities around,
which ones are you implicitly applying by using the epsilon prescription at all?

Very good question. I have asked that myself since the electrodynamics lecture I took years ago.

These epsilon-prescriptions seem like writing the software for an Automated Teller Machine without being able to say how much money the customer will get if he requests $100. If he gets at least some money, it has to be okay for him...

 

[Micha]

What still perplexes me is, that all the QFT textbooks I know do it the other way. Hans seems to say that for QFT it doesn't matter which propagator you take, the differences cancel anyway (and it's hard to believe that anyone has done successful calculations if it did matter). But then why not take the advanced propagator ? Strange...

Don't forget, that this Kleiss in the Field theory lecture at Cern in the link, I posted, is selling epsilon as the decay rate of the particle. So it seems, there is not only mathematics, but also Physics in it.

Edit: Hans is suspicously calm. He is figuring this all out quietly it seems. Or our discussion is just too trivial for him.

 

[OOO]

In 1+1 dimensions it would be:

harm(k0,k1) = delta(k0-p0)*delta(p0^2-*k1^2-m^2)

for some arbitrary p0

Seems to work...

 

[OOO]

Don't forget, that this Kleiss in the Field theory lecture at Cern in the link, I posted, is selling epsilon as the decay rate of the particle. So it seems, there is not only mathematics, but also Physics in it.

I'm not sure. An electron (which is, of course, no KG particle) has no decay rate. But probably unstable particles can be dealt with this way (without sending epsilon to zero). But then they don't obey Klein-Gordon but some dissipative equation.

Edit: Hans is suspicously calm. He is figuring this all out quietly it seems. Or our discussion is just too trivial for him.

As a wise man he doesn't work on weekends...

 

[Micha]

I'm not sure. An electron (which is, of course, no KG particle) has no decay rate. But probably unstable particles can be dealt with this way (without sending epsilon to zero). But then they don't obey Klein-Gordon but some dissipative equation.

The fact, that we haven't seen an electron decaying experimentally just means, that the decay time of the electron is bigger than some limit of 10^35 years or so. Remember the experiments about proton decay. It would still be clear, which sign of epsilon would be the "right" one. Kleiss goes so far to connect the sign of epsilon to the arrow in time in QFT.
Don't worry about Dirac vs. KG by the way. The denominator of the propagator is just the same.

 

[Haelfix]

Exactly what do you want the electron to decay into?

Keep in mind i epsilon really has to do with Wick rotation, with epsilon --> zero. Its justified mathematically in that sense ultimately. Typically we denote something that looks like i sigma for the decay rate in cases where the resonance is so thin that its hard to make sense off. Sigma is not a limit though, but a small and positive number.

Depending on definitions they can be used interchangeably, but ultimately one is utilized as a mnemonic for a trick, and the other is an actual physical thing.

 

[Micha]

I think, I understand now, in which direction to search for the source of our confusion.
We should remind ourselves, that, while we are calculation in natural units, where hbar = 1, there is an 1/hbar^2 sitting in front of the rest mast term.
This means, that the KG equation with finite rest mass does know about quantum mechanics, whereas the massless equation does not!
So for the massless equation we will of course rediscover the classical behaviour of the photon traveling along the light cone.
If we now look at the leaking out of the lightcone for the massive case, it goes with
exp(-m/hbar*abs(x)). From there we see, that it is clearly a quantum mechanical effect! If hbar goes to zero, there is no leaking out of the light cone! We know from things as the tunnel effect, that a quantum mechanics state can leak into classical forbidden regions with an exponentially decaying amplitude.
We can see the same thing also in a different way.
A strictly causal propagator would be completely concentrated in the point x=0 for t=0. On the other hand, the propagtor is described by 1/(p^2-m^2), so its momentum is also completely sharp.
We know from the uncertainty principle, that such states do not exist in quantum mechnacis. I think, Hans solution for the massive case should be looked at with this in mind.
By the way, in the classical electrodynmacis book of Jelitto, it is said, that the epsilon prescription is used exactly because it automatically ensures causality. I didn't follow the argument so far. Maybe this statement can be generalized to say, that the epsilon prescription is used in QFT, because it respects causality as much as possible under the laws of quantum mechancis.

 

[Micha]

Exactly what do you want the electron to decay into?

Keep in mind i epsilon really has to do with Wick rotation, with epsilon --> zero. Its justified mathematically in that sense ultimately. Typically we denote something that looks like i sigma for the decay rate in cases where the resonance is so thin that its hard to make sense off. Sigma is not a limit though, but a small and positive number.

Depending on definitions they can be used interchangeably, but ultimately one is utilized as a mnemonic for a trick, and the other is an actual physical thing.

Maybe my argument is far fetched. But I don't see how the epsilon in
1/(p^2-m^2+i*epsilon) is coming from a wick rotation.

 

[OOO]

A strictly causal propagator would be completely concentrated in the point x=0 for t=0. On the other hand, the propagtor is described by 1/(p^2-m^2), so its momentum is also completely sharp.

Momentum isn't sharp (why should it be ). I think we agree that the propagator is not a plane wave.

And remember, my numerics (as well as Hans') show no acausality if the boundary condition is Phi(t<0)=0. Sometimes it seems to me that hbar has been invented just for the purpose of having an excuse whenever one doesn't know exactly what is going on...

 

[Micha]

And remember, my numerics (as well as Hans') show no acausality if the boundary condition is Phi(t<0)=0.

Weren't you calculating the massless case only?

 

[Micha]

Momentum isn't sharp (why should it be ). I think we agree that the propagator is not a plane wave.

True. Only the modulus of the momentum is sharp.

 

[Haelfix]

Hi Micha, its a bit of a long story to see where i epsilon ultimately comes from. On one hand it seems like a bit of a hack in Feynman's prescription, and on the other it sort of appears naturally in canonical quantization (See A. Zee for this point).

Either way I think its clear by now that you guys can see that i.epsilon is related to causal structure in a certain sense (it is).

From there, enter mathematical rigor with things like the Osterwalder Schrader theorem, and its a hop leap and a jump to Wick rotation. Sorry to be vague, but its just totally nontrivial to expose this in a way that makes good sense (I only have it from theory discussion notes).

 

[OOO]

Weren't you calculating the massless case only?

You seem to refer to my analytical confirmation of Hans' result, that's true (although I have little doubt that the calculation yields a similar result for the massive case). But I was referring to my simulations (and I guess Hans has done something similar). The finite difference approximation of the massive KG equation does give a propagator that's exactly zero outside the lightcone (with the appropriate initial conditions).

True. Only the modulus of the momentum is sharp.

Neither. 1/(p^2-m^2) is nonzero for all p, not just the ones on mass shell.

 

[OOO]

Either way I think its clear by now that you guys can see that i.epsilon is related to causal structure in a certain sense (it is).

Haelfix, what (I think) we are discussing about is, whether there is one propagator that is exactly zero outside the light cone and why the textbooks don't just take this one to prove causality.

From there, enter mathematical rigor with things like the Osterwalder Schrader theorem, and its a hop leap and a jump to Wick rotation. Sorry to be vague, but its just totally nontrivial to expose this in a way that makes good sense (I only have it from theory discussion notes).

I have looked for the Osterwalder Schrader theorem before (in a different context) but I haven't found something useful.

Could you please tell me the reference to a publication or textbook that deals with the OS theorem.

 

[Haelfix]

I like Hans argument for the vanishing of the propagator offshell, but its not necessary really.

The commutator argument works fine either as a consequence of the Smatrix satisfying certain properties, or even just defining it as such (alla Weinberg who just imposes microcausality).

Constructive field theory makes all of this painfully rigorous as they go to great lengths to spell out the requirements and axioms necessary for a field to be causal. I don't have a good introductory link, but Streeter-Wightman probably can get you started.

This whole business is really a very old debate, that was done 40-50 years ago, i'd imagine it would be quite hard to track down the appropriate papers absent some old proffessor who remembers things.

 

[OOO]

I like Hans argument for the vanishing of the propagator offshell, but its not necessary really.

The commutator argument works fine either as a consequence of the Smatrix satisfying certain properties, or even just defining it as such (alla Weinberg who just imposes microcausality).

I think there must be something wrong with QFT (at least pedagogically) if Peskin & Schroeder are forced to introduce a concept on page 30 of their book, that, as you say, works fine, but the reason of which can't be understood without axiomatic quantum field theory.

I mean there is not even the slightest hint of a justification for why we should calculate the commutator by shifting the poles by ++ and not by +- epsilon. The first possibility yields the retarded propagator (which is zero outside the lightcone) and the second possibility yields the Feynman propagator (which is nonzero outside the cone, although being exponentially damped for larger separation). As P&S say:

"The p^0 integral of (2.58) can be evaluated according to four different contours, of which that used in (2.54) is only one."

So we are expected to learn by heart that the retarded propagator is used for calculating the commutator and the Feynman propagator is used for Feynman graphs. That's like learning zoology ! So what is all this jabbering for, that P&S involve into ?

Constructive field theory makes all of this painfully rigorous as they go to great lengths to spell out the requirements and axioms necessary for a field to be causal. I don't have a good introductory link, but Streeter-Wightman probably can get you started.

Thanks. I have looked into the Streater-Wightman book at Amazon but it doesn't seem to mention the Osterwalder-Schrader theorem in the index.

 

[Micha]

Neither. 1/(p^2-m^2) is nonzero for all p, not just the ones on mass shell.

True.

 

[Micha]

@OOO
I am confused by post 167.
I don't have the book of P&S at hand. I never saw two different propagators advocated.
From checking the integral I think, that the plus plus propagator (1/(p^2-m^2+i*epsilon)) is leaking out of the lightcone. And the reason is, it is doing this already at t=0.
This is, why there is no contraction with your numerical simulations.

 

[OOO]

@OOO
I am confused by post 167.
I don't have the book of P&S at hand. I never saw two different propagators advocated.
From checking the integral I think, that the plus plus propagator (1/(p^2-m^2+i*epsilon)) is leaking out of the lightcone. And the reason is, it is doing this already at t=0.
This is, why there is no contraction with your numerical simulations.

I have reread that part of P&S and to me it's confusing as well. P&S first calculate the vacuum expectation value of two field operators: <0|Phi(x)Phi(y)|0>. For this they show that there is the "leakage outside the lightcone", i.e. it does not vanish outside the light cone.

Then they say: this is void of meaning because it's not what we measure, so consider commutators instead. For simplicity they calculate the commutator as an expectation value again (valid because commutator is a c-number): <0| [Phi(x)Phi(y)] |0>. They show that this yields the 4D Fourier transform of what we have been discussing here all the time: 1/(p^2-m^2). Then they say that, by moving both poles into the lower half plane (so I should have rather said -- instead of ++, sorry), one gets a propagator (retarded) that vanishes outside the lightcone, which is what they seemed to expect.

Finally, they knock our socks off by saying that the (--) prescription is not the only one, but the Feynamn propagator is obtained by the (+-) prescription, or simply by p^2-m^2+iepsilon, and, again, this one does not vanish outside the lightcone.

 

[Micha]

I see. I think, it is ok to be confused.

Let me ask another question. Do you think, the leaking of the lightcone is physical? Can it be measured?

 

[OOO]

I see. I think, it is ok to be confused.

Let me ask another question. Do you think, the leaking of the lightcone is physical? Can it be measured?

As this thread has shown, many people seem to say, that it can't, because causality has to be preserved if special relativity is to make any sense. On the other hand there are advocats of superluminal propagation. I'm not in the position to question either side, because I'm still trying to learn that stuff too.

My personal suspicion is, that the idea of superluminal propagation could well be grounded in a psychological motivation to fuel esotericism and/or science fiction movies.

 

[Avodyne]

1) The Feynman propagator does not vanish outside the lightcone. Explicit expressions (in four spacetime dimensions) are given in Appendix C of Relativistic Quantum Fields by Bjoken and Drell.

2) The i-epsilon prescription that leads to the Feynman propagator corresponds to taking the vacuum expection value of the time-ordered product of two free fields. Time-ordered products of fields are relevant because they are related (by the LSZ reduction formula) to scattering amplitudes.

3) Causality is related to the commutator of two fields; this should vanish outside the lightcone, so that a measurement of the field at one point does not affect the measurement at a spacelike separated point.

 

[jostpuur]

I must say I'm surprised by this debate about the propagators. It seems to be always going on in some thread.

If physicists used more rigor mathematics to justify their conclusions about this propagator problem, we probably wouldn't have this debate. The physicists always have the policy, that they don't need to understand the math, as long as their calculations work. Now, as a consequence, there is no agreement about the behaviour of the relativistic propagator.

 

[OOO]

1) The Feynman propagator does not vanish outside the lightcone. Explicit expressions (in four spacetime dimensions) are given in Appendix C of Relativistic Quantum Fields by Bjoken and Drell.

Edit: this post is obsolete. Avodyne's statement is compatible with what Peskin & Schroeder say. Sorry.

Welcome to this delicate discussion Avodyne. What you say is interesting because it adds a little to my confusion. In Peskin & Schroeder, eqs. 2.51 and 2.52, the authors calculate the quantity

&lt;0|\phi(x)\phi(y)|0&gt; =: D(x-y)

and afterwards they explicitely say: "So again we find that outside the light-cone, the propagation amplitude is exponentially vanishing but nonzero.". I've hacked their intermediate result into maple and it seems to me they are right.

Because the above D(x-y) reduces to the time-ordered product (i.e. the Feynman propagator) in the special case x^0&gt;y^0, it seems that the Feynman propagator is nonzero outside the lightcone too. At least if one believes in what they have done with D(x-y).

Anyway I'll have a look at Bjorken-Drell. Meanwhile I'm at a point where nothing comes as a surprise...

 

[Micha]

Why is Avodyne's remark confusing you? What you say is in agreement with what he said, isn't it?

 

[Micha]

Could somebody say then in one sentence, what is the exact physical meaning of the Feynman propagator?

Edit: I'd say, it is the amplitude to find a particle at spacetime point y, when you have found one at earlier spacetime point x.

 

[OOO]

Why is Avodyne's remark confusing you? What you say is in agreement with what he said, isn't it?

Thanks for pointing that out, Micha. My brain has become a knot. I'm not even able to read properly.

 

[jostpuur]

Could somebody say then in one sentence, what is the exact physical meaning of the Feynman propagator?

Edit: I'd say, it is the amplitude to find a particle at spacetime point y, when you have found one at earlier spacetime point x.

I tried to ask about this here https://www.physicsforums.com/showthread.php?t=176563

I don't feel like I got satisfying answer. The answer seems to be, that the propagator doesn't necessarily mean anything. It just works when used correctly to compute scattering amplitudes.

 

[Hans de Vries]

Thanks for pointing that out, Micha. My brain has become a knot.

At the end all the confusion is unlikely to have any influence on real calculations since these are all done in momentum space, not in position space, and non physical results are massaged away with other prescriptions.

Leaking outside the light cone with exp-m at t=0 would mean instantaneous propagation at infinite speed over micron size distances in the case of neutrinos. The size of living species.Regards, Hans

 

[Avodyne]

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.

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.)

 

[meopemuk]

Hi 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.

 

[Hans de Vries]

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

 

[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.

 

[Hans de Vries]

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

 

[Micha]

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?

 

[meopemuk]

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.

 

[Avodyne]

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&#039;)]=f(k)\delta(k-k&#039;),
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&#039;|k\rangle = f(k)\delta(k&#039;-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&#039;|x\rangle = h(x)\delta(x&#039;-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&#039;|x\rangle = \delta(x&#039;-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&#039;-x)=\langle x&#039;|x\rangle<br /> =\int {dk\over f(k)}\langle x&#039;|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}|g(k)|^2 e^{ik(x&#039;-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&#039;|e^{-iHt}|x\rangle<br /> =\int {dk\over f(k)}\langle x&#039;|e^{-iHt}|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}e^{-iE(k)t}\langle x&#039;|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x&#039;-x)}<br /> =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x&#039;-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.

 

[meopemuk]

Now let's compute the propagation amplitude. This is given by
\langle x&#039;|e^{-iHt}|x\rangle<br /> =\int {dk\over f(k)}\langle x&#039;|e^{-iHt}|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}e^{-iE(k)t}\langle x&#039;|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x&#039;-x)}<br /> =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x&#039;-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.

 

[OOO]

Now let's compute the propagation amplitude. This is given by
\langle x&#039;|e^{-iHt}|x\rangle<br /> =\int {dk\over f(k)}\langle x&#039;|e^{-iHt}|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}e^{-iE(k)t}\langle x&#039;|k\rangle\langle k|x\rangle<br /> =\int {dk\over f(k)}|g(k)|^2 e^{-iE(k)t}e^{ik(x&#039;-x)}<br /> =\int {dk\over 2\pi}e^{-iE(k)t}e^{ik(x&#039;-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 ?

 

[meopemuk]

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.

 

[OOO]

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.

 

[meopemuk]

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.

 

[OOO]

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.

Have you ever considered Google ads

 

[Avodyne]

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.

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.

 

[Micha]

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?

 

[Avodyne]

Avodyne, I think your result breaks Lorentz invariance.

That depends on what you mean by "Lorentz invariance". The real problem is defining what we mean by a particle at a definite position. My states of definite position are eigenstates of the Newton-Wigner position operator, as explained by Eugene in http://www.arxiv.org/abs/physics/0504062, but there is no comparable "time" operator. So I am treating space and time differently at the very beginning. Lorentz transformations do not connect these "space" and "time" labels.

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

 

[Avodyne]

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.

 

[meopemuk]

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.

I don't think it is necessary to model particle detectors and measuring apparatuses. Once we have the position-space wave function (e.g., the one that you calculated) we already have by definition the probability amplitude for measuring particle position at a given point in space. It doesn't matter what kind of physical device will be actually used for this measurement. It is only important that this is a position-measuring device, and this fact is already incorporated in our choice of the position-space representation for the wave function.

It is true that in reality physical observables cannot be measured with unlimited precision. So, the quantum-mechanical assumption about precise measurements of observables is, of course, an idealization. But I guess that without such an idealization the theory would be a complete mess.

Eugene.

 

[meopemuk]

Avodyne, I think your result breaks Lorentz invariance.

It is possible to prove that the inner product of Avodyne's wave functions

\langle \psi | \phi \rangle = \int d^3r \psi^*(\mathbf{r}) \phi(\mathbf{r})

is the same in all moving reference frames. Boost transformations of position-space wave functions are rather tricky, but this can be done. So, the Lorentz invariance is not violated.

Eugene.