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

[Micha]

It's is a non-physical artifact of perturbation theory where there are
pairs of diagrams which have a physical meaning. Independently
they have no physical meaning because they can be converted into
each other via a Lorentz transform, and only together they respect
causality and special relativity.
Regards, Hans

The term "propagation outside the lightcone" seems to be loaded with mystery.
So let's avoid it and talk about the Fourier transform of the function
1/(p^2-m^2)

We are just talking about the Green's function of the classical Klein-Gordon
equation. Why would we need pertubation theory for this?

Making a distinction of cases for t>0 and t<0 in solving the integral over energy
is a completely valid mathematical step. If you see any problems, please
indicate them on mathematical grounds.

@OOO
Interesting argument.
Do you think, it will hold for an infinite Dirac delta peek?

 

[Micha]

If 1+1 dimensions is too hard to do, why not going to 1 dimension for a while.
I claim the propagator here is
1/(2m)*exp(-m*abs(x)).

I think this also indicates, that the massless propagator really gets singular here.

I should mention, that I talk about the Fourier transform of the function 1/(p^2+m^2) here. This
means, I talk about one spacelike dimension here. You can go to the timelike case
by substituting m=i*m. This means, you get oscillatory behaviour in a timelike dimension,
and exponential decay in a spacelike dimension.

 

[Micha]

@OOO
Another thought:
You might as well solve your equation for
Psi(x+1) and then you get a spacelike correlation instead
of a timelike.

 

[OOO]

Interesting argument.
Do you think, it will hold for an infinite Dirac delta peek?

You can't do an infinite delta peak in lattice numerics. But as an approximation you can use boundary conditions where Psi does not vanish at exactly one site (with finite but arbitrarily large value however; this doesn't really matter since the KG equation is linear).

What you get then is two peaks running left and right (1+1 dimension) or a circular wavefront (1+2 dimension). If there's dispersion (m != 0) then there are oscillations lagging behind the wavefront(s), but not before it (which would be outside the light cone). In the case of the massless equation m=0 (e.g. photons, apart from polarization) then there are no such oscillations (within numerical accuracy), one gets a sharp "retarded propagator".

So yes, this also holds for an approximate delta peak, and that's really the test case, which is why you mentioned it.

 

[OOO]

@OOO
Another thought:
You might as well solve your equation for
Psi(x+1) and then you get a spacelike correlation instead
of a timelike.

What for ?

 

[Micha]

What for ?

You are solving for Psi(t+1), because you know, that t is the time variable.
On mathematical grounds, x and t are just symbols, so why not solving
for Psi(x+1) and see a propagation in space?

I think, that you get only timelike propagation and not spacelike, because
you are already putting this in as an assumption..

 

[OOO]

You are solving for Psi(t+1), because you know, that t is the time variable.
On mathematical grounds, x and t are just symbols, so why not solving
for Psi(x+1) and see a propagation in space?

I think, that you get only timelike propagation and not spacelike, because
you are already putting this in as an assumption..

I think I don't know what you mean. The difference between space and time is not just one between mathematical symbols. It's a difference in sign ! You may, of course, try to solve the equations that way. But, to make a long story short, this scheme becomes unstable. You actually said it in one of your posts above: exchanging t<->x amounts to m<->i*m and this exchanges oscillatory solutions with exponentially damped and (more importantly !) excited ones. You'd have to select your boundary conditions very carefully in order to have only damped solutions.

Anyway, this does only make sense from a mathematical POV.

 

[reilly]

Hans -- your discussion in post 5 is one of the best I've ever read.

In my usual simple minded way, I'll suggest that the standard retarded Green's Function, like in E&M, prohibits propagation outside the light cone almost by definition -- it all has to do with those tricky i epsilons and contours of integration -- beloved by Electrical Engineers.

A personal note: a few years ago I became interested in the spreading of wave packets for the Dirac Equation. Initially I thought that the edges of the packet spread more quickly than Special Relativity allowed. But after extensive calculations, I rediscovered the retarded propagator and decided that my initial hunch was incorrect, which it is.
Regards, Reilly Atkinson

 

[Micha]

I think I don't know what you mean. The difference between space and time is not just one between mathematical symbols. It's a difference in sign ! You may, of course, try to solve the equations that way. But, to make a long story short, this scheme becomes unstable. You actually said it in one of your posts above: exchanging t<->x amounts to m<->i*m and this exchanges oscillatory solutions with exponentially damped and (more importantly !) excited ones. You'd have to select your initial conditions very carefully in order to have only damped solutions.

Anyway, this does only make sense from a mathematical POV.

Yes, I know about the sign difference and I know there is no propagation in space, but
only exponential damping. My whole point is, there is correlation along the spacelike dimension which accounts for the exponential decay. Remember, the discussion is,
whether we have exponential decay in the spacelike dimension or if the propagtor is exactly zero for spacelike dimensions.

 

[OOO]

Yes, I know about the sign difference and I know there is no propagation in space, but
only exponential damping. My whole point is, there is correlation along the spacelike dimension which accounts for the exponential decay. Remember, the discussion is,
whether we have exponential decay in the spacelike dimension or if the propagtor is exactly zero for spacelike dimensions.

Hmm, still not sure what you mean. If there was correlation along spacelike separation then you couldn't specify initial conditions. But you can. Choose any Psi(t,x) and Psi(t-1,x) for all x, that you like, and you will get perfectly definite time evolution.

 

[Micha]

Hmm, still not sure what you mean.

Me neither. Maybe correlation is even the wrong word. After all we know, that wave packets and information can only travel inside the light cone, right?

Remember, I was only claiming, that the Fourier transform of the function 1/(p^2-m^2) in two dimensions with Minkowski metric is not exactly zero at (x,0).
You help me please on why you think, you can proof me (and Feynman and the textbooks) wrong without doing the integral, or if I am right, what this exactly means for the time evolution of the KG equation.

 

[Anonym]

Hans -- your discussion in post 5 is one of the best I've ever read.

I am confused. Do you mean post #6 in this session by Hans de Vries?

a few years ago I became interested in the spreading of wave packets for the Dirac Equation. Initially I thought that the edges of the packet spread more quickly than Special Relativity allowed. But after extensive calculations, I rediscovered the retarded propagator and decided that my initial hunch was incorrect, which it is.

Provided you mean post #6, Hans de Vries discuss KG and not Dirac. Please reproduce the relevant summary of your calculations that allowed your decision. Notice that it means that V.A. Fock was wrong.

Regards, Dany.

 

[OOO]

Me neither. Maybe correlation is even the wrong word. After all we know, that wave packets and information can only travel inside the light cone, right?

Remember, I was only claiming, that the Fourier transform of the function 1/(p^2-m^2) in two dimensions with Minkowski metric is not exactly zero at (x,0).
You help me please on why you think, you can proof me (and Feynman and the textbooks) wrong without doing the integral, or if I am right, what this exactly means for the time evolution of the KG equation.

I didn't say anything about calculating the propagator in position representation from its momentum representation, did I ? If you feel the urge to do that, I won't stop you, but I won't try to find your respective mistakes either.

I did say that, on lattice-discretizing the KG equation, the impulse response, which is nothing but the retarded propagator of the difference equations, shows no sign of having values outside the cone.

Of course, we could go on and on and on, talking about different things, but I was thinking that you referred to what I was saying. If this is not the case then I apologize for that misconception.

 

[Micha]

I didn't say anything about calculating the propagator in position representation from its momentum representation, did I ? If you feel the urge to do that, I won't stop you, but I won't try to find your respective mistakes either.

I did say that, on lattice-discretizing the KG equation, the impulse response, which is nothing but the retarded propagator of the difference equations, shows no sign of having values outside the cone.

Of course, we could go on and on and on, talking about different things, but I was thinking that you referred to what I was saying. If this is not the case then I apologize for that misconception.

Ok, let me explain, where I am coming from.

I am trying to learn quantum field theory by the book of Zee.

Meanwhile I read post #6 from Hans, which says, that the propagator of KG in spacetime representation is exactly zero for spacelike intervals. This contracticts the book of Zee, which Hans himself mentions in his discussion.
So I tried to check the issue myself and now I am convinced, that Zee is right and Hans is wrong. If this is true, I think this might be interesting for other people as well, who are reading this thread.

Now you came in saying
"I have also done numerical simulations of KG and it seems quite obvious to me that there is no propagation outside the light cone."
So I thought you were supporting Hans result. I guess this is just a misunderstanding
about mixing up the two statements

1. "The propagator in spacetime representation is strictly zero outside the light cone."

and

2. "There is no propagation outside the lightcone."

I now think, you mean, that no wave packets can travel outside the light cone, and if so, I think, you are completely right.

To sum up, I think, that 1 is wrong and 2 is right.

 

[OOO]

I am trying to learn quantum field theory by the book of Zee.

I love that extraordinary book. Very much fun to read.

So I thought you were supporting Hans result. I guess this is just a misunderstanding
about mixing up the two statements

"The propagator in spacetime representation is zero outside the light cone."

and

"There is no propagation outside the lightcone."

I now think, you mean, that no wave packets can travel outside the light cone, and if so, I think, you are completely right.

Indeed I don't know the difference between these two statements. So this must be the problem. What I was trying to say is the first statement (derived from numerical evidence). So how can there be propagation outside the lightcone when the propagator is zero there ?

Edit: As you say, you think 1 is wrong, I say numerical simulation gives no indication for that. Thus, actually, I wanted to support what Hans said. I didn't claim anything about wave packets or group velocity, but about the maximum speed of interactions represented by the behaviour of the wavefront of the KG propagator.

 

[OOO]

Meanwhile I read post #6 from Hans, which says, that the propagator of KG in spacetime representation is exactly zero for spacelike intervals. This contracticts the book of Zee, which Hans himself mentions in his discussion.
So I tried to check the issue myself and now I am convinced, that Zee is right and Hans is wrong.

I'm too tired to check Zee's calculation at the moment. But how about this:

By integrating in various ways around singularities in the complex frequency plane you get different propagators. In fact there is an infinite multitude of such propagators. If you have found one propagator, you may always add a homogeneous (source free) solution to this propagator, and you get a valid propagator again. By choosing a different path in the complex omega-plane, it's as if you effectively added a homoheneous solution to the former propagator.

So what Zee and other authors found is probably a valid propagator, but one that does not meet the boundary conditions of Psi(t0,x)=delta(x-x0).

 

[Hans de Vries]

I'll suggest that the standard retarded Green's Function, like in E&M, prohibits propagation outside the light cone almost by definition -- it all has to do with those tricky i epsilons and contours of integration --

I tend to agree, That's my feeling too. Regards, Hans

 

[Hans de Vries]

Hans de Vries discuss KG and not Dirac.

In this case it shouldn't make any difference Dany. The Green's function of the
Dirac propagator is just a sum of first order derivatives of the Klein Gordon
Green's function. If the latter is zero outside the light cone then the one for
the Dirac propagator is zero as well and visa versa.Regards, Hans

 

[Micha]

I'm too tired to check Zee's calculation at the moment. But how about this:

By integrating in various ways around singularities in the complex frequency plane you get different propagators. In fact there is an infinite multitude of such propagators. If you have found one propagator, you may always add a homogeneous (source free) solution to this propagator, and you get a valid propagator again. By choosing a different path in the complex omega-plane, it's as if you effectively added a homoheneous solution to the former propagator.

So what Zee and other authors found is probably a valid propagator, but one that does not meet the boundary conditions of Psi(t0,x)=delta(x-x0).

The uniqueness problem is a good point. But actually, in the calculation we are just going from momentum space to real space via a Fourier transform, and this is a unique operation. Also, when you do the integral, you see, that there is no choice involved about the integration path. You want to close the loop in such a way, that your exponential goes to zero, so you get no additional contribution from closing the loop.

I think the boundary condition making the propagator unique is, that the propagator goes to zero for infinite spacelike separation, which the Fourier transform of 1/(p^2-m^2) does (while not being exactly zero).

 

[Micha]

I tend to agree, That's my feeling too. Regards, Hans

Hans, I saw your paper, where you calculated all these propagators in different dimensions.
Sure, a lot of effort went into this.

I really ask you to do the calculation of Zee in 1+1 dimensions and tell me, where the error is.
It is only a few lines of algebra and the steps are in the book, until you end up with a one dimensional integral,
which you can put on a computer easily.

Also, do you agree with the result, I put here for the 1 dimensional case?

 

[Hans de Vries]

Hans, I saw your paper, where you calculated all these propagators in different dimensions.
Sure, a lot of effort went into this.

I really ask you to do the calculation of Zee in 1+1 dimensions and tell me, where the error is.
It is only a few lines of algebra and the steps are in the book, until you end up with a one dimensional integral,
which you can put on a computer easily.

Also, do you agree with the result, I put here for the 1 dimensional case?

It's just not the same Green's function anymore because of the pole prescription.
As Feynman remarks himself (in chapter 17 and 18 of Fundamental processes) what
you get is two different Green's functions depending on what pole you pick up, both
violate special relativity but the two diagrams they represent together become
Lorentz invariant again.

The diagrams he shows are spin 0 diagrams, where a spin 0 particle in one diagram
is replaced by its anti-particle in the other. He mentions that both are just two
representations of the same thing because a Lorentz transform can change one
into the other. Of course, changing a particle into its anti-particle via Lorentz
transform is impossible as Feynman remarks himself, except if two subsequent
interactions of the same particle are outside each others light cone...
(Actually, I don't see these kind of diagrams used in practice)

The question is. Are these only mathematical constructs, handy to do calculations?
Do they have a real physical meaning or not? Now, since the physical effects
always cancel, I would say not.


Regards, Hans

 

[OOO]

The uniqueness problem is a good point. But actually, in the calculation we are just going from momentum space to real space via a Fourier transform, and this is a unique operation. Also, when you do the integral, you see, that there is no choice involved about the integration path. You want to close the loop in such a way, that your exponential goes to zero, so you get no additional contribution from closing the loop.

I disagree. This is not just a harmless Fourier transform as it might seem at first. The KG operator is singular so its Green's function in momentum space has (two) poles. There's nothing unique in adding (+i epsilon) to the denominator. One could add (-i epsilon) as well. In the end adding a small imaginary part amounts to integrating around the original singularities along a small half circle. So one could even integrate around the negative pole in the upper half plane and around the positive pole in the lower half plane or vice versa. And of all these propagators one gets, one may take any weighted average to get a valid propagator again.

No, I remain confident that this integral is by no means unique.

I think the boundary condition making the propagator unique is, that the propagator goes to zero for infinite spacelike separation, which the Fourier transform of 1/(p^2-m^2) does (while not being exactly zero).

I can't see that this should be a complete boundary condition that determines the solution uniquely. Talking again about the continuum (not my discrete simulation), one may specify boundary conditions where $\Psi(t_0,x)$ and $\partial_t\Psi(t_0,x)$ are given for all x.

You will understand that there are already lots of functions $\Psi_0(x)=\Psi(t_0,x)$ that tend to zero for infinite spacelike separation. One of them is $\Psi_0(x)=\delta(x)$. Let alone the time derivative of $\Psi(t,x)$ at t0...

 

[Micha]

I disagree. This is not just a harmless Fourier transform as it might seem at first. The KG operator is singular so its Green's function in momentum space has (two) poles. There's nothing unique in adding (+i epsilon) to the denominator. One could add (-i epsilon) as well. In the end adding a small imaginary part amounts to integrating around the original singularities along a small half circle. So one could even integrate around the negative pole in the upper half plane and around the positive pole in the lower half plane or vice versa. And of all these propagators one gets, one may take any weighted average to get a valid propagator again.

No, I remain confident that this integral is by no means unique.

At the end, we are sending epsilon to zero, right? Do the integral and see, that you get the same result, if you use either -i*epsilon or +i*epsilon.
I think, there is a problem with your integration contour.
We want to integrate from minus infinity to plus infinity along the real axis.
We start with an integral from -r to r for r some (big) real number.
Now we are closing the integral via a big (not small!) half circle around the complex plane.
Do we use the upper or lower half circle? This depends on sign of our exponential, which in the end is just definition
of the Fourier transform. (Either the Fourier transform or its inverse get a minus.)
If we use exp(i*k*x), we want to use the upper half circle, because then in the limit, that r goes to infinity, the integrand around the half circle goes to zero. Notice, that the sign of epsilon was not involved here.

 

[Micha]

I admit, I have to think more about what makes the propagator unique.
Maybe we need to demand translation invariance to make the propagator unique.

But I am very sure, there is no uniqueness problem in going from momentum
space to real space.

Let me approach the uniqueness question from another perspective.
We know the solutions of the homogeneous equation.
These are the harmonic waves, right?
Now, for them, it is easy to go from momentum space to real space, so
if you add a harmonic wave to the propagator in real space, you have to do the
same in momentum space and vice versa.

 

[Micha]

It's just not the same Green's function anymore because of the pole prescription.
As Feynman remarks himself (in chapter 17 and 18 of Fundamental processes) what
you get is two different Green's functions depending on what pole you pick up, both
violate special relativity but the two diagrams they represent together become
Lorentz invariant again.

The diagrams he shows are spin 0 diagrams, where a spin 0 particle in one diagram
is replaced by its anti-particle in the other. He mentions that both are just two
representations of the same thing because a Lorentz transform can change one
into the other. Of course, changing a particle into its anti-particle via Lorentz
transform is impossible as Feynman remarks himself, except if two subsequent
interactions of the same particle are outside each others light cone...
(Actually, I don't see these kind of diagrams used in practice)

The question is. Are these only mathematical constructs, handy to do calculations?
Do they have a real physical meaning or not? Now, since the physical effects
always cancel, I would say not.


Regards, Hans

I know, that Feynman decribes antiparticles as particles going backward in time and that any consistent QFT needs both branches of the propagator, which is why, antimatter must exist.
These topics confuse me still and are far ahead of the current simple question, which is, how the propagator looks in real space.

 

[Micha]

It's just not the same Green's function anymore because of the pole prescription.
As Feynman remarks himself (in chapter 17 and 18 of Fundamental processes) what
you get is two different Green's functions depending on what pole you pick up, both
violate special relativity but the two diagrams they represent together become
Lorentz invariant again.

The diagrams he shows are spin 0 diagrams, where a spin 0 particle in one diagram
is replaced by its anti-particle in the other. He mentions that both are just two
representations of the same thing because a Lorentz transform can change one
into the other. Of course, changing a particle into its anti-particle via Lorentz
transform is impossible as Feynman remarks himself, except if two subsequent
interactions of the same particle are outside each others light cone...
(Actually, I don't see these kind of diagrams used in practice)

The question is. Are these only mathematical constructs, handy to do calculations?
Do they have a real physical meaning or not? Now, since the physical effects
always cancel, I would say not.


Regards, Hans

I know, that Feynman decribes antiparticles as particles going backward in time and that any consistent QFT needs both branches of the propagator, which is why, antimatter must exist.
These topics confuse me still and are far ahead of the current simple question, which is, how the propagator looks in real space. This is a question of classical field theory and not of QFT.
Which is why this thread should be renamed to:
"Very simple question of classical field theory"

 

[OOO]

At the end, we are sending epsilon to zero, right? Do the integral and see, that you get the same result, if you use either -i*epsilon or +i*epsilon.
I think, there is a problem with your integration contour.
We want to integrate from minus infinity to plus infinity along the real axis.
We start with an integral from -r to r for r some (big) real number.
Now we are closing the integral via a big (not small!) half circle around the complex plane.
Do we use the upper or lower half circle? This depends on sign of our exponential, which in the end is just definition
of the Fourier transform. (Either the Fourier transform or its inverse get a minus.)
If we use exp(i*k*x), we want to use the upper half circle, because then in the limit, that r goes to infinity, the integrand around the half circle goes to zero. Notice, that the sign of epsilon was not involved here.

You misunderstood what I was saying. There is no problem with my integration contour. Of course there is always that big half circle the contribution of which should vanish for r to infinity.

What I mean is: instead of moving the singularities into the upper or lower half plane and integrating along the real axis we could also leave the singularities where they are and integrate along parts of the real axis until we are just before a singularity, then we integrate around that singularity (thus leaving the real axis for a moment) along a small half circle and immediately return back to the real axis. In the end we treat these small circles in the limit of zero radius. That's the way it is explained in some textbooks on electrodynamics (for the massless propagator of course).

I don't know where your mistake is, but probably you didn't take the residue theorem into account.

 

[Micha]

You misunderstood what I was saying. There is no problem with my integration contour. Of course there is always that big half circle the contribution of which should vanish for r to infinity.

What I mean is: instead of moving the singularities into the upper or lower half plane and integrating along the real axis we could also leave the singularities where they are and integrate along parts of the real axis until we are just before a singularity, then we integrate around that singularity (thus leaving the real axis for a moment) along a small half circle and immediately return back to the real axis. In the end we treat these small circles in the limit of zero radius. That's the way it is explained in some textbooks on electrodynamics (for the massless propagator of course).

I don't know where your mistake is, but probably you didn't take the residue theorem into account.

I understand your point now. I will think about it.

 

[OOO]

Let me approach the uniqueness question from another perspective.
We know the solutions of the homogeneous equation.
These are the harmonic waves, right?
Now, for them, it is easy to go from momentum space to real space, so
if you add a harmonic wave to the propagator in real space, you have to do the
same in momentum space and vice versa.

Yes, indeed. So this shows that the propagator can't be unique in momentum space too. That's why you have to do the epsilon trick. If you do the trick in two different ways you get two different propagators (yes, I know, you don't believe me, but that's the well-known way you switch between retarded and advanced propagators for example).

There is simply no unique meaning of a Fourier integral over first order singularities. It's not defined and so, in my opinion, the problem lies already in the Ansatz.

 

[Micha]

Yes, indeed. So this shows that the propagator can't be unique in momentum space too. That's why you have to do the epsilon trick. If you do the trick in two different ways you get two different propagators (yes, I know, you don't believe me, but that's the well-known way you switch between retarded and advanced propagators for example).

There is simply no unique meaning of a Fourier integral over first order singularities. It's not defined and so, in my opinion, the problem lies already in the Ansatz.

Your argument suggests, that the two propagators with plus and minus epsilon might differ.
I will do the integral with minus epsilon again carefully.
Notice, that we are still far away from having an infinite number of propagtors in real space, even if we allow epsilon to change its sign.

By the way, passing one singularity from above and one from below is not allowed, because we should choose epsilon once.

So my modified claim is:

The Fourier transform of the function 1/(p^2-m^2+i*epsilon) with epsilon > 0 is unique.
(It is not my claim, it is all well known physics from the textbooks.)

By the way, a finite epsilon makes an unstable particle with epsilon being the decay rate.
As we know that particles are decaying and not appearing, we have an arrow of time built-in in QFT
and we should choose epsilon > 0.

For the further discussion about the propagator exponentially decaying, I suggest, we take
1/(p^2-m^2+i*epsilon) with epsilon > 0.

 

[OOO]

Your argument suggests, that the two propagators with plus and minus epsilon might differ.
I will do the integral with minus epsilon again carefully.
Notice, that we are still far away from having an infinite number of propagtors in real space, even if we allow epsilon to change its sign.

You're right. I don't dare to say that we are able to generate all homogeneous solutions by moving the singularities around by some infinitesimal displacement. But it is obvious that the difference between two propagators must be a homogeneous solution. It would be interesting to see how one of these homogeneous solution is connected to Fourier space. If I consider the difference between the retarded and advanced propagator of the massless case, it's clear that it must be a spherical delta/r-wave coming from infinity, contracting to the location of the event at (t0,x0), and inflating again as a delta/r-wave for t->infinity. So the question is: what representation has this homogeneous solution in Fourier space ? Are we sure that this solution has got a well-defined Fourier transform at all ? I don't know.

By the way, passing one singularity from above and one from below is not allowed, because we should choose epsilon once.

Says who ? Instead of the prescription $\omega^2-\omega_k^2 \to \omega^2-\omega_k^2+i\epsilon$ (where $\omega_k^2:=k^2+m^2$) you could also use the prescription

$$\omega^2-\omega_k^2 = (\omega-\omega_k)(\omega+\omega_k)\to (\omega-\omega_k\pm i\epsilon)(\omega+\omega_k\pm i\epsilon)$$

That's the way the massless propagator is usually obtained in electrodynamics textbooks. So why shouldn't this be allowed here ?

The Fourier transform of the function 1/(p^2-m^2+i*epsilon) with epsilon > 0 is unique.

I tend to agree.

By the way, a finite epsilon makes an unstable particle with epsilon being the decay rate.
As we know that particles are decaying and not appearing, we have an arrow of time built-in in QFT
and we should choose epsilon > 0.

For the further discussion about the propagator exponentially decaying, I suggest, we take
1/(p^2-m^2+i*epsilon) with epsilon > 0.

I think you're going too far if you assign to this decay any physical meaning a priori (probably there is one but this doesn't seem to be justified for now). It's just a calculational aid to resolve the indeterminacy of the integral over the singular argument.

 

[Micha]

Says who ? Instead of the prescription $\omega^2-\omega_k^2 \to \omega^2-\omega_k^2+i\epsilon$ (where $\omega_k^2:=k^2+m^2$) you could also use the prescription

$$\omega^2-\omega_k^2 = (\omega-\omega_k)(\omega+\omega_k)\to (\omega-\omega_k\pm i\epsilon)(\omega+\omega_k\pm i\epsilon)$$

That's the way the massless propagator is usually obtained in electrodynamics textbooks. So why shouldn't this be allowed here ?

That's right. I do not have any a priori arguments to exclude those cases.
My only point is, once we write the propagator as
1(p^2-m^2+i*epsilon), we have excluded those cases.

 

[Micha]

I think you're going too far if you assign to this decay any physical meaning a priori (probably there is one but this doesn't seem to be justified for now). It's just a calculational aid to resolve the indeterminacy of the integral over the singular argument.

Correct. At this level it is just a mathematical trick. The statement has to be justified.
See this lecture series.
http://indico.cern.ch/conferenceDisplay.py?confId=a032459

 

[Micha]

You're right. I don't dare to say that we are able to generate all homogeneous solutions by moving the singularities around by some infinitesimal displacement. But it is obvious that the difference between two propagators must be a homogeneous solution. It would be interesting to see how one of these homogeneous solution is connected to Fourier space. If I consider the difference between the retarded and advanced propagator of the massless case, it's clear that it must be a spherical delta/r-wave coming from infinity, contracting to the location of the event at (t0,x0), and inflating again as a delta/r-wave for t->infinity. So the question is: what representation has this homogeneous solution in Fourier space ? Are we sure that this solution has got a well-defined Fourier transform at all ? I don't know.

I think, the homogeneous solutions in Fourier space are just a product of delta functions for the components of the momentum vector, where you should choose only momentum vectors on mass shell: p^2-m^2 = 0

 

[OOO]

That's right. I do not have any a priori arguments to exclude those cases.
My only point is, once we write the propagator as
1(p^2-m^2+i*epsilon), we have excluded those cases.

Yes, partly. But as Zee remarks, in the limit epsilon to zero this amounts to $\omega_k\to\omega_k-i\epsilon$ (expanding a square root) and this is just one of those cases that you are trying to exclude.

I admit that I have always wondered why it is done this way for the KG propagator as opposed to the massless propagator. There seems to be no good reason for it, besides being slightly easier to write down (only one epsilon term, no binomial decomposition).