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

[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).

 

[OOO]

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

That sounds resonable because then the two Fourier profiles differ only at the location of the singularities (which are on mass shell). That would explain why we think that the propagator in momentum space is unique. It's because we tend to think only of its values off mass shell.

Edit: I have got some objections against the term product of delta functions, not sure what you mean by that. This would give discrete singularities whereas they actually form a manifold.

 

[Micha]

I worked out the propagator in one spatial dimension and this is, what I get:

For epsilon > 0:

exp(-m*abs(x))/(-2m)For epsilon < 0:

exp(m*abs(x))/(2m)It seems you get the propagator for epsilon < 0 by putting in a negative rest mass
in the propagator for epsilon >0. Mathematically, it is not surprising, that this is
possible, because only m^2 is appearing in KG.

Maybe the reason for the differences in the solutions of Hans and Zee lie in different choices of epsilon. Zee explicitly works out the properties of the +i*epsilon propagator.
Maybe Hans works implicitly or explicitly with a sum of the plus and minus epsilon solution, and so the contributions out of the light cone cancel exactly to zero. But I don't know.
Edit: Maybe this is, what Hans was trying to tell us with post #126.

 

[Micha]

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).

I don't understand.

Zee can take the epsilon out of the square root, because it is small. It is just a linear approximation in epsilon.

 

[OOO]

I don't understand.

Zee can take the epsilon out of the square root, because it is small. It is just a linear approximation in epsilon.

Yes, exactly. There is no problem with this argument at all. But by joining the two epsilons into one (or vice versa, depending where you start from) one restricts generality more than one needs to.

 

[OOO]

Maybe the reason for the differences in the solutions of Hans and Zee lie in different choices of epsilon. Zee explicitly works out the properties of the +i*epsilon propagator.
Maybe Hans works implicitly or explicitly with a sum of the plus and minus epsilon solution, and so the contributions out of the light cone cancel exactly to zero. But I don't know.

As I have said above, I am being sceptical about the ability to put all boundary conditions into moving the singularities. But if you modify your statement to "Maybe the reason for the differences in the solutions of Hans and Zee lie in different boundary conditions" I'd definitely agree.

Edit: Maybe this is, what Hans was trying to tell us with post #126.

I have just done Hans' calculation he explained in

http://chip-architect.com/physics/Higher_dimensional_EM_radiation.pdf

for the massless propagator (I think generalization to the massive one is straightforward) in 1+1 dimension. He uses lightcone coordinates \xi=\omega-k and \zeta=\omega+k for doing the double Fourier integral. With this trick the integral decomposes into two independent single integrals and no epsilon-tricks have to be applied. Although I don't quite understand how he used the convolution theorem in this case, I got the same result by a different method (change of variables).

I think the crucial point is that this method is not unique as well, like any other method to compute the propagator. If you do it that way, you get two "integration constants" c1(t-x) for the special solution Theta(t+x) and c2(t+x) for the special solution Theta(t-x), as far as I can see. But, in my opinion, Hans correctly assumes that the causal propagator should be zero for t<0 which rules out both "constants" and you get

\phi(t,x) = \Theta(t-x)\Theta(t+x) = \Theta(t)\Theta(t^2-x^2)

As Hans has stressed in one of his earlier posts, the textbook authors don't seem to be concerned about the past, which is why they neglect the Heaviside factor. And probably because they are historically blind the don't notice that values outside the lightcone must have their origin somewhere in the past.

 

[Micha]

Edit: I have got some objections against the term product of delta functions, not sure what you mean by that. This would give discrete singularities whereas they actually form a manifold.

Can you explain. I think, a product of delta functions for different coordinates
under a multidimensional integral is just fine. Only if you multiply delta functions for
the same integration variable, you get problems.

 

[Micha]

As I have said above, I am being sceptical about the ability to put all boundary conditions into moving the singularities. But if you modify your statement to "Maybe the reason for the differences in the solutions of Hans and Zee lie in different boundary conditions" I'd definitely agree.

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?

 

[Micha]

But, in my opinion, Hans correctly assumes that the causal propagator should be zero for t<0 which rules out both "constants" and you get

\phi(t,x) = \Theta(t-x)\Theta(t+x) = \Theta(t)\Theta(t^2-x^2)

As Hans has stressed in one of his earlier posts, the textbook authors don't seem to be concerned about the past, which is why they neglect the Heaviside factor. And probably because they are historically blind the don't notice that values outside the lightcone must have their origin somewhere in the past.

Very interesting. I need some time to look into this.

Edit: Maybe we should write a paper about this, once we understand it all. :-)

 

[OOO]

Can you explain. I think, a product of delta functions for different coordinates
under a multidimensional integral is just fine. Only if you multiply delta functions for
the same integration variable, you get problems.

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

 

[OOO]

Very interesting. I need some time to look into this.

Edit: Maybe we should write a paper about this, once we understand it all. :-)

That would probably be a bit too little for a paper...

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...