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

[Hans de Vries]

OK, They can indeed be considered as being equivalent, as it should be. Klein Gordon Forward-in-time propagator

<br /> \frac{-1}{(E-i\epsilon)^2 -p^2-m^2} ~~~~\Leftrightarrow~~~~\frac{-1}{E^2-p^2-m^2}\ \ +\ \ \frac{\pi}{2i\sqrt{p^2+m^2}}\bigg( \delta(E-\sqrt{p^2+m^2})-\delta(E+\sqrt{p^2+m^2}) \bigg)<br />

Klein Gordon Backward-in-time propagator

<br /> \frac{-1}{(E+i\epsilon)^2 -p^2-m^2} ~~~~\Leftrightarrow~~~~\frac{-1}{E^2-p^2-m^2}\ \ -\ \ \frac{\pi}{2i\sqrt{p^2+m^2}}\bigg( \delta(E-\sqrt{p^2+m^2})-\delta(E+\sqrt{p^2+m^2}) \bigg)<br />

Both sides give imaginary delta-pulses in the poles with the right signs.
More convincingly, The "pole-pick-prescriptions" at the left side give
the same formula as the direct Fourier transforms of the right side:Fourier transform (time <--> energy)

<br /> \frac{\sin\left(\sqrt{p^2+m^2}~~ |t|\right)}{2\sqrt{p^2+m^2}}~~~~\Leftrightarrow~~~~\frac{-1}{E^2-p^2-m^2}<br /><br /> \frac{\sin\left(\sqrt{p^2+m^2}~~~ t~~\right)}{2\sqrt{p^2+m^2}}~~~~~~\Leftrightarrow~~~~\frac{\pi}{2i\sqrt{p^2+m^2}}\bigg( \delta(E-\sqrt{p^2+m^2})-\delta(E+\sqrt{p^2+m^2}) \bigg)<br />So, adding the two selects the forward light-cone, subtracting the
two selects the backward light-cone. The two are a Hilbert pair.
They respect the Kramers, Kronig relation.Regards, Hans

 

[Micha]

We have discussed the single pole Feynman propagator here before. It violates
special relativity as you say. In practice one doesn't use the pole prescription
when evaluating Feynman diagrams. The expression without the prescription,
simply: 1/(p^2-m^2) supports both positive and negative solutions as well.

Regards, Hans.

Here exactly is our disagreement.

The Feynman propagator does not correspond to a directly observable quantity in QFT. This is why QFT does not violate special relativity and this is what matters.

In cases, where epsilon does not fall out of the formulas and you end up with an integral of 1/(p^2-m^2) over p (eg. when calculating loop diagrams), then 1/(p^2-m^2) is just an ill-defined mathematical quantity. The only right choice is the Feynman propagator then.

 

[reilly]

I've got time problems. But I want to point out that the issues of "i epsilons" and Green's functions go back a very long time. A key feature of all this is doing the standard 2nd order equations in the complex plane(s) -- see, for example, Ince's classic Ordinary Differential Equations((1926),Whittaker and Watson's Modern Analysis(1902), Watson's formidable A Treatise on the Theory of Bessel Functions, Courant and Hilbert, Sommerfeld's PDE and .. When I have more time I'll go through this, including Weyl's important work on system's with continuous spectra.

It's all about using contour integration for Green's functions -- resolvants -- and integral representations of special functions.
Regards,
Reilly Atkinson

 

[Hans de Vries]

It's all about using contour integration for Green's functions -- resolvants -- and integral representations of special functions.
Regards,
Reilly Atkinson

Exactly, The contour integral is generally not well understood, works only in a few
cases and is resulting in basically incorrect "lingo" in many books

For instance, the contour integral for higher order poles is zero! (Link 1 at bottom)
The method doesn't work in conjunction with for instance the Fourier transform of
the solid light-cone: (forward+backward lightcone)

\frac{1}{(E^2-p^2)^2}The correct and general method to constrain a propagator to either the forward
or backward light-cone is to find the pair of real and imaginary functions which are
related to each other via the Hilbert transform, the convolution:Im\{\ f(E)\ \} = \frac{1}{i\pi E}\ *\ Re\{\ f(E)\ \}
Re\{\ f(E)\ \} = \frac{1}{i\pi E}\ *\ Im\{\ f(E)\ \}The convolution with the function 1/i\pi E, which is a first order pole, is the reason
that Cauchy's integral theorem gives useful results for propagators which are first
order poles or linear combinations of first order poles. The elementary Hilbert pair is:\frac{1}{E-p} \qquad \qquad = \qquad \frac{1}{i\pi E}\ *\ i\pi\ \delta(E-p)\

i\pi\ \delta(E-p)\ \quad \ = \qquad \frac{1}{i\pi E}\ *\ \frac{1}{E-p}\

This pair has the (inverse) Fourier transforms. (energy --> time)\mbox{\Huge $\mathcal{F}$}\ \bigg\{~~~~~\frac{1}{E-p}~~~~\bigg\} \quad = \quad \exp(-ipt)~\mbox{sgn}(t)

\mbox{\Huge $\mathcal{F}$}\ \bigg\{~i\pi\ \delta(E-p)~\bigg\} \quad = \quad \exp(-ipt)Note: A convolution with 1/i\pi E becomes a multiplication with sgn(t). The latter
is the (inverse) Fourier transform of 1/i\pi E. (See link 2 at bottom, rule 309)
Incorrect in the standard lingo are expressions like:
"For t>0 we close the contour at the upper half plane to pick up the pole"

The Fourier transform is an integral over a straight line from from -\infty to +\infty
It is not a contour integral! This lingo gives the right answer for the wrong reason.

What happens is that if we move the pole "up" or "down" like in 1/(E-p\pm\i\epsilon) then
we pick up the imaginary delta functions when E=p The Fourier integral straight through the pole:

\mbox{\Huge $\mathcal{F}$}\ \bigg\{~~~~~~~\frac{1}{E-p} \qquad \bigg\} \qquad \qquad \Rightarrow \qquad \qquad ~~~~\exp(-ip~t~)~\mbox{sgn}(t)

The Fourier integral i\epsilon "above" the pole:

\mbox{\Huge $\mathcal{F}$}\ \bigg\{~\frac{1}{E-p}\ +\ i\pi\ \delta(E-p) \bigg\} ~~~~~~ \Rightarrow \qquad \qquad 2\ \exp(-ipt)~\theta(t)


The Fourier integral i\epsilon "below" the pole:

\mbox{\Huge $\mathcal{F}$}\ \bigg\{~\frac{1}{E-p}\ -\ i\pi\ \delta(E-p) \bigg\} ~~~~~~ \Rightarrow \qquad \qquad 2\ \exp(-ipt)~\theta(-t)The last two expressions are constrained to the forward light-cone and the
backward light-cone respectively via the Heaviside step-function \theta(t).Regards, Hans

Links:
http://en.wikipedia.org/wiki/Residue_(complex_analysis)
http://en.wikipedia.org/wiki/Fourier_transform#Distributions
http://en.wikipedia.org/wiki/Cauchy's_integral_formula
http://en.wikipedia.org/wiki/Hilbert_transform

 

[Micha]

Incorrect in the standard lingo are expressions like:
"For t>0 we close the contour at the upper half plane to pick up the pole"

The Fourier transform is an integral over a straight line from from -\infty to +\infty
It is not a contour integral! This lingo gives the right answer for the wrong reason.

Hans, using contour integration to calculate integrals along the real axis, is a well established and basic piece of mathematics. Please go back to any basic textbook and see, why and how it works.

 

[Hans de Vries]

Here exactly is our disagreement.

The Feynman propagator does not correspond to a directly observable quantity in QFT.

This is just something you throw in here. The propagator is the amplitude for a particle
to go from A to B, hence it leads to an observable probability. At least it does so in the
opinion of Feynman. You can read his book here:

http://chip-architect.com/physics/KG_propagator_Feynman.jpg"

Some physicist agree while others disagree and use postulated commutator arguments.

Next you can see that Feynman's position space propagator in his book is very different
from the Wiki page you are referring to.

Hans, Please go back to any basic textbook and see, why and how it works.

Oh, well. I read these books 30 years ago.

Micha, The last 20 posts of you on this thread are virtually absent of any mathematical
contents. Further. you do not shows signs of grasping the math I'm talking about, nor
do you show any signs of doing an effort to understand it.

I can assure you of the extreme cautiousness I observe to make sure that what I'm
presenting is mathematically correct. There is some very serious effort from me going
in here.

Please refrain from these type of contentless insultsRegards, Hans

 

[kdv]

This is just something you throw in here. The propagator is the amplitude for a particle
to go from A to B, hence it leads to an observable probability. At least it does so in the
opinion of Feynman. You can read his book here:

http://chip-architect.com/physics/KG_propagator_Feynman.jpg"

Some physicist agree while others disagree and use postulated commutator arguments.

Next you can see that Feynman's position space propagator in his book is very different
from the Wiki page you are referring to.



Oh, well. I read these books 30 years ago.

Micha, The last 20 posts of you on this thread are virtually absent of any mathematical
contents. Further. you do not shows signs of grasping the math I'm talking about, nor
do you show any signs of doing an effort to understand it.

I can assure you of the extreme cautiousness I observe to make sure that what I'm
presenting is mathematically correct. There is some very serious effort from me going
in here.

Please refrain from these type of contentless insults


Regards, Hans

Hans,

I appreciate the work and thoughfulness you have put in your posts and in your calculations.

Before I try to understand your comments, can you summarize your position?
If I understand you correctly, you you are saying that the propagator does represents the amplitude for a particule to go from A to B, right? And you are saying that if one is careful, it does vanish putside of the lightcone. Is that your position?

That would make me feel much more comfortable about my understanding of what the propagator is but at the same time it confuses the heck out of me to see that no QFT book explains that correctly!

Regards

 

[Hans de Vries]

Hans,

I appreciate the work and thoughfulness you have put in your posts and in your calculations.

Before I try to understand your comments, can you summarize your position?
If I understand you correctly, you you are saying that the propagator does represents the amplitude for a particule to go from A to B, right? And you are saying that if one is careful, it does vanish outside of the lightcone. Is that your position?

Yes Indeed,

For those who want to be interpretation-agnostic one could state:
If the wave function has a certain amplitude in A, then the propagator tells
you how much of that amplitude ends up in B.

That would make me feel much more comfortable about my understanding of what the propagator is but at the same time it confuses the heck out of me to see that no QFT book explains that correctly!

Regards

Yes, There is confusion, this is why you see these discussions. This is why I put so
much effort in this. Of course, the intend is to reduce the confusion by double
and triple checking the math.

People like Micha want peace of mind and do not want confusion. However if one
studies at graduate level then one should be prepared to live with the fact that not
everything is that clear cut.

Progress in physics often works in the ant-colony's way. The ants find their food
3 meters south of the colony, but to gather the food they all follow the established
chemical pathway which may go up and down the tree in the east, go through the
dog's house in the north and the hollow tree in the west.

QFT produces incredible results but this doesn't mean that the whole process wasn't
a process of trial and error which took decades. It's my desire to understand the
basics, the foundations, rather then to wander off to some far, far away physical
theory as you see so often these days.

Well, I hope that people at least appreciate the intention.
Regards, HansFrom my book: http://chip-architect.com/physics/Book_Chapter_Klein_Gordon.pdf"

 

[Micha]

Micha, The last 20 posts of you on this thread are virtually absent of any mathematical
contents. Further. you do not shows signs of grasping the math I'm talking about, nor
do you show any signs of doing an effort to understand it.

I can assure you of the extreme cautiousness I observe to make sure that what I'm
presenting is mathematically correct. There is some very serious effort from me going
in here.Regards, Hans

Hans, you are right, that I am not checking or trying to understand in detail the math you are presenting. I can see and I do respect, that you put some serious effort into this.

I do understand these things well enough though, that I can see, when you get basic stuff wrong. And when I do, my willingness to dig deeper into your math is limited.

The contour integration is a good example because it is simple mathematics and not physics. It is no "lingo" that you close the integration contour by a half circle across the lower or upper complex half plane, it is a short version of a rock solid mathematical proof. Of course, for a finite halfcircle you get another contribution to the integral. But when the integrand falls off fast enough, this contribution is zero in the limit of an infinite half circle. I do not see the point of being more explicit here, because you can really read this in many books.

From the solid ground of contour integrals it is clear, that it is not unimportant, which epsilon prescription you are using when calculating Feynman diagrams and that you can not simply omit epsilon. This is another statement of yours, which goes against established mathematical and physical theory and which holds me off looking further into your math.

 

[Hans de Vries]

Of course, for a finite halfcircle you get another contribution to the integral. But when the integrand falls off fast enough, this contribution is zero in the limit of an infinite half circle. I do not see the point of being more explicit here, because you can really read this in many books.

That's the point. It does not become zero. The contribution of a half-circle is always
i\pi times the residue, independent of the radius.

If the contribution would be zero then it would not make any difference if you
close the contour with a half-circle in the upper or the lower half.

What I say is that if you do the integral along the straight line above the pole then
you go through +180 degrees around the pole and if you go below the pole then you
go through -180 degrees around the pole. The difference between the two is
a full circle.

From the solid ground of contour integrals it is clear, that it is not unimportant, which epsilon prescription you are using when calculating Feynman diagrams and that you can not simply omit epsilon.

This is true. The expression you get by by ignoring the epsilon is correct everywhere,
except for on-shell propagating virtual particles within the poles This is what I'm talking
about in post #301Regards, Hans

 

[Micha]

That's the point. It does not become zero. The contribution of a half-circle is always
i\pi times the residue, independent of the radius.

If the contribution would be zero then it would not make any difference if you
close the half-circle in the upper or the lower half.Regards, Hans

Now we might come closer to settle this.

You can not choose, how you close the circle, when you want to get a zero contribution.
Usually in the integrand there is some function say exp(i*x*...). This one you want to close with an upper half circle to get a zero contribution, because only then the e function goes to zero (assume x > 0)

For exp(-i*x*...) or when x<0 it is the opposite of course.

The residue theorem does not make any statement on the contribution of a half circle
or any other part of the integration path. You only now from it, what the result for a full closed loop will be.

 

[Hans de Vries]

Now we might come closer to settle this.

The residue theorem does not make any statement on the contribution of a half circle
or any other part of the integration path. You only now from it, what the result for a full closed loop will be.

OK

But in our case of a half circle you may use symmetry considerations for the
180 degrees contribution. If the pole is only dependent on E as a single pole
in the denominator like:

\oint^{\phi_1}_{\phi_2} \frac{1}{E-p}\ d\phi

Then the contribution should be simply proportional to the total angle \phi_1-\phi_2
One can derive this from lemma 2.1 in.

The Hilbert Transform: http://w3.msi.vxu.se/exarb/mj_ex.pdf

This Master Thesis describes the relation between the Hilbert transform, Cauchy's
integral and the Fourier transform. Quite appropriate for the discussion here.


Regards, Hans.

 

[Micha]

Symmetry arguments require, that you are integrating over a radially symmetric function.
When doing Fourier transforms, there is a factor exp(i*k*x) around, which is clearly not radially symmetric. Just look at in on the imaginary axis x' = i*x. You get exp(-x'). This is clearly not symmetric when you go from x' to -x'. This is why it makes a difference how you close the contour. The integration along one half circle gives zero, the other not.
Edit: Rather then radially symmetric, I should say not of the simple form 1/(z-a).
Notice, that lemma 2.1 in the link you presented does only make a claim about an infinitely small circle, where every simple pole looks like 1/(z-a).

 

[Hans de Vries]

Notice, that lemma 2.1 in the link you presented does only make a claim about an infinitely small circle, where every simple pole looks like 1/(z-a).

Yes,

Note that the poles of the Klein Gordon propagator are of this simple form 1(z-a)<br /> \frac{-1}{E^2-p^2-m^2}\ =\ <br /> \frac{-1}{2\sqrt{p^2+m^2}}\left(\frac{1}{E-\sqrt{p^2+m^2}}\ -\ \frac{1}{E+\sqrt{p^2+m^2}} \right)<br />

We are talking about the "energy --> time" Fourier transform and it is custom
to define \sqrt{p^2+m^2} as \omega_p or E_p which is a constant for the Fourier transform.<br /> \frac{-1}{E^2-p^2-m^2}\ =\ <br /> \frac{-1}{2\omega_p}\left(\frac{1}{E-\omega_p}\ -\ \frac{1}{E+\omega_p} \right)<br />
Regards, Hans

 

[Hans de Vries]

For the explicit Feynman propagator you have to add the Hilbert partner of the
positive pole to constrain it to the forward light-cone, and you have subtract
the Hilbert partner of the negative pole to constrain it to the backward light-cone.<br /> \frac{-1}{E^2-p^2-m^2 -i\epsilon}\ =\ <br /> \frac{-1}{2\omega_p}\left(~~\frac{1}{E-\omega_p}\ +\ i\pi\delta(E-\omega_p)~~~~ - ~~~~ \frac{1}{E+\omega_p} + \ i\pi\delta(E+\omega_p)~~\right)<br />

Where the right hand side is the limit of epsilon going to zero. The two delta functions
can be combined into one so can write for the Feynman propagator with epsilon --> 0:


Feynman propagator

<br /> \lim_{\epsilon \rightarrow 0}~~\frac{-1}{E^2-p^2-m^2 -i\epsilon}~~~~ \Rightarrow ~~~~<br /> \frac{-1}{E^2-p^2-m^2 }~~ - ~~\ i\pi\ \delta(E^2-p^2-m^2)<br />
Regards, Hans

 

[Micha]

Yes,

Note that the poles of the Klein Gordon propagator are of this simple form 1(z-a)


Regards, Hans

Yes, but as I pointed out earlier, doing a Fourier transform you have a factor exp(...) around, so the whole integrand is not of the simple form.

 

[Hans de Vries]

Yes, but as I pointed out earlier, doing a Fourier transform you have a factor exp(...) around, so the whole integrand is not of the simple form.

entirely correct.

 

[Hans de Vries]

To go back to the original question:Why does the Cauchy integral produce the same result as the Fourier integral ?
I'm using the latter. Both arguments are exactly the same, only the paths differ.\mbox{ $\frac{1}{2\pi}$} \int^{\infty}_{-\infty}dE\ e^{iEt} \bigg\{\ \frac{1}{E-\omega_p}~~~~~~~~+~~~~i\pi\ \delta(E-\omega_p)\ \bigg\} ~~~~ = ~~~~ \mbox{ $\frac{1}{2\pi}$} \oint dE\ e^{iEt}\bigg\{\ \frac{1}{E-\omega_p}~~~~~~~~+~~~~i\pi\ \delta(E-\omega_p)\ \bigg\}

~\Rightarrow ~~~~~~~~~~~~~~~~~~~~~~~~~~~~\mbox{$\frac{i sgn(t)}{2}$}\ e^{-i\omega_p t}~~ + ~~~~~~\mbox{ $\frac{i}{2}$}\ e^{-i\omega_p t}<br /> ~~~~~~~~~~~~~~ = ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~ie^{-i\omega_p t}~~~~~~~~~~+~~~~~~~~~~~~0


Fourier integral

Both Hilbert partners provide half of the result at (t>0) while they cancel
at (t<0). This automatically provides the result i\theta(t)\exp(-i\omega_p t)

Cachy integral

The pole provides the whole result. The contour has to be "hand-picked" to
include the pole at (t>0) and exclude the pole at (t<0) to give i\theta(t)\exp(-i\omega_p t)
The Hilbert partner (the delta) has zero contribution. In fact it is generally
not even considered in the calculation at all.

But the Hilbert partner has to be considered. It shows up trivially from the
Fourier transform back:i\mbox{\Huge $\mathcal{F}$}\ \bigg\{\ \theta(t)\ e^{-i\omega_p t}\ \bigg\} ~~~~=~~~~\frac{i}{2\pi} \mbox{\Huge $\mathcal{F}$}\bigg\{\ \theta(t)\ \bigg\}\ *\ \mbox{\Huge $\mathcal{F}$}\bigg\{\ e^{-i\omega_p t} \bigg\}~~~~=


\frac{i}{2\pi}\bigg\{\ 2\pi\mbox{ $\frac{1}{2}$}\left(\frac{1}{i\pi E}+\delta(E)\right) \bigg\}\ *\ \bigg\{\ 2\pi\ \delta(E-\omega_p) \bigg\}\ ~~~~=~~~~\left(~\frac{1}{E}~~+~~i\pi\ \delta(E)~\right)~*~\delta(E-\omega_p)~~~~=

\frac{1}{E-\omega_p}~~+~~i\pi\ \delta(E-\omega_p)The convolution is just a shift by \omega_p applied on the Fourier transform of the
Heaviside step-function \theta(t) given here:

http://en.wikipedia.org/wiki/Fourier_transform#Distributions (formula 310, with f=2\pi E)

Regards, Hans