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

[Micha]

Small addition: The concept of causality behind $[\phi(x), \phi(y)] = 0$ obviously is: Two measurements at spacelike distances can not influence each other. This is for me a respectable concept of causality, too. I am only not sure, whether the two definitions of causality, we then have, are necessarily equivalent.

 

[strangerep]

A technical point:
$$\lim_{m\to 0}mK_1(mr)={1\over r}$$
not zero.

Oops! OK, thanks.

This means the over-simplified UTK formula is not adequate to illustrate
that massless photons propagate on the lightcone. One must use the more general
formula from Scharf, which has a $\Theta(-x^2)$ multiplying that term,
not to mention a $\delta(x^2)$ elsewhere. So it all gets horribly messy
near the lightcone.

 

[strangerep]

How can Reeh-Schlieder proove such a theorem, while Peskin
& Schroeder is providing a proof for causality?

I'm not familiar with the details of the Reeh-Schlieder proof, but others have
suggested the roots lie in the non-locality of the vacuum state itself.

But in order to say, that $[\phi(x), \phi(y)] = 0.$ means, that causality is
preserved, you need to have a concept of causality first, to which you can connect.
And the best definition of causality in physics I know, is, that you can not send a signal
from A to B with a speed greater than the speed of light.

Small addition: The concept of causality behind $[\phi(x), \phi(y)] = 0.$ obviously is:
Two measurements at spacelike distances can not influence each other. This is for me a
respectable concept of causality, too. I am only not sure, whether the two definitions of
causality, we then have, are necessarily equivalent.

They're equivalent - because tachyons have never been physically observed.

[and yes, this thread is getting too long]

 

[Hans de Vries]

Micha,

OK, the UTK web document you quoted is essentially a course version
that combines material covered in Peskin & Schroeder, ch2, with some
extra stuff.

In particular, UTK's eq(1.4.6) is the equal-time (ie special) case of D(x-y):

$$D(x-y) = \frac{m}{4\pi^2\sqrt{-(x-y)^2}} ~ K_1(m \sqrt{-(x-y)^2})$$

This equal-time expression, involving the $K_1$ Bessel
function, is a special case of the more general expression I mentioned
in Scharf.

This expression is also used in Weinberg's treatment of the subject:
Volume one, chapter 5.2: Causal scalar fields.

Weinberg refrains from an explicit picture of outside the lightcone
propagating particles and counter propagating anti particles which
cancel each other in his treatment.


Regards, Hans

 

[Micha]

They're equivalent - because tachyons have never been physically observed.

I am too stupid to clearly see this equivalence.

Imagine I had a source of radioactive particles and the source was shielded by some lead. Now in order to send a signal I would remove the lead. Is removing the lead a measurement? Mhm, probably yes. But is it a measurement of the radioactive particles I am sending? Is it nonsense to think about this altogether?

 

[geoffc]

For me, it made matters much worse when I had to "unlearn" everything about the measurement process and the notion of "particles" when looking at quantum field theory in curved spacetime.

A great explanation of the difficult is described in the introduction of:
http://arxiv.org/abs/gr-qc/9707062

An article that discusses of the notion of "particle" and "measurement" as it pertains to Unruh radiation is here:
http://xxx.lanl.gov/abs/0710.5373

 

[strangerep]

I am too stupid [...]

Trust me... everyone feels stupid now and then. Only the totally ignorant and braindead
never feel stupid. Don't worry about it.

Imagine I had a source of radioactive particles and the source was shielded by some lead. Now in order to send a signal I would remove the lead. Is removing the lead a measurement? Mhm, probably yes. But is it a measurement of the radioactive particles I am sending? Is it nonsense to think about this altogether?

You're over-complicating it. The crux is that any type of event at x cannot affect any
type of event at y, if x-y is spacelike. That is, information cannot be propagated between
x and y.

(A measurement at x is just a type of event at x.)

 

[Micha]

Trust me... everyone feels stupid now and then. Only the totally ignorant and braindead
never feel stupid. Don't worry about it.You're over-complicating it. The crux is that any type of event at x cannot affect any
type of event at y, if x-y is spacelike. That is, information cannot be propagated between
x and y.

(A measurement at x is just a type of event at x.)

Now you replaced the word measurement by the word event. While this might give us a better feeling, because the word event seems innocent and is surely less mysterious then the word measurement in the context of quantum mechanics, I don't see, what this really explains. Especially I am not able to make a clear connection between events and creation and annihilation operators.

 

[strangerep]

Now you replaced the word measurement by the word event. While this might give us a better feeling, because the word event seems innocent and is surely less mysterious then the word measurement in the context of quantum mechanics, I don't see, what this really explains.

I was just using "event at x" as a generic term for anything that happens at x,
ie: any change of configuration of the quantum field at x. To perform a measurement
within a region A of spacetime involves the creation and annihilation of initial/final
states within A. (Think of preparation and detection.) The point is that any side effects
of these operations within A cannot affect the result of some other measurement off in
another spacetime region B, if A and B are spacelike-separated.

Especially I am not able to make a clear connection between events and creation and annihilation operators.

Think of each operator as a mapping from one state to another. Ie: a change in the
configuration of the quantum field.

 

[Micha]

Thanks, strangerep,
I think I just forgot the basic starting point of quantum mechanics: Every observable is associated with an operator.

But I want to say another thing:
Look what wikipedia has now on the propagator:
http://en.wikipedia.org/wiki/Propagator
It looks like a very nice summary of our thread with some details added.

 

[Haelfix]

Since this thread has resurfaced, i'd like to point out a neat little paper I read on this subject (with regards to curved spacetimes)

arXiv:0709.1483. In it they try to show that causality of the classical theory implies microcausality of the quantum picture.

Now, there is something a little fishy going on in at least part of their derivation, but i'll leave it to the reader to see if they can spot my objections. I do agree with the premise though, at least in part.

 

[jostpuur]

Look what wikipedia has now on the propagator:
http://en.wikipedia.org/wiki/Propagator
It looks like a very nice summary of our thread with some details added.

What does $\delta(t,t')$ mean in the first equation?

 

[strangerep]

What does $\delta(t,t')$ mean in the first equation?

It's just a Dirac delta function. (Look at the next line, where it says that $\delta(x,x')$
is a Dirac delta fn, even though they've used "$\delta(x-x')$" just before that.)
Their notation is a bit inconsistent.

 

[tiny-tim]

thread approaching mass of black hole … HELP!

I started to read this thread - but it's RIDICULOUSLY LONG!

Themes and sub-themes have come and gone, and there could be almost anything in the middle pages.

I can't plough through all that just in case.

Shouldn't it be someone's job to split it up every so often?

Or perhaps once a thread gets a certain size, there should be an inhibitor encouraging new contributors to start new threads?

I don't mind short threads that wander off-topic, or long threads that stay on-topic.

But, please, no long threads that meander all over the place!

Am I the only one who gets turned off by threads like this one?

 

[Hans de Vries]

I made some nice progress here, writing the chapter on the Klein Gordon equation
of my textbook.

http://chip-architect.com/physics/Book_Chapter_Klein_Gordon.pdf" (see 8.9 and further) For instance I found out why we can get away with 'sloppy' math like using.

$$\frac{-1}{E^2-p^2}~~~~ \mbox{and}~~~~\frac{-1}{E^2-p^2-m^2}$$

as the propagators for the photon and KG equation. They are symmetric in
E(nergy) and thus symmetric in time. This means that the electromagnetic
field would have advanced potentials as well as retarded potentials. Clearly
in conflict with experiment.

It turns out that the difference between the two-sided propagators (forward
and backward in time) and the fully causal, forward-in-time-only propagators
is only in the poles. For instance:

Causal, forward in time Photon propagator

$$D^{\triangledown}(E,p)\ =\ \frac{-1}{E^2-p^2}\ \ +\ \ \frac{\pi}{2ip}\bigg( \delta(E-p)-\delta(E+p) \bigg)$$

The extra term only propagates on-shell. It modifies the poles. It does not influence
first order Feynman diagrams since the virtual photons of those diagrams can not be
real. (real electrons can not emit a single real photon)
The same is true for the Klein Gordon equation:

Causal, forward in time Klein Gordon propagator

$$D^{\triangledown}(E,p)\ =\ \frac{-1}{E^2-p^2-m^2}\ \ +\ \ \frac{\pi}{2i\sqrt{p^2+m^2}}\bigg( \delta(E-\sqrt{p^2+m^2})-\delta(E+\sqrt{p^2+m^2}) \bigg)$$

Again, only the poles are different. First order diagrams are unaffected.
The extra deltas have an amplitude in the poles which is infinitely much
higher as the normal poles. This is proved with Rayleigh's theorem.

We can reorganize the forward propagator on a pole-by-pole base to
study the behavior at the poles, for instance for the photon

$$D^{\triangledown}(E,p)\ =\ -\frac{1}{2p} \left( \frac{1}{E-p}+i\pi\delta(E-p)\right)\ \ \ +\ \ \ \frac{1}{2p}\left( \frac{1}{E+p}+i\pi\delta(E+p)\right)$$

Going slowly through a pole gives an amplitude which can be symbolically
written as:

$$-\infty ~~~~ \rightarrow ~~~~ i\infty^2 ~~~~ \rightarrow ~~~~ +\infty$$

Where the middle term $i\infty^2$ symbolizes the contribution from the delta.

A compelling argument can be made that the addition of the delta functions
is a physical requirement. Any real life photon has a finite lifetime and is therefor
not exactly on-shell. Its frequency spectrum extends to both sides of the pole
which propagate with opposite sign, and therefor, would destructively interfere.

The destructive interference would be 100% in the exactly symmetric case.
Minimal changes in frequency would move the spectrum to either side of the
pole and the destructive interference would disappear. The on-shell propagation
would be ill-defined.

Only the addition of the delta functions at the poles makes the on-shell propagation
well-behaved since the magnitude of the deltas is much higher as that of the
reciprocal functions.Regards, Hans

 

[Micha]

Hans, maybe this sloppy mathematics works for you.
But I will certainly not use it, when the "real" mathematics is so clear and easy
as presented in wikipedia.

Note also, that the advanced propagator, which is surely the causal propagator in classical physics, is not the one to be used in quantum field theory. Instead the Feynman propagator has to be used, because it contains both the positive and negative energy solutions (particles and antiparticles). As explained in wikipedia, while it is nonzero outside the light cone, causality is ensured by the commutators of space-like separated field operators being zero.

 

[Hans de Vries]

Hans, maybe this sloppy mathematics works for you.
But I will certainly not use it, when the "real" mathematics is so clear and easy
as presented in wikipedia.

1) The article in Wikipedia has many errors. It's not for nothing that it says in the
header that the article needs an expert.
http://en.wikipedia.org/wiki/Talk:Propagator

2) Sloppy mathematics certainly doesn't work for me. That's exactly the reason
for me to go to the bottom of things.

Note also, that the advanced propagator, which is surely the causal propagator in classical physics, is not the one to be used in quantum field theory.

The advanced potentials come from the backward propagator which goes from the future
to the past. This is the part you want to remove from the propagator and that is what I did.

Instead the Feynman propagator has to be used, because it contains both the positive and negative energy solutions (particles and antiparticles). As explained in wikipedia, while it is nonzero outside the light cone,

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.

causality is ensured by the commutators of space-like separated field operators being zero.

One can postulate this as Pauli did in his 1940 paper on spin-statistics, but that's
hardly satisfying. Regards, Hans.

 

[Micha]

1) The article in Wikipedia has many errors. It's not for nothing that it says in the
header that the article needs an expert.
http://en.wikipedia.org/wiki/Talk:Propagator

I know this article from before. It already had this comment on it, but since then it has been completely reworked. And now it contains nicely everything, what I learned from this thread.
I think, they just didn't remove the comment yet.

I do not want to discuss the whole thing all over again here. But since a contribution of yours already caused a lot of confusion in this thread earlier (if for nobody than certainly for me), I felt I had to make this remark. Nothing personal.

Edit: It seems I confused the advanced with the retarded propagator in my earlier post. I mean a propagator being non zero in the future of course.

 

[Hans de Vries]

I know this article from before. It already had this comment on it, but since then it has been completely reworked. And now it contains nicely everything, what I learned from this thread.
I think, they just didn't remove the comment yet.

The article has many errors

The use a Klein Gordon propagator in position space which is totally wrong,
it has the wrong Bessel function (Bessel Y1 instead of J1). The Greens function
has the wrong argument. It misses out on the delta function..What is presented as a causal propagator violates the basic laws of the Fourier
transform. A causal propagator has to respect the Kramers Kronig relation. A
real forward-only propagator has a Fourier transform which has an even real
part and an odd imaginary part, the two are related via the Hilbert transform.

I do not want to discuss the whole thing all over again here. But since a contribution of yours already caused a lot of confusion in this thread earlier (if for nobody than certainly for me), I felt I had to make this remark. Nothing personal.

This just is a confusing subject. There are so many different versions going
around of the Klein Gordon propagator in position space that textbook authors
avoid writing about it because they are not sure which one to choose.

That alone is a reason to discuss these things here.Regards, Hans

 

[Hans de Vries]

What is presented as a causal propagator violates the basic laws of the Fourier
transform. A causal propagator has to respect the Kramers Kronig relation. A
real forward-only propagator has a Fourier transform which has an even real
part and an odd imaginary part, the two are related via the Hilbert transform.

I may have been a bit too impulsive here, one would have to prove that
the causal propagator:

$$\frac{-1}{(E-i\epsilon)^2 -p^2-m^2}$$

respects the Kramers Kronig relation. It would have to be equal to my
expression for which I'm quite sure that it does:

$$\frac{-1}{E^2-p^2-m^2}\ \ +\ \ \frac{\pi}{2i\sqrt{p^2+m^2}}\bigg( \delta(E-\sqrt{p^2+m^2})-\delta(E+\sqrt{p^2+m^2}) \bigg)$$

Note that both are always real except in the poles where they become
imaginary delta functions.

I would actually be quite happy if they turn out to be the same. It's
late now, tomorrow more.Regards, Hans

 

[Hans de Vries]

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

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

Klein Gordon Backward-in-time propagator

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

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)

$$\frac{\sin\left(\sqrt{p^2+m^2}~~ |t|\right)}{2\sqrt{p^2+m^2}}~~~~\Leftrightarrow~~~~\frac{-1}{E^2-p^2-m^2}$$$$\frac{\sin\left(\sqrt{p^2+m^2}~~~ t~~\right)}{2\sqrt{p^2+m^2}}~~~~~~\Leftrightarrow~~~~\frac{\pi}{2i\sqrt{p^2+m^2}}\bigg( \delta(E-\sqrt{p^2+m^2})-\delta(E+\sqrt{p^2+m^2}) \bigg)$$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)$$\frac{-1}{E^2-p^2-m^2}\ =\ \frac{-1}{2\sqrt{p^2+m^2}}\left(\frac{1}{E-\sqrt{p^2+m^2}}\ -\ \frac{1}{E+\sqrt{p^2+m^2}} \right)$$

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.$$\frac{-1}{E^2-p^2-m^2}\ =\ \frac{-1}{2\omega_p}\left(\frac{1}{E-\omega_p}\ -\ \frac{1}{E+\omega_p} \right)$$
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.$$\frac{-1}{E^2-p^2-m^2 -i\epsilon}\ =\ \frac{-1}{2\omega_p}\left(~~\frac{1}{E-\omega_p}\ +\ i\pi\delta(E-\omega_p)~~~~ - ~~~~ \frac{1}{E+\omega_p} + \ i\pi\delta(E+\omega_p)~~\right)$$

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

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