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Peskin & Schroeder Problem 4.1

by ChrisVer
Tags: peskin, schroeder
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ChrisVer
#1
May22-14, 07:34 AM
P: 895
I am having some problems in evaluating the current problem's question (b)...

I have reached the point (writing only the term which I have problem with):

[itex]A_{3}= - \frac{1}{2} \int_{t_{0}}^{t} dt' dt'' \int d^{3}x \int d^{3}y j(x,t') j(y,t'') D_{F}(x-y)[/itex]

So for some unknown reasons, I cannot see that I can find the Fourier Transform of the classical source to bring it in the needed form...
eg.
[itex] D_{F}= \int \frac{d^{4}p}{(2 \pi)^{4}} \frac{i}{p^{2}-m^{2}} e^{i p (x-y)}[/itex]

[itex]A_{3}=- \frac{1}{2} \int d^{4}x \int d^{4}y j(x) j(y) \int \frac{d^{4}p}{(2 \pi)^{4}} \frac{1}{p^{2}-m^{2}} e^{i p x} e^{-ipy}[/itex]


[itex]A_{3}=- \frac{1}{2} \int \frac{d^{4}p}{(2 \pi)^{4}} \frac{1}{p^{2}-m^{2}} \int d^{4}x j(x) e^{i p x}\int d^{4}y j(y) e^{-ipy}[/itex]

However I don't know how this [itex]p^{0}=E_{p}[/itex] integration can be performed...
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Einj
#2
May22-14, 10:04 AM
P: 323
I'm not sure I understand what your question is. However the p0 integration is usually performed as follows:
$$
\int \frac{d^4p}{(2\pi)^4}\frac{e^{ip(x-y)}}{p_0^2-(\vec p^2+m^2)}=\int \frac{d^4p}{(2\pi)^4}\frac{e^{ip_0(x-y)_0}}{(p_0-E_p)(p_0+E_p)}e^{-i\vec p\cdot(\vec x-\vec y)},
$$
where [itex]E_p=\vec p^2+m^2[/itex]. In order for the integral to converge you need to close in the upper-half plane, i.e. you get the residue at [itex]p_0=E_p[/itex] and hence the final result is:
$$
i\int \frac{d^3p}{(2\pi)^32E_p}e^{ip(x-y)},
$$
where now [itex]p_0=E_p=\vec p^2+m^2[/itex].

Is this what you were looking for?
ChrisVer
#3
May22-14, 12:34 PM
P: 895
yes but unfortunately i also have the [itex]j(p)=j(p^{0},\vec{p})[/itex] in the integrand...

Einj
#4
May22-14, 12:47 PM
P: 323
Peskin & Schroeder Problem 4.1

Why do you say so? From what I see your sources depend on x and y not on p.
ChrisVer
#5
May22-14, 12:49 PM
P: 895
[itex] \int d^{4}x j(x) e^{i p x}= j(p)[/itex]
it;s the fourier tranf of j
Einj
#6
May22-14, 12:52 PM
P: 323
Yes, exactly. But once you write that all the p dependence is in the exponential [itex]e^{ip(x-y)}[/itex] which I considered in my solution.
strangerep
#7
May22-14, 09:19 PM
Sci Advisor
P: 1,924
ChrisVer,

I slogged through these problems ages ago, so I've pulled out my old workbook to refresh my memory. I approached it similarly to you, but I now realize it was subtly wrong.

At least I now know the fully correct solution. It's easy to fudge a correct-looking solution, but a truly correct solution is rather more difficult.
There are some tricky bits involving Feynman contours and real parts, etc, which are easy to overlook (as I discovered the hard way).

I'm not sure whether you want step-by-step help like a homework question, or just the final solution. I'll assume the former unless you say otherwise...

Einj's hints were in the right direction, but to see it clearly we might need to go back a couple of steps and write out ##P(0)## in terms of contracted ##\phi(x)## and ##\phi(y)##. The next step is to replace the contraction with the Feynman propagator, which (I guess) you've tried to do in your ##A_3## expression. But you didn't say how your ##A_3## relates to the desired ##P(0)## in the original question. (Getting that right makes an important difference.) A review of the material on p31 of P+S might also be helpful.

HTH.
Einj
#8
May22-14, 09:31 PM
P: 323
Quote Quote by strangerep View Post
But at least I now know the fully correct solution. It's easy to fudge a correct-looking solution, but a truly correct solution is rather more difficult.
There are some tricky bits involving Feynman contours and real parts, etc, which are easy to overlook (as I discovered the hard way).
Yes, you are right. As you can see I took the pole on the upper-half plane even though the zeros of my denominator both lies on the real axis. I haven't explicitly said that I was taking the right Feynman path, I am sorry I didn't mention that.
ChrisVer
#9
May23-14, 06:38 AM
P: 895
my problem with Einj is that:
if you firstly do the integration of the Feynman propagator, without caring for the sources [itex]j(x)[/itex], then you will lose/have integrated the [itex]p^{0}[/itex] from the exponential and so you are left with the exponential with the 3-momentum on it. On the other hand, my sources also depend on the [itex]x^{0}=t[/itex] variable, and missing that I can't bring them to the Fourier transformed expression....
If I try to write them as:
[itex] \int dt j(\vec{x},t) [/itex]
then I'll probably get (I am not sure) after FT:
[itex] \int dt j(\vec{p},t)[/itex]
which is weird- I don't want the integral over time...

But I think I found a way out- Probably I can do the integral:
[itex]\int d^{4}p |j(p)|^{2} [propagator] [/itex]
The only problem with that is I don't know what will happen to [itex]j[/itex] when I look for the Residue.

(of course I did that strangerep- I just gave the final expression I'm trying to compute... I am not yet able to see through what kind of contractions happen without first writing the Wick's theorem on them)
strangerep
#10
May23-14, 07:27 PM
Sci Advisor
P: 1,924
Quote Quote by ChrisVer View Post
my problem with Einj is that:
if you firstly do the integration of the Feynman propagator, without caring for the sources [itex]j(x)[/itex], then you will lose/have integrated the [itex]p^{0}[/itex] from the exponential and so you are left with the exponential with the 3-momentum on it. On the other hand, my sources also depend on the [itex]x^{0}=t[/itex] variable, and missing that I can't bring them to the Fourier transformed expression....
But did you actually attempt what Einj suggested, or are you just guessing about what might go wrong?

If you do the integration of the Feynman propagator carefully, you should still have factor(s) like $$e^{\pm i E_p (x^0 - y^0)}$$which should help with that perceived problem.

If I try to write them as:
[itex] \int dt j(\vec{x},t) [/itex]
then I'll probably get (I am not sure) after FT:
[itex] \int dt j(\vec{p},t)[/itex]
which is weird- I don't want the integral over time...

But I think I found a way out- Probably I can do the integral:
[itex]\int d^{4}p |j(p)|^{2} [propagator] [/itex]
The only problem with that is I don't know what will happen to [itex]j[/itex] when I look for the Residue.
It's possible to do it this way (that was my first method). One must make certain assumptions about convergence of ##j## anyway (i.e., that it vanishes fast enough on the large semicircle part of the contour). So one can perform that ##p^0## contour integral by standard methods.

Personally, I think it's easier to do the ##D_F## integration first -- provided you do it carefully.

(of course I did that strangerep- [...]
Hmm. Sounds like I'm not the right person to help you. (I'm not a mind reader.)
ChrisVer
#11
May23-14, 08:50 PM
P: 895
Quote Quote by strangerep View Post
Hmm. Sounds like I'm not the right person to help you. (I'm not a mind reader.)
Sorry I didn't mean it like that... I just explained that there was no other way for me to write [itex]A_{3}[/itex] as I gave it, but by having followed the way you indicated before doing the contraction of the field etc :) nothing more/less...

"but to see it clearly we might need to go back a couple of steps and write out P(0) in terms of contracted ϕ(x) and ϕ(y). The next step is to replace the contraction with the Feynman propagator, which (I guess) you've tried to do in your A3 expression. But you didn't say how your A3 relates to the desired P(0) in the original question. (Getting that right makes an important difference.)"

Sorry if it didn't sound "right"/ sounded impolite ...

So, to make it more clear, the [itex]A_{3}[/itex] is just a part of the expansion of the amplitude...
[itex]A= 1+ A_{3} [/itex]
the 1 middle term which appears, dies because the field has zero vev.
Then the probability [itex]P(0)[/itex] will be just the module square of the amplitude:
[itex]P(0)= |1+A_{3}|^{2}[/itex]
strangerep
#12
May23-14, 09:14 PM
Sci Advisor
P: 1,924
Quote Quote by ChrisVer View Post
Sorry I didn't mean it like that...
OK. Let us know if you need further clues.


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