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Peskin & Schroeder Problem 4.1 
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#1
May2214, 07:34 AM

P: 754

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}(xy)[/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 (xy)}[/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... 


#2
May2214, 10:04 AM

P: 305

I'm not sure I understand what your question is. However the p0 integration is usually performed as follows:
$$ \int \frac{d^4p}{(2\pi)^4}\frac{e^{ip(xy)}}{p_0^2(\vec p^2+m^2)}=\int \frac{d^4p}{(2\pi)^4}\frac{e^{ip_0(xy)_0}}{(p_0E_p)(p_0+E_p)}e^{i\vec p\cdot(\vec x\vec y)}, $$ where [itex]E_p=\vec p^2+m^2[/itex]. In order for the integral to converge you need to close in the upperhalf plane, i.e. you get the residue at [itex]p_0=E_p[/itex] and hence the final result is: $$ i\int \frac{d^3p}{(2\pi)^32E_p}e^{ip(xy)}, $$ where now [itex]p_0=E_p=\vec p^2+m^2[/itex]. Is this what you were looking for? 


#3
May2214, 12:34 PM

P: 754

yes but unfortunately i also have the [itex]j(p)=j(p^{0},\vec{p})[/itex] in the integrand...



#4
May2214, 12:47 PM

P: 305

Peskin & Schroeder Problem 4.1
Why do you say so? From what I see your sources depend on x and y not on p.



#5
May2214, 12:49 PM

P: 754

[itex] \int d^{4}x j(x) e^{i p x}= j(p)[/itex]
it;s the fourier tranf of j 


#6
May2214, 12:52 PM

P: 305

Yes, exactly. But once you write that all the p dependence is in the exponential [itex]e^{ip(xy)}[/itex] which I considered in my solution.



#7
May2214, 09:19 PM

Sci Advisor
P: 1,901

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 correctlooking 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 stepbystep 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. 


#8
May2214, 09:31 PM

P: 305




#9
May2314, 06:38 AM

P: 754

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 3momentum 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) 


#10
May2314, 07:27 PM

Sci Advisor
P: 1,901

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


#11
May2314, 08:50 PM

P: 754

"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] 


#12
May2314, 09:14 PM

Sci Advisor
P: 1,901




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