Peskin and Schroeder eq. (2.52)

[sizzleiah]

Hi there.

I've just finished reading chapter 2 of Peskin and Schroeder, and I managed to follow all of their calculations - with one exception:

Homework Statement

I'm not sure how P&S arrive at the integral in equation (2.52) (page 27) from the previous step in the calculation of D(x-y).

Homework Equations

We're trying to calculate $$D(x-y)=<0|\phi (x) \phi (y) |0>$$ for a real Klein-Gordon scalar field $$\phi$$, where $$x-y$$ is purely spatial.

The Attempt at a Solution

Getting to the step right before eq. (2.52) is easy enough - it's just a standard integration in spherical coordinates. Then P&S make branch cuts to create a simply connected domain, so that they can apply path independence to the contour integration. I'm ok with all of that, but then they lose me when they write down the integral in eq. (2.52). It's confusing to me for a couple of reasons. One is that I'm not entirely sure how to deal with a contour that goes off to infinity in this way - where we can't restrict the variable of integration to be real (doesn't the complex plane only have one infinity?). Another is that it seems that for the lower limit of the integration to be valid, P&S are claiming that we have $$p=i m$$. Are they implying that we should be integrating along the branch cut? This seems very strange to me. I'm obviously no complex analyst, but I knew enough to be able to understand fairly easily what they did on the next few pages with the Feynman propagator. So...what am I missing?

Thanks!

 

[Ben Niehoff]

There IS only one infinity in the complex plane, which is precisely why this contour deformation works. The original contour is along the real axis. Instead, we push it up so that it is along the branch cut. That is, the first part of the contour comes down on the left side of the branch cut, then goes around the pole, and then back up the right side of the branch cut. So, splitting the contour up piecewise, there are really three integrals:

$$\int_{i\infty - \varepsilon}^{im - \varepsilon} f(p) \; dp\; + \int_{\pi/4}^{9\pi/4} f(im + \varepsilon e^{i\theta}) \; d\theta \; + \int_{im + \varepsilon}^{i\infty + \varepsilon} f(p) \; dp$$

The middle integral vanishes. The third integral is equal to the first integral times a phase (this phase is incurred by going around the pole to get to the other side of the branch cut). Evidently, the phase is -1; thus giving us +1 when we reverse the limits of integration to match the first integral. Notice in P&S 2.52 that a factor of 2 has been canceled from the previous line; this is due to taking the sum of the first and third integral in my expression above.

 

[sizzleiah]

Hey, thanks a lot for responding.

It looks like my complex analysis is more rusty than I thought.

Why are the limits of integration on the 2nd integral not $$-\pi$$ to 0? Naively, I would think that we'd just integrate from the left part of the contour over to the right part along a semicircle in this way.

For the 3rd integral, after we've reversed the limits of integration, you're saying that we can change the $$+\epsilon$$ in the limits to $$-\epsilon$$ at the cost of a -1 phase? I don't see why that is.

 

[Ben Niehoff]

Why are the limits of integration on the 2nd integral not $$-\pi$$ to 0? Naively, I would think that we'd just integrate from the left part of the contour over to the right part along a semicircle in this way.

You can, but it's easier to make a full circle, because then all the dependence on the pole is isolated in the middle integral.

For the 3rd integral, after we've reversed the limits of integration, you're saying that we can change the $$+\epsilon$$ in the limits to $$-\epsilon$$ at the cost of a -1 phase? I don't see why that is.

Look at f(p). Put in $p = I am + \varepsilon e^{i\pi/4}$ and $p = I am + \varepsilon e^{i9\pi/4}$. You should get a phase difference of -1.

Also, note that we're taking the limit as $\varepsilon \rightarrow 0$.

 

[sizzleiah]

Ok, I get it. Thanks a lot for the help.