2
$\begingroup$

In Matthew Schwartz's QFT text, on page 39, he has the following contour integral:

$$\int_{-\infty}^{\infty}dk\frac{e^{ikr}-e^{-ikr}}{k+i\delta }.\tag{3.63}$$

This can be split into two terms, one for the positive exponent and one for the negative exponent.

What I am confused on is why does he choose to close the contour up for the first term, and down for the second term. The contour is a semicircle.

Specifically, he says "for $e^{ikr}$ we must close the contour integral up to get exponential decay at large k".

How do you get "exponential decay" when $e^{i*something*x}$ is usually a cyclical function of x?

$\endgroup$
  • 1
    $\begingroup$ Hint: $k$ has an imaginary part when we deform the contour in the complex $k$-plane. $\endgroup$ – Qmechanic Feb 17 '19 at 23:59
  • 1
    $\begingroup$ Well, when you do contour integration $k$ will become complex and $k = k_1 + ik_2$. Hence $$e^{ikX}=e^{-k_2X}e^{ik_1X}$$ Hence you have exponential decay as $k_2\to\infty$ when $X = r> 0$ and as $k_2\to -\infty$ when $X =- r < 0$. The first case happens precisely when you are integrating over a semicircle on the upper half-plane and the second case happens exactly when you are integrating over a semicircle on the lower half-plane. $\endgroup$ – user1620696 Feb 18 '19 at 0:04
  • $\begingroup$ Thank you both, this did the trick. In retrospect, guess it should have been obvious! $\endgroup$ – EthanT Feb 18 '19 at 4:48
1
$\begingroup$

I will consider only one of the integrals and leave the other for you as an exercise.

We wish to evaluate the integral

$$ I = \int_{-\infty}^{\infty}dk\frac{e^{-ikr}}{k+i\delta }. $$

To do this we consider first a different integral

$$I'= \oint_\gamma dz\frac{e^{-izr}}{z+i\delta }$$

where $z$ is complex-- you may think of $z = k + iy$, if you like. The goal is to find a closed contour $\gamma$ we can integrate over that is along the real axis (returning our original integral) and goes to zero at every other point along the contour. Here's how we do it.


Solution:

The key is to first realize that this integral has a pole at $z = -i\delta$ and so any contour that doesn't enclose this point will be zero by cauchy's theorem. This suggests we take the simplest curve we can think of that encloses the point $z= -i\delta$ and runs along the real line.

Namely, we should take a semicircle of infinite radius that closes in the Lower Half Plane. Note that if we enclosed it in the Upper Half Plane the integral is just zero as it would not contain the pole.

Continuing through the LHP, we can split our integral into two contours as

$$I' = I'_{real\ axis} + I'_{semicircle,\ LHP} = I + I'_{semicircle,\ LHP} = 2\pi i \mathrm{Res}(z= -i\delta) $$

That is ,

our desired integral is given by

$$ I + I'_{semicircle,\ LHP} = 2\pi i \mathrm{Res}(z= -i\delta) $$

Our goal is now to show that the integral over the semi circle goes to zero as $R\to \infty$. This is easily shown by writing in polar form, making the substitution $z = Re^{i\theta}$

$$I'_{semicircle, LHP} = \lim_{R\to\infty} \int_{0}^{-\pi} (d\theta\ iRe^{i\theta}) \frac{e^{-irRe^{i\theta}}}{Re^{i\theta} + i\delta} = -\lim_{R\to\infty} \int_{0}^{-\pi} (d\theta\ iRe^{i\theta}) \frac{e^{-irR\cos\theta } e^{Rr\sin\theta}}{Re^{i\theta} + i\delta} $$

Now we see the importance of choosing to enclose our semicircle in the LHP. Namely, the fact that $\sin\theta\leq 0$ for $\theta \in [0,-\pi]$ is what allows us to say that this integral goes to zero as $R\to \infty$.

All that is left is to calculate the resides, which is straight forward.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.