homework and exercises - Three integrals in Peskin's Textbook

Peskin's QFT textbook

1.page 14

$$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}$$

when $x^2\gg t^2$, how do I apply the method of stationary phase to get the book's answer.

2.page 27

$$\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}$$

where $r>0$

3.page 27

$$\int_{m}^{\infty}\mathrm{d}E \sqrt{E^2-m^2}e^{-iEt}$$

where $m>0$

I'm crazy about these integrals, but the textbook doesn't give the progress.

Answer

1. Since $x\gg p$, we see that $\sin(px)$ is highly oscillatory. In fact, the integral becomes

$$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}\sim \int_{-\infty} ^\infty \mathrm{d}p\ p\ e^{ipx-it\sqrt{p^2 +m^2}}$$

modulo some factor of $\pm2/i$. Observe now this integral resembles $\int f(p)\exp(g(p))\,\mathrm{d}p$. We find the point $\tilde{p}$ such that

$$g'(\tilde{p})=0.$$

Then just replace $g(p)$ with $g(\tilde{p})+\frac{1}{2}g''(\tilde{p})(p-\tilde{p})^{2}$ and carry out the integral as a moment of a Gaussian. For more on this approximation, see e.g. the relevant chapter of Hunter and Nachtergaele's book (freely and legally available).

2. This is just a Fourier transform of $p\,(p^{2}+m^{2})^{-1/2}$.

3. I assume you are referring to Eq (2.51) on page 27. We write the integral as

$$I(t) = \int^{\infty}_{m}\sqrt{E^{2}-m^{2}}e^{-iEt}\,\mathrm{d}E.$$

Peskin and Schroeder consider this integral as $t\to\infty$. If we consider a change of variables to

$$E^{2}-m^{2}=\mu^{2}\quad\Longrightarrow\quad \mathrm{d}E = \frac{\mu}{\sqrt{m^{2}+\mu^{2}}}\mathrm{d}\mu$$

We have

\begin{align} I(t) &= \int^{\infty}_{0} \frac{\mu^{2}}{\sqrt{m^{2}+\mu^{2}}}e^{-it\sqrt{m^{2}+\mu^{2}}}\mathrm{d}\mu \\ &=\frac{1}{2}\int^{\infty}_{-\infty} \frac{\mu^{2}}{\sqrt{m^{2}+\mu^{2}}}e^{-it\sqrt{m^{2}+\mu^{2}}}\mathrm{d}\mu \end{align}

Let

$$f(\mu) = \frac{\mu^{2}}{\sqrt{m^{2}+\mu^{2}}},\quad\mbox{and}\quad \phi(\mu) = \sqrt{m^{2}+\mu^{2}}$$

so

$$I(t)=\frac{1}{2}\int^{\infty}_{-\infty}f(\mu)e^{-it\phi(\mu)}\mathrm{d}\mu.$$

Observe

$$f(\mu)=\mu\phi'(\mu).$$

As $t\to\infty$, the integral becomes highly oscillatory.

There are two ways to approach the problem from here. The first, unforgivably handwavy but faster: take the stationary phase approximation, and pretend that $f(\mu_{\text{crit}})$ is some arbitrary constant.

The critical points for $\phi$ are $\mu_{0}=0$ and $\mu_{\pm}=\pm im$. We only care about the real $\mu$, so we Taylor expand about $\mu_0$ to second order:

\begin{align} \phi(\mu)&=\phi(0)+\frac{1}{2!}\phi''(0)\mu^{2}\\ &=m + \frac{1}{2m}\mu^{2} \end{align}

We now approximate the integral as

$$I(t) \sim \int^{\infty}_{-\infty}f(c)e^{-itm}e^{-it\mu^{2}/2m}\mathrm{d}\mu \approx f(c)e^{-itm}\sqrt{\frac{4\pi m}{t}}.\tag{1}$$

The other approximation doesn't fix $f$. Observe $f(\mu)\sim|\mu|$, so we have

$$I(t) \sim e^{-itm} \int^{\infty}_{0}\mu e^{-it\mu^{2}/2m}\mathrm{d}\mu.$$

We have (using Fresnel integrals)

$$\int^{\infty}_{0}\mu e^{-it\mu^{2}/2m}\mathrm{d}\mu\sim \frac{im}{t}.$$

Hence

$$I(t)\sim\frac{im}{t}e^{-imt}.\tag{2}$$