Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 23

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

where $r>0$

3.page 23

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

where $m>0$

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

share|improve this question
    
Can you define all the quantities above. What is $r$? –  Prahar Mar 25 at 13:28
6  
Hey, what about Mr. Schroeder? –  Love Learning Mar 25 at 13:29

1 Answer 1

up vote 5 down vote accepted

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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