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I have a quick calculational question.

In Peskin and Schroeder, Chapter 2, they want to look at the amplitude for a particle to propagate between two arbitrary points, $x$ and $x_0$, in an arbitrary amount of time t given the Hamiltonian $\sqrt{p^2c^2+m^2c^4}$. To do this, we look at the inner product of the time-evolved particle that was at $x_0$ with the $x$ eigenket: $\langle \vec{x}|\exp(-i\hat{H}t/\hbar)|\vec{x_0}\rangle$, correct?

If we insert two complete sets of momenta, the integral becomes (give or take factors of $\hbar$):

$\frac 1 {(2\pi)^3}\int d^3p\; \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar)\exp(i\vec{p}\cdot(\vec{x}-\vec{x_0}))$

I'm not quite sure where to proceed from here... Peskin and Schroeder somehow reach:

$\frac 1 {2\pi^2 |x-x_0|}\int_0^\infty dp\;p\sin(p|x-x_0|)\exp((-it\sqrt{p^2c^2+m^2c^4}/\hbar)$.

To get here, however, it seems that you would have to assume that $p$ points in the same direction as $x-x_0$ to get spherical symmetry or something. Why can we do this? Aren't we summing over all possible momenta?


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I'm not sure but it seems that they are using spherical coordinates? – DaniH Apr 12 '12 at 20:28
Well yeah, definitely - that would explain the vanished $4\pi$ and the $dp\;p$. But how did the exponential become a sine and how did the $\theta,\phi$ integrals integrate out trivially? – Nilay Kumar Apr 12 '12 at 20:35

2 Answers 2

up vote 4 down vote accepted

Work with spherical coordinates in momentum space and choose $z$-axis in $p$-space along $\vec{x}-\vec{x_0}$. Then $$\vec{p}\cdot(\vec{x}-\vec{x_0})=p|x-x_0| \cos\theta$$

Then your integral $$ \frac 1 {(2\pi)^3}\int d^3p \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar)\exp(i\vec{p}\cdot(\vec{x}-\vec{x_0})) \\ = \frac 1 {(2\pi)^3}\int_0^{\infty} p^2 dp \int_0^{2\pi} d\phi \int_{-1}^{1} d(\cos\theta) \exp(ip|x-x_0| \cos\theta) \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar) $$

Now do the $\phi$ and $\cos\theta$ integrals to get $$ \frac 1 {4\pi^2 \, i \, |x-x_0|}\int_0^{\infty} p dp \Big[ \exp(ip|x-x_0|) - \exp(-ip|x-x_0|) \Big] \exp(-it\sqrt{p^2c^2+m^2c^4}/\hbar) $$

Recognize that the quantity in the square brackets is $2i\sin(p|x-x_0|)$ to get your answer.

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Got it, thank you very much! – Nilay Kumar Apr 12 '12 at 20:56

What Peskin and Schroeder start here is a 3D Fourier transform from the (mixed) time-momentum domain to the time-space domain. Now you can calculate the steps from 3D integral to 1D integral explicitly (as in the first answer) but you can also directly apply the standard formula for the radial Fourier transform, which is presumably what Peskin & Schroeder did.

Radial Fourier transforms

There are simplified expressions for radial Fourier transforms. These expressions are in principle one dimensional integrals. Specifically for three spatial dimensions we have the radial transform and its inverse.

\begin{equation}%\label{} F (p) \ \stackrel{\mathrm{def}}{\mbox{ $=$}} \ ~~~4\pi ~~\frac{1}{p}~ \int_0^{\infty} ~r~\mbox{ f}(r)~ \sin(pr) \ dr \end{equation}

\begin{equation}%\label{} ~\mbox{ f}(r) ~~\stackrel{\mathrm{def}}{\mbox{ $=$}} ~~\frac{4\pi}{(2 \pi)^3} ~\frac{1}{r}~\int_0^{\infty} p~\mbox{ F}(p)~ \sin(pr) \ dp \end{equation}

If the functions are symmetric, that is, $f(r)=f(-r)$ or $F(p)=-F(p)$ then we can express the three dimensional radial transforms as one dimensional normal Fourier transforms so that we can use the standard Fourier transform tables.

\begin{equation}%\label{} \mbox{ F} (p) \ \stackrel{\mathrm{def}}{\mbox{ $=$}} ~~-2\pi ~~\frac{1}{ip}~ \int_{-\infty}^{\infty} ~r~\mbox{ f}(r)~ e^{-ipr} \ dr \end{equation}

\begin{equation}%\label{} ~~\mbox{ f}(r) ~~\stackrel{\mathrm{def}}{\mbox{ $=$}} ~~\frac{2\pi}{(2 \pi)^3} ~\frac{1}{ir}~\int_{-\infty}^{\infty} p~\mbox{F}(p)~ e^{+ipr} \ dp \end{equation}


The n dimensional radial Fourier transforms

The generalized radial Fourier transforms for n-dimensions expressed as one dimensional integrals with the help of integer and half-integer order Bessel functions. The general radial transform and its inverse are

\begin{equation}%\label{} \mbox{ F} (p) \ \stackrel{\mathrm{def}}{\mbox{ $=$}} ~~~~4\pi ~~~\frac{1}{p^{(n/2-1)}}~ \int_0^{\infty} r^{n/2}~~\mbox{ f}(r)~~ \mbox{ J}_{(n/2-1)}(pr) \ dr \end{equation}

\begin{equation}%\label{} ~~\mbox{ f}(r) ~~\stackrel{\mathrm{def}}{\mbox{ $=$}} ~~\frac{4\pi}{(2 \pi)^n} ~\frac{1}{r^{(n/2-1)}}~\int_0^{\infty} p^{n/2}~\mbox{ F}(p)~ \mbox{ J}_{(n/2-1)}(pr) \ dp \end{equation}

For odd number of spatial dimensions the half integer order Bessel function can be expressed in series of sine and cosine functions. The series are the spherical Bessel functions generated by a repeated differentiation.

The spherical Bessel functions of the first kind are related to the Bessel function by. \begin{equation}j_n(x)~~ =~~\sqrt{\frac{\pi}{2x}}~J_{n+\frac{1}{2}} \end{equation}

which are generated by repeatedly differentiating the sync function.

\begin{equation}j_n(x)~~ =~~ (-x)^n\left(\frac{1}{x}~\frac{d}{dx}\right)^n\,\frac{\sin x}{x}~~~~~~\end{equation}


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Good to know - thanks! – Nilay Kumar Apr 18 '12 at 2:33

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