Finding momentum distribution for particle in square well

[perishingtardi]

Homework Statement

A particle of mass $m$ is trapped between two walls in an infinite square well with potential energy
V(x) = \left\{ \begin{array}{cc} +\infty & (x < -a), \\ 0 & (-a \leq x \leq a), \\ +\infty & (x > a).\end{array} \right.
Suppose the wavefuntion of the particle at time $t=0$ is \psi(x) = \frac{1}{\sqrt{2a}}.
Show that the probability density for the particle having momentum $p$ is g(p) = \frac{\sin^2 (pa)}{(\pi a p)^2} \qquad (-\infty < p < +\infty).

Homework Equations

The Attempt at a Solution

I know that to transform the wave function to momentum representation I use \phi(p) = \frac{1}{\sqrt{2\pi}} \int_{-a}^a \psi(x) e^{-ipx/\hbar}\,dx. From this I got \phi(p) = \frac{\hbar \sin(pa/\hbar)}{p\sqrt{\pi a}}. Squaring this isn't going to give the required answer though. Any help appreciated. Thanks.

 

[TSny]

It could be that the person who wrote the problem is assuming units where $\hbar = 1$. Even so, the answer for g(p) as stated in the problem appears to have a typographical error. Only the p should be squared in the denominator.

Your work looks good except I think the numerical factor in front of the integral for $\phi(p)$ should have $\hbar$ in it. This will modify your answer a little. Consider the dimensions of $\phi(p)$ and see if your answer has the correct dimensions.

 

[vanhees71]

Yes, it should be
\phi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{\mathbb{R}} \mathrm{d} x \psi(x) \exp(\mathrm{i} p x/\hbar).
This must be so, because of (a) the correct units, because \phi should be of dimension 1/\sqrt{[p]} and \psi is of dimension 1/\sqrt{[x]} and (b) to make the Fourier transform a unitary transformation, changing from the position representation to the momentum representation of states. It's a change from one generalized basis to another and thus should be a unitary mapping of \mathrm{L}^2(\mathbb{R},\mathbb{C}) \to \mathrm{L}^2(\mathbb{R},\mathbb{C}).