# Finding momentum distribution for particle in square well

1. Oct 23, 2013

### perishingtardi

1. The problem statement, all variables and given/known data
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 $$g(p) = \frac{\sin^2 (pa)}{(\pi a p)^2} \qquad (-\infty < p < +\infty).$$

2. Relevant equations

3. 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.

2. Oct 23, 2013

### 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.

3. Oct 23, 2013

### 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 $\mathrm{L}^2(\mathbb{R},\mathbb{C}) \to \mathrm{L}^2(\mathbb{R},\mathbb{C})$.