Finding momentum distribution for particle in square well

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perishingtardi
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Homework Statement


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

Homework Equations




The Attempt at a Solution


I know that to transform the wave function to momentum representation I use [tex]\phi(p) = \frac{1}{\sqrt{2\pi}} \int_{-a}^a \psi(x) e^{-ipx/\hbar}\,dx.[/tex] From this I got [tex]\phi(p) = \frac{\hbar \sin(pa/\hbar)}{p\sqrt{\pi a}}.[/tex] Squaring this isn't going to give the required answer though. Any help appreciated. Thanks.
 
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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.
 
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Yes, it should be
[tex]\phi(p)=\frac{1}{\sqrt{2 \pi \hbar}} \int_{\mathbb{R}} \mathrm{d} x \psi(x) \exp(\mathrm{i} p x/\hbar).[/tex]
This must be so, because of (a) the correct units, because [itex]\phi[/itex] should be of dimension [itex]1/\sqrt{[p]}[/itex] and [itex]\psi[/itex] is of dimension [itex]1/\sqrt{[x]}[/itex] 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 [itex]\mathrm{L}^2(\mathbb{R},\mathbb{C}) \to \mathrm{L}^2(\mathbb{R},\mathbb{C})[/itex].
 
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