1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding momentum distribution for particle in square well

  1. Oct 23, 2013 #1
    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
    [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]

    2. Relevant equations


    3. 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.
     
  2. jcsd
  3. Oct 23, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    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.
     
  4. Oct 23, 2013 #3

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted