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Homework Help: Probability for a momentum value for particle in a box

  1. May 11, 2008 #1
    [SOLVED] Probability for a momentum value for particle in a box

    1. The problem statement, all variables and given/known data

    This is the problem 7.37 from Levine's Quantum Chemistry Textbook.

    Show that, for a particle in a one-dimensional box of length [tex]l[/tex], the probability of observing a value of [tex]p_{x}[/tex] between [tex]p[/tex] and [tex]p+dp[/tex] is:

    [tex]\frac{4|N|^{2}s^{2}}{l(s^{2}-b_{2})}\left(1-(-1)^{n}cos(bl)\right)dp[/tex]​

    Where [tex]s\equiv n\pi l^{-1}[/tex] and [tex]b\equiv p\hbar^{-1} [/tex]. The constant [tex]N[/tex] is to be chosen so that the integral from minus infinity to infinity of the previous equation is unity.

    2. Relevant equations

    Since [tex]p[/tex] has a continue range of eigenvalues, the probability of finding a value of [tex]p[/tex] beetween [tex]p[/tex] and [tex]p+dp[/tex] for a system in the state [tex]\Psi[/tex] is

    [tex]|\left\langle g_{p}(x)|\Psi (x,t) \right\rangle |^{2}dp [/tex]​

    where [tex]g_{p}[/tex] are the eigenfunctions of [tex]\hat{p}[/tex].

    For a particle in a box:

    [tex]\Psi = \left(\frac{2}{l}\right)^{1/2}sin\left(\frac{n\pi x}{l}\right)[/tex]​

    and the eigenfunctions of the momentum operator are:

    [tex]g=Ne^{ipx/\hbar}[/tex]​

    3. The attempt at a solution

    I made the corresponding substitutions to obtain the integral:

    [tex]\int ^{0}_{l}Ne^{-ipx/\hbar}\left(\frac{2}{l}\right)^{1/2}sin\left(\frac{n\pi x}{l}\right)dx[/tex]​

    Which after computing and using some identities gives:

    [tex]\int ^{0}_{l}Ne^{-ipx/\hbar}\left(\frac{2}{l}\right)^{1/2}sin\left(\frac{n\pi x}{l}\right)dx = \frac{\sqrt{2}Ns\left[1-(-1)^{n}e^{-ibl}\right] }{\sqrt{l}\left(s^{2}-b^{2}\right)} [/tex]​

    Which looks a lot like the square root of the expression that I seek, however I'm stuck at this point. Any help will be appreciated, if more informations is needed please tell me.

    Thanks in advance
     
  2. jcsd
  3. May 12, 2008 #2
    Sorry, I made a mistake in the first equation. Instead of

    [tex]\frac{4|N|^{2}s^{2}}{l(s^{2}-b_{2})}\left(1-(-1)^{n}cos(bl)\right)dp[/tex]​

    Must be:

    [tex]\frac{4|N|^{2}s^{2}}{l(s^{2}-b^{2})^{2}}\left(1-(-1)^{n}cos(bl)\right)dp[/tex]​
     
  4. May 12, 2008 #3
    Ok thanks anyway, I already solved it. I forgot to take the absolute value.
     
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