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Finding the momentum-space wave function for the infinite square well

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the momentum-space wave function for the nth stationary state of the infinite square well.

    2. Relevant equations

    Nth state position-space wavefunction:

    [itex]\Psi_n(x,t) = \sqrt(\frac{2}{a})sin(\frac{n\pi}{a}x)e^{-iE_nt/\hbar}[/itex].

    Momentum operator in position space:

    [itex]\hat{p} = -i\hbar\frac{d}{dx}[/itex].

    3. The attempt at a solution

    As a result of an earlier question I know that you can write

    [itex]\Psi_n(x,t) = \sqrt(\frac{2}{a})sin(\frac{n\pi}{a}x)e^{-iE_nt/\hbar}[/itex]
    [itex]= \sqrt(\frac{2}{a})[\frac{e^{i n\pi x/a}-e^{-i n\pi x/a}}{2i}]e^{-iE_nt/\hbar}[/itex]
    [itex]= \frac{-i}{\sqrt(2a)}[\sqrt(a)f_{n\pi\hbar/a}(x) - \sqrt(a)f_{-n\pi\hbar/a}(x)]e^{-iE_nt/\hbar}[/itex]
    [itex]= \frac{-i}{\sqrt(2)}[f_{n\pi\hbar/a}(x) - f_{-n\pi\hbar/a}(x)]e^{-iE_nt/\hbar}[/itex](*)

    where [itex]f_{p}(x) = \frac{1}{\sqrt(a)}e^{ipx/\hbar}[/itex] are the momentum eigenfunctions (when we're in position space) normalized over 0≤x≤a.

    Now when you write [itex]\Psi_n(x,t)[/itex] as a sum (or integral) of momentum eigenfunctions then [itex]\Phi_n(p,t)[/itex] is the coefficient of these eigenfunctions right? So combining this with (*) do we not get that

    [itex]\Phi(p,t) = \frac{-i}{\sqrt(2)}e^{-iE_nt/\hbar}[/itex] when [itex]p=n\pi\hbar/a[/itex];
    [itex]\Phi(p,t) = \frac{i}{\sqrt(2)}e^{-iE_nt/\hbar}[/itex] when [itex]p=-n\pi\hbar/a[/itex];
    [itex]\Phi(p,t) = 0[/itex] otherwise .

    And note that the sum of the squares of these adds up to 1, as it should.

    But now if I just did it without thinking so much, and did

    [itex]\Phi_n(p,t) = \frac{1}{\sqrt(2\pi\hbar)}\int^{\infty}_{-\infty}\Psi_n(x,t)e^{-ipx/\hbar}dx[/itex]

    then I get the result

    [itex]\Phi_n(p,t) = \frac{\sqrt(a\pi)}{\sqrt(\hbar)}\frac{ne^{-iE_nt/\hbar}}{(n\pi)^2 - (ap/\hbar)^2}[1-(-1)^ne^{-ipa/\hbar}][/itex].

    Now this function does peak at those "special" momentum values [itex]p=\pm n\pi\hbar/a[/itex], but it is non-zero for the other values.

    So how come the two methods do not match up? The first says that I am certain to measure one of the two momentum values [itex]p=\pm n\pi\hbar/a[/itex], but the second says I can also get others. Which is correct?

    Thanks for any help! :smile:
     
    Last edited: Feb 6, 2012
  2. jcsd
  3. Feb 6, 2012 #2

    vela

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    It's because the ##\psi(x) = 0## outside the well, but the linear combination of fpn(x) and f-pn(x) doesn't vanish. In other words, you can't express the spatial wave function as a linear combination of just the two f's.
     
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