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Quantum Mechanics, Momentum Space

  1. Mar 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Show that

    [tex] <x> = \int \Phi^* \left(-\frac{\hbar}{i}\frac{\partial}{\partial p} \right) \Phi dp [/tex]


    2. Relevant equations

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

    [tex] \Psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \Phi(p,t)dp [/tex]

    3. The attempt at a solution

    I started out with

    [tex] <x> = \int^{\infty}_{-\infty} \Psi^* x \Psi dx [/tex]

    Using the above equation for [itex] \Psi(x,t) [/itex] (and it's conjugate) gives:

    [tex] \Psi^* (x,t) = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{-ipx}{\hbar}} \Phi(p,t)dp [/tex]

    and

    [tex] x\Psi(x,t) = \frac{x}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \Phi(p,t)dp[/tex]

    [tex] = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} x e^{\frac{ipx}{\hbar}} \Phi(p,t)dp[/tex]

    [tex] = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} \frac{\partial}{\partial p} e^{\frac{ipx}{\hbar}} \Phi(p,t)dp[/tex]

    [tex] = \frac{1}{\sqrt{2\pi\hbar}} \left[ \left( e^{\frac{ipx}{\hbar}} \Phi(p,t) \right) \bigg|^{\infty}_{-\infty} -\int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \frac{\partial}{\partial p}\Phi(p,t)dp \right][/tex]

    [tex] = -\frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \frac{\partial}{\partial p} \Phi(p,t) dp [/tex]

    Substituting into the original equation for [itex] <x> [/itex] then gives

    [tex]<x> = \int^{\infty}_{-\infty}\left( \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty}e^{\frac{-ipx}{\hbar}} \Phi^* (p,t) dp \right) x\Psi(x,t) dx [/tex]

    [tex] = \frac{1}{\sqrt{2\pi\hbar}} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} e^\frac{-ipx}{\hbar} (x\Psi(x,t))dx dp [/tex]

    [tex] = -\frac{1}{2\pi\hbar} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} e^{\frac{-ipx}{\hbar}} \int^{\infty}_{-\infty} e^{\frac{ipx}{\hbar}} \frac{\partial}{\partial p} \Phi(p,t) dp dx dp[/tex]

    [tex] = -\frac{1}{2\pi\hbar} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} \frac{\partial}{\partial p} \Phi(p,t) \int^{\infty}_{-\infty} e^{\frac{-ipx}{\hbar}} e^{\frac{ipx}{\hbar}} dx dp dp[/tex]

    [tex] = -\frac{1}{2\pi\hbar} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} \frac{\partial}{\partial p} \Phi(p,t) \int^{\infty}_{-\infty} dx dp dp [/tex]

    I'm pretty sure I messed up somewhere, since that integral is infinite...

    Any help would be appreciated.
     
    Last edited: Mar 2, 2007
  2. jcsd
  3. Mar 2, 2007 #2
    I also tried the reverse, starting with the expression you're supposed to get for <x>, and working back from there using similar methods... but it gives me the same problem.
     
    Last edited: Mar 2, 2007
  4. Mar 2, 2007 #3
    Your calculation seems fine up until the point where you substitute your expression for [tex]x\Psi(x,t)[/tex]. You should be integrating over two dummy variables in your final expression, say [itex]p[/itex] and [itex]p^{\prime}[/itex], but you have written both dummy variables as the same variable [itex]p[/itex]. The expression you derived for [tex]x\Psi(x,t)[/tex] in terms of an integral over [itex]p[/itex], change [itex]p[/itex] to [itex]p^{\prime}[/itex] and everything should work out.
     
  5. Mar 2, 2007 #4
    So then for the second last line we end up with
    [tex] <x> = -\frac{1}{2\pi\hbar} \int^{\infty}_{-\infty} \Phi^* (p,t) \int^{\infty}_{-\infty} \frac{\partial}{\partial p'} \Phi(p',t) \int^{\infty}_{-\infty} e^{\frac{-i(p-p')x}{\hbar}} dx \ dp' \ dp [/tex]

    Is that integral doable?
     
  6. Mar 3, 2007 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You should recognize the x integral as a delta function.
     
  7. Mar 3, 2007 #6
    Oh, right... Duh. :blushing: Got it now
     
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