Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]


    [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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook