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Schroedinger Equation in Momentum Space

  1. Dec 23, 2008 #1
    Derive the Schroedinger equation (for harmonic oscillator) in momentum space.

    The attempt at a solution
    We have

    [tex]ih \frac{\partial}{\partial t} \langle p' | \alpha \rangle = \langle p' | \frac{p^2}{2m} | \alpha \rangle + \langle p' | V(x) | \alpha \rangle[/tex]
    [tex]\iff ih \frac{\partial}{\partial t} \langle p' | \alpha \rangle = \frac{p'^2}{2m}\langle p' | \alpha \rangle + \left(i\hbar\frac{\partial}{\partial p'}\right)^2 \langle p' | \alpha \rangle[/tex]
    \iff ih\frac{\partial \Phi(p)}{\partial t} = \frac{p^2}{2m}\Phi(p) - \hbar^2 \frac{\partial^2 \Phi(p)}{\partial p^2}

    Is the above correctly done? If so, is this the generalized Schroedinger equation (without electrodynamics, etc.. just p^2/2m kinetic energy)?:

    \iff ih\frac{\partial \Phi(p)}{\partial t} = \frac{p^2}{2m}\Phi +\displaystyle\sum_{n = 0}^\infty \frac{V^{(n)}(0)}{n!}\left(i\hbar \frac{\partial}{\partial p}}\right)^n \Phi

    Thanks very much for your help :)
    Last edited: Dec 23, 2008
  2. jcsd
  3. Dec 23, 2008 #2


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    Science Advisor

    Yes and yes (but V and its derivatives should be evaluated at 0, not x).
  4. Dec 23, 2008 #3
    Yes, sorry that was a typo. :) Thanks very much!
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