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Quantum: Finding time derivative of momentum operator in terms of V(x)

  1. Jan 30, 2006 #1


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    Hi all, got a little caught up in tonights hw assignment; which basically reads:

    Find an expression for [tex]\frac {d<\hat{p}>} {dt} [/tex] in terms of V(x).

    I had a few ideas from my math methods for theoretical phys. class i took a few quarters ago, involving applying the derivative operator to the already derived momentum operator, but so far this class hasn't really tackled any kind of math involving working inside hilbert spaces, so i think a strictly calc based approach may be necessary here. Any hints would be appreciated here, ty you all for the help.
  2. jcsd
  3. Jan 30, 2006 #2
    Yes this is usually done using just calc. So for starters write down the integral for <p> and take it's time derivative. If you've done the proof on how the time derivative of psi squared is zero you'll know what to do after that.
  4. Jan 30, 2006 #3


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    So you're trying to find one of Ehrenfest's theorems, right...? From the form of the potential, you've already chosen the Hilbert space to be [itex] L^{2}\left(\mathbb{R}\rihgt) [/itex] and now remember the definition of the average of an observable on the quantum state [itex] \psi (x) [/itex].

  5. Jan 30, 2006 #4


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    I've actually worked out the the derivative for psi squared (in the derivation of the momentum operator using calc), and it turned out not to be zero; I got this:

    \newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } }
    \pd{(\Psi^*\Psi)}{t}{} = \frac{i\hbar^2}{2 m}\left( \Psi^*\pd{\Psi}{x}{2} - \pd{\Psi^*}{x}{2}\Psi \right) [/tex]

    Taking this into the integral for the computation of the derivative of the expectation value of position (classically the velocity,) we end up recognizing the momentum operator we all know and love (multiplying by an m of course). Now the problem i have is with the actual time derivative of <p>, and how to relate it to V(x). I'm not exactly sure how to start it. Thanks for the help again.
    Last edited: Jan 30, 2006
  6. Jan 30, 2006 #5
    I expressed myself poorly. That's indeed nonzero. I meant to referr to conservation of probability or that the that the wave function stays normalized as time passes. So if you integrate that over all space it goes to zero.

    Anyways, as in doing that you'll have now to solve the time derivatives of psi from Schrödingers equation and plug those into the integral for <p> and then use the approperiate boundary conditions.
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