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Expectation of Momentum Operator

  1. Feb 2, 2006 #1


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    I need help getting started on this problem:

    A free particle moving in one dimension is in the initial state [itex] \Psi(x,0) [/itex]. Prove that [itex] <\hat{p}> [/itex] is constant in time by direct calculation (i.e. without recourse to the commutator theorem regarding constants of the motion).

    Our professor strongly advised against doing it this way:

    [tex] <\hat{p}> = \int{\Psi^* \hat{p} \Psi \, dx} [/tex]

    instead saying that we should try this:

    [tex] <\hat{p}> = \int{p |\phi(k,t)|^2 \, dk} [/tex]

    Since we know that:

    [tex] \Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\phi(k)e^{ikx} \, dk} [/tex]

    and therefore

    [tex] \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}{\Psi(x,0)e^{-ikx} \, dx} [/tex]

    Yay Fourier transforms :yuck:

    My questions are:

    1. Practically speaking, how does one go about expressing phi(k) using this method?

    2. Even if I get a reasonable expression for phi(k), what about the time dependence? How do I then get phi(k,t)? I'm going to need that in order to find <p> and show that d<p>/dt = 0.
    Last edited: Feb 2, 2006
  2. jcsd
  3. Feb 2, 2006 #2


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    Since the particle is free, then the whole time dependence of the state vector (it's actually a functional) can be factorized and, by virtue of being expressed in terms of a complex exponential, vanishes when you consider this baby

    [tex] \langle \Psi (t) |\hat{p}| \Psi (t) \rangle [/tex]

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