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Momentum of a state vector

  1. Feb 3, 2012 #1
    If we assume a system (pure for now) is in a state described by a single state vector, how can you determine the momentum? The momentum of a wavefunction is simply -i times the gradient, but that's for a continuous function. In the hilbert space representation of psi as a ket vector, what does the momentum operator look like?
     
  2. jcsd
  3. Feb 3, 2012 #2

    Ben Niehoff

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    In the position basis, it still looks like [itex]-i \hbar \vec \nabla[/itex]. In the momentum basis, it's simply [itex]p[/itex]. To obtain it in some other basis, you change basis in the standard way.

    The best way to find the result of acting with [itex]\hat p[/itex] is to insert identity operators like so:

    [tex]\begin{align}
    \hat{p} | \psi \rangle &= \int dp \; \hat{p} | p \rangle \langle p | \psi \rangle = \int dp \; p | p \rangle \langle p | \psi \rangle \\
    &= \int dp \; p | p \rangle \int dx \; \langle p | x \rangle \langle x | \psi \rangle \\
    &= \int dp \; | p \rangle \int dx \; p e^{-ipx} \psi(x) \\
    &= \int dp \; | p \rangle \int dx \; i \frac{d}{dx} \big( e^{-ipx} \big) \psi(x) \\
    &= \int dp \; | p \rangle \int dx \; e^{-ipx} \bigg( -i \frac{d}{dx} \psi(x) \bigg) \\
    &= \int dx \; \int dp \; | p \rangle \langle p | x \rangle \bigg( -i \frac{d}{dx} \psi(x) \bigg) \\
    &= \int dx \; | x \rangle \bigg( -i \frac{d}{dx} \psi(x) \bigg)
    \end{align}[/tex]
    Here I've done these steps:

    1. Insert the resolution of the identity in the p basis.
    2. Since we're in the p basis, the p operator can be replaced by its eigenvalue on each basis ket.
    3. Insert the resolution of the identity in the x basis (since our goal is to get the p operator in the x basis).
    4. Use the fact that [itex]\langle p | x \rangle = e^{-ipx}[/itex].
    5. Do standard calculus tricks, integrate by parts.
    6. Observe that we have an identity in the p basis we can remove.
    7. Finally, we see the action of the p operator in the x basis.
     
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