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QM, propagator of particle

  1. Oct 30, 2007 #1
    1. The problem statement, all variables and given/known data
    (this is from R. Shankar, Principles of Quantum Mechanics, 2nd ed, exercise 5.4.3)
    Consider a particle subject to a constant force f in one dimension. Solve for the propagator in momentum space and get

    [tex] U(p,t;p',0) = \delta (p-p'-ft) e^{ i(p'^3-p^3)/6m\hbar f } [/tex]

    2. Relevant equations
    3. The attempt at a solution

    I write a hamiltonian H = p^2/2m + fx, plug that into H|p>=E|p>, with the x operator in momentum space being ih d/dp, it's all nice and seperable and I get

    [tex] \psi(p) = A exp \left( i \frac{p^3-6mEp}{6m\hbar f} \right) [/tex]

    but what do I do now? I'm not sure how to go about normalizing this.
  2. jcsd
  3. Oct 30, 2007 #2


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    First of all, you don't really mean H|p>=E|p>; |p> is an eigenstate of the momentum operator, but not of H; you mean H|E>=E|E>, where <p|E> is your momentum-space wave function.

    You want to normalize it so that you can write a completeness statement in the form
    [tex]\int_{-\infty}^{+\infty}dE\;|E\rangle\langle E| = I,[/tex]
    where I is the identity operator.
  4. Oct 31, 2007 #3
    It's a bit much too write everything as tex but I ended up with
    [tex] A = \left( \frac{1}{2 \pi \hbar f} \right)^{1/2} [/tex]
    Now, how do I get a propagator out of this!
    Last edited: Oct 31, 2007
  5. Oct 31, 2007 #4


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    The propagator is <p'|e^(-iHt/hbar)|p>. Can you think of a use for the completeness statement?
  6. Oct 31, 2007 #5
    \langle p'| e^{-iHt/ \hbar} |p\rangle = \int_{-\infty}^{+\infty}dE \langle p'|E \rangle \langle E| e^{-iHt/ \hbar} |p\rangle = \int_{-\infty}^{+\infty}dE \langle p'|E \rangle \langle E|p-ft\rangle
    and I can plug in my expression for psi, do the integral and out comes the given answer.
    But how come e^(-iHt/hbar)|p> = |p-ft>? I just guessed it from experience of doing lots of integrals like these.
    Last edited: Oct 31, 2007
  7. Oct 31, 2007 #6


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    You don't need to guess it, and in fact it's not correct; evaluating your final expression does not yield the given result (so you made a mistake somewhere when you evaluated it).

    In your middle expression, you can replace H with E, since H is sitting next to one of its eigenstates. Then the integral over E will generate the given answer.
  8. Nov 22, 2011 #7
    Hi, i have a quick question on middle part of the integral. How do you evaluate <E|e[itex]^{-iEt/h}[/itex]|p> ?
    Last edited: Nov 22, 2011
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