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Homework Help: QM - Position/Momentum representation problem

  1. Dec 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Write down the time independant Schrodinger eqn in the momentum representation for a particle with mass m when the potential is given by [itex]V(x) = \frac{1}{2} \gamma x^2[/itex]
    Given that a possible solution is given by [itex]\Phi(p) = e^{\frac{-Bp^2}{2}}[/itex]

    determine B and the corresponding energy eigenvalue.


    2. Relevant equations

    position / momentum

    [itex]x \rightarrow i\hbar \frac{d}{dp}[/itex]
    [itex]-i\hbar \frac{d}{dx} \rightarrow p[/itex]



    3. The attempt at a solution

    in position representation the full time independant schrodinger eqn is:

    [itex]- \frac{\hbar^2}{2m}\frac{d^2}{dx^2} \Psi(x) + \frac{1}{2}\gamma x^2 \Psi(x) = E\Psi(x)[/itex]

    becomes

    [itex]\frac{p^2}{2m}\Phi(p) - \frac{\hbar^2 \gamma}{2}\frac{d^2}{dp^2}\Phi(p) = E\Phi(p)[/itex]
    in the momentum representation, where Phi is the FT of Psi.

    After plugging in the Trial Solution I get:

    [itex]E = \frac{p^2}{2m} - \frac{\hbar^2 \gamma}{2}\left( B^2 p^2 - B\right)[/itex]

    Not sure what to do after this bit, I tried to normalise the wavefunction [itex]A = \left(\frac{B}{\pi}\right)^{\frac{1}{4}}[/itex]

    But I dont think that helps.

    Any ideas on where i could get a second equation to find B and E?

    Thanks
     
  2. jcsd
  3. Dec 15, 2011 #2

    vela

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    The energy E needs to be a constant, so it can't depend on p. For what value of B will the dependence of E on p vanish?
     
  4. Dec 15, 2011 #3
    ok, the only way i can see of doing this is, first rearrange to get:

    [itex]p^2 \left(\frac{1}{2m} - \frac{B^2 \hbar^2 \gamma}{2}\right) + \frac{B\hbar^2 \gamma}{2} = E[/itex]

    choose B such that:

    [itex]B^2 \frac{\hbar^2 \gamma}{2} = \frac{1}{2m}[/itex]

    which gives:

    [itex]B = \frac{1}{\hbar}\sqrt{\frac{1}{\gamma m}}[/itex]

    So the energy eigenvalue would be:

    [itex]\frac{\hbar}{2}\sqrt{\frac{\gamma}{m}}[/itex]
     
  5. Dec 15, 2011 #4

    vela

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    Looks good.
     
  6. Dec 15, 2011 #5
    thanks very much
     
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