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Ground State of the Simple Harmonic Oscillator in p-space

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data
    A particle is in the ground state of a simple harmonic oscillator, potential → V(x)=[itex]\frac{1}{2}[/itex]mω[itex]^{2}[/itex]x[itex]^{2}[/itex]
    Imagine that you are in the ground state |0⟩ of the 1DSHO, and you operate on it with the momentum operator p, in terms of the a and a† operators. What is the eigenstate? Then, how would you represent this operation in terms of a probability density in momentum space?

    2. Relevant equations
    p=i[itex]\sqrt{\frac{\hbar m ω}{2}}[/itex](a†-a)
    Fourier Transform, x→p-space

    3. The attempt at a solution
    p|0⟩=i[itex]\sqrt{\frac{\hbar m ω}{2}}[/itex]|1⟩
    So now the particle is in the first excited state of the SHO

    But I don't understand the next part of the problem. How can I represent this terms of probability density in p-space?

    I can perform the Fourier Transform and find that,

    [itex]\Phi[/itex](p,t)=[itex]\sqrt{2}[/itex]([itex]\frac{\pi \hbar}{m ω}[/itex])[itex]^{\frac{1}{4}}[/itex] e[itex]^{\frac{-p^{2}}{2 \hbar m ω} - \frac{i t ω}{2}}[/itex]

    And the probability density is,

    |[itex]\Phi[/itex](p,t)|[itex]^{2}[/itex]=2[itex]\sqrt{\frac{\pi \hbar}{m ω}}[/itex] e[itex]^{\frac{-p^{2}}{\hbar m ω}}[/itex]

    I don't believe I understand what this part of the question is asking. Any suggestions/ideas would be greatly appreciated.
     
  2. jcsd
  3. Nov 3, 2013 #2

    vela

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    It doesn't make sense to me either. Is that the problem statement exactly as it was given to you?
     
  4. Nov 3, 2013 #3
    Yes. Word for word.

    Maybe it is supposed to be something like; "Find the expectation value of p...and then do the same thing in p-space."
    But that seems too simple and we've already covered that. The p operator in p-space is just p.
     
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