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Harmonic Oscillator, overlap in states

  1. May 23, 2014 #1
    1. The problem statement, all variables and given/known data

    33makpj.png

    Particle originally sits in ground state about x=0. Equilibrium is suddenly shifted to x=s. Find probability of particle being in new first excited state.

    2. Relevant equations



    3. The attempt at a solution

    Shifted wavefunctions are for ground state: ##\phi'_0 = \frac{\alpha^{\frac{1}{2}}}{\pi^{\frac{1}{4}}} e^{ -\frac{\alpha^2}{2}(x-s)^2}## and for first excited state: ##\phi'_1 = \frac{2\alpha^{\frac{3}{2}}}{\pi^{\frac{1}{4}}} (x-s) e^{-\frac{\alpha^2}{2} (x-s)^2}##.

    When the equilibrium is shifted to x=s, the particle's ground state at x=0 wavefunction becomes:

    [tex]\phi'_{0(0)} = \frac{\alpha^{\frac{1}{2}}}{\pi^{\frac{1}{4}}} e^{-\frac{\alpha^2}{2}s^2}[/tex]

    To find probability of it being in first excited state, we overlap it with the first excited state:

    [tex]\langle \phi'_1|\phi'_{0(0)}\rangle [/tex]
    [tex]= \frac{2\alpha^2}{\pi^{\frac{1}{2}}} e^{-\frac{(\alpha s)^2}{2}} \int_{-\infty}^{\infty}(x-s) e^{-\frac{\alpha^2}{2}(x-s)^2 } dx[/tex]
    [tex] = \frac{2\alpha^2}{\pi^{\frac{1}{2}}} e^{-\frac{(\alpha s)^2}{2}} \left[ \int_{-\infty}^{\infty} (x-2s)e^{-\frac{\alpha^2}{2}(x-s)^2} dx + \int_{-\infty}^{\infty}s \space e^{-\frac{\alpha^2}{2}(x-s)^2} \right] dx[/tex]

    Now, we let ##2s = b## and ##a = \frac{\alpha}{\sqrt 2}## and change ##x## to ##x-s## for the second integral:

    [tex]=\frac{2\alpha^2}{\pi^{\frac{1}{2}}} e^{-\frac{(\alpha s)^2}{2}} \left[ -\frac{(2s)\pi^{\frac{1}{2}}}{\sqrt 2 \alpha} dx + s \int_{-\infty}^{\infty} e^{-\frac{\alpha^2}{2}x'^2} dx' \right][/tex]

    [tex]=\frac{2\alpha^2}{\pi^{\frac{1}{2}}} e^{-\frac{(\alpha s)^2}{2}} \left[ -\frac{\sqrt 2 \pi^{\frac{1}{2}}s}{\alpha} + s\pi^{\frac{1}{2}} \frac{\sqrt 2}{\alpha} \right] [/tex]

    [tex] = 0[/tex]
     
    Last edited: May 23, 2014
  2. jcsd
  3. May 23, 2014 #2
    What you're looking for is [itex]\langle \phi'_1|\phi_{0}\rangle^2 [/itex]
     
  4. May 23, 2014 #3
    Why is it ##\phi_0## and not ##\phi_0'##?
     
  5. May 23, 2014 #4
    Because the particle was in the state [itex]\phi_0[/itex], not [itex]\phi_0'[/itex]
     
  6. May 24, 2014 #5
    Got it, thanks!
     
  7. May 25, 2014 #6
    I did that, and I got an answer of ##2(\alpha s)^2##..
     
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