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Quantum Mechanical Harmonic Oscillator Problem Variation

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data
    At time t < 0 there is an infinite potential for x<0 and for x>0 the potential is 1/2m*w^2*x^2 (harmonic oscillator potential. Then at time t = 0 the potential is 1/2*m*w^2*x^2 for all x.
    The particle is in the ground state.
    Assume t = 0+ = 0-
    a) what is the probability that a measurement will give the value hbar*w/2?
    b) what is the probability that a measurement will give the value 3*hbar*w/2?


    2. Relevant equations
    This seems more conceptual than anything. The eigenenergies for the harmonic oscillator might help \hbar*ω(n+1/2)


    3. The attempt at a solution

    I know that at t <0 the half harmonic oscillator only allows the odd eigenfunctions to survive and then when the infinite potential is removed and a full harmonic oscillator exists for t>0 all of the eigenfunctions can exist but I don't know how to get probability from this knowledge.

    I apologize for my lack of latex. I couldn't find an hbar in the latex reference and didn't know how to do it on my own.
     
  2. jcsd
  3. Nov 27, 2012 #2

    TSny

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    Suppose you had mathematical expressions for the following:

    1. The wavefunction ##\psi(x)## at time t = 0 for -∞< x ∞

    2. The eigenfunctions ##\phi_0(x)## and ##\phi_1(x)##of the full harmonic well corresponding to ##E = \hbar ω/2## and ##E = 3\hbar ω/2##, respectively.

    How would you use these functions to determine the probabilities asked for?

    If you need some help, see page 2 of http://www.umich.edu/~gevalab/Geva/lecture/chem461/Chapter3.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  4. Nov 27, 2012 #3
    If I was given the wavefunction ψ(x). The probability is just calculated <[itex]\phi_0[/itex]|ψ(x)>.

    For t > 0, Ʃ<[itex]\phi_n[/itex]|ψ(x)>|ψ(x)> will be the expanded wavefunction. This means that to get the different values I would just integrate and multiply by the [itex]\phi_n^*[/itex] to get the probabilities where n = 0,1.
     
  5. Nov 27, 2012 #4

    TSny

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    Good. But keep in mind that <[itex]\phi_0[/itex]|ψ(x)> is the "probability amplitude" rather than the probability (and the functions need to be normalized).
     
  6. Nov 27, 2012 #5
    If that's the probability amplitude than the probability would just be the magnitude of that squared correct? So now all I need is to figure out the form of the wave function before and after.

    t < 0 the wave function is in the groundstate of the half harmonic oscillator so for x >0 it looks like the 1st excited state of the full harmonic oscillator and for x < 0, it goes to 0.

    Then when it's switched on it's the full harmonic oscillator but I'm not sure what eigenstates make up the wave function is it just the ground state 1st and 2nd excited state?
     
  7. Nov 27, 2012 #6

    TSny

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    Yes. That all sounds good. You'll need to look up the wavefunctions for the eigenstates of the full harmonic oscillator potential that correspond to the ground state and the first excited state. You won't need the 2nd excited state.
     
  8. Nov 27, 2012 #7
    I got the probability for hbarω/2 = 1/(2*∏) and then 3hbarω/2 = 1/4. What did you get?
     
  9. Nov 27, 2012 #8

    TSny

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    I haven't worked out the integrals. Maybe I will if I get a chance.

    Off hand, though, the answer for 3hbarω/2 doesn't look right to me. Did you normalize the initial wave function properly?
     
  10. Nov 27, 2012 #9
    I believe so. The normalized initial wavefunction is just A_0*(1/√2)*2*x'*exp(-x'^2/2) where x' = (xmω/hbar)^(1/4) and A_0 = (1/∏)^(1/4). Am I missing something?
     
  11. Nov 27, 2012 #10

    TSny

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    Remember, that the initial wave function is zero for negative x. So, you need to choose the nomalization factor with that in mind.
     
  12. Nov 28, 2012 #11
    Ah yes of course. So that means I would have a √2 out front for the normalization. So I would multiply each of my probabilities by two. I get 1/∏ for the ground state and 1/2 for the first excited state for t > 0.
     
  13. Nov 28, 2012 #12

    TSny

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    Good. That's what I get, too.
     
  14. Nov 28, 2012 #13
    Good. Thanks for your help.
     
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