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QM: Harmonic oscillator

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all.

    At time t<0 a particle is in the stationary state [itex]\left| {\psi _0 } \right\rangle [/itex] of the harmonic oscillator with frequency omega1 (i.e. the ground state of the H.O.).

    At t=0 the Hamiltonian changes in such a way that the new angular frequency is omega = omega1/2. We assume that at t=0 the particle is still in the state [itex]\left| {\psi _0 } \right\rangle [/itex], but for t>0 the state evolves according to the new Hamiltonian with omega = omega1/2.

    I wish to find the expectation value of position, x, for all times t.


    3. The attempt at a solution
    For t<0 it is easy, since the particle is in the stationary state, so using raising/lowering operators, one findes that <x> = 0 for t<0.

    For t>0 I don't think it is so trivial, since the particle is now in a superposition of all the possible states. If I use raising/lowering operators to find <x> for t>0, I get an infinite sum (of course it might converge, but still ..).

    Do you have any suggestions for the case t>0?

    Best regards,
    Niles.
     
  2. jcsd
  3. Mar 24, 2009 #2

    turin

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    I don't understand this part. Do you mean that the Hamiltonian changes adiabatically? Wouldn't you then need to know the details in order to determine the observables as a function of t?
     
  4. Mar 24, 2009 #3
    I'm not quite sure I understand what you mean by 'adiabatically'. But it is a harmonic oscillator potential, so for t<0 we have:

    [tex]
    \hat H = \frac{\hat p^2}{2m} + \frac{1}{2}m\omega_1^2\hat x^2,
    [/tex]

    and for t>0 we have:

    [tex]
    \hat H = \frac{\hat p^2}{2m} + \frac{1}{2}m\omega^2\hat x^2,
    [/tex]

    where [itex]2\omega= \omega_1[/itex].
     
  5. Mar 26, 2009 #4
    I'm giving this a final shot: Anybody has any good ideas about this problem?

    EDIT: I understand now what you meant about the adiabatic change of the Hamiltonian. It is not adiabatic, since I am told that the probability of finding the particle in the ground state of the new Hamiltonian is √[8/9].
     
    Last edited: Mar 26, 2009
  6. Mar 26, 2009 #5

    turin

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    I tried to do this strictly in terms of the creation and annihilation operators, but I also got an infinite series. hmm... I'll continue to think about it.

    BTW, thanks for clarifying those Hamiltonians; that helped.
     
  7. Mar 26, 2009 #6
    That was no problem; I really appreciate that you are looking at this.

    I've also thought about this problem the last couple of days, and I still haven't reached a conclusion.
     
  8. Mar 27, 2009 #7
    Shouldn't <x> = 0? Looking on those Hamiltonians I can't figure out why <x> should be positive or negative =)
     
  9. Mar 27, 2009 #8
    I think that too (since the <x>=0 for the stationary states of the Hamiltonian for the H.O.), but my problem is to show it.
     
  10. Mar 27, 2009 #9
    Are you currently studying perturbation theory? This might be the way to approach the problem if so. However I have a sneaking suspicion that it is simpler than we are making out to be. What I tried is as follows:

    [tex]\langle \psi(t)|x|\psi(t) \rangle = \langle 0|e^{\frac{iHt}{\hbar}} x e^{\frac{-iHt}{\hbar}}|0 \rangle = \sum_{n,m}\langle 0|n\rangle \langle n|e^{\frac{iHt}{\hbar}} x e^{\frac{-iHt}{\hbar}} |m \rangle \langle m|0 \rangle[/tex]

    [tex]= \sum_{n,m}e^{i \frac{\omega}{2}(n-m)t}\langle 0|n\rangle \langle n|x|m \rangle \langle m|0 \rangle = 4 cos(\frac{\omega}{2}t) \sqrt{\frac{\hbar}{m \omega}}{\sum_n \sqrt{n+1} \langle 0|n+1 \rangle[/tex]

    Here the n,m represent the "new" states. I had to do a bit of gymnastics to reduce the two sums (two surviving terms from the bracket with [tex]x[/tex]) to one. The only issue is that, as of yet, I can't quite see how the old and new states are related. Staring at the position-space wavefunctions might do the trick. I like this one. I'll think about it some more.
     
  11. Mar 27, 2009 #10
    this may be dumb... but here is my idea =)
    you know that for any operator [tex] \hat{f} [/tex] and it's expectation value [tex]\bar{f}[/tex]
    [tex] \frac{d\bar{f}}{dt} = \frac{\partial \hat{f}}{\partial t} + \frac{i}{\hbar}[\hat{H}, \hat{f}] [/tex]
    If [tex]\hat{f} = \hat{x}[/tex] then [tex]\partial \hat{x}/\partial t = 0[/tex] and both Hamiltonians commute with [tex]\hat{x}[/tex] so [tex] d\bar{f}/dt = 0 [/tex] and since it was zero in the beginning it would stay zero.
    This idea should be checked on some simple case like harmonic oscillator with the same frequency but with different position of minimum... i'll try to look at it later =)
     
  12. Mar 27, 2009 #11
    Thanks for the suggestions - I'll check them out in detail soon.
     
  13. Mar 27, 2009 #12

    turin

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    No they don't. As a simple excercise, determine the commutator [x,H] for the harmonic oscillator. It is just as easy to due this calculation with x and p or with a and a+, and the result does not vanish. BTW, I don't know of any Hamiltonian that commutes with x.

    Actually, I think I figured out the way to show that the expecation value is always zero, and it amounts to determining how the commutators [x,H] and [p,H] appear in the expectation value in the state exp(Ht/ihbar)|ψ0>, and then recognizing that the expectation values vanish. I'll throw it out there, and you can make what you will of it.

    I started with the expression as physicsboy started. Then, I expanded in powers of t. So, I got a power series with coefficients that contain expectation values. These expectation values can be written as expectation values of commutators, and they vanish (not the commutators themselves, but their expectation values).

    EDIT:

    Ah, there's an even better way to do it. Thanks to quzz for reminding me. The following is in the Schroedinger picture.

    [tex]
    \frac{d}{dt}\Bigg(\langle{}x\rangle{}\Bigg)
    =
    \frac{d}{dt}\Bigg(\langle{}\psi{}|\Bigg)x|\psi{}\rangle{}+\langle{}\psi{}|x\frac{d}{dt}\Bigg(|\psi{}\rangle{}\Bigg)
    [/tex]

    Then, you just have to deal will a commutator and an initial value problem.
     
    Last edited: Mar 27, 2009
  14. Mar 27, 2009 #13
    You're right =) that idea of mine was silly
     
  15. Mar 27, 2009 #14

    turin

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    Actually, I think your idea basically solves the problem. However, you wrote an expectation value on one side and then operators on the other, so the expression didn't quite make sense. This is kind of like saying something like

    [tex]
    \int_{x_1}^{x_2}dx\ y(x)=\frac{d}{dx}
    [/tex]

    But, I knew what you meant. It was a good suggestion.
     
    Last edited: Mar 27, 2009
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