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Homework Help: Harmonic oscillator expectation values

  1. Nov 18, 2006 #1
    I'm given that there is a harmonic oscillator in a state that is a superposition of the ground and first excited stationary states given by [tex]\psi = \frac{1}{\sqrt{2}}\abs{\psi_0(x,t) + \psi_1(x,t)}[/tex], where [tex]\psi_0 = \psi_0(x)e^{\frac{-iE_0t}{\hbar}}[/tex] and [tex]\psi_1 = \psi_1(x)e^{\frac{-iE_1t}{\hbar}}[/tex]. I need to calculate expectation values for position, momentum, and total energy.

    Here's what I've done: I'm assuming this is a simple harmonic oscillator, and for the x operator I have [tex]\frac{i}{\sqrt{2m\omega}}(a_- - a_+)[/tex]. I think that a- operating on Psi0 would be zero, and on Psi1 would be Psi0, and a+ operating on Psi0 would be Psi1, and on Psi1 would be Psi2. So... I have for <x>:

    [tex] <x> = \frac{i}{2\sqrt{2m\omega}}<\psi_0 + \psi_1 | \psi_0 - \psi_1 - \psi_2>[/tex].. but.. uh.. how do you do this? I feel like I must have done something wrong. A push in the right direction would be much appreciated.

    Thanks so much!
     
  2. jcsd
  3. Nov 18, 2006 #2
    There's no way that expectation value could be zero, right? because [tex]<\psi_n | \psi_{n+1}> = 0[/tex], I think.. so all that stuff inside the <> would be zero. No?
     
  4. Nov 19, 2006 #3

    George Jones

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    Is that supposed to be

    [tex]\Psi = \frac{1}{\sqrt{2}} \left( \abs{\Psi_0(x,t) + \Psi_1(x,t)} \right)?[/tex]

    Brackets are important, as are distinctions between upper case and lower case. For example,

    doesn't make sense without such a distinction. If fact, in this case, it doesn't make sense to write

    [tex]\psi = \psi_{0} + \psi_{1},[/tex]

    since the sum of two energy eigenstates is not an energy eigenstate.

    No - don't forget the numerical factor.

    Now calculate

    [tex]\left< \Psi \left| x \right | \Psi \right>[/tex]

    by, as you did, replacing [itex]x[/itex] with the appropriate, but don't to forget to include all the time dependences and other numerical factors.

    Also, use the orthogonality condition that you gave in your last post.

    I don't mean to sound so critical, but, in this calculation, it's important to get the details right.
     
  5. Nov 19, 2006 #4
    I did mean to put brackets around it. I'm not sure why I didn't. I have trouble getting LaTeX to display things correctly sometimes.

    Edit: nevermind, I'm an idiot and figured out what you were talking about. See next post for confusion.
     
    Last edited: Nov 19, 2006
  6. Nov 19, 2006 #5
    OK. I tried to go back through and put in the terms you were talking about, and I've gotten something.. but I'm still confused. I've gotten down to [tex]\frac{i}{2\sqrt{2m\omega}}<\psi_0 + \psi_1 | \sqrt{\hbar\omega}\psi_0 - \sqrt{\hbar\omega}\psi_1 - \sqrt{2\hbar\omega}\psi_2>[/tex].


    [tex]<\psi_0 | \psi_0> [/tex] I know how to do, but [tex] <\psi_0 | \psi_1>[/tex] I don't. So you said to use the orthoganility condition I brought up in my last post.. so is this just zero since it's Psi0 and Psi1? If not, here's what I'm getting for that, plugging in the x and t dependences: [tex]-\sqrt{\hbar\omega}<\psi_0(x)e^{\frac{-iE_0t}{\hbar}} | \psi_1(x)e^{\frac{-iE_1t}{\hbar}}>[/tex]. I have no idea how to do that. So I hope it's zero.

    So if it is, then the <Psi0 | Psi0> part of that would just be [tex]\sqrt{\hbar\omega}[/tex]. That seems way too easy though.
     
    Last edited: Nov 19, 2006
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