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Quantum mechanics: Find the wave function given the conditions

  1. Nov 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a 1-dimensional linear harmonic oscilator. Any measurement of it's energy can either return the value of ħw/2 or 3ħw/2, with equal probability. The mean value of the momentum <P> at the instant t=0 is
    <P> = (mħw/2)1/2

    Find the wave function ψ(x,0) for this state.

    2. Relevant equations



    3. The attempt at a solution

    I started by assuming that the wave function would be something of the kind:

    ψ(x,0) = α|ϕ0> + β|ϕ1>

    Since the probability of measuring the energy for n=0 and n=1 is the same I concluded that α=β

    Image with my resolution:

    http://oi41.tinypic.com/2e64ykz.jpg

    I think my way of solving the problem is correct, however the result doesn't make sense... If someone could give me a hint about where I might have gone wrong I'd appreciate. Thanks.
     
  2. jcsd
  3. Nov 2, 2013 #2

    TSny

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    You can't assume that α = β follows from the equal probabilities. What is the relationship between α and the probability of measuring the energy to be [itex]\hbar \omega/2[/itex]?
     
    Last edited: Nov 2, 2013
  4. Nov 2, 2013 #3

    TSny

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    It looks like you dropped a negative sign in your calculations. See the third line of your calculation of <Px> where you have a negative sign in front of α that you drop in the last equality of the third line.

    [Edit: Also, I think your expression for the operator Px has the wrong overall sign.]
     
    Last edited: Nov 2, 2013
  5. Nov 2, 2013 #4
    Ok, so I reviewed the problem.
    The probability of measuring the energy to be ħw/2 is given by the expression:

    [itex]\frac{|<ϕ_{0}|ψ>|^{2}}{<ψ|ψ>}[/itex]




    where ϕ0 is the eigenfunction associated with the eigenvalue of the first level of energy. Developing this expression I got:

    [itex]\frac{|<ϕ_{0}|ψ>|^{2}}{<ψ|ψ>}[/itex] = [itex]\frac{|α|^{2}}{1}[/itex] = [itex]\frac{1}{2}[/itex]

    |α| = ±[itex]\sqrt{1/2}[/itex]

    Using the same process I also concluded that:

    |β| = [itex]\sqrt{1/2}[/itex]

    Correcting the sign mistake I had I arrived at the following result, regarding the mean value of the momentum operator P:

    i[itex]\sqrt{\frac{mwħ}{2}}[/itex] ([itex]α^{*}[/itex]β - αβ[itex]^{*}[/itex]) = [itex]\sqrt{\frac{mwħ}{2}}[/itex]

    i (α[itex]^{*}[/itex]β - αβ[itex]^{*}[/itex]) = 1

    I'm not sure where I am supposed to go next. Have I made any other error?
     
    Last edited: Nov 2, 2013
  6. Nov 2, 2013 #5

    TSny

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    Since you can multiply the wavefunction by any complex number of magnitude 1 without changing the state or the normalization, you may assume ##\alpha## is real and positive. But you will need to allow for a phase factor in ##\beta##: ##\beta =e^{i\phi} \sqrt{1/2}##.

    I think you might still have an overall sign error here. Did you correct the sign error in your expression for the Px operator in terms of the creation and destruction operators?
     
    Last edited: Nov 2, 2013
  7. Nov 2, 2013 #6
    My bad! I corrected it just now. It will result in a change of sign in the final expression.

    [itex]i\left(αβ^{*}-α^{*}β\right) = 1[/itex]

    I assumed that both [itex]α[/itex] and [itex]β[/itex] were complex numbers, given by

    [itex]α = |α|^{i\theta}[/itex]

    [itex]β = |β|^{i\phi}[/itex]

    Substituting in my expression, and developing it in a sine, I arrived at the expression:

    [itex]sin(\theta-\phi) = -1[/itex]

    [itex]\theta-\phi = \frac{3\pi}{2}+2n\pi[/itex]

    If I force [itex]\theta[/itex] to be 0, as you said, I'll get the value of [itex]\phi[/itex]. Am I correct now? (I don't have the solution of the problem, so I'm not 100% sure)
     
  8. Nov 2, 2013 #7

    TSny

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    Yes, that looks good. You should get a simple expression for ##\beta##.
     
  9. Nov 2, 2013 #8
    Thank you very much for your help!
     
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