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Harmonic oscillator superposition amplitude evaluation

  1. Mar 16, 2013 #1
    Hi all

    1. The problem statement, all variables and given/known data
    I have the first three states of the harmonic oscillator, and I need to know the amplitudes for the states after the potential is dropped.


    2. Relevant equations
    [itex]u_{0}=(\frac{1}{\pi a^{2}})^{\frac{1}{4}} e^{{\frac{-x^2}{2a^2}}}[/itex]

    [itex]u_{1}=(\frac{4}{\pi})^{\frac{1}{4}} (\frac{1}{a^2})^\frac{3}{4} x e^{({\frac{-x^2}{2a^2}})}[/itex]

    [itex]u_{2}=(\frac{1}{4\pi a^2})^{\frac{1}{4}} (\frac{2x^2}{a^2}-1) e^{{\frac{-x^2}{2a^2}}}[/itex]

    Then, the potential is lowered suddenly, by setting [itex]\omega ' = \frac{\omega}{2}[/itex].

    The system is then modelled with the expansion postulate, as a superposition of these three new states, with [itex]\omega'[/itex] in them.

    The particle was originally in the ground state before the potential was lowered, so I have that:

    [itex]C_{i}=\int u_{i}^* \psi dx[/itex]

    Where [itex]\psi[/itex] represents the ground state function, as stated above, with [itex]\omega[/itex]. The states are real though, so the complex conjugate is unchanged from the normal state.


    3. The attempt at a solution
    When I try to evaluate these integrals, I find that I can't actually do it. The problem is that the different symbols keep tripping me up... The ω and ω' values don't seem to gel.

    Am I doing the physics wrong, or am I simple falling short on the maths?
    Thanks, all.
     
  2. jcsd
  3. Mar 16, 2013 #2
    Ah! Solved the maths problems...

    I have one more question, though. For a one-dimensional quantum harmonic oscillator, when determining the amplitudes,[itex]C_{i}[/itex], should I integrate from 0 to infinity, or from -infinity to infinity? I'm guessing it's the latter.
     
  4. Mar 16, 2013 #3

    vela

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    You're right, the latter.
     
  5. Mar 16, 2013 #4
    Thanks!

    To that end, I have a standard integral that gives the answer for the limits 0 and infinity. Is it simply a matter of multiplying by two or have I missed something?
     
  6. Mar 16, 2013 #5

    vela

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    If the integrand is an even function, you can simply multiply by 2. If it's odd, the integral will vanish.
     
  7. Mar 16, 2013 #6

    It's an exponential, similar in form to:

    [itex]\int e^{-ax^2} dx = 0.5[\frac{\pi}{a}]^\frac{1}{2}[/itex]

    Where the limits are taken over 0 and infinity for this identity.

    So, am I okay to multiply by two?
     
  8. Mar 16, 2013 #7

    vela

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    You should be able to answer that on your own. A function f(x) is even if f(x)=f(-x) and odd if f(x)=-f(x). Which one do you have on your hands here?
     
  9. Mar 16, 2013 #8
    I've been having a duh moment here...(I blame late night maths). it's clearly even, because the x has an even exponent. So multiplying by two is permitted.
     
  10. Mar 17, 2013 #9
    Okay, so now I have a superposition of the 3 states, complete with amplitudes. I've introduced a time dependence by setting
    [itex]\psi = \Sigma C_{i} U_{i} e^{\frac {-iE_{i}t} {h}} [/itex]

    Which gives me a superposition state with 3 terms. Now, I want to find the probability of finding this particle in an area -a to a, so I've multiplied the above by its complex conjugate, which gives an expression with 9 terms. If I now integrate over -a to a, am I right in thinking that the terms that contain two of the original eigenstates (eg U0 and U1) will reduce to the kronecker delta, or is that only valid for an integral over all space?
     
  11. Mar 17, 2013 #10

    vela

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    To rely on orthogonality, you have to integrate over all space.
     
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