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Harmonic Oscillator Potential Approximation

  1. Dec 31, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle is in a region with the potential
    V(x) = κ(x2-l2)2
    What is the approximate ground state energy approximation for small oscillations about the location of the potential's stable equilibrium?

    2. Relevant equations

    ground state harmonic oscillator ~ AeC*x2


    3. The attempt at a solution

    My plan is to find the value of x for which V(x) is at stable equilibrium, then expand about that point, and the expansion should look something like the harmonic oscillator potential (1/2*mωx2). Once the potential is known, I know H = p2/2m + V, and I can use <0|H|0> to find the ground state energy, where |0> is the GS energy eigenstate for the harmonic oscillator.

    I have found that the stable equilibrium of V(x) is at x = l.
    Next, I expanded V(x) about this point and found:
    V = 4κl2(x2-2lx+l2) + O(x-l)3

    But what can I do about the 2lx term? I need it to get lost in order to get the HO potential, right?
     
  2. jcsd
  3. Dec 31, 2011 #2

    vela

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    You don't want to do anything. You want the potential in terms of powers of (x-l), which is exactly what you have.
    $$V \cong \frac{1}{2}(8\kappa l)(x-l)^2$$
     
  4. Dec 31, 2011 #3
    So, in calculating <0|H|0>, I will have to integrate ∫(x-l)2e-mωx2/[itex]\hbar[/itex]dx from l-[itex]\epsilon[/itex] to l+[itex]\epsilon[/itex]
    Right?
     
  5. Dec 31, 2011 #4

    vela

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    You don't need to calculate that expectation value, and no, that isn't how you'd calculate the expectation value either. The idea is you can use the approximation that you have a simple harmonic oscillator. What's the ground-state energy of a harmonic oscillator?
     
  6. Dec 31, 2011 #5
    So E=[itex]\hbar[/itex]ω/2, and V=1/2*kx2=1/2*8κl2(x-l)2.
    ω=√(8kl2/m), so E0=[itex]\hbar[/itex]/2*√(8kl2/m)
     
  7. Dec 31, 2011 #6

    vela

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    Right.
     
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