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Quantum harmonic question

  1. Mar 19, 2008 #1
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  2. jcsd
  3. Mar 20, 2008 #2

    malawi_glenn

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    Show work done and the formulas /relations you know, then someone will try to help you.
     
  4. Mar 20, 2008 #3
    My first thought about this problem is trying to set up a boundary condition for the wavefunction. For 0<x<A, the wavefunction takes the form of simple harmonic oscillators, i.e. N exp(-1/2 alpha x^2) . With boundary condition where psi(x) = 0 at x=0, I'm stuck with the expression of N=0. Maybe, I didn't think it right...
     
  5. Mar 21, 2008 #4

    malawi_glenn

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    N is just a normalization factor, leave that now for the moment.

    But you also have Hermite polynomials in the harmonoc wave functions right?
     
  6. Mar 21, 2008 #5
    so, the harmonic wave function can be written as : N H(y) exp(-1/2 alpha x^2) . With the mentioned boundary condition, do I get N*H(y) = 0 ...how would I go from here on to solve the problem?
     
  7. Mar 22, 2008 #6
    The energy is calculated as the expectation value of the Hamiltonian i.e.:
    [tex]E_{n}=\int{{\phi_{n}{(x)H\phi_{n}(x)dx}}[/tex]
    where [tex]\phi_{n}(x)[/tex] is the eigenfunction for the energy-level [tex]E_{n}[/tex]
    In the case of an ordinary one-dimensional harmonic oscillator
    [tex]H=\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}[/tex]
    And therefore the energy is given by
    [tex]E_{n}=\int_{-\infty}^{\infty}{{(\phi_{n}{(x)\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\phi_{n}(x))dx}}[/tex]
    In your case you instead integrate from 0 to [tex]\infty[/tex].
    So the energy is:
    [tex]E_{n}=\int_{0}^{\infty}{{(\phi_{n}{(x)\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\phi_{n}(x))dx}}[/tex]
     
  8. Mar 22, 2008 #7
    but then how do you actually carry out the intergration?
     
  9. Mar 22, 2008 #8

    malawi_glenn

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    have you tried? Where do you get stuck?
     
  10. Mar 23, 2008 #9
    Hi again

    I did only give you a clue so you could continue by yourself. But maybe you need some more help. As you know the integral over the even interval [tex]-\infty[/tex] to [tex]\infty[/tex] you can use the fact that the integrand is an even function (why?), i.e.
    [tex]f(-x)=f(x)[/tex]

    Know you want to integrate over the half interval and therefore use the rule (I suppose you have seen it in some course in mathematics):
    [tex]\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx[/tex]

    I will give you a detailed solution because I want you to try by yourself, but kmow I suppose you have pathway at least.
     
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