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3D Quantum harmonic Oscillator

  1. Sep 4, 2011 #1
    1. The problem statement, all variables and given/known data
    What are the stationary states of an isotropic 3D quantum harmonic oscillator in a potential [tex]U(x,y,z) = {1\over2}m\omega^2 (x^2+y^2+z^2)[/tex] in the form [tex]\psi(x,y,z)=f(x)g(y)h(z)[/tex] and how many linearly independent states have energy [tex]E=({3\over 2}+n)\hbar\omega[/tex]?


    2. Relevant equations

    See above.

    3. The attempt at a solution
    The solution of a HO in 1D, say the x-direction, is [tex]c H_n e^{\sqrt{m\omega\over\hbar}x}[/tex] where [tex]H_n[/tex] is the [tex]nth[/tex] Hermite polynomial. So I am guessing [tex]\psi = c^3 H_n^3 e^{{m\omega\over\hbar}(x^2+y^2+z^2)}[/tex]. I don't quit understand how to count L.I. states, though. I am guessing the number of combinations of [tex]n_x,n_y,n_z[/tex] such that [tex]n_x+n_y+n_z=n[/tex]? Please help! Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 4, 2011 #2

    vela

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    That's not quite right.
    Decent guess. Try plugging [itex]\psi(x,y,z)[/itex] into the Schrodinger equation. You should find the equation separates, so you can solve it.
    Yes, that's correct.
     
  4. Sep 4, 2011 #3
    Thanks, vela. So is [tex]\psi=cH_{n_x}H_{n_y}H_{n_z}e^{{m\omega\over\hbar}{(x^2+y^2+z^2)}}[/tex], where c is some normalization constant?
     
  5. Sep 4, 2011 #4

    vela

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    Close. You're missing a factor of -1/2 in the exponent. What's the argument of the Hermite polynomials?
     
  6. Sep 4, 2011 #5
    Ah, thanks. Is the argument [tex]\sqrt{m\omega\over\hbar}x_i[/tex], where [tex]x_i\in\{x,y,z\}[/tex]?
     
  7. Sep 4, 2011 #6

    vela

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    Yup!
     
  8. Sep 4, 2011 #7
    Thanks! :-)
     
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