1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A 3D Harmonic Oscillator

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a particle of mass [itex]m[/itex] moving in a 3D potential

    [tex]V(\vec{r}) = 1/2m\omega^2z^2,~0<x<a,~0<y<a[/tex].

    [tex]V(\vec{r}) = \inf[/tex], elsewhere.

    2. The attempt at a solution

    Given that I know the solutions already for a 1D harmonic oscillator and 1D infinite potential well I'm going to combine them [itex]E_x + E_y + E_z = E[/itex], and [itex]\psi=\psi(z)\psi(x)\psi(y)[/itex] as for separation of variables of 3D Schrodinger.

    Therefore,

    [tex]E = (n_z+1/2)\hbar\omega + \frac{\pi^2\hbar^2}{2ma^2}(n_x^2+n_y^2)[/tex]

    [tex]\psi=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^{n_z}n_z!}}H_{n_z}(\zeta)e^{-\zeta^2/2}\sqrt{\frac{2}{a}}\sin\left(\frac{n_x\pi}{a}x\right)\sqrt{\frac{2}{a}}\sin\left(\frac{n_y\pi}{a}y\right)[/tex]

    where [itex]H[/itex] are the Hermite polynomials and [tex]\zeta=\sqrt{m\omega/\hbar}z[/tex].

    Is this a correct approach? I couldn't see how going all through the separation of variables for 3D Schrodinger would give me a different answer.
     
    Last edited: Jun 2, 2009
  2. jcsd
  3. Jun 1, 2009 #2

    Avodyne

    User Avatar
    Science Advisor

    You are correct (modulo some minor typos in your expression for psi).
     
  4. Jun 2, 2009 #3
    I believe I fixed the typo. Did I miss anything?

    My OP was because the second part asks: Assuming [tex]\hbar\omega>3\pi^2\hbar^2/(2ma^2)[/tex] find the energies of the ground state and the first excited state, labeling by each state by its three quantum numbers.

    And I don't see why this assumption needs to be made. Wouldn't I just sub into [itex]E[/itex],

    [tex](n_x, n_y, n_z) = (1, 1, 0)[/tex] for the ground state and,

    [tex](n_x, n_y, n_z) = (1, 1, 1)[/tex] for the first excited state?
     
  5. Jun 2, 2009 #4

    Hao

    User Avatar

    For the ground state, the assumption makes no difference. However, it has an effect for the first excited state.

    The first excited state is defined as the state that is second lowest in energy.

    If [tex]\hbar \omega[/tex] were to be very small, [tex](n_x, n_y, n_z) = (1, 1, 1)[/tex] would be second lowest in energy, and hence the first excited state.
    However, if [tex]\hbar \omega[/tex] were to be very large, it can be possible for [tex](n_x, n_y, n_z) = (2, 1, 0)[/tex] to be lower in energy than [tex](n_x, n_y, n_z) = (1, 1, 1)[/tex] instead.
     
  6. Jun 2, 2009 #5
    Thanks, this makes sense now. I was too caught up thinking it had something to do with keeping the energy positive/negative.
     
  7. Jun 2, 2009 #6

    Avodyne

    User Avatar
    Science Advisor

    Nope! It all looks correct now.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A 3D Harmonic Oscillator
Loading...