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Particle in harmonic oscillator potential

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle with mass m moves in 3-dimensions in the potential [tex]V(x,y,z)=\frac{1}{2}m\omega^{2}x^{2}[/tex]. What are the allowed energy eigenvalues?


    2. Relevant equations



    3. The attempt at a solution
    The Hamiltonian is given by [tex]H=\frac{P^{2}}{2m}+\frac{1}{2}m\omega^{2}X^{2}[/tex] where P is the momentum operator in three dimensions. Projecting this into the coordinate basis, we have [tex]\left( -\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\right)\psi = \left(E + \frac{\hbar^{2}}{2m}\left( \frac{d^{2}}{dy^{2}}+\frac{d^{2}}{dz^{2}}\right) \right)\psi [/tex]
    The left side is now in the form of a simple harmonic oscillator. However, I am stuck after this point.
     
  2. jcsd
  3. Dec 4, 2011 #2

    vela

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    Try assuming a solution of the form [itex]\psi(x,y,z)=X(x)Y(y)Z(z)[/itex].
     
  4. Dec 5, 2011 #3
    Hmm so if we assume that, then we find that [itex]E=E_{x}+E_{y}+E_{z}[/itex]. But [itex]E_{y}[/itex] and [itex]E_{z}[/itex] correspond to the eigenvalues of a free particle, since the particle is free in the y and z directions. So does this just mean that the eigenvalues are continuous, since the energy is only quantized along one direction?
     
  5. Dec 5, 2011 #4
    Why do you say it's only quantized in one direction? And no, the energy is not continuous in this case.
     
  6. Dec 5, 2011 #5
    Don't we just have [tex]E=E_{x}+E_{y}+E_{z}=(n+\frac{1}{2})\hbar\omega+\frac{p_{y}^{2}}{2m}+\frac{p_{z}^{2}}{2m}[/tex]Then, since p_y and p_z are continuous, wouldn't we just have that the ground state energy is [itex]\frac{\hbar\omega}{2}[/itex] and that above that, the eigenvalues are continuous?
     
  7. Dec 5, 2011 #6

    vela

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    Yes.
    Because the potential only depends on x. The particle isn't bound when moving in the y and z directions, so Ey and Ez can take on any value. Ex is quantized, but the total energy isn't.
     
  8. Dec 5, 2011 #7
    Why do you treat y and z differently from x? your solution should be the same for X(x), Y(y) and Z(z). I would say [itex]n=n_x+n_y+n_z[/itex]
     
  9. Dec 5, 2011 #8
    OMG, I'm sorry. I didn't read the OP carefully enough. Yes, then you are right. I was thinking of [itex]\frac{1}{2}m\omega r^2[/itex] where [itex]r^2=x^2+y^2+z^2[/itex]
     
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