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3DAnisotropic oscillator in Spherical Harmonic basis-States with L_z=0

  1. Sep 17, 2014 #1
    I've been trying to prove a rather simple looking concept. I have a code that calculates states of a 3D anisotropic oscillator in spherical coordinates. The spherical harmonics basis used to expand it's solutions in radial coordinate constraint the spectrum such that when the Hamiltonian is diagonalized it calculates only states with Lz=0, because the potential has a spherical harmonic (Y10)2. i.e.

    VHO=1/2[itex]\hbar[/itex]m [[itex]\omega[/itex]xy2(x2+y2) + [itex]\omega[/itex]z2z2]
    VHO=1/2[itex]\hbar[/itex]m [[itex]\omega[/itex]xy2(x2+y2+z2) + ([itex]\omega[/itex]z2-[itex]\omega[/itex]xy2)z2]
    Since z=rCos(θ)
    VHO=1/2[itex]\hbar[/itex]m [r2([itex]\omega[/itex]xy2 + ([itex]\omega[/itex]z2-[itex]\omega[/itex]xy2)2[itex]\pi[/itex]/3(Y10)2]


    Now, we know the system spectrum in Cartesian ENx,Ny,Nz = 1/2[itex]\hbar[/itex][[itex]\omega[/itex]xy(Nx+Ny+1) + [itex]\omega[/itex]z(Nz+1/2)]. So to calculate this spectrum on paper for verification one can either
    (a) calculate spectrum for 3D anisotropic oscillator in spherical coordinates directly OR
    (b) look for states with Lz=0 in terms of Nx, Ny, Nz by introducing constraints on Nx, Ny & Nz -> like Nx=Ny, Nz=0 OR Nx=2Ny, Nz always even or some such rules..
    Does anyone have advice on how to derive the anisotropic oscillator spectrum in spherical coordinates (using regular spherical harmonics Ymlas basis for solutions).. if not any advice on how to derive constraints on Nx, Ny, Nz to give Lz=0 states only?? Any help will be greatly appreciated.. thanks!!
     
    Last edited: Sep 18, 2014
  2. jcsd
  3. Sep 17, 2014 #2

    Avodyne

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    Is it isotropic in the x-y plane? I can't tell from what you wrote ...
     
  4. Sep 18, 2014 #3
    Yes, it is isotropic in x-y plane.. But sorry for the confusion, I had written the wrong Hamiltonian.. I've corrected the main question now..
     
  5. Sep 18, 2014 #4

    Avodyne

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    You can find combinations of Nx and Ny that give Lz=0 by writing Lz = x py - y px and expressing x, px, y, py in terms of ax and ax+; you should find something like ##L_z \sim a_x^\dagger a_y-a_y^\dagger a_x##. From this you can see that when Lz acts on |Nx,Ny>, Nx+Ny is unchanged. So fix Nx+Ny to some particular value, and find the linear combinations of |Nx,Ny> that are annihilated by Lz. This is an eigenvalue problem for a finite-dimensional matrix.

    I suspect a spherical harmonic expansion of an exact eigenstate |Nx,Ny,Nz> requires an infinite number of terms, except at special values of the ratios of the frequencies.
     
  6. Sep 19, 2014 #5

    Avodyne

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    Addendum: you always need an infinite number of spherical harmonics whenever ##\omega_{xy}\ne\omega_z##. This is because the gaussian factor of the wave function will have some ##\cos\theta## dependence in the exponent, and this can't be expressed as a finite linear combination of ##Y_{\ell m}##'s.
     
  7. Sep 21, 2014 #6
    @Avodyne: That is correct. The basis is a "spherical" harmonic basis.. so any anisotropy will have to be an infinite series expansion, though the degree of anisotropy i.e ##|\omega_{xy}/\omega_z-1|## will decide the numerical convergence.. your earlier suggestion was helpful, thanks!
     
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