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For an isotropic 3D QHO in a potential $V(x,y,z)={1\over 2}m\omega^2(x^2+y^2+z^2)$. I can see by independence of the potential in the $x,y,z$ coordinates that the solution to the Schrodinger equation would be of the form $\psi(x,y,z)=f(x)g(y)h(z)$. Explicitly, what would it be? Is it $\psi(x,y,z) = cH_{n_x}H_{n_y}H_{n_z}e^{-{m\omega\over2\hbar}(x^2+y^2+z^2)}$, where $H_{n_i}$ are the $ith$ Hermite polynomial? (A side query, surely since the potential is radial, there is a polar coordinate form of solution which might be better? But this is not asked for in the question. Also, does isotropic just mean that the potential is spherically symmetric?) 

How many linearly independent states have energy $E=\hbar\omega({3\over2}+N)$? Am I supposed to be counting the number of combinations of $n_x,n_y,n_z$ s.t. $n_x+n_y+n_z = N$? I vaguely remember some notion $(n,l)$ mentioned once, but I can't remember what it is nor find the bit of notes on this.

Thanks in advance.

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Much as I'd like to answer the question, I think it belongs on physics.SE. The answers are well-known to physicists. See e.g. en.wikipedia.org/wiki/…. –  joriki Sep 3 '11 at 12:36

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up vote 6 down vote accepted

  2. Since the potential is radially symmetric - it commutes with with angular momentum operator (L^2 and Lz for instance). Hence you may build a solution of the form |nlm> where n states for the radial state description and lm - the angular. Is it better? Depends on the problem. It's just the other basis in which you may represent the solution.


  4. Yes, you have to count the number of combinations where nx+ny+nz=N.

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