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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.

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@J.M.: If you see it as a question about the Ornstein-Uhlenbeck operator, it fits here ;-). – Jonas Teuwen Sep 3 '11 at 12:22
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.…. – joriki Sep 3 '11 at 12:36
@joriki: Thanks for the link! – Walter W. Sep 3 '11 at 14:02
All your questions are answered here, see in particular section on the N-dimensional harmonic oscillator. – Tomáš Brauner Sep 4 '11 at 8:50
Thanks, Tomas. There is still something I don't quite understand. Does the ground state of the system correspond to $N=1$ in $E=\hbar\omega\left({3\over2}+N\right)=\hbar\omega\left({3\over2}+n_x+n_y+n_z\ri‌​ght)$? But then I think the $n_i$'s must be $\geq1$? And I don't quite get what linearly independent states mean in this context. What do I have to check to show that they are L.I.? – walter w Sep 4 '11 at 11:18
up vote 6 down vote accepted
  1. Your solution is correct (multiplication of 1D QHO solutions).

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

  3. Isotropic - probably means what you suggest - the potential is spherically symmetric. Depends on the context.

  4. Yes, you have to count the number of combinations where $n_x+n_y+n_z=N$.

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