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.

  • $\begingroup$ @J.M.: If you see it as a question about the Ornstein-Uhlenbeck operator, it fits here ;-). $\endgroup$
    – JT_NL
    Sep 3, 2011 at 12:22
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    $\begingroup$ 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/…. $\endgroup$
    – joriki
    Sep 3, 2011 at 12:36
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    $\begingroup$ All your questions are answered here, see in particular section on the N-dimensional harmonic oscillator. $\endgroup$ Sep 4, 2011 at 8:50
  • $\begingroup$ Thanks, Tomas. There is still something I don't quite understand. Does the ground state of the system correspond to $N=1$ in $E=\hbar\omega\left({3\over2}+N\right)=\hbar\omega\left({3\over2}+n_x+n_y+n_z\right)$? 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.? $\endgroup$
    – walter w
    Sep 4, 2011 at 11:18
  • $\begingroup$ The ground state corresponds to $N=0$ as all $n_i$'s must be $\geq0$. The occupation numbers $n_i$ are eigenvalues of the mutually commuting Hermitian operators $a_i^\dagger a_i$. Any two states with differing sets of $n_i$ are therefore orthogonal, thus also linearly independent. As you say, in order to find the degeneracy of the energy levels, you then just need to find the number of solutions to the equation $n_x+n_y+n_z=N$ with non-negative integers $n_i$. You can also derive the same result using the polar coordinate quantum numbers $n,l$ - for that see Wikipedia. $\endgroup$ Sep 4, 2011 at 11:41

1 Answer 1

  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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