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In our QM class, the prof said:

"We are ready to begin constructing the individual states of the 3D isotropic harmonic oscillator system. The key property is that the states must organize themselves into representations of angular momentum. Since angular momentum commutes with the Hamiltonian, angular momentum multiplets represent degenerate states."

Why do they "must"? Is that because there is rotational invariance and therefore there "must" be a conserved angular momentum, hence an algebra of angular momentum operators, which necessarily lead to $|\ell,m\rangle$ eigenkets spanning the space of states for the 3D oscillator?

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Frank
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1 Answers1

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I) Perhaps it is helpful to point out that even if the physical system $S$ has no rotational symmetry (e.g. if the system $S$ is a 3D an-isotropic harmonic oscillator), then the Lie group $G=SO(3)$ of rotations still has a group action $G \times S \to S$ on the system. See also e.g. this Phys.SE post.

In particular the Hilbert space ${\cal H}$ of the system still becomes (a possibly infinite-dimensional, possibly reducible) representation$^1$ of $G$.

And the Hilbert space ${\cal H}=\oplus_j{\cal H}_j$ can be decomposed into finite-dimensional $G$-irreps ${\cal H}_j$.

Moreover, the angular momentum $J_i$, $i\in\{1,2,3\}$, are the generators of the corresponding Lie algebra $so(3)$.

II) Now, if the $J_i$ , $i\in\{1,2,3\}$, happen to commute with the Hamiltonian $H$, then one can say more along the lines of what OP's professor mentions. In particular, the aforementioned $G$-irreps ${\cal H}_j$ become (degenerate) energy-eigenspaces.

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$^1$ Concerning the single-valuedness of the wave-function, see also e.g. this Phys.SE question.

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