operators - Symmetry and degeneracy in quantum mechanics

If an operator commutes with the Hamiltonian of a problem, must it always admit degeneracy? It appears that not necessarily. For example, the parity operator commutes with the Hamiltonian of a free particle as well as that of the 1-D linear harmonic oscillator. But in the former case, we have two-fold degeneracy for a given energy and no degeneracy for the latter. Does it mean that $[P, H]=0$ is not the sufficient condition to have degeneracy in the energy eigenstates?

Answer

If an operator commutes with the Hamiltonian of a problem, must it always admit degeneracy?

No. The identity always commutes with the Hamiltonian, yet we don't generally find all energy levels to be degenerate.

Does it mean that $[P,H]=0$ is not the sufficient condition to have degeneracy in the energy eigenstates?

Yes. It is not a sufficient condition.

Let $A,B$ be any two self-adjoint operators that commute, $[A,B]=0$. Define the eigenvectors of $A$ through

$$A|a\rangle=a\ |a\rangle$$

Now we prove that $B|a\rangle$ is also an eigenvector of $A$: $$ A(B|a\rangle)=BA|a\rangle=a(B|a\rangle) $$

Welp, that was easy.

We could naively say that $a$ is degenerate, because both $|a\rangle$ and $B|a\rangle$ have the same eigenvalue, but this breaks down if $|a\rangle$ is an eigenvector of $B$, because in that case $B|a\rangle\propto |a\rangle$ and there is no degeneracy!

Therefore, if $|a\rangle$ is not an eigenvector of $B$ we find that $a$ is degenerate, with at least two eigenvectors, $|a\rangle$ and $B|a\rangle$. If $|a\rangle$ is an eigenvector of $B$ we can't conclude anything about the degeneracy of $a$.

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In the case $B=\mathbb I$, we obviously have $[H,B]=0$, but as any vector is an eigenvector of $\mathbb I$, the fact that these operators commute gives no information about energy-degeneracies.

In the case of a free particle, the parity operators acts on kets by changing the sign, $P|\boldsymbol p\rangle=|-\boldsymbol p\rangle$, which is not proportional to $|\boldsymbol p\rangle$ (a.e.), and therefore we do have the two-fold degeneracy you said.