Let me remind you about the following classical examples in quantum mechanics.
Example 1. Bound states in 1-dim potential V(x).
Let $V(x)$ be a symmetric potential i.e. $$V(x) = V(-x)$$ Let us introduce the parity operator $\hat\Pi$ in the following way: $$\hat\Pi f(x) = f(-x).$$ It is obvious that $$[\hat H,\hat\Pi] = 0.$$ Therefore, for any eigenfunction of $\hat H$ we have: $$\hat H|\psi_E(x)\rangle = E|\psi_E(x)\rangle = E\hat\Pi|\psi_E(x)\rangle,$$ i.e. state $\hat\Pi|\psi_E(x)\rangle$ is eigenfunction with the same eigenvalue. Is $E$ a degenerate level? No, because of linear dependence of $|\psi\rangle$ and $\hat\Pi|\psi\rangle.$
Consider the second example.
Example 2. Bound states in 3-dim a potential $V(r)$. Where $V(r)$ possesses central symmetry, i.e. depends only on distance to center.
In that potential we can choose eigenfunction of angular momentum $\hat L^2$ for basis $$|l,m\rangle,$$ where $l$ is total angular momentum and $m$ - its projection on chosen axis (usually $z$). Because of isotropy eigenfunction with different $m$ but the same $l$ correspond to one energy level and linearly independent. Therefore, $E_l$ is a degenerate level.
My question is if there is some connection between symmetries and degeneracy of energy levels. Two cases are possible at the first sight:
- Existence of symmetry $\Rightarrow$ Existence of degeneracy
- Existence of degeneracy $\Rightarrow$ Existence of symmetry
It seems like the first case is not always fulfilled as shown in the first example. I think case 1 may be fulfilled if there is continuous symmetry. I think the second case is always true.