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.
Best Answer
This material seems to be poorly covered in most introductory QM books, so here's the logic:
Below are some examples.
In summary, your second point is true (generally, degeneracy implies symmetry), but your first point is false. Continuous symmetries guarantee you get conserved quantities, not degeneracy.