What is the relationship between symmetry and degeneracy in quantum mechanics?

Question

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:

1. Existence of symmetry $\Rightarrow$ Existence of degeneracy
2. Existence of degeneracy $\Rightarrow$ Existence of symmetry

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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.

Answer 1

This material seems to be poorly covered in most introductory QM books, so here's the logic:

• Suppose there is a group of transformations $G$. Then it acts on the Hilbert space by some set of unitary transformations $\mathcal{O}$.
• The Hilbert space is therefore a representation of the group $G$, and it splits up into subspaces of irreducible representations (irreps). The important thing is that if $|\psi\rangle$ and $|\phi \rangle$ are in the same irrep iff you can get from one to the other by applying operators $\mathcal{O}$.
• If the transformations are symmetries of the Hamiltonian, then the operators $\mathcal{O}$ commute with the Hamiltonian. Then if $|\psi\rangle$ is an energy eigenstate, then $\mathcal{O}|\psi \rangle$ is an energy eigenstate with the same energy.
• Therefore, all states in an irrep have the same energy. So if there are nontrivial irreps of dimension greater than one present, then there will be degenerate states.
• Conversely, if there is any degeneracy at all, we typically think of it as being caused by some symmetry, which may be well hidden. Ideally, there should be no 'accidental' degeneracy.
• If $G$ is an abelian group, then all irreps are one-dimensional, and hence yield no degeneracy.

Below are some examples.

• Particle in a 1D symmetric potential. The group is $\mathbb{Z}_2$ and it is generated by the parity operator. The group is abelian, so there's no degeneracy.
• The free particle in 1D. There are two symmetries: translational symmetry and parity symmetry. As a result, the group is not abelian and can have nontrivial irreps. There are irreps of dimension two, and these correspond to the degeneracy of the plane wave states $e^{\pm ikx}$.
• A particle in 1D with $H = p^3$. There's no degeneracy; the argument for the free particle fails because we don't have parity symmetry, only translational symmetry. This shows that a continuous symmetry (translations) doesn't guarantee degeneracy. It does guarantee a conserved quantity (here, momentum), but that's a different issue.
• Particle in a 3D central potential. The group is $SU(2)$, which is nonabelian. The degenerate sets of states $\{l, m\}_{-l \leq m \leq l}$ are just the irreps of $SU(2)$.
• Hydrogen atom. There is an additional degeneracy between states with the same $n$ but different $l$ quantum numbers. This comes from a hidden $SO(4)$ symmetry of the Hamiltonian.

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.

Answer 2

knzhou's answer is very well-explained, but it's perhaps worth mentioning that the energy gaps between different symmetry sectors typically decrease with system size, and formally vanish in the thermodynamic limit. So an infinite-size system can indeed (but doesn't have to) have symmetry-induced degeneracy, even if the symmetry is abelian (regardless of whether the symmetry is discrete or continuous - e.g. the quantum transverse Ising model, which has $\mathbb{Z}_2$ symmetry, has twofold ground-state degeneracy in the thermodynamic limit, and the $X$-$Y$ model, which has $U(1)$ symmetry, has infinite GS degeneracy). If there is a symmetry-induced degenerate GS manifold in the thermodynamic limit, then the symmetry is typically broken: the physically realistic ground states are not invariant under the symmetry.

Also, even in absence of symmetry, an infinitely large system in a topologically ordered phase can have a finite degeneracy. This degeneracy is extremely robust because unlike in the symmetry-induced case, no possible perturbation can lift it.