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

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

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.

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.

• Degeneracy can also imply topological order 0:) Then again that in its turn can be linked to gauge symmetries... – Ruben Verresen Mar 16 '17 at 22:15
• Thank you for your answer. I have to ask. Firstly, when you say that Hilbert space is a representation of group what do you mean under it? I thought it should be matrices in Hilbert space. Secondly, sorry for my ignorance but I do not understand why If group is abelian then there is no degeneracy. Could you give some hint? – LRDPRDX Mar 17 '17 at 7:30
• Mathematically, a representation is the vector space, but sometimes in physics we say that the operators that act on this vector space are "the representation". In this answer, I mean the first. – knzhou Mar 17 '17 at 15:17
• If the group is abelian you can diagonalize all the operators at once. That means that acting with a symmetry operator in this basis never gives you another state, so you get no degeneracy. It works just like your example of parity. – knzhou Mar 17 '17 at 15:18
• @Wolfgang whoops, forgot to tag you. – knzhou Mar 17 '17 at 15:18

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.