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.


2 Answers 2


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.

  • $\begingroup$ Degeneracy can also imply topological order 0:) Then again that in its turn can be linked to gauge symmetries... $\endgroup$ Commented Mar 16, 2017 at 22:15
  • $\begingroup$ 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? $\endgroup$
    Commented Mar 17, 2017 at 7:30
  • $\begingroup$ 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. $\endgroup$
    – knzhou
    Commented Mar 17, 2017 at 15:17
  • $\begingroup$ 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. $\endgroup$
    – knzhou
    Commented Mar 17, 2017 at 15:18
  • $\begingroup$ @knzhou, When you say diagonalize all the operators at once you mean reduce them to $\lambda I$, when $I$ is identity and $\lambda$ is a number (this number depends on operator)? $\endgroup$
    Commented Mar 18, 2017 at 14:05

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.


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