A symmetry carries each energy eigenstate to another state of the same energy. If the copy is a genuinely different state, the level is degenerate, and the symmetry has forced it. The cleanest version of the argument: two symmetries that do not commute with each other cannot both act as mere numbers on every level, so some level must hold several states. Break the symmetry and the degenerate level splits; the pattern of the splitting tells which symmetry was lost.

A square drum has pairs of vibrations with the same pitch — a pattern and the same pattern turned by a right angle. Stretch the drum into a rectangle and the pairs split into two slightly different notes.

On the left, a level of a rotationally symmetric system holding three states of one energy. On the right, the same states after a perturbation that keeps only rotations about one axis: nothing now forces them to share an energy, and the level splits into three.

If $[\hat{U},\hat{H}]=0$, then $\hat{U}$ maps each eigenspace of $\hat{H}$ to itself. A level whose states are not all mapped to multiples of themselves must hold more than one state; two symmetries that do not commute guarantee such levels.

• A free particle on a line: translations and reflection do not commute, and each energy $\hbar^{2}k^{2}/2m$ holds $e^{ikx}$ and .

Suppose every level held a single state. Each symmetry maps a state of energy $E$ to a state of energy $E$, which must then be a multiple of it: both symmetries act on every eigenstate as numbers. Numbers commute, so $\hat{A}$ and $\hat{B}$ would commute on a basis, and so everywhere — a contradiction.

Argue by contradiction.

Each maps $\ket{E}$ to a state of energy $E$, which can only be a multiple of $\ket{E}$.

On each eigenstate both act as numbers, and numbers commute.

The eigenstates of $\hat{H}$ form a basis.

This contradicts $[\hat{A},\hat{B}]\neq0$.

A symmetry that commutes with the Hamiltonian carries each energy eigenstate to another eigenstate of the same energy. When the copy is a different state, the level is degenerate.

If two symmetries of $\hat{H}$ do not commute with each other, some level must be degenerate: on non-degenerate levels both would act as numbers, and numbers commute.

For a free particle, translations and reflection both commute with the Hamiltonian but not with each other. Each energy is shared by $e^{ikx}$ and $e^{-ikx}$.

A perturbation that breaks a symmetry can split a degenerate level into several. The pattern of the splitting tells which symmetry was broken.

Great work! You now know how to:

• explain how a symmetry carries states within a level
• prove that two clashing symmetries force a degeneracy
• read a splitting as the breaking of a symmetry