In a general central potential the energy depends on $n_{r}$ and $\ell$ separately. In hydrogen it depends only on their sum: 2s and 2p have exactly the same energy, and so do 3s, 3p and 3d. Rotations explain why the $2\ell+1$ values of $m$ share a level; they do not explain why different $\ell$ do. Hydrogen has an extra conserved quantity, the Runge–Lenz vector, the quantum version of the fact that a planet’s orbit does not precess, and its hidden symmetry makes each level hold $n^{2}$ states.

Keplerian orbits of the same energy all have the same long axis, whether round or thin: the thin and round ones are equally cheap. Hydrogen’s $\ell$ values are its round and thin orbits, and they cost the same.

Hydrogen’s levels for $n=1$ to 4 in columns by $\ell$, in units of 13.6 eV. Levels of the same $n$ sit at exactly the same height across the columns: 2s with 2p, 3s with 3p and 3d. In a general central potential they would all differ.

For each $n$, $\ell$ runs from 0 to $n-1$ and $m$ from $-\ell$ to $$, all with the same energy. The count is — twice that when the electron’s spin is included, as a later chapter does.

• The $2\ell+1$ values of $m$ are forced by rotations, as in every central potential.
• The equality of different $\ell$ is forced by a further conserved vector, the Runge–Lenz vector $\hat{\mathbf{A}}$, which commutes with but not with .

Each $\ell$ contributes its $2\ell+1$ values of $m$. Adding over $\ell$ from 0 to $n-1$ is adding the first odd numbers, which make a square.

All $m$ for each allowed $\ell$ share the energy $E_{n}$.

Split the sum.

The sum of the first $n-1$ whole numbers.

The first $n$ odd numbers add to $n^{2}$.

Level $n$ of hydrogen holds every $\ell$ from 0 to $n-1$, each with its $2\ell+1$ values of $m$: states in all, leaving spin aside.

In hydrogen the energy depends on $n$ alone, so 2s and 2p, or 3s, 3p and 3d, have exactly the same energy. In a general central potential they would differ.

Rotations force the $2\ell+1$ values of $m$ to share an energy. The sharing between different $\ell$ needs a further symmetry, carried by the Runge–Lenz vector.

The first $n$ odd numbers add to $n^{2}$, so no sum needs doing: level $n$ holds $n^{2}$ states.

Great work! You now know how to:

• count the $n^{2}$ states of a hydrogen level
• separate the degeneracy rotations explain from the one they do not
• name the hidden symmetry behind it