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intermediate · Physics · The Hydrogen Atom in Depth

Degeneracy and Quantum Numbers

Several distinct quantum states of hydrogen share the same energy. Counting them — the degeneracy of each level — explains the structure of the periodic table and the capacities of electron shells.

The three spatial quantum numbers

A bound state is specified by three integers:

The constraint n1\ell \le n-1 comes from the node count nr=n10n_r = n - \ell - 1 \ge 0, and m|m| \le \ell comes from the spherical harmonics.

Counting orbital states

For a fixed \ell there are 2+12\ell + 1 values of mm. Summing over all allowed \ell at a given nn:

=0n1(2+1)=n2.\sum_{\ell=0}^{n-1}(2\ell + 1) = n^2.

So the level nn has an orbital degeneracy of n2n^2. The n=1n=1 level has 11 orbital state, n=2n=2 has 44 (2s2s plus three 2p2p), n=3n=3 has 99, and so on. The energy is independent of \ell (an "accidental" Coulomb degeneracy) and independent of mm (a consequence of rotational symmetry, true for any central potential).

Adding spin

The electron also carries spin s=12s = \tfrac{1}{2}, with two states ms=±12m_s = \pm\tfrac{1}{2}. Each spatial orbital therefore holds two electrons, doubling the count. The total degeneracy of level nn is

gn=2n2.g_n = 2n^2.

This is exactly the pattern of the periodic table: the KK, LL, MM shells hold 2,8,182, 8, 18 electrons — that is 2122 \cdot 1^2, 2222 \cdot 2^2, 2322 \cdot 3^2.

The takeaway

Hydrogen states are labelled by (n,,m)(n, \ell, m) with 0n10 \le \ell \le n-1 and m|m| \le \ell. The orbital degeneracy of level nn is (2+1)=n2\sum_\ell (2\ell+1) = n^2, and including the electron's two spin states gives a total degeneracy gn=2n2g_n = 2n^2.

Try it

This is a numerical exercise — return a number. Count the total number of states (including spin) for the n=3n = 3 hydrogen level by summing 2+12\ell+1 over the allowed \ell and then multiplying by the two spin states.

Run your code to see the quantum state.

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