Degeneracy of Levels

Energy depends only on n

In earlier lessons we found that the hydrogen atom has three quantum numbers: $n$ , $\ell$ , and $m_\ell$ . Each one restricts the allowed wavefunctions differently, yet the energy formula

$E_n = -\frac{13.6\,\text{eV}}{n^2}$

contains only $n$ . Two states with the same $n$ but different $\ell$ or $m_\ell$ have identical energies — they are called degenerate states.

Counting the states at shell n

The allowed values of the quantum numbers are:

$n = 1, 2, 3, \ldots$
$\ell = 0, 1, 2, \ldots, n-1$ (so there are $n$ distinct values of $\ell$ )
$m_\ell = -\ell, -\ell+1, \ldots, 0, \ldots, \ell-1, \ell$ (so there are $2\ell + 1$ values for each $\ell$ )

For a fixed $n$ , the total number of states is obtained by summing $2\ell + 1$ over all allowed $\ell$ :

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

This is the sum of the first $n$ odd integers. Since odd integers $1, 3, 5, \ldots$ satisfy $\sum_{k=0}^{n-1}(2k+1) = n^2$ (a standard identity), the result is

$\boxed{d(n) = n^2.}$

Worked example: n = 3

Let us verify the formula explicitly by listing every $(n, \ell, m_\ell)$ triplet at $n = 3$ :

| $\ell$ | $m_\ell$ values | States | |--------|-----------------|--------| | 0 | 0 | 1 | | 1 | $-1, 0, +1$ | 3 | | 2 | $-2, -1, 0, +1, +2$ | 5 |

Total: $1 + 3 + 5 = 9 = 3^2$ . All nine states sit at exactly the same energy, $E_3 = -13.6/9\,\text{eV} \approx -1.51\,\text{eV}$ .

A table for small n

| $n$ | Subshells ( $\ell$ ) | States per subshell ( $2\ell+1$ ) | Total $d(n) = n^2$ | |-----|-------------------|----------------------------------|---------------------| | 1 | $s$ ( $\ell=0$ ) | 1 | 1 | | 2 | $s$ , $p$ | 1, 3 | 4 | | 3 | $s$ , $p$ , $d$ | 1, 3, 5 | 9 | | 4 | $s$ , $p$ , $d$ , $f$ | 1, 3, 5, 7 | 16 |

The growing degeneracy explains why more electrons can be accommodated in higher shells — each shell has room for $n^2$ distinct spatial states (and $2n^2$ when electron spin is included).

Try it

This is a numerical exercise — return a number. Count all degenerate quantum states for $n = 3$ in hydrogen by summing $2\ell + 1$ over $\ell = 0, 1, 2$ .

Run your code to see the quantum state.