Solving the Schrödinger equation for hydrogen in three dimensions reproduces Bohr's energies, but each energy level now holds several different states. Each state is labeled by a set of quantum numbers. Like the whole numbers in a particle-in-a-box, they come from the boundary conditions, now around a sphere.
Key idea
- $n = 1, 2, 3, \ldots$ (principal: sets the energy $-13.6\ \text{eV}/n^2$)
- $l = 0, 1, \ldots, n - 1$ (orbital angular momentum, $L = \sqrt{l(l + 1)}\,\hbar$; letters s, p, d, f for $l = 0, 1, 2, 3$)
- $m_l = -l, \ldots, +l$ (the component $L_z = m_l\hbar$)
- $m_s = \pm\frac{1}{2}$ (spin up or down)
Why it works
| $n$ | allowed $l$ | states |
|---|---|---|
| 1 | 0 (1s) | 2 |
| 2 | 0, 1 (2s, 2p) | 2 + 6 = 8 |
| 3 | 0, 1, 2 (3s, 3p, 3d) | 2 + 6 + 10 = 18 |
| 4 | 0, 1, 2, 3 | 32 |
Worked examples
Worked example: Count the n = 3 states
- $l = 0, 1, 2$ (3s, 3p, 3d).
- $m_l$ values: 1, 3 and 5. Times 2 spins: 2 + 6 + 10 = 18.
Answer: 18 states, which is $2n^2$
Worked example: The size of L
- d means $l = 2$: $L = \sqrt{2 \cdot 3}\,\hbar = \sqrt{6}\,\hbar = 2.45\hbar$.
- Smallest angle when $m_l = 2$: $\cos\theta = \frac{2}{\sqrt{6}} = 0.816$, so $\theta = 35.3^\circ$.
Answer: $L = 2.45\hbar$; the angle is never less than $35.3^\circ$
Worked example: Spot the impossible set
- $(2, 2, 0, +\frac{1}{2})$ has $l = n$, but $l$ can be at most $n - 1 = 1$.
- The others satisfy $l \le n - 1$ and $|m_l| \le l$.
Answer: $(2, 2, 0, +\frac{1}{2})$
States with the same $n$ and $l$ but different $m_l$ have the same energy, until you switch on a magnetic field. Then each $m_l$ shifts a little, because the electron's orbit acts like a tiny current loop, and a single spectral line splits into several. This Zeeman effect, seen in 1896, was the first evidence for $m_l$; astronomers use it today to measure the magnetic fields in sunspots.
Common mistake