Energy levels, degeneracy, and quantum numbers

The derivation of the last section bore a whole family tree of states: each state is labelled by three integers $(n,l,m)$, and the energy acknowledges only the first. In this section we count that tree, read it properly, and then hook it up to one more thing — the real spectrum. After all, this whole subject began with a few spectral lines (section 1.4); now it is our turn not only to compute where the lines are, but to answer the question Bohr could not: why some transitions never light up.

The division of labour among the three quantum numbers

First lay out the rules. Each bound state $\psi_{nlm}$ we solved for is labelled by three quantum numbers:

Quantum number	Name	Values	Determines
$n$	principal	$1,2,3,\dots$	energy $E_n=-13.6\ \text{eV}/n^2$; size $\sim n^2a_0$
$l$	orbital (azimuthal)	$0,1,\dots,n-1$	magnitude of angular momentum $\lvert\vec L\rvert=\sqrt{l(l+1)}\,\hbar$; shape of the radial distribution
$m$	magnetic	$-l,\dots,+l$	$z$-component $L_z=m\hbar$; orientation of the angular distribution

Spectroscopic tradition gives $l$ letter aliases: $l=0,1,2,3$ are called s, p, d, f (from nineteenth-century descriptions of how the lines looked — sharp, principal, diffuse, fundamental; the names are older than the theory). So "$n=2,l=1$" is written 2p, and "$n=3,l=2$" is 3d. The notation you memorised in chemistry class gets its birth certificate here.

The electron also has spin $s=1/2$ (section 5.3), with two orientations $m_s=\pm1/2$. The Coulomb Hamiltonian contains no spin, so in this section we do the books “spinless” first and multiply by 2 at the end.

Counting the degeneracy: exactly n² states per level

Degeneracy = n² in three linesbasic~4 minexpand

The energy depends only on $n$. Fix $n$: $l$ runs from $0$ to $n-1$, and each $l$ offers $2l+1$ values of $m$. Total:

$$d(n)=\sum_{l=0}^{n-1}(2l+1) =2\cdot\frac{(n-1)n}{2}+n=n^2\tag{6.4.1}$$

Roll call, level by level: $n=1$ has only 1s — one state; $n=2$ has 2s plus three orientations of 2p, $1+3=4$; $n=3$ has 3s, 3p, 3d, $1+3+5=9$. Double for spin: $2n^2=2,8,18,\dots$ — the capacity sequence of the periodic table’s rows shows its face here. (Why the electrons don’t all pile into 1s must wait for the Pauli principle in Chapter 8.)

This $n^2$-fold degeneracy splits into two accounts of entirely different character:

Connecting to the spectrum: selection rules

We have the level diagram — so the spectral lines are just the level differences… right? Not quite. If any two levels could interconvert, the spectrum would be far denser than what is observed. Experimentally, a great many lines that “ought” to exist simply never appear.

The rule is: electric-dipole radiation (the dominant mechanism by which atoms emit light) requires

$$\Delta l=\pm1,\qquad \Delta m=0,\pm1\tag{6.4.3}$$

with no restriction on $n$.

Where selection rules come from: the photon carries off one unit of angular momentumadvanced~6 minexpand

The transition rate is proportional to the matrix element $\lvert\braket{\psi_{n'l'm'}}{\hat{\vec r}\,\psi_{nlm}}\rvert^2$ (rigorous derivation in Chapter 7’s perturbation theory; here we only ask when it vanishes).

The parity argument. Under spatial inversion $\vec r\to-\vec r$ the spherical harmonics transform as $Y_l^m\to(-1)^lY_l^m$, so $\psi_{nlm}$ has parity $(-1)^l$. The operator $\hat{\vec r}$ has odd parity. If $l'=l$, the integrand as a whole is odd and the integral vanishes — $\Delta l=0$ is forbidden.

The angular-momentum argument. The three components of the position operator $\hat{\vec r}$ can be recombined into the form $rY_1^m$ — it “carries $l=1$” all by itself. The matrix element is a coupling integral of three angular-momentum objects ($l'$, $1$, $l$); by the addition rules of section 5.6, it is non-zero only if $l'$ lies between $\lvert l-1\rvert$ and $l+1$. Combined with parity ruling out $l'=l$, only $\Delta l=\pm1$ survives.

In physical language: the photon carries one unit of angular momentum, so when an atom emits one, its own $l$ must change by exactly 1 to balance the books. The $\Delta m$ rule is the same statement for the $z$-component.

One corollary: a 2s state wanting to fall to 1s finds the single-photon channel blocked by $\Delta l=0$; it can only take the far slower two-photon route — a lifetime of 0.12 seconds, eight orders of magnitude longer than an ordinary excited state (nanoseconds). Such “metastable states” are important players both in nebular spectra and in precision measurements of hydrogen.

The spectral series: an experimental photograph of the level diagram

Apply the selection rules to the level diagram, group transitions by their landing level $n_1$, and each group is a spectral series:

Series	Landing level	Wavelength range	Band	Notes
Lyman	$n\to1$	91.2–121.6 nm	vacuum UV	Ly$\alpha$ 121.6 nm is the brightest hydrogen line in the universe
Balmer	$n\to2$	364.6–656.3 nm	visible	$\mathrm H\alpha$ 656 nm red, $\mathrm H\beta$ 486 nm blue-green
Paschen	$n\to3$	820.4–1875 nm	infrared
Brackett	$n\to4$	1458–4051 nm	infrared

The series limit at each series’ short-wavelength end (the $n\to\infty$ limiting wavelength, e.g. 91.2 nm for Lyman) corresponds to an electron dropping from the ionisation threshold to the landing level; at wavelengths shorter than the limit the spectrum becomes a continuous band — free electrons being captured, their initial energies continuous. The observed positions of the series limits are a direct test of the energy-level formula.

∎Key formulas

• Quantum-number rules

  $$n=1,2,\dots;\quad l=0,\dots,n-1;\quad m=-l,\dots,+l$$

  n runs the energy, l the angular-momentum magnitude, m the orientation

• Degeneracy

  $$d(n)=\sum_{l=0}^{n-1}(2l+1)=n^2\ \ (\text{with spin}\ 2n^2)$$

  2n² = 2, 8, 18, … the embryo of the periodic table's row capacities

• Runge–Lenz vector

  $$\hat{\vec A}=\frac{1}{2\mu}(\hat{\vec p}\times\hat{\vec L}-\hat{\vec L}\times\hat{\vec p})-\frac{e^2}{4\pi\varepsilon_0}\frac{\hat{\vec r}}{r}$$

  Conserved only for the 1/r potential; the root of the l (accidental) degeneracy

• Selection rules

  $$\Delta l=\pm1,\quad\Delta m=0,\pm1$$

  The photon carries one unit of angular momentum; Δn unrestricted

• Rydberg formula

  $$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),\quad R_\infty=1.0974\times10^7\ \text{m}^{-1}$$

  Bohr's formula stands unchanged — but now every term has a pedigree

?Self-check4 questions

1. 1.

   What is the degeneracy of the hydrogen n = 4 level (ignoring spin), and which part of it is "accidental"?

   • 16-fold; that different l (4s, 4p, 4d, 4f) share the same energy at all is the accidental degeneracy peculiar to the Coulomb potential
   • 16-fold; different m sharing the same energy is the accidental degeneracy
   • 8-fold; spin up and spin down sharing an energy is the accidental degeneracy
   • 4-fold; there is no accidental component

   Submit
2. 2.

   Why is the lifetime of the 2s state (about 0.12 s) eight orders of magnitude longer than that of 2p (about 1.6 ns)?

   • 2s→1s is a Δl = 0 transition, forbidden by the electric-dipole selection rule; it can only take the much slower two-photon channel
   • The 2s energy is much lower than 2p
   • The 2s electron is too far from the nucleus
   • The 2s state has no such thing as a transition matrix element

   Submit
3. 3.

   Which statements about the hydrogen spectral series are correct? (Select all that apply.)

   Select all that apply

   • The Balmer series is the family of transitions landing on n = 2, and happens to fall in the visible band
   • Each series has a series limit at its short-wavelength end, corresponding to a drop from the ionisation threshold
   • Beyond the series limit (shorter wavelengths) the spectrum is continuous, because free-electron energies are continuous
   • The Lyman series lies in the infrared

   Submit
4. 4.

   Compute the wavelength of the Lyman α line (n = 2 → 1) in nm. (E₁ = −13.606 eV, hc = 1240 eV·nm)

   nm100% relative tolerance

   Submit

What comes next

Levels, degeneracies, spectral lines — all in hand. One thing remains unseen: what the wavefunctions themselves look like. What exactly are those spherical, dumbbell, and cloverleaf “electron cloud” pictures in chemistry books depicting? Is the place where $\lvert\psi\rvert^2$ peaks the radius where the electron is most likely to be found? (Hint: no — and almost everyone falls for it.) The next section is devoted to teaching you how to “read” an orbital correctly.