Oscillator Strengths

The name is a fossil, and a useful one. Before quantum mechanics, Lorentz modeled each absorption line as an electron on a spring resonating at that line's frequency — and the model fits dispersion data beautifully, except that each line behaved like a fraction of an electron. Spectroscopists in the 1920s bookkept this with a fudge factor $f$ per line: this line responds like 0.44 of a classical oscillating electron, that one like 0.01.

Quantum mechanics then explained the fudge factor instead of discarding it: a transition with energy $\omega$ and transition moment $⟨f|\hat{\mu }|i⟩$ responds to light exactly like $f$ classical oscillators, with $f$ computable from the matrix element. That heritage is why $f$ is dimensionless and why it is the right currency for comparing transitions across molecules: an $f$ of 1 means "as bright as one whole classical electron," whether the molecule is H₂ or a dye with 300 atoms.

In atomic units, for a transition from $|i⟩$ up to $|f⟩$:

Each factor has a job. The squared transition moment is the coupling to light — the antenna strength. The factor $\omega$ weights fast oscillations more (a classical oscillator at higher frequency carries more energy for the same amplitude). The 2/3 averages a 3D vector moment over the orientations of a randomly tumbling molecule — in a strict 1D world the formula is $f=2\omega |⟨f|x|i⟩{|}^2$ with no 3.

Smallest case, by hand: the harmonic oscillator with $m=\omega =\hslash =1$. The only allowed absorption from the ground state is 0 → 1, with ${\omega }_{10}=1$ and $|⟨0|x|1⟩{|}^2=1/2$, so $f=2\cdot 1\cdot \frac{1}{2}=1$. One electron, one line, strength exactly one — the harmonic oscillator is the classical oscillator, and the definition is calibrated so it says so. The grid check below reproduces this to seven digits.

The Thomas-Reiche-Kuhn sum rule (1925 — it predates the Schrödinger equation) says the strengths from any state sum to the number of electrons:

The proof is two lines of commutator algebra: evaluate $⟨i|[\hat{x},[\hat{H},\hat{x}]]|i⟩$ once directly — for any potential that depends only on position the double commutator is a constant, $1/m$ per electron — and once by inserting a complete set of states, which produces exactly the sum of $f$-values. No details of the potential survive. Hydrogen, water, a protein: the total brightness is fixed the moment you count electrons.

So oscillator strength is a budget. A molecule never creates or destroys brightness; it only distributes it among its transitions. The harmonic oscillator posts the entire budget on one line ($f=1$, everything else zero — forced, since only 0 → 1 is allowed and the total must be 1). The particle in a box spreads it: 0.9607 on the first line, 0.0307 on the next allowed one, a thinning tail above. When a dye's absorption grows, some other transition of that dye got dimmer. And when a computed sum comes out at 3.95 for a 10-electron molecule — see below — the missing 6 is a diagnosis, not a rounding error.

In Casida's equation you diagonalize once and get every excitation of the molecule — but the eigenvector $(X,Y)$ of state $n$ is a list of occupied→virtual amplitudes, not yet a spectrum. The bridge is one contraction: assemble the one-particle dipole integrals $⟨{\varphi }_i|\mathbf{r}|{\varphi }_a⟩$ in the MO basis, dot them with $X+Y$, and you have the many-electron transition dipole,

(The $\sqrt{2}$ is spin adaptation for a closed shell; the response combination is $X+Y$ because the dipole perturbation drives excitation and de-excitation coherently.) That's the entire distance from "diagonalized a matrix" to "predicted a UV-vis spectrum": five lines of einsum, run against water below and validated against PySCF to six digits.

Some states have $f=0$ exactly — not small, zero — because the transition density has a symmetry that no component of the dipole vector can match. Water's second excited state (11.29 eV in the calculation below) is the textbook case: its transition density transforms as $A_2$ in the C_2v point group, and none of $x,y,z$ does, so every component of the transition moment vanishes identically. The state is there — the eigenvalue is real, the wavefunction is real — but one-photon absorption cannot reach it. It is dark.

Dark states are not a curiosity; they run photochemistry. A molecule pumped to a bright state often slides into a nearby dark state and parks there, invisible to your spectrometer, doing chemistry — fluorescence quenching, photoprotection in DNA and sunscreen, the slow phosphorescence of triplets (dark for the stronger reason that a spin flip is required). An absorption spectrum shows what couples to light, not what exists. The $f$-column of a Casida calculation is often the fastest symmetry assignment you can get: zeros label the representations.