Fermi's golden rule

Remarks

First-order time-dependent perturbation theory gives the amplitude $-\frac{i}{\hbar}\int_0^t e^{i\omega s} \ip{\psi_f}{V\psi_i}\, ds$ with $\omega = (E_f - E_i)/\hbar$. Hence the transition probability to one final state is $$\abs{\ip{\psi_f}{V\psi_i}}^2 \, \frac{4 \sin^2(\omega t/2)}{\hbar^2 \omega^2}.$$ As a function of $\omega$, the factor $4\sin^2(\omega t/2)/\omega^2$ has integral $2\pi t$ and concentrates near $\omega = 0$ as $t$ grows, so it behaves like $2\pi t\, \delta(\omega) = 2\pi\hbar t\, \delta(E_f - E_i)$. Summing over final states with the density $\rho$ gives a probability $\Gamma t$, which is the rule.

Example: the rule gives the rate at which an atom absorbs light from a broadband beam, and the rate of decay of a discrete state coupled to a continuum. Without a continuum of final states the rule does not apply. For a single discrete final state the probability oscillates in time instead of growing linearly, as in a two-state system.

Going deeper

For a student

The rule comes from the first term of the Dyson series in the interaction picture. It is an approximation, valid for weak coupling and intermediate times, not an exact theorem; a rigorous version is obtained in the weak-coupling (van Hove) limit. Despite its name, the formula goes back to Dirac; Fermi called it a golden rule. Its validity requires a smooth density of final states; near thresholds or band edges corrections are needed.