Fermi's golden rule

The last section left a crack. For a single final state, first-order theory gives a probability that either grows madly as $t^2$ (on resonance) or oscillates back and forth (off resonance). Yet excited atoms in the laboratory wear an entirely different face: a population of hydrogen atoms in the 2p state loses a fixed fraction every nanosecond, its number sliding smoothly down as $\ee^{-t/\tau}$ with $\tau=1.6\ \text{ns}$ — no oscillation, no acceleration. “A fixed fraction per unit time” means there exists a constant transition rate $\Gamma$. Where does this constant rate spring from?

The difference hides in the nature of the final states. Last section we tacitly assumed the transition’s destination was one discrete state; but when an atom emits light, the destination is “the atom drops to 1s plus a photon flies away” — and the photon can fly in any direction, carry either of two polarisations, and take any frequency within the resonance width. The final state is not a point; it is a continuum. When we measure “has the atom decayed yet?”, we do not care which way the photon went, so the probability must be summed over the whole continuum. It is this summation that flattens the oscillations into a rate.

Counting the final states: the density of states

To sum over a continuum, we first need to answer “how many states are there per unit energy interval near $E$?” That quantity is the density of states, written $\rho(E)$: there are $\rho(E)\,\dd E$ states with energy between $E$ and $E+\dd E$. It is a pure counting problem, with no dynamics in it.

The most common example: a particle (or photon) placed in a large box of volume $V=L^3$, with momentum quantised onto the lattice $\vec k=\frac{2\pi}{L}(n_x,n_y,n_z)$. Each state occupies a volume $(2\pi/L)^3$ of $k$-space; count the lattice points in a spherical shell of radius $k$, convert to energy, and out comes $\rho(E)$. For photons ($E=\hbar ck$, two polarisations):

$$\rho_{\text{photon}}(E)=\frac{V E^2}{\pi^2\hbar^3c^3}\tag{7.6.1}$$

The box is only scaffolding: in any physical answer, $V$ always cancels against the $1/V$ normalisation factor in the matrix element, and the final rate is independent of the box size.

From sinc² to a rate

Deriving the golden rulebasic~10 minexpand

Step 1: sum the single-state probability over the continuum of final states.

Take the cleanest case first: a perturbation $\hat V$ switched on suddenly at $t=0$ and held constant thereafter (the periodic perturbation $\hat V\cos\omega t$, after the rotating-wave approximation, is completely isomorphic — just replace $V_{fi}\to V_{fi}/2$ and the detuning by $\Delta=\omega_{fi}-\omega$). The last section’s first-order amplitude integral gives (writing $\Delta\equiv\omega_{fi}$):

$$P_{i\to f}(t)=\frac{|V_{fi}|^2}{\hbar^2}\,t^2\, \mathrm{sinc}^2\!\left(\frac{\Delta t}{2}\right), \qquad \mathrm{sinc}\,x\equiv\frac{\sin x}{x}\tag{7.6.2}$$

The total transition probability is the sum over all final states; use the density of states to write it as an integral ($\dd E_f=\hbar\,\dd\Delta$):

$$P(t)=\int P_{i\to f}(t)\,\rho(E_f)\,\dd E_f =\frac{t^2}{\hbar}\int|V_{fi}|^2\rho\, \mathrm{sinc}^2\!\left(\frac{\Delta t}{2}\right)\dd\Delta\tag{7.6.3}$$

Step 2: see what sinc² looks like at long times.

As a function of $\Delta$, $\mathrm{sinc}^2(\Delta t/2)$ has a main peak of height 1 and width about $4\pi/t$. The larger $t$, the narrower the peak — and once it is narrower than any scale over which $|V_{fi}|^2\rho$ varies with energy, the factor $|V_{fi}|^2\rho$ is essentially constant across the peak and can be pulled out at its central value:

$$P(t)\approx\frac{t^2}{\hbar}\,|V_{fi}|^2\rho\,\Big|_{\Delta=0} \int_{-\infty}^{\infty}\mathrm{sinc}^2\!\left(\frac{\Delta t}{2}\right)\dd\Delta\tag{7.6.4}$$

The remaining integral is the standard result $\int\mathrm{sinc}^2u\,\dd u=\pi$, which after the change of variable equals $2\pi/t$.

Step 3: the miracle.

$$P(t)\approx\frac{t^2}{\hbar}\cdot\frac{2\pi}{t}\cdot|V_{fi}|^2\rho =\frac{2\pi}{\hbar}\,|V_{fi}|^2\rho\;t\tag{7.6.5}$$

The peak width’s $1/t$ shaves one power off the $t^2$: the total probability grows linearly in time. Linear growth is the very definition of “constant rate”, so

$$\boxed{\ \Gamma=\frac{2\pi}{\hbar}\,|V_{fi}|^2\,\rho(E_f)\Big|_{E_f=E_i+\hbar\omega}\ }\tag{7.6.6}$$

This is Fermi’s golden rule (it was actually Dirac who derived it; Fermi called it “golden rule number two” in his lecture notes, and the name stuck — yet another miscarriage of justice in the history of physics naming).

What happened, retold in pictures: the probability for each single final state oscillates, but final states at different detunings oscillate with different periods. In the sum, the states sitting exactly on the resonance shell grow steadily, while those slightly off rise and fall out of phase with one another — summing over the continuum averages the oscillations away, leaving only the net outflow through the resonance shell. Mathematically this is the limit

$$\lim_{t\to\infty}\frac{t}{2\pi\hbar}\,\mathrm{sinc}^2\!\left(\frac{(E_f-E_i-\hbar\omega)t}{2\hbar}\right) =\delta(E_f-E_i-\hbar\omega)\tag{7.6.7}$$

The sinc² peak sharpens into a δ function: in the long-time limit energy conservation is strictly enforced, which is why the golden rule is often written with an explicit δ function (after integrating over final states the two forms agree).

An order-of-magnitude calculation: why the atom lives 1.6 nanoseconds

The golden rule’s most famous client: spontaneous emission. A hydrogen atom falls from 2p back to 1s, releasing a 10.2 eV photon. The matrix element comes from the electric-dipole interaction ($\hat V\sim e\vec r\cdot\vec{\mathcal E}$, dipole moment of order $ea_0$); the density of states is the photon formula above. Feed both into the golden rule, sum over photon directions and polarisations, and the rate tidies up to

$$\Gamma\approx\frac{4\alpha}{3}\,\frac{\omega^3a_0^2}{c^2}\times|\text{matrix element}/ea_0|^2\tag{7.6.10}$$

($\alpha\approx1/137$ is the fine-structure constant.) The arithmetic: $\omega=1.55\times10^{16}\ \text{rad/s}$, taking the matrix-element factor as 1 for now:

$$\Gamma\approx\frac{4}{3\times137}\times\frac{(1.55\times10^{16})^3\times(5.29\times10^{-11})^2}{(3\times10^8)^2} \approx1.1\times10^{9}\ \text{s}^{-1}\tag{7.6.11}$$

The exact matrix element $|\bra{1s}z\ket{2p_0}|=0.745\,a_0$ squeezes in a further factor of 0.55, giving $\Gamma=6.3\times10^8\ \text{s}^{-1}$, i.e.

$$\tau=\frac{1}{\Gamma}\approx1.6\ \text{ns}\tag{7.6.12}$$

in exact agreement with experiment. Take a moment to feel what this number means: the light wave’s period is $2\pi/\omega\sim0.4\ \text{fs}$, so the atom oscillates through four million periods before emitting one photon — the coupling of atom to light field (powers of $\alpha$) is genuinely feeble, which in turn explains why first-order perturbation theory is so trustworthy here.

Push the same rule outward and you get the whole of spectroscopy and scattering physics: photoabsorption cross-sections, beta-decay rates (Fermi’s own use for it), electron–phonon scattering in solids, the resistance of metals — for any problem where “an initial state flows through a weak coupling into a continuum of final states”, step one is always to write down the golden rule.

∎Key formulas

• Golden rule

  $$\Gamma=\frac{2\pi}{\hbar}\,|V_{fi}|^2\,\rho(E_f)$$

  Rate = coupling strength × number of final states on the energy shell

• δ-function limit

  $$\lim_{t\to\infty}\frac{t}{2\pi\hbar}\,\mathrm{sinc}^2\!\left(\tfrac{(E_f-E_i-\hbar\omega)t}{2\hbar}\right)=\delta(E_f-E_i-\hbar\omega)$$

  Energy conservation strictly enforced in the long-time limit

• Photon density of states

  $$\rho_{\text{photon}}(E)=\frac{VE^2}{\pi^2\hbar^3c^3}$$

  The box volume V eventually cancels against the matrix element’s 1/V

• Exponential decay

  $$N(t)=N_0\,\ee^{-\Gamma t},\qquad\tau=1/\Gamma$$

  Hydrogen 2p→1s: Γ=6.3×10⁸ s⁻¹, τ≈1.6 ns

?Self-check4 questions

1. 1.

   The transition probability to a single final state oscillates. Why does summing over a continuum of final states produce a constant rate?

   • Final states at different detunings oscillate with different periods; the sum averages the oscillations away, leaving only the net outflow on the resonance shell
   • Energy is not conserved for continuum final states, so the oscillation loses meaning
   • The summation makes the matrix elements larger
   • The continuum final states collide with each other and destroy the coherence

   Submit
2. 2.

   Put an excited atom inside the band gap of a photonic crystal (where the photon density of states at the relevant frequency is zero). What happens?

   • Spontaneous emission is suppressed — ρ(E_f)=0 means zero rate, no matter how large the matrix element
   • The atom radiates faster because the energy is pent up
   • No effect; spontaneous emission depends only on the atom itself
   • The atom absorbs instead of emitting

   Submit
3. 3.

   Which statements about the golden rule’s window of validity in time are correct? (Select all that apply.)

   Select all that apply

   • At very short t the δ-function approximation fails and the decay behaves as t²
   • The derivation requires the initial state not yet visibly drained (Γt ≪ 1)
   • In real atomic systems this window often spans many orders of magnitude
   • The golden rule holds strictly for times of any length

   Submit
4. 4.

   The spontaneous-emission rate of hydrogen 2p→1s is Γ = 6.3×10⁸ s⁻¹. Find the lifetime τ of the 2p state, in ns.

   ns10% relative tolerance

   Submit

What comes next

Time-dependent perturbation theory handles disturbances that are “fast but weak”: the light field oscillates millions of periods before triggering a single transition. Now twist the dial to the opposite end — what if the disturbance is extremely slow? Compress the well gradually, rotate the magnetic field gently: will the system jump away, or follow along obediently? The answer (the adiabatic theorem) is useful in its own right, but in 1984 Berry discovered a character in the story that had been overlooked for half a century: a system carried slowly around a closed loop comes back with a phase imprinted on its wavefunction that depends only on the geometry of the path. This “geometric phase” went on to thread together the Aharonov–Bohm effect and the entire field of topological matter.