Fermi’s golden rule

Intuition

Often the final levels are not one but a continuum — an electron knocked out of an atom can leave with any energy, a photon can go off in any direction. Then the resonance peak of the last lesson sweeps over many final levels at once. Its height grows as the square of the time and its width shrinks as one over the time, so the area under it, the probability summed over the levels, grows in proportion to the time: a constant rate. That rate is Fermi’s golden rule. A constant rate of leaving empties a level exponentially, and a level that empties in a time $\tau$ has an energy blurred by $\hbar/\tau$ .

Water leaks from a tank through a sieve with thousands of holes of every size. Each hole alone would dribble unsteadily, but together they drain it at a steady rate — and a steady fraction lost each second empties the tank exponentially.

The resonance curve of a constant perturbation against the Bohr frequency, divided by $2\pi t$ so that the area under it is 1, after the times 2, 4 and 8. It grows in proportion to $t$ and narrows as $1/t$ : in the limit it is the delta function $\delta(\omega_{fi})$ , which picks out the final levels of the initial energy.

Let the final levels form a continuum with $\rho(E)$ levels per unit energy. A constant perturbation empties the initial level into them at the constant rate:

\Gamma_{i\to f}=\frac{2\pi}{\hbar}\,|V_{fi}|^{2}\,\rho(E_{f}),\qquad E_{f}=E_{i}

Properties

A periodic perturbation gives the same rate with $F_{fi}$ in place of $V_{fi}$ and $E_{f} =$ for absorption, for stimulated emission.

The golden rule

\frac{d}{dt}\sum_{f}P_{i\to f}=\frac{2\pi}{\hbar}\,|V_{fi}|^{2}\,\rho(E_{i})

Add up the constant-perturbation probability over a continuum of final energies. For long times the resonance curve is a narrow peak of area $2\pi t$ at the initial energy, so the sum picks out the final levels there and grows in proportion to the time.

Proof steps

P_{i\to f}=\frac{|V_{fi}|^{2}}{\hbar^{2}}\,\frac{\sin^{2}(\omega_{fi}t/2)}{(\omega_{fi}/2)^{2}}

The constant-perturbation result of the last lesson.

\sum_{f}\ \to\ \int\rho(E_{f})\,dE_{f}=\hbar\int\rho\,d\omega_{fi}

A continuum of final levels, $\rho$ of them per unit energy.

\frac{\sin^{2}(\omega t/2)}{(\omega/2)^{2}}\ \to\ 2\pi t\,\delta(\omega)\qquad(t\to\infty)

The peak narrows as $1/t$ while its area stays $2\pi t$ .

\sum_{f}P_{i\to f}=\frac{2\pi}{\hbar}\,|V_{fi}|^{2}\,\rho(E_{i})\,t

$V_{fi}$ and $\rho$ change little across the narrow peak, so they leave the integral at $E_{f}=E_{i}$ .

\Gamma=\frac{d}{dt}\sum_{f}P_{i\to f}=\frac{2\pi}{\hbar}\,|V_{fi}|^{2}\,\rho(E_{i})

A probability that grows in proportion to the time is a constant rate.

Applications

Practice

The Golden Rule

A transition into a continuum goes at a constant rate: $2\pi/\hbar$ times the squared matrix element, times the number of final levels per unit energy at the energy the jump conserves.

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

Try it

Which quantities set the rate of a transition into a continuum?