Time-Dependent Perturbation Theory & Fermi's Golden Rule

Strike a match, switch on a sodium lamp, watch a laser pointer glow: in every case atoms are trading energy with light — soaking up a photon to jump upstairs, or dropping down and spitting one back out. And they do it at astonishingly sharp colours. Sodium's flame is a particular, unmistakable yellow; hydrogen glows a specific red; a helium-neon laser is locked to one precise wavelength. Behind every one of these facts sits a single number you can actually calculate: a transition rate — the probability per unit time that a quantum system, nudged by a time-varying influence, hops from one state to another.

The time-independent Schrödinger equation gave us the ladder of stationary energy levels, and time-independent perturbation theory told us how a static nudge shifts those levels. But a stationary state, left alone, sits on its rung forever — it never jumps. To make an atom change state, something has to shake it in time: an oscillating electromagnetic field, a passing collision, a switched-on interaction. This page is about that shaking, and about the one formula it culminates in — Fermi's Golden Rule — which turns "how strongly does the perturbation connect these two states?" into "how many times per second does the jump happen?"

That rate is the quiet engine under an enormous amount of physics: the intensity of every spectral line, the lifetime of every excited atom and unstable nucleus, the gain of every laser, and the cross section of nearly every scattering and decay process. Learn to compute it once, here, and you have the master tool for how fast quantum things happen.

The setup: a perturbation switched on in time

Start with a system whose stationary states we already know — the eigenstates |n\rangle of an unperturbed Hamiltonian \hat{H}_0, with \hat{H}_0|n\rangle = E_n|n\rangle. At t=0 we turn on a small, time-dependent extra piece:

\hat{H}(t) = \hat{H}_0 + \hat{H}'(t),

with \hat{H}' weak enough that the old eigenstates are still a good language to describe what happens. Because the |n\rangle form a complete basis, the true evolving state can always be written as a superposition of them, but now with time-dependent coefficients that carry the free phase explicitly:

|\psi(t)\rangle = \sum_n c_n(t)\,e^{-iE_n t/\hbar}\,|n\rangle.

All the interesting physics lives in how the c_n(t) drift away from their starting values once \hat{H}' is on. Suppose the system starts cleanly in a single initial state |i\rangle, so c_i(0) = 1 and every other c_n(0) = 0. The question of the whole page is: what is the probability of later finding it in some different final state |f\rangle?

P_{i\to f}(t) = |c_f(t)|^2.

Feeding the ansatz into the time-dependent Schrödinger equation and projecting onto \langle f| gives an exact but tangled set of coupled equations for the c_n(t). Perturbation theory cuts the knot: since \hat{H}' is small, we solve it in successive orders, keeping the leading one.

First-order transition amplitude

To lowest order we set the coefficient on the right-hand side to its initial value (c_i \approx 1, all others small) and simply integrate. The result is the workhorse formula of the whole subject — the first-order transition amplitude:

c_f(t) = -\frac{i}{\hbar}\int_0^t \langle f|\hat{H}'(t')|i\rangle\; e^{\,i\omega_{fi}t'}\,dt', \qquad \omega_{fi} \equiv \frac{E_f - E_i}{\hbar}.

Read it slowly, because everything else on the page is a special case of it. The matrix element \langle f|\hat{H}'|i\rangle asks how strongly the perturbation connects the two states — if the perturbation cannot link them, this is zero and nothing happens. The exponential e^{i\omega_{fi}t'} is a stopwatch ticking at the Bohr frequency \omega_{fi}, the natural frequency of the i\to f gap. The integral adds up the little kicks the perturbation delivers over time, each weighted by that spinning phase.

The phase factor is the crux. If \hat{H}' pushes in step with \omega_{fi}, the contributions pile up in phase and the amplitude grows — a resonance. If it pushes at the wrong tempo, the spinning phase makes successive kicks cancel, and almost nothing accumulates. That single tension — constructive build-up at resonance, destructive cancellation off it — is the origin of sharp spectral lines.

P_{i\to f}(t) = \left|c_f(t)\right|^2 = \frac{1}{\hbar^2}\left|\int_0^t \langle f|\hat{H}'(t')|i\rangle\,e^{i\omega_{fi}t'}\,dt'\right|^2.

Everything now depends on the time profile of the perturbation. The single most important profile — the one that describes light hitting an atom — is a sinusoid, and that is where we go next.

A sinusoidal perturbation: resonance and the birth of spectral lines

Light is an oscillating electromagnetic field, so shine it on an atom and the perturbation oscillates:

\hat{H}'(t) = 2\hat{V}\cos(\omega t) = \hat{V}\big(e^{i\omega t} + e^{-i\omega t}\big).

Drop this into the first-order integral. The cosine splits into two counter-rotating exponentials, and combined with the stopwatch e^{i\omega_{fi}t'} they produce two terms oscillating at (\omega_{fi}+\omega) and (\omega_{fi}-\omega). One of them can go resonant:

Keeping the resonant term, the transition probability works out to the celebrated lineshape:

P_{i\to f}(t) = \frac{|V_{fi}|^2}{\hbar^2}\; \frac{\sin^2\!\big[(\omega - \omega_{fi})\,t/2\big]}{\big[(\omega - \omega_{fi})/2\big]^2}, \qquad V_{fi} = \langle f|\hat{V}|i\rangle.

This is a sinc² curve in the detuning \omega - \omega_{fi}: a tall central peak sitting exactly at resonance, flanked by small ripples. Two features matter enormously. As the interaction time t grows, the peak height at exact resonance climbs as t^2 (take the limit \omega \to \omega_{fi} and the fraction becomes t^2), while the peak width shrinks as 1/t. The peak becomes taller and narrower — the system grows ever pickier about matching the exact resonant frequency the longer you watch. Drag the time slider below and see it happen.

That sharpening is the whole reason spectral lines are sharp. A long-lived interaction admits only frequencies extremely close to \omega_{fi}; everything else cancels away. Push t to infinity and the peak becomes infinitely tall and infinitely thin — a Dirac delta function enforcing \hbar\omega = |E_f - E_i| exactly. Energy conservation for the atom-plus-photon is not put in by hand; it emerges from letting the perturbation act for a long time.

Into a continuum: Fermi's Golden Rule

A jump between two isolated levels at exact resonance actually oscillates — probability sloshes up to |f\rangle and back down again (we meet that as Rabi flopping in the Watch-out below). To get a steady rate, the final state must not be a lone level but one of a continuum — a dense band of nearly degenerate states, described by a density of states \rho(E_f) (the number of available final states per unit energy). A decaying atom emits into the continuum of photon modes; a scattered particle flies off into a continuum of momenta. Sum the sinc² probability over that band, take the long-time limit so the delta function bites, and the sharp lineshape integrates to a probability that grows linearly in time. A probability linear in time is a constant rate — and that rate is Fermi's Golden Rule.

Dirac derived it in 1927; Fermi called it "golden" because he used it so often it deserved a nickname — and the name stuck to the wrong person, in the grand tradition of physics. Its power is that it reduces an entire dynamical calculation to two ingredients you can often compute by hand: a matrix element and a density of states.

What the rule buys you: selection rules, lifetimes, lasers

Selection rules. If the matrix element \langle f|\hat{H}'|i\rangle = 0, the rate is exactly zero: the transition is forbidden, no matter how much light you shine. Very often the matrix element vanishes by symmetry — a parity or angular-momentum mismatch between |i\rangle, \hat{H}' and |f\rangle makes the integral cancel. This is why only certain atomic transitions light up: the periodic table's spectra are a giant catalogue of which matrix elements happen to be nonzero.

Lifetimes. An excited atom left alone still decays — spontaneous emission — at a rate \Gamma given by (a fully quantised-field version of) the golden rule. If each atom jumps with constant probability per unit time \Gamma, a population decays exponentially, N(t) = N_0 e^{-\Gamma t}, and the characteristic lifetime is simply

\tau = \frac{1}{\Gamma}.

A strongly-coupled transition (big matrix element) is fast and short-lived; a nearly-forbidden one can linger for seconds — the "metastable" states that make phosphorescence glow in the dark.

Spontaneous vs stimulated. Our sinusoidal treatment gave stimulated emission, driven by an applied field; a full theory also has spontaneous emission into empty modes. Einstein tied the two together before quantum electrodynamics existed with his A and B coefficients — the fixed ratio between spontaneous decay and stimulated up/down rates that any working laser exploits by engineering a population inversion so stimulated emission wins.

And it does not stop at atoms. Replace \hat{H}' by a scattering potential and the same rule gives collision cross sections; replace it by the weak interaction and it gives nuclear beta-decay rates. Fermi's Golden Rule is the universal answer to "how fast?"

Worked examples

Example 1 — a rate straight from the rule. A perturbation connects an initial state to a continuum with matrix element |M| = 2\times10^{-3}\,\text{eV} and the final-state density is \rho = 5\times10^{18}\,\text{J}^{-1}. Fermi's rule gives \Gamma = (2\pi/\hbar)\,|M|^2\,\rho. The lesson is structural: the rate is linear in \rho and quadratic in the coupling. Double the density of final states and the rate doubles; double the matrix element and the rate goes up four-fold. Everything else is just plugging in \hbar = 1.055\times10^{-34}\,\text{J·s} with the units made consistent.

Example 2 — a forbidden transition. Suppose |i\rangle and |f\rangle are both even functions of x and the perturbation is \hat{H}' \propto x (an odd function — the usual dipole coupling to a field). Then the integrand \langle f|x|i\rangle = \int \psi_f^*(x)\,x\,\psi_i(x)\,dx is even × odd × even = odd, so it integrates to zero over a symmetric range. The matrix element vanishes, the rate is zero, and the transition is forbidden by parity — exactly why, for example, the 2s \to 1s transition in hydrogen is so spectacularly slow. You did not need to compute a single number; symmetry killed it.

Example 3 — lifetime from a rate. An excited state decays with rate \Gamma = 2.5\times10^{8}\,\text{s}^{-1} (a typical allowed optical transition). Its lifetime is

\tau = \frac{1}{\Gamma} = \frac{1}{2.5\times10^{8}\,\text{s}^{-1}} = 4.0\times10^{-9}\,\text{s} = 4\,\text{ns}.

A few nanoseconds — the ballpark for ordinary allowed atomic transitions, and why fluorescent things flash off almost the instant you kill the excitation. A metastable state with \Gamma a billion times smaller would glow for seconds instead.

A fair worry: the first-order formula gave a probability P_{i\to f}(t), yet the golden rule is a rate. The bridge is that, once you sum the sinc² lineshape over a continuum of final states, all the wiggly time-dependence collapses and the surviving probability is simply proportional to t: P(t) = \Gamma t. Differentiate and the rate \Gamma = dP/dt is a constant — no time dependence left. That linear growth is what makes a constant rate meaningful at all: a fixed slice of probability accrues each second, so a fixed fraction of a population jumps per second, which is exactly what "rate" means.

There is a catch worth naming: because P = \Gamma t grows without bound, it would eventually exceed 1, which is nonsense. First-order theory is only valid in the window \Gamma t \ll 1 — long enough for the delta function to sharpen, short enough that the initial state is not appreciably depleted. Inside that window, the golden rule reigns.

The classic trap. Fermi's Golden Rule gives a steady, constant rate — but that only happens when the system decays into a dense set of final states (that is what the density of states \rho(E_f) is doing in the formula). If instead you drive a transition between two isolated discrete levels on exact resonance, there is no rate at all in the golden-rule sense. The probability does not grow linearly forever; it oscillates sinusoidally — sloshing fully up to the excited state and all the way back down again. These are Rabi oscillations, and the population cycles at the Rabi frequency \Omega = |V_{fi}|/\hbar:

P_{i\to f}(t) = \sin^2\!\big(\Omega t\big) \quad\text{(two levels, on resonance)} \qquad\text{vs.}\qquad P_{i\to f}(t) = \Gamma\,t \quad\text{(into a continuum).}

So before you reach for the golden rule, ask: is there somewhere for the system to decay into a continuum? If yes — a photon field, a band of momenta, a dense spectrum — the rule applies and you get a lifetime. If the final states are just a couple of sharp levels, you get coherent flopping instead, and you must solve the two-state problem directly. Same physics, utterly different behaviour.