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Transition amplitudes from a changing Hamiltonian, resonance and its width, Fermi's golden rule for decay into a continuum, and the sudden and adiabatic limits.
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By the end of this lesson you will be able to compute first-order transition probabilities, predict the width of a resonance, apply Fermi's golden rule, and use the sudden and adiabatic approximations.
You know how stationary states evolve by phases, how a two-state system oscillates when its states are coupled, and time-independent perturbation theory. You know that the Hamiltonian determines time evolution. This lesson lets the perturbation itself depend on time — a light wave, a pulse, a switch — and asks how it moves the system between states.
| Term | What it means |
|---|---|
| Transition amplitude | $c_b(t)$, the coefficient of the final state $\psi_b$; its square is the transition probability. |
| Bohr frequency | $\omega_0 = (E_b - E_a)/\hbar$, the natural frequency of a transition. |
| Detuning | $\Delta = \omega - \omega_0$, how far a drive is from resonance. |
| Fermi's golden rule | $\Gamma = \tfrac{2\pi}{\hbar}\vert V\vert ^2\rho(E)$, the rate of transitions into a continuum. |
| Density of states | $\rho(E)$, the number of final states per unit energy. |
| Sudden approximation | For a very fast change the state is unchanged and must be re-expanded in the new eigenstates. |
| Adiabatic theorem | For a very slow change a system stays in the corresponding instantaneous eigenstate. |
Let $\hat{H} = \hat{H}^0 + \hat{H}'(t)$ and write the state as $\Psi = \sum_n c_n(t)\psi_ne^{-iE_nt/\hbar}$. Without $\hat{H}'$ the coefficients would be constant; with it they change. Starting in state $a$ and working to first order,
$$c_b^{(1)}(t) = -\frac{i}{\hbar}\int_0^t H'_{ba}(t')\,e^{i\omega_0t'}\,dt', \qquad \omega_0 = \frac{E_b - E_a}{\hbar}.$$
For a perturbation oscillating at frequency $\omega$, the integrand has a phase $e^{i(\omega_0 - \omega)t}$ that averages away unless $\omega \approx \omega_0$. The result is a probability sharply peaked at resonance,
$$P_{a \to b}(t) = \frac{|V_{ba}|^2}{\hbar^2}\frac{\sin^2(\Delta t/2)}{(\Delta/2)^2}, \qquad \Delta = \omega - \omega_0,$$
with height growing as $t^2$ and width shrinking as $1/t$. If instead the final states form a continuum with density $\rho(E)$, summing over them turns the peak into a delta function and the probability grows linearly in time, at a constant rate — Fermi's golden rule:
$$\Gamma = \frac{2\pi}{\hbar}|V|^2\rho(E).$$
This rate governs spontaneous emission, beta decay, scattering and almost every other process in which a quantum system decays.
Another way: picture
Picture pushing a child on a swing. Push at random moments and the pushes mostly cancel; push in step with the swing and each adds to the last, so the amplitude grows steadily. A quantum transition works the same way: the perturbation's phase must keep step with the phase difference between the two states, $e^{i\omega_0t}$. The longer you push, the more precisely in step you must be, which is why a long drive picks out a sharp frequency.
Another way: steps
Checks. First-order probabilities must be small, or the calculation is outside its range. Rates must scale as $|V|^2$. The resonance width in angular frequency must be about $2\pi/t$, the uncertainty relation between time and energy. And a golden-rule rate must be much slower than the frequency spread of the continuum, or the continuum is not really continuous on that time scale.
For a single final state, $P(t) \propto \sin^2(\Delta t/2)/(\Delta/2)^2$ oscillates. For a continuum, each final energy has its own detuning, and the total probability is the integral of $P$ over final energies weighted by $\rho$. As $t$ grows, the function $\sin^2(\Delta t/2)/(\Delta/2)^2$ becomes a tall, narrow peak of area $2\pi t$, so the integral picks out $\rho$ at the resonant energy and grows in proportion to $t$.
A linear growth of probability means a constant rate, and a constant rate means exponential decay of the initial state when the argument is carried beyond first order: $P_{\text{survive}} = e^{-\Gamma t}$. This is why radioactive nuclei and excited atoms decay exponentially with a well-defined lifetime. Enrico Fermi, who named the rule "golden" for its usefulness, applied it to beta decay in 1934, where the density of states of the electron and neutrino fixes the shape of the emitted electrons' energy spectrum.
The peak in $P(\Delta)$ has its first zeros at $\Delta = \pm 2\pi/t$. A drive lasting one microsecond selects frequencies within about a megahertz of resonance; one lasting a second selects within a hertz. This is the energy–time uncertainty relation in action: a measurement lasting $t$ cannot resolve energies finer than about $h/t$.
Atomic clocks exploit this directly. The longer the atoms interact with the probe field, the narrower the resonance and the more precisely the frequency can be set. Norman Ramsey's method of separated fields, recognized with the 1989 Nobel Prize, applies two short pulses separated by a long free interval, producing fringes whose width is set by the interval. The cesium fountain clocks that define the second use Ramsey's method with an interval of about a second, giving a resonance about a hertz wide on a $9.2$ GHz transition.
If the Hamiltonian changes much faster than any of the system's natural periods, the wave function has no time to respond. Just after the change it is the same function as before, now expanded in the new Hamiltonian's eigenstates, and the probability of each new state is $|\langle\psi_n^{\text{new}}|\psi^{\text{old}}\rangle|^2$.
Beta decay is a clean example. When tritium, a hydrogen isotope, decays to helium-3, the nuclear charge jumps from $1$ to $2$ as a fast electron leaves in about $10^{-18}$ s, far shorter than the orbital period of about $10^{-16}$ s. The atomic electron, left in a hydrogen $1s$ orbital around a helium nucleus, ends in the helium ion's ground state with probability $64 \times 8/729 = 0.70$ and is otherwise excited or ejected. The KATRIN experiment in Germany, which measures the tritium beta spectrum to weigh the neutrino, must account for these final-state probabilities precisely.
The opposite limit is a change so slow that the system can follow it. The adiabatic theorem says that a system starting in the $n$-th eigenstate of $\hat{H}(0)$ ends in the $n$-th eigenstate of $\hat{H}(T)$, provided the change is slow compared with $\hbar$ divided by the gap to neighboring levels, and the levels never cross. A particle in a box whose wall is moved slowly stays in the ground state of the slowly changing box, its energy falling as the box widens.
Adiabatic evolution is a practical tool. Rapid adiabatic passage sweeps a laser's frequency through resonance slowly enough that every atom is transferred from one state to another, regardless of small differences in intensity. Adiabatic quantum computing, the principle behind D-Wave's annealing machines, prepares a system in the easy ground state of one Hamiltonian and slowly changes it to a Hamiltonian whose ground state encodes the answer to an optimization problem. The speed limit is set by the smallest gap encountered along the way.
First-order theory assumes the initial state is barely depleted. For a resonant drive of a single transition, that fails once $P$ is no longer small, and the exact two-state result, $P = \sin^2(|V|t/\hbar)$ on resonance, takes over: the population oscillates instead of growing without bound. That is the Rabi oscillation of the two-state lesson, and first order is its short-time limit.
Some transitions have $V_{ba} = 0$ by symmetry. Then the leading process is second order, passing through an intermediate state: two-photon absorption, Raman scattering, and the virtual transitions responsible for van der Waals forces. Their amplitudes contain sums over intermediate states divided by energy mismatches, exactly as in second-order time-independent theory, and they are typically much weaker, which is why forbidden lines are faint.
The KATRIN experiment at the Karlsruhe Institute of Technology measures the energies of electrons from tritium beta decay with a spectrometer $24$ m long. If neutrinos have mass, the spectrum's endpoint is pulled down and its shape distorted by a tiny amount. In 2025 KATRIN reported that the neutrino's mass is below $0.45$ eV, the tightest direct laboratory limit.
Both ideas of this lesson are essential. Fermi's golden rule gives the spectrum's shape, through the density of states of the outgoing electron and neutrino. The sudden approximation gives the final states of the daughter molecule: when the nuclear charge jumps, the molecule's electrons are left in excited states roughly $40$ percent of the time, taking energy that the beta electron no longer has. The experiment's sensitivity depends on calculating those probabilities to better than a percent — a textbook projection of old wave functions onto new ones, done with care.
D-Wave Systems builds quantum annealers with thousands of superconducting qubits. Each run starts with all qubits in the easy ground state of a strong transverse field, then slowly turns that field off while turning on couplings that encode an optimization problem, such as scheduling or routing. If the change is slow enough, the adiabatic theorem guarantees that the system ends in the ground state of the final Hamiltonian, the problem's best solution.
How slow is slow enough depends on the smallest energy gap between the ground and first excited states along the way: the required time grows roughly as $\hbar$ over the gap squared. For hard problems the gap can shrink rapidly with the number of qubits, and a run that is too fast drives transitions into excited states — wrong answers — exactly as time-dependent perturbation theory predicts. Users therefore repeat each anneal many times and keep the best result.
It is natural to think that shaking a system harder always excites it. But the transition amplitude integrates the perturbation against the phase $e^{i\omega_0t}$, and off resonance the contributions cancel. The probability is peaked at $\omega = \omega_0$ with a width of about $2\pi/t$; far from resonance, even a strong drive does little. That selectivity is what makes spectroscopy possible.
A second misconception is that first-order probabilities keep growing as $t^2$ forever. They cannot exceed one. For a single final state, the exact answer oscillates; for a continuum, the linear growth of the golden rule becomes exponential decay of the initial state once it is appreciably depleted.
Write the first-order amplitude for a constant coupling on resonance.
$c_b(t) = -\dfrac{i}{\hbar}\displaystyle\int_0^tV\,dt' = -\dfrac{iVt}{\hbar}$
In the rotating frame the coupling is constant.
Square it for the probability.
$P = \dfrac{|V|^2t^2}{\hbar^2}$
Quadratic growth.
Evaluate for $|V| = 0.2$ μeV and $t = 1$ ns.
$\dfrac{|V|t}{\hbar} = \dfrac{0.2}{0.6582} = 0.304, \qquad P = 0.092$
Nine percent.
Compare with the exact result.
$\sin^2(0.304) = 0.089$
First order is three percent high.
Find when first order fails badly.
$\dfrac{|V|t}{\hbar} = 1: \quad P_{(1)} = 1, \quad \sin^2 1 = 0.71$
By then the initial state is heavily depleted.
Write the probability for a detuned drive.
$P(\Delta) = \dfrac{|V|^2}{\hbar^2}\dfrac{\sin^2(\Delta t/2)}{(\Delta/2)^2}$
From the first-order integral with $e^{i(\omega_0 - \omega)t}$.
Find the peak value.
$\Delta \to 0: \quad P = \dfrac{|V|^2t^2}{\hbar^2}$
Since $\sin x/x \to 1$.
Find the first zeros.
$\dfrac{\Delta t}{2} = \pm\pi \quad\Rightarrow\quad \Delta = \pm\dfrac{2\pi}{t}$
The line's half-width.
Evaluate the width for a $1$ μs pulse.
$\dfrac{\Delta}{2\pi} = \pm\dfrac{1}{t} = \pm 1\ \text{MHz}$
In ordinary frequency.
Find the area under the curve.
$\displaystyle\int_{-\infty}^{\infty}\dfrac{\sin^2(\Delta t/2)}{(\Delta/2)^2}\,d\Delta = 2\pi t$
Grows linearly with $t$, while the width shrinks as $1/t$.
Connect to the golden rule.
$\dfrac{\sin^2(\Delta t/2)}{(\Delta/2)^2} \to 2\pi t\,\delta(\Delta)$
At long times the peak acts like a delta function.
A state couples to a continuum with $|V| = 0.5$ μeV and $\rho = 2$ per μeV. Write the golden rule.
$\Gamma = \dfrac{2\pi}{\hbar}|V|^2\rho$
The rate for many closely spaced final states.
Evaluate the product $|V|^2\rho$.
$|V|^2\rho = 0.25 \times 2 = 0.5\ \mu\text{eV}$
Units of energy.
Evaluate the rate.
$\Gamma = \dfrac{2\pi \times 0.5}{0.6582}\ \text{ns}^{-1} = 4.77\ \text{ns}^{-1}$
With $\hbar = 0.6582$ μeV ns.
Find the lifetime.
$\tau = \dfrac{1}{\Gamma} = 0.21\ \text{ns}$
About two hundred picoseconds.
Find the energy width of the decaying state.
$\hbar\Gamma = 0.6582 \times 4.77 = 3.14\ \mu\text{eV}$
The level is broadened by its own decay.
Check the continuum assumption.
$\rho\,\hbar\Gamma = 2 \times 3.14 \approx 6 \text{ states}$
Several final states lie within the width, marginally continuous.
Write the survival probability.
$P_a(t) = e^{-\Gamma t} = e^{-t/0.21\ \text{ns}}$
Exponential decay.
Evaluate after $1$ ns.
$P_a = e^{-4.77} = 0.0085$
Less than one percent survive.
Write the short-time formula.
$P = \left(\dfrac{|V|t}{\hbar}\right)^2$
On resonance.
Substitute the values.
$\dfrac{|V|t}{\hbar} = \dfrac{0.1 \times 2}{0.6582} = 0.304$
The same product as $0.2$ μeV for $1$ ns.
Evaluate the probability.
A state decays into a continuum at a rate of $35 \times 10^{6}$ s⁻¹. If the coupling matrix element to the continuum were doubled, what would the rate be?
Complete the worked solution: a decay rate is $16$ μs⁻¹. Find the rate if the coupling is doubled, if it is tripled, and the difference between those two rates, in μs⁻¹.
Scale by the square of two.
$\Gamma_{2V} = 4 \times 16 =$ a
Twice the coupling.
Scale by the square of three.
$\Gamma_{3V} = 9 \times 16 =$ b
Three times the coupling.
Subtract the two rates.
$\Gamma_{3V} - \Gamma_{2V} =$ c
Five times the original rate.
Match each result of time-dependent perturbation theory to its statement.
| $-\frac{i}{\hbar}\int H'_{ba}e^{i\omega_0t}dt$ | $(|V|t/\hbar)^2$ | $\frac{2\pi}{\hbar}|V|^2\rho$ | follows the instantaneous eigenstate | |
|---|---|---|---|---|
| the first-order amplitude | ||||
| the short-time resonant probability | ||||
| the golden rule | ||||
| the adiabatic theorem |
A weak drive acts for $7$ μs with detuning $\Delta$ from resonance. Fill in the transition probability relative to its value on resonance for each value of $\Delta t$.
| $P(\Delta)/P(0)$ | |
|---|---|
| $\Delta t = 0$ | |
| $\Delta t = \pi$ | |
| $\Delta t = 2\pi$ | |
| $\Delta t = 3\pi$ |
A resonant drive couples two states with matrix element $|V| = 0.25$ μeV. Starting in the lower state, what is the probability of finding the upper one after $1$ ns, to first order? Use $\hbar = 0.6582$ μeV ns.
Answer:
A discrete state is coupled to a continuum of final states by a matrix element $|V| = 0.3$ μeV, and the continuum has $\rho = 3$ states per μeV near the initial energy. Using Fermi's golden rule, what is the state's lifetime, in ns? Use $\hbar = 0.6582$ μeV ns.
Answer: ns
When a hydrogen-like lithium-8 ion decays to beryllium, the nuclear charge jumps from $Z_1 = 3$ to $Z_2 = 4$ in far less time than the electron's orbital period. What is the probability that the electron, initially in the ground state, ends in the new ground state?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A weak drive acts for $4$ μs with detuning $\Delta$ from resonance. Fill in the transition probability relative to its value on resonance for each value of $\Delta t$.
| $P(\Delta)/P(0)$ | |
|---|---|
| $\Delta t = 0$ | |
| $\Delta t = \pi$ | |
| $\Delta t = 2\pi$ | |
| $\Delta t = 3\pi$ |
You can calculate transitions driven by time-dependent perturbations. Explain to someone why a longer pulse picks out a sharper frequency.
17. Your turn: a resonant coupling of $0.1$ μeV acts for $2$ ns. What is the first-order transition probability?, step 3
$P = 0.092$
Small enough for first order to hold.