Categories: Optics, Physics, Quantum mechanics.

Fermi’s golden rule

In quantum mechanics, Fermi’s golden rule expresses the transition rate between two states of a system, when a sinusoidal perturbation is applied at the resonance frequency \(\omega = E_g / \hbar\) of the energy gap \(E_g\). The main conclusion is that the rate is independent of time.

From time-dependent perturbation theory, we know that the transition probability for a particle in state \(\ket{a}\) to go to \(\ket{b}\) is as follows for a periodic perturbation at frequency \(\omega\):

\[\begin{aligned} P_{ab} = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big((\omega_{ba} - \omega) t / 2\big)}{(\omega_{ba} - \omega)^2} \end{aligned}\]

Where \(\omega_{ba} \equiv (E_b - E_a) / \hbar\). If we assume that \(\ket{b}\) irreversibly absorbs an unlimited number of particles, then we can interpret \(P_{ab}\) as the “amount” of the current particle that has transitioned since the last period \(2 \pi n / (\omega_{ba} \!-\! \omega)\).

For generality, let \(E_b\) be the center of a state continuum with width \(\Delta E\). In that case, \(P_{ab}\) must be modified as follows, where \(\rho(E_x)\) is the destination’s density of states:

\[\begin{aligned} P_{ab} &= \frac{|V_{ba}|^2}{\hbar^2} \int_{E_b - \Delta E / 2}^{E_b + \Delta E / 2} \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x} \end{aligned}\]

If \(E_b\) is not in a continuum, then \(\rho(E_x) = \delta(E_x - E_b)\). The integrand is a sharp sinc-function around \(E_x\). For large \(t\), it is so sharp that we can take out \(\rho(E_x)\). In that case, we also simplify the integration limits. Then we substitute \(x \equiv (\omega_{xa}\!-\!\omega) / 2\) to get:

\[\begin{aligned} P_{ab} &\approx \frac{2}{\hbar} |V_{ba}|^2 \rho(E_b) \int_{-\infty}^\infty \frac{\sin^2(x t)}{x^2} \:dx \end{aligned}\]

This definite integral turns out to be \(\pi |t|\), so we find, because clearly \(t > 0\):

\[\begin{aligned} P_{ab} &= \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) \: t \end{aligned}\]

The transition rate \(R_{ab}\), i.e. the number of particles per unit time, then takes this form:

\[\begin{aligned} \boxed{ R_{ab} = \pdv{P_{ab}}{t} = \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) } \end{aligned}\]

Note that the \(t\)-dependence has disappeared, and all that remains is a constant factor involving \(E_b = E_a \!+\! \hbar \omega\), where \(\omega\) is the resonance frequency.


  1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.

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