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.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.