Categories: Optics, Physics, Quantum mechanics, Two-level system.

# 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 sinusoidal 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.

## References

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