---
title: "Fermi's golden rule"
date: 2021-07-10
categories:
- Physics
- Quantum mechanics
- Two-level system
- Optics
layout: "concept"
---

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](/know/concept/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](/know/concept/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
1.  D.J. Griffiths, D.F. Schroeter,
    *Introduction to quantum mechanics*, 3rd edition,
    Cambridge.