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---
title: "Fermi's golden rule"
sort_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.
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