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+---
+title: "Fermi's golden rule"
+firstLetter: "F"
+publishDate: 2021-07-10
+categories:
+- Physics
+- Quantum mechanics
+- Optics
+
+date: 2021-07-03T14:41:11+02:00
+draft: false
+markup: pandoc
+---
+
+# 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](/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.