---
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"
---

Consider a general quantum system that is being affected by a sinusoidal potential.
Specifically, the system's time-independent base Hamiltonian $$\hat{H}_0$$
is perturbed by an $$\hat{H}_1(t)$$ given by:

$$\begin{aligned}
    \hat{H}_1(\vb{r}, t)
    \equiv \hat{V}(\vb{r}) \cos(\omega t)
    = \frac{1}{2} \hat{V}(\vb{r}) \: (e^{i \omega t} + e^{-i \omega t})
\end{aligned}$$

This potential is switched on at $$t = 0$$.
Let us study the system's response using
[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
by expressing the perturbed wavefunction $$\ket{\Psi(t)}$$
as a superposition of known eigenstates $$\ket{n}$$ of $$\hat{H}_0$$
with time-dependent coefficient $$c_n(t)$$:

$$\begin{aligned}
    \ket{\Psi(t)}
    = \sum_n c_n(t) \ket{n} e^{-i E_n t / h}
\end{aligned}$$

In the absence of $$\hat{H}_1$$,
the coefficients would simply be constants $$c_n^{(0)}$$.
Perturbation theory then states that the first-order correction
$$c_n^{(1)}(t) \approx c_n(t) - c_n^{(0)}$$ is as follows,
with $$V_{mn} \equiv \matrixel{m}{\hat{V}}{n}$$:

$$\begin{aligned}
    c_m^{(1)}(t)
    &= - \frac{i}{2 \hbar} \sum_{n} c_n^{(0)} \: V_{mn}
    \int_0^t e^{i \tau (\omega_{mn} + \omega)} + e^{i \tau (\omega_{mn} - \omega)} \dd{\tau}
    \\
    &= - \frac{1}{2 \hbar} \sum_{n} c_n^{(0)} \: V_{mn}
    \bigg( \frac{e^{i t (\omega_{mn} + \omega)} - 1}{\omega_{mn} + \omega}
    + \frac{e^{i t (\omega_{mn} - \omega)} - 1}{\omega_{mn} - \omega} \bigg)
\end{aligned}$$

For simplicity, we let the system start in a known state $$\ket{a}$$,
such that $$c_n^{(0)} = \delta_{na}$$,
and we focus on a single destination state $$m = b$$,
effectively turning this into a two-level system.
Let us assume that the driving frequency is close
to that transition's resonance $$\omega \approx \omega_{ba}$$,
such that the second term is dominant over the first, which can then be neglected.
We therefore get:

$$\begin{aligned}
    c_b^{(1)}(t)
    &= - \frac{V_{ba}}{2 \hbar} \frac{e^{i (\omega_{ba} - \omega) t} - 1}{\omega_{ba} - \omega}
    \\
    &= - i \frac{V_{ba}}{2 i \hbar} \frac{e^{i (\omega_{ba} - \omega) t / 2} - e^{- i (\omega_{ba} - \omega) t / 2}}{\omega_{ba} - \omega} \: e^{i (\omega_{ba} - \omega) t / 2}
    \\
    &= - i \frac{V_{ba}}{\hbar} \frac{\sin\!\big( (\omega_{ba} - \omega) t / 2 \big)}{\omega_{ba} - \omega} \: e^{i t (\omega_{ba} - \omega) / 2}
\end{aligned}$$

Taking the norm squared yields the **transition probability**:
the probability that a particle starting in state $$\ket{a}$$
will be found in $$\ket{b}$$ at time $$t > 0$$:

$$\begin{aligned}
    \boxed{
        P_{a \to b}(t)
        = |c_b^{(1)}(t)|^2
        = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ba} - \omega) t / 2 \big)}{(\omega_{ba} - \omega)^2}
    }
\end{aligned}$$

This would be the same if $$\hat{H}_1 \equiv \hat{V} \sin(\omega t)$$.
However, many authors define $$\hat{H}_1 \equiv \hat{V} e^{- i \omega t}$$,
and then $$P_{a \to b}$$ is larger by a factor of $$4$$,
which can cause confusion when comparing literature.

In any case, the probability oscillates as a function of $$t$$
with period $$T = 2 \pi / (\omega_{ba} - \omega)$$,
so after one full period the particle is back in $$\ket{a}$$,
and after $$T/2$$ the particle is fully in $$\ket{b}$$.
See [Rabi oscillation](/know/concept/rabi-oscillation/)
for a more accurate treatment of this "flopping" behaviour.

However, when regarded as a function of $$\omega$$, the probability
takes the form of a sinc-function centred around $$(\omega_{ba} - \omega)$$,
so it is maximized for transitions with energy
$$\hbar \omega = E_b - E_a$$.
Also notice that the sinc-function becomes narrower over time:
this has no effect if $$\omega = \omega_{ba}$$,
however, if we are slightly off-resonance,
then this means that $$P_{a \to b}$$ decreases over time.

In fact, for $$t \to \infty$$, the sinc-function
becomes so sharp that it can be related to
the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$.
Consider this general identity (yes, the notation is sloppy, but it is clear):

$$\begin{aligned}
    \lim_{t \to \infty} \frac{\sin^2(x t / 2)}{x^2}
    = \frac{\pi}{2} \: \delta(x) \: t
\end{aligned}$$


{% include proof/start.html id="proof-relation" -%}
First, observe that we can rewrite the fraction using an integral:

$$\begin{aligned}
    \frac{\sin(x t / 2)}{x}
    = \frac{e^{i x t / 2} - e^{-i x t / 2}}{2 i x}
    = \frac{i}{2 i} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau}
\end{aligned}$$

By taking the limit $$t \to \infty$$,
it can be turned into a nascent Dirac delta function:

$$\begin{aligned}
    \lim_{t \to \infty} \frac{\sin(x t / 2)}{x}
    = \lim_{t \to \infty} \frac{\pi}{2 \pi} \int_{-\infty}^{\infty} e^{i x \tau} \dd{\tau}
    = \lim_{t \to \infty} \pi \delta(x)
\end{aligned}$$

On the topic of the delta function, we also show that its square is given by:

$$\begin{aligned}
    \delta^2(x)
    = \delta(x) \lim_{t \to \infty} \frac{1}{2 \pi} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau}
    = \delta(x) \lim_{t \to \infty} \frac{t}{2 \pi}
\end{aligned}$$

Where we have set $$x = 0$$ according to the first delta function.
This gives the target:

$$\begin{aligned}
    \lim_{t \to \infty} \frac{\sin^2(x t / 2)}{x^2}
    = \pi^2 \delta^2(x)
    = \frac{\pi}{2} \: \delta(x) \: t
\end{aligned}$$
{% include proof/end.html id="proof-relation" %}


Inserting this into $$P_{a \to b}$$
and using the general property that $$\delta(x) / c = \delta(c x)$$:

$$\begin{aligned}
    P_{a \to b}
    &= \frac{\pi t}{2 \hbar^2} |V_{ba}|^2 \: \delta(\omega_{ba} - \omega)
    \\
    &= \frac{\pi t}{2 \hbar} |V_{ba}|^2 \: \delta(E_b - E_a - \hbar \omega)
\end{aligned}$$

By taking the $$t$$-derivative, we obtain the **transition rate** $$R_{a \to b}$$,
i.e. the number of particles per unit time,
which is given by **Fermi's golden rule**:

$$\begin{aligned}
    \pdv{P_{a \to b}}{t} =
    \boxed{
        R_{a \to b}
        = \frac{\pi}{2 \hbar} |V_{ba}|^2 \: \delta(E_b - E_a - \hbar \omega)
    }
\end{aligned}$$

Once again, beware that many authors define
$$\hat{H}_1 \equiv \hat{V} e^{- i \omega t}$$,
in which case the result is larger by a factor of $$4$$,
which can cause confusion when comparing literature.
For that alternative derivation, see the one-photon section of
[multi-photon absorption](/know/concept/multi-photon-absorption/).

What is most noteworthy about this result is
that the $$t$$-dependence has disappeared,
and that the rate is proportional to $$|V_{ba}|^2$$,
i.e. the overlap between $$\ket{a}$$ and $$\ket{b}$$ with respect to $$\hat{V}$$.
In practice, this result is usually used inside an integral over $$\omega$$,
hence that "naked" Dirac delta function.

For generality, let us take $$E_b$$ to be the center
of a state continuum with width $$2 W$$.
In that case, $$P_{a \to b}$$ must be modified as follows,
where $$\rho(E_x)$$ is the destination's
[density of states](/know/concept/density-of-states/):

$$\begin{aligned}
    P_{a \to b}
    &= \frac{|V_{ba}|^2}{\hbar^2} \int_{E_b - W}^{E_b + W}
    \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x}
    \\
    &= \frac{\pi t}{2 \hbar} |V_{ba}|^2
    \int_{E_b - W}^{E_b + W} \delta(E_x - E_a - \hbar \omega) \: \rho(E_x) \dd{E_x}
    \\
    &= \frac{\pi t}{2 \hbar} |V_{ba}|^2 \rho(E_a + \hbar \omega)
\end{aligned}$$

Where we used the aforementioned relation for $$t \to \infty$$ again.
The transition rate is therefore:

$$\begin{aligned}
    \pdv{P_{a \to b}}{t} =
    \boxed{
        R_{a \to b}
        = \frac{\pi}{2 \hbar} |V_{ba}|^2 \, \rho(E_b)
    }
\end{aligned}$$

This is an alternative version of Fermi's golden rule.
Unsurprisingly, if $$E_b$$ is not in a continuum,
then we simply have $$\rho(E_x) = \delta(E_b - E_x)$$ and recover the previous version.



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