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

Fermi’s golden rule

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, 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 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 $\delta$ . 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}

Proof

Proof. 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}

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.

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:

\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

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