Fermi's Golden Rule

Previosly, we have looked at the quantum TD-LRT, now let us put that into action by consider an perturbation that has a harmonic time-dependence (typical of interaction with radiation):

$$V(t) = V e^{-i\omega t} + V^\dagger e^{i\omega t}$$

Then the relevant transition matrix element is:

$$V_{fi}(t) = \braket{f|V(t)|i}e^{-i\omega t} + \braket{f|V^\dagger(t)|i} e^{i\omega t}$$

Eq. 5 in quantum TD-LRT becomes

$$\begin{aligned} {c}_f(t) = {c}_f(t)^{(1)} &= \frac{V_{fi}}{i\hbar} \int_{t_0}^t \delta t' e^{i(\omega_{fi}-\omega)t'} + \int_{t_0}^t\delta t' e^{i(\omega_{fi}+\omega)t'} \\ &=\frac{V_{fi}}{i\hbar} \frac{e^{i(\omega_{fi}-\omega)t}-1}{i(\omega_{fi}-\omega)} + \frac{V_{fi}}{i\hbar} \frac{e^{i(\omega_{fi}+\omega)t}-1}{i(\omega_{fi}+\omega)} \end{aligned}$$

Here, let’s make an assumption that $\omega$ is close to $\omega_fi$ (which is called the Bohr angular frequency of $\psi_f$ and $\psi_i$). In this case, the first term dominates and we can safely ignore the second term.

$$\begin{aligned} {c}_f(t) = {c}_f(t)^{(1)}&= \frac{V_{fi}}{i\hbar} \frac{e^{i(\omega_{fi}-\omega)t}-1}{i(\omega_{fi}-\omega)} \\ &= \frac{V_{fi}}{i\hbar} \frac{e^{i(\omega_{fi}-\omega)t/2}(e^{i(\omega_{fi}-\omega)t/2}-e^{-i(\omega_{fi}-\omega)t/2})}{i(\omega_{fi}-\omega)}\\ &= \frac{V_{fi}}{i\hbar} \frac{e^{i(\omega_{fi}-\omega)t/2}(2i\sin(\frac{\omega_{fi}-\omega t}{2}))}{i(\omega_{fi}-\omega)}\\ &= \frac{V_{fi}}{i\hbar} \frac{e^{i(\omega_{fi}-\omega)t/2}\sin(\frac{(\omega_{fi}-\omega) t}{2})}{(\omega_{fi}-\omega)/2}\\ \end{aligned}$$

The corresponding transition probability is

$$P_{i\to f}(t) = |{c}_f(t)^{(1)}|^2 = \frac{|V_{fi}|^2}{\hbar^2} \left[\frac{\sin(\frac{(\omega_{fi}-\omega) t}{2})}{(\omega_{fi}-\omega)/2}\right]^2$$

!!! note Nascent Dirac delta function

The equation

$$F(\omega_{fi}-\omega,t)=\left[\frac{\sin(\frac{(\omega_{fi}-\omega) t}{2})}{(\omega_{fi}-\omega)/2}\right]^2$$

behaves peaks at $\omega_{fi}$ with a width of $\Delta \omega$: where $\Delta \omega \approx \frac{4pi}{t}$. We see that as $t \to \infty$, this becomes a delta function.

However, this function has a different normalization comparing to a delta function:

$$\int \left[\frac{\sin(\frac{(\omega_{fi}-\omega) t}{2})}{(\omega_{fi}-\omega)/2}\right]^2 \hbar d \omega = \hbar (\frac{2}{t}) t^2 \int \frac{\sin^2(x)}{x^2} d x =2 \pi \hbar t$$

!!!

Converting using Nascent Dirac delta function, we get

$$P_{i\to f}(t) = |{c}_f(t)^{(1)}|^2 = \frac{|V_{fi}|^2}{\hbar^2} 2 \pi \hbar t \delta(\omega_{fi}-\omega) = \frac{2 \pi}{\hbar} |V_{fi}|^2 t \delta(\omega_{fi}-\omega)$$

And the transition rate is

$$W_{i\to f} = P_{i\to f}(t)/t= |{c}_f(t)^{(1)}|^2 = \frac{2 \pi}{\hbar} |V_{fi}|^2 \delta(\omega_{f} - \omega_i-\omega)$$

Which is the Fermi’s Golden Rule

Continuum

Assuming that the final state lives in a continuum of states, we need to account for all states the system can jump to using

$$\begin{aligned} P_{i}(t) &= \int P_{i\to f}(t) \rho(E) dE\\ &= \int \frac{|V_{fi}|^2}{\hbar^2} \left[\frac{\sin(\frac{(\omega_{fi}-\omega) t}{2})}{(\omega_{fi}-\omega)/2}\right]^2 \rho_f(E) dE\\ \end{aligned}$$

since $F(\omega_{fi}-\omega,t)$ acts as a delta function, we can extract $\rho(E)$ as $\rho(E_{fi})$

$$\begin{aligned} P_{i}(t) &= \int \frac{|V_{fi}|^2}{\hbar^2} \left[\frac{\sin(\frac{(\omega_{fi}-\omega) t}{2})}{(\omega_{fi}-\omega)/2}\right]^2 \rho_f(E) dE\\ &= \frac{2\pi}{\hbar}|V_{fi}|^2 \rho_f(E_{fi}) t \end{aligned}$$

And the transition rate is:

$$W = P_{i}(t) /t = \frac{2\pi}{\hbar}|V_{fi}|^2 \rho_f(E_{fi})$$

Which is the Fermi’s Golden Rule in continuum.