Fermi's golden rule

Perturbation of a quantum system

Consider a quantum system that is characterized by a Hamiltonian \(\hat{H}_0\). Note that its eigenstates are represented by \(\Phi_{n}=\phi_{n}\exp(-iE_n t/\hbar), n=1,2,\dots\), thus the time-independent Schrödinger equation is given by

\[ \hat{H}_0\phi_{n} = E_{n}\phi_{n} \]

Now, we perturb the system by adding a perturbation \(\hat{\mathcal{V}}(t)\), by which the new Hamiltonian becomes \(\hat{H} = \hat{H}_0 + \hat{\mathcal{V}}(t)\), and we try to solve the new Schrödinger equation. Note that those eigenstates constitute a complete set, thus,

\[ \begin{aligned} \underbrace{\left( \hat{H}_0 + \hat{\mathcal{V}}(t)\right)}_{\bar{H}} \underbrace{\sum_{j} c_{j}(t) \Phi_{j}}_{\Phi} = i\hbar \frac{\partial}{\partial t} \underbrace{\sum_{j} c_{j}(t) \Phi_{j}}_{\Phi} \end{aligned} \]

Due to that for any \(j\), \(\hat{H}\Phi_j = i\hbar\frac{\partial}{\partial t}\Phi_j\), the above equation can be rewritten by cancelling stuff related with the original Hamiltonian \(\hat{H}_0\),

\[ \hat{\mathcal{V}}(t) \sum_{j} c_j (t) \Phi_{j}= i\hbar \sum_j \Phi_{j} \frac{\mathrm{d} c_j (t)}{\mathrm{d} t} \]

Mathematically, for any \(f\), we can try to multiply \(\Phi_f^*\) on both sides of the above equation, and then integrate over the whole space, thus,

\[ \sum_{j} c_j (t) \Braket{\Phi_f | \hat{\mathcal{V}}(t) | \Phi_j} = i\hbar \frac{\mathrm{d} c_f (t)}{\mathrm{d} t} \]

The braket on the left side can be further expressed as,

\[ \Braket{\Phi_f | \hat{\mathcal{V}}(t) | \Phi_j} = \underbrace{\Braket{\phi_f | \hat{\mathcal{V}}(t) | \phi_j}}_{\equiv \mathcal{V}_{fj}} \exp\left( \frac{i(E_f - E_j)t}{\hbar} \right) \]

Therefore, our task to study the perturbated system is then relying on the following equations:

\[ i\hbar \frac{\mathrm{d}c_f (t)}{\mathrm{d}t} = \sum_{j} \mathcal{V}_{fj} c_j (t) e^{\frac{i (E_f-E_j)t}{\hbar}},\quad \forall f \]

Notice that the above equation is a set of coupled differential equations that depict the transitions from different states \(\phi_{n}\). It is then clear to say, if we want to study the transition from state \(i\) to state \(f\), we will give the initial conditions:

\[ c_j(t)\Big|_{t=0} = \delta_{ij} \]

1st-order approximation of time-independent perturbation

We now consider a simple case where the perturbation is time-independent, i.e., \(\hat{\mathcal{V}}(t) = \hat{\mathcal{V}}\). Also, we only consider the transition from state \(i\) to state \(f\), that is to set \(c_j(t)|_{t=0}=\delta_{ij}\). The first-order approximation of such problem is given by,

\[ i\hbar \frac{\mathrm{d}c_f (t)}{\mathrm{d}t} = \mathcal{V}_{fi} \cancelto{1}{c_i (t)} e^{\frac{i (E_f-E_i)t}{\hbar}} \]

\[ \Rightarrow c_f(t) = -\frac{\mathcal{V}_{fi}}{E_f-E_i} e^{\frac{i (E_f-E_i)t'}{\hbar}} \Big|_{t'=0}^{t'=t} = -\frac{\mathcal{V}_{fi}t}{\hbar} \frac{\sin ((E_j-E_i)t/2\hbar)}{(E_j-E_i)t/2\hbar} i e^{i(E_j-E_i)t/2\hbar} \]

The module of the final state \(f\) is then given by,

\[ \begin{aligned} \left|c_f (t)\right|^2 &= |\mathcal{V}_{fi}|^2 \left(\frac{t}{\hbar}\frac{\sin ((E_j-E_i)t/2\hbar)}{(E_j-E_i)t/2\hbar} \right)^2\\ &= \frac{2\pi}{\hbar} |\mathcal{V}_{fi}|^2 t \times\delta(E_f-E_i) \end{aligned} \]

When we are talking about a quantum mode that has a frequency \(\sim\omega\), the time scale should be much larger than \(1/\omega\). It is normal that the scale of energy difference is \(\sim \hbar\omega\). Thus, except for the case where \(E_f=E_i\), the term \(\frac{(E_j-E_i)t}{2\hbar} \gg \frac{(E_j-E_i)}{\hbar\omega}\sim 1\), making the term \(\sin ((E_j-E_i)t/2\hbar)/((E_j-E_i)t/2\hbar)\) to be either \(1\) or \(0\). Also, for t much larger than \(1/\omega\), with \(|\mathcal{V}|_{fi}\) the same magnitude as \(\hbar\omega\), the whole term is then proportional to the dirac delta function \(\delta(E_f-E_i)\).

After careful examination, it is written as,

\[ \left(\frac{t}{\hbar}\frac{\sin ((E_j-E_i)t/2\hbar)}{(E_j-E_i)t/2\hbar} \right)^2 = \frac{2\pi}{\hbar} t \times\delta(E_f-E_i) \]

due to that \(\int_{-\infty}^{\infty} (\sin x)^2/x^2 \mathrm{d}x = \pi\).

The transition rate can be read from \(|c_f (t)|^2\):

\[ W_{fi} = \frac{2\pi}{\hbar} |\mathcal{V}_{fi}|^2 \delta(E_f-E_i) \]

This is the very golden rule that we are looking for. It is the basis of the derivation of the scattering rate in the previous post, where we use the anharmonic potential as the perturbation with respect to the harmonic Hamiltonian.

Reference

1. Davies, John H., Scattering Rates: the Golden Rule (A chapter from the book The Physics of Low-dimensional Semiconductors: An Introduction).
2. Griffths, David J., Introduction to Quantum Mechanics.