The Bethe Ansatz

Fermi's Golden Rule d.b.F

Introducing some perturbation into a system generates some generally very complex time-dependent behaviour. One way to picture this time dependence is to still consider the original unperturbed basis of states, but have time-dependent state amplitudes. The probability of finding the system in a given state thus becomes time-dependent. For a small perturbation, the rates at which this probability flows from one state to another is given by Fermi's Golden rule.

Let us consider an exactly-solvable, time-independent Hamiltonian \(H_0\) for which we know a basis of eigenstates

\begin{equation*} H_0 |\alpha^0 \rangle = E_{\alpha^0} |\alpha^0 \rangle. \end{equation*}

Let us consider perturbing this theory with a time-dependent operator which is adiabatically turned on from \(t = -\infty\) onwards:

\begin{equation*} V(t) = V e^{-i \omega t + \eta t}, \hspace{20mm} \eta \rightarrow 0^+ \end{equation*}

(we will evaluate this \(t\) for times much less than \(1/\eta\)). Here, \(V\) is some perturbing operator in the Schrödinger representation.

We now address the following question. If the initial state is

\begin{equation*} |\psi^S (t = t_0) \rangle = |\alpha^0_i \rangle \end{equation*}

for some initial state \(|\alpha^0_i\rangle\), what is the probability amplitude for finding the system in state \(|\alpha^0_f \rangle\) (with \(i \neq f\)) at time \(t\)?

In the interaction representation, we had \(|\psi^I (t) \rangle = U^I (t, t_0) |\psi^I (t_0) \rangle.\) Since \(|\psi^I(t) \rangle = e^{\frac{i}{\hbar} H_0 t} |\psi^S(t) \rangle\), we have \(|\psi^I(t_0)\rangle = e^{\frac{i}{\hbar} H_0 t_0} |\alpha^0_i \rangle\) and thus

\begin{equation*} |\psi^S(t) \rangle = e^{-\frac{i}{\hbar} H_0 t} U^I (t, t_0) e^{\frac{i}{\hbar} H_0 t_0} |\alpha^0_i\rangle. \end{equation*}

Consider now calculating the amplitude for being in state \(|\alpha^0_f\rangle\), \(f \neq i\), at time \(t\):

\begin{equation*} \langle \alpha^0_f | \psi^S(t) \rangle = \langle \alpha^0_f | e^{-\frac{i}{\hbar} H_0 t} U^I (t, t_0) e^{\frac{i}{\hbar} H_0 t_0} | \alpha^0_i \rangle = e^{-\frac{i}{\hbar} E_{\alpha^0_f} t + \frac{i}{\hbar} E_{\alpha^0_i} t_0} \langle \alpha^0_f | U^I (t, t_0) | \alpha^0_i \rangle. \end{equation*}

Using the series expansion for the propagator and keeping only the linear (in \(V\)) response, the matrix element of the propagator can be written

\begin{equation*} \langle \alpha^0_f | U^I (t, t_0) | \alpha^0_i \rangle = \langle \alpha^0_f | \alpha^0_i \rangle - \frac{i}{\hbar} \int_{t_0}^t dt' \langle \alpha^0_f | V^I(t') | \alpha^0_i \rangle + ... \end{equation*}

The first term vanishes (we are looking for transition rates, so the initial and final states are different, \(f \neq i\)). The second term gives (substituting the explicit form of the perturbation)

\begin{equation*} -\frac{i}{\hbar} \int_{t_0}^t dt' \langle \alpha^0_f| e^{\frac{i}{\hbar} H_0 t'} V (t') e^{-\frac{i}{\hbar} H_0 t'} |\alpha^0_i \rangle = -\frac{i}{\hbar} \int_{t_0}^t dt' e^{\frac{i}{\hbar} [E_{\alpha^0_f} - E_{\alpha^0_i} - \hbar \omega - i \hbar \eta] t'} \langle \alpha_f^0 | V | \alpha_i^0 \rangle. \end{equation*}

Since the matrix element is time-independent, we can perform the time integral, giving us

\begin{equation*} \langle \alpha^0_f | U^I (t, t_0) | \alpha^0_i \rangle = -\frac{\langle \alpha_f^0 | V | \alpha_i^0 \rangle}{E_{\alpha^0_f} - E_{\alpha^0_i} - \hbar \omega - i \hbar \eta} e^{\frac{i}{\hbar} [E_{\alpha^0_f} - E_{\alpha^0_i} - \hbar \omega - i \hbar \eta] t'}|_{t_0}^t + \mbox{O} (V^2). \end{equation*}

Let us now take the limit \(t_0 \rightarrow -\infty\), with \(\eta \rightarrow 0^+\) but \(\eta t_0 \rightarrow -\infty\) while keeping \(t\) finite (that is, the perturbation is turned on at a vanishingly slow rate from the infinite past). This gives us

\begin{equation*} \langle \alpha^0_f | \psi^S(t) \rangle = \frac{\langle \alpha_f^0 | V | \alpha_i^0 \rangle}{\hbar \omega - (E_{\alpha^0_f} - E_{\alpha^0_i}) + i\eta \hbar} e^{-\frac{i}{\hbar} E_{\alpha_i^0} (t-t_0)} e^{-i\omega t + \eta t} + O(V^2). \end{equation*}

The probability to be in state \(|\alpha^0_f\rangle\) at time \(t\), given that we initially started in state \(|\alpha^0_i\rangle\) is thus

\begin{equation*} P_{f \leftarrow i} (t) = |\langle \alpha^0_f | \psi^S(t) \rangle|^2 = \frac{|\langle \alpha_f^0 | V | \alpha_i^0 \rangle|^2 e^{2\eta t}}{(\hbar \omega - (E_{\alpha^0_f} - E_{\alpha^0_i}))^2 + \eta^2 \hbar^2}. \end{equation*}

The rate at which this probability changes is thus

\begin{equation*} \frac{d}{dt} P_{f \leftarrow i} (t) = |\langle \alpha_f^0 | V | \alpha_i^0 \rangle|^2 \lim_{\eta \rightarrow 0^+} \frac{2\eta}{(\hbar \omega - (E_{\alpha^0_f} - E_{\alpha^0_i}))^2 + \eta^2 \hbar^2}. \end{equation*}

Using the representation of the Dirac delta function \(\delta(x) = \lim_{\eta \rightarrow 0^+} \frac{1}{\pi} \frac{\eta}{x^2 + \eta^2}\) then yields Fermi's Golden rule

\begin{equation} \frac{d}{dt} P_{f \leftarrow i} (t) = \frac{2\pi}{\hbar} |\langle \alpha_f^0 | V | \alpha_i^0 \rangle|^2 \delta (\hbar \omega - (E_{\alpha^0_f} - E_{\alpha^0_i})) \tag{Fgr}\label{Fgr} \end{equation}

Creative Commons License Except where otherwise noted, all content is licensed under a Creative Commons Attribution 4.0 International License.

Author: Jean-Sébastien Caux

Created: 2024-01-18 Thu 14:24