# The Bethe Ansatz

#### Fermi's Golden Ruled.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

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