E Fermi's Golden Rule

   The time dependent Schrödinger Equation can be written as
 $$\begin{gather} (H_0 + H_1)\psi = \text{i}\hbar\frac{\partial\psi}{\partial t}, \end{gather}$$
where H₁ is a small perturbation to the Hamiltonian H₀. The solution of the unperturbed case is
 $$\begin{gather} H_0\psi_n=\hbar\omega_n\psi_n \quad\text{with}\quad \int\vert\psi_n\vert^2\,d^3\mathbf{r} = 1 \quad\text{and}\quad n=0,1,2,3,\ldots \end{gather}$$
The eigenvalues are $\hbar\omega_n$, and the orthonormal eigenfunctions are $\psi_n$. Under the assumption that the perturbation H₁ is small, the wave function for the perturbed case can be expanded in a series of the orthonormal wave functions of the unperturbed case [29].
 $$\begin{gather} \psi=\sum_n c_n(t)\psi_n e^{-\text{i}\omega_n t} \end{gather}$$
Inserting (E.3) into the time dependent Schrödinger equation (E.1), and using (E.2) gives
$$\begin{gather}\text{i}\hbar\sum_n\frac{dc_n(t)}{dt}\psi_n e^{-\text{i}\omega_n t} = \sum_n c_n(t)H_1\psi_n e^{-\text{i}\omega_n t}. \end{gather}$$
If this is multiplied by $\psi_m^*\exp(\text{i}\omega_m t)$, and integrated over the volume one obtains
 $$\begin{gather} \text{i}\hbar\frac{dc_m(t)}{dt} = \sum_n c_n(t)\int\psi_m^*H_1\psi_n\, d^3\boldsymbol{r}\, e^{\text{i}t(\omega_m-\omega_n)}, \end{gather}$$
since orthonormality makes
$$\begin{gather}\int\psi_m^*\psi_n\,d^3\mathbf{r} = \delta_{mn}, \end{gather}$$
which vanishes for $m \neq n$ and equals unity for m=n. With the definitions
$$\begin{gather}\omega_{mn} = \omega_m - \omega_n \quad\mbox{and}\quad H_{mn} = \int\psi_m^*H_1\psi_n\,d^3\mathbf{r} \end{gather}$$
one can write (E.5) as
 $$\begin{gather} \text{i}\hbar\frac{dc_m(t)}{dt}=\sum_n c_n(t)H_{mn}e^{\text{i}\omega_{mn}t}. \end{gather}$$
In general, when the perturbation is turned on at t=0, it may be assumed that the system is in state m at t=0, and c_m(0)=1, while c_n(0)=0for all other states. Furthermore, it is assumed that the scattering out of the initial state is small, so that c_m(t)=1 for all time. This assumption neglects the  conservation of particles. Considering these assumptions, one can solve (E.8).
$$\begin{gather}c_n(t)= -\frac{\text{i}}{\hbar}\int_0^t H_{mn}e^{\text{i}\omega_{m... ...c{H_{mn}}{\hbar\omega_{mn}}\left(1-e^{\text{i}\omega_{mn}t}\right). \end{gather}$$
The probability of occupying state n is then
$$\begin{gather}\vert c_n(t)\vert^2=\frac{4\vert H_{mn}\vert^2t^2}{\hbar^2}\frac{\sin^2(\frac{\omega_{mn}t}{2})}% {\omega_{mn}^2t^2}. \end{gather}$$
For times $t\rightarrow\infty$, which are long enough such that the scattering process has been completed, the function
$$\begin{gather}\frac{\sin^2\left(\frac{\omega_{mn}t}{2}\right)}% {\left(\frac{\om... ...\right)}{\left(\frac{\omega_{mn}}{2}\right)^2}\,d\omega_{mn}=2\pi t \end{gather}$$
has the property of a $\delta$-function (see Fig. E.1).

  

Figure: The function $\protect\sin^2(\protect\omega_{mn}t/2)/(\protect\omega_{mn}/2)^2$ has the property of a $\protect\delta$-function.

$$\begin{gather}2\pi t\delta(\omega_{mn})=2\pi\hbar t\delta(\hbar\omega_{mn}). \end{gather}$$
The transition probability per unit time, the   transmission rate, is then given by the so called Fermi Golden Rule
$$\begin{gather}\Gamma = \frac{d\vert c_n(t)\vert^2}{dt}=\frac{2\pi}{\hbar}\vert H_{mn}\vert^2\delta(E_m-E_n). \end{gather}$$
This means that a scattering process, in our case a tunnel event, only takes place if the energy of the particle is conserved, E_m = E_n.