Time-Independent Hamiltonian $V(t) = V$

In this case

$$\langle n\vert V_I(t')\vert m\rangle = \langle n\vert e^{iH_0t'} V e^{-iH_0t'}\vert m\rangle = e^{-i (\varepsilon_n-\varepsilon_m)t'} \langle n\vert V\vert m\rangle$$

$$\leadsto P_{m\to n}(t) = \left\vert\langle n\vert V\vert m\rangle \right\vert^2 \left\vert \int_{0}^t dt' e^{-i (\varepsilon_n-\varepsilon_m)t'} \right\vert^2$$

$$= \left\vert\langle n\vert V\vert m\rangle \right\vert^2 \left\vert... ...-i (\varepsilon_n-\varepsilon_m)t}-1}{\varepsilon_n-\varepsilon_m}\right\vert^2$$

$$= \left\vert\langle n\vert V\vert m\rangle \right\vert^2 4 \frac{\sin^2 \frac{\varepsilon_n-\varepsilon_m}{2} t}{(\varepsilon_n-\varepsilon_m)^2 }$$ (3.11)

As for the $\sin^2$ function, we now use the representation of the Dirac Delta-function,

Theorem:

For any integrable, normalised function $f(x)$ with $\int_{-\infty}^{\infty}dx f(x)=1$,

$$\lim_{\varepsilon\to 0 } \frac{1}{\varepsilon}f\left( \frac{x}{\varepsilon}\right)=\delta(x).$$ (3.12)

Here, we apply it with $f(x) = \frac{1}{\pi}\frac{\sin^2(x)}{x^2}$

$$\lim_{\varepsilon\to 0 } \frac{1}{\varepsilon} \frac{1}{\pi}\frac{\sin^2(x/\varepsilon)}{(x/\varepsilon)^2} = \delta(x),\quad \lim_{t\to \infty } \frac{t}{2\pi}\frac{\sin^2( \Delta E t/2 )}{(\Delta E t/2)^2 } = \delta(\Delta E)$$

$$\leadsto \lim_{t\to \infty } \frac{1}{t}P_{m\to n}(t) = \lim_{t\to \infty } \left\vert\langle n\vert V\vert m\rangle \rig... ... \frac{\varepsilon_n-\varepsilon_m}{2} t}{[(\varepsilon_n-\varepsilon_m)t/2]^2}$$

$$= 2\pi \left\vert\langle n\vert V\vert m\rangle \right\vert^2 \delta(\varepsilon_n-\varepsilon_m).$$ (3.13)

This is an extremely important result, and we therefore highlight it here again, introducing the transition rate $\Gamma_{m\to n}$,

$$\Gamma_{m\to n}\equiv \lim_{t\to \infty } \frac{1}{t}P_{m\to n}(t... ...langle n\vert V\vert m\rangle \right\vert^2 \delta(\varepsilon_n-\varepsilon_m)$$ (3.14)

The total transition rate into any final state $\vert n\rangle$ is, within first order perturbationtheory in $V$, given by the sum over all $n$,

$$\Gamma_{m} \equiv \frac{2\pi}{\hbar^2}\sum_n \left\vert\langle n\vert V\vert m\rangle \right\vert^2 \delta(\varepsilon_n-\varepsilon_m)$$

$$\mbox {\rm\bf Fermi's Golden Rule}.$$ (3.15)