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Time-Independent Hamiltonian \bgroup\color{col1}$ V(t) = V$\egroup

In this case
$\displaystyle \langle n\vert V_I(t')\vert m\rangle$ $\displaystyle =$ $\displaystyle \langle n\vert e^{iH_0t'} V e^{-iH_0t'}\vert m\rangle
= e^{-i (\varepsilon_n-\varepsilon_m)t'} \langle n\vert V\vert m\rangle$  
$\displaystyle \leadsto P_{m\to n}(t)$ $\displaystyle =$ $\displaystyle \left\vert\langle n\vert V\vert m\rangle \right\vert^2
\left\vert \int_{0}^t dt' e^{-i (\varepsilon_n-\varepsilon_m)t'} \right\vert^2$  
  $\displaystyle =$ $\displaystyle \left\vert\langle n\vert V\vert m\rangle \right\vert^2
\left\vert...
...-i (\varepsilon_n-\varepsilon_m)t}-1}{\varepsilon_n-\varepsilon_m}\right\vert^2$  
  $\displaystyle =$ $\displaystyle \left\vert\langle n\vert V\vert m\rangle \right\vert^2 4 \frac{\sin^2 \frac{\varepsilon_n-\varepsilon_m}{2} t}{(\varepsilon_n-\varepsilon_m)^2 }$ (3.11)

As for the \bgroup\color{col1}$ \sin^2$\egroup function, we now use the representation of the Dirac Delta-function,

Theorem:

For any integrable, normalised function \bgroup\color{col1}$ f(x)$\egroup with \bgroup\color{col1}$ \int_{-\infty}^{\infty}dx f(x)=1$\egroup,

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

Here, we apply it with \bgroup\color{col1}$ f(x) = \frac{1}{\pi}\frac{\sin^2(x)}{x^2}$\egroup

$\displaystyle \lim_{\varepsilon\to 0 } \frac{1}{\varepsilon}
\frac{1}{\pi}\frac{\sin^2(x/\varepsilon)}{(x/\varepsilon)^2}$ $\displaystyle =$ $\displaystyle \delta(x),\quad
\lim_{t\to \infty } \frac{t}{2\pi}\frac{\sin^2( \Delta E t/2 )}{(\Delta E t/2)^2 }
= \delta(\Delta E)$  
$\displaystyle \leadsto \lim_{t\to \infty } \frac{1}{t}P_{m\to n}(t)$ $\displaystyle =$ $\displaystyle \lim_{t\to \infty } \left\vert\langle n\vert V\vert m\rangle \rig...
... \frac{\varepsilon_n-\varepsilon_m}{2} t}{[(\varepsilon_n-\varepsilon_m)t/2]^2}$  
  $\displaystyle =$ $\displaystyle 2\pi \left\vert\langle n\vert V\vert m\rangle \right\vert^2 \delta(\varepsilon_n-\varepsilon_m).$ (3.13)

This is an extremely important result, and we therefore highlight it here again, introducing the transition rate \bgroup\color{col1}$ \Gamma_{m\to n}$\egroup,
$\displaystyle \Gamma_{m\to n}\equiv \lim_{t\to \infty } \frac{1}{t}P_{m\to n}(t...
...langle n\vert V\vert m\rangle \right\vert^2 \delta(\varepsilon_n-\varepsilon_m)$     (3.14)

The total transition rate into any final state \bgroup\color{col1}$ \vert n\rangle$\egroup is, within first order perturbationtheory in \bgroup\color{col1}$ V$\egroup, given by the sum over all \bgroup\color{col1}$ n$\egroup,
$\displaystyle \Gamma_{m}$ $\displaystyle \equiv$ $\displaystyle \frac{2\pi}{\hbar^2}\sum_n \left\vert\langle n\vert V\vert m\rangle \right\vert^2 \delta(\varepsilon_n-\varepsilon_m)$  
    $\displaystyle \mbox {\rm\bf Fermi's Golden Rule}.$ (3.15)


next up previous contents index
Next: Higher Order Perturbation Theory Up: First Order Perturbation Theory Previous: First Order Perturbation Theory   Contents   Index
Tobias Brandes 2005-04-26