Perturbation Theory transmission probability

[raintrek]

I'm trying to bridge the gap between several expressions describing the insertion of a constant perturbation:

a_{f}(t) = \frac{1}{i\hbar} V_{fi} \int^{t}_{0} e^{i(E_{f}-E_{i})t'/\hbar}dt' = \frac{1}{i\hbar}V_{fi}\frac{e^{i(E_{f}-E_{i})t/\hbar} - 1}{i(E_{f}-E_{i})/\hbar}

so:

|a_{f}(t)|^{2} = \frac{1}{\hbar^{2}}|V_{fi}|^{2} 2 \frac{1 - cos(E_{f} - E_{i})t/\hbar}{(E_{f} - E_{i})^{2}/\hbar^{2}}

^ I'm not sure how they arrive at that step... I assume it's something to do with the Euler equation, but I can't see how if it is where the sin terms disappear to...

Furthermore, they state that:

P_{if} = \frac{d}{dt}|a_{f}(t)|^{2} = \frac{2\pi}{\hbar}|V_{fi}|^{2}\frac{sin(E_{f} - E_{i})t/\hbar}{\pi(E_{f} - E_{i})} \stackrel{}{\rightarrow}\frac{2\pi}{\hbar}|V_{fi}|^2\delta(E_{f} - E_{i}) for large t

Which baffles me further - factors of \pi have krept in somehow!

Would be more than thankful is someone would be able to explain these steps! Thanks in advance

 

[Rainbow Child]

I'm trying to bridge the gap between several expressions describing the insertion of a constant perturbation:

a_{f}(t) = \frac{1}{i\hbar} V_{fi} \int^{t}_{0} e^{i(E_{f}-E_{i})t'/\hbar}dt' = \frac{1}{i\hbar}V_{fi}\frac{e^{i(E_{f}-E_{i})t/\hbar} - 1}{i(E_{f}-E_{i})/\hbar}

The numerator is of the form

e^{i\,x}-1=e^{i\,x/2}(e^{i\,x/2}-e^{-i\,x/2})=e^{i\,x/2}\,2\,i\,\sin\frac{x}{2}\Rightarrow |e^{i\,x}-1|=2\,\sin\frac{x}{2}

so squaring that

|e^{i\,x}-1|^2=4\,\sin^2\frac{x}{2}=4\,\frac{1-\cos x}{2}=2\,(1-\cos x)

Furthermore, they state that:

P_{if} = \frac{d}{dt}|a_{f}(t)|^{2} = \frac{2\pi}{\hbar}|V_{fi}|^{2}\frac{sin(E_{f} - E_{i})t/\hbar}{\pi(E_{f} - E_{i})} \stackrel{}{\rightarrow}\frac{2\pi}{\hbar}|V_{fi}|^2\delta(E_{f} - E_{i}) for large t

Which baffles me further - factors of \pi have krept in somehow!

For a proper \phi(x), holds

\lim_{k\rightarrow \infty}\int_{-\infty}^{+\infty}\frac{\sin(k\,x)}{\pi\,x} \phi(x)\,d\,x=\phi (0)\Rightarrow \lim_{k\rightarrow \infty}\int_{-\infty}^{+\infty}\frac{\sin(k\,x)}{\pi\,x}\,d\,x=\delta (0)

that's why the \pi disappears!