Perturbation Theory transmission probability

1. Jan 5, 2008

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

2. Jan 5, 2008

Rainbow Child

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

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

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!