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Perturbation Theory transmission probability

  1. Jan 5, 2008 #1
    I'm trying to bridge the gap between several expressions describing the insertion of a constant perturbation:

    [tex]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}[/tex]

    so:

    [tex]|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}}[/tex]

    ^ 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:

    [tex]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})[/tex] for large t

    Which baffles me further - factors of [tex]\pi[/tex] have krept in somehow!

    Would be more than thankful is someone would be able to explain these steps! Thanks in advance :smile:
     
  2. jcsd
  3. Jan 5, 2008 #2
    The numerator is of the form

    [tex]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}[/tex]

    so squaring that

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

    For a proper [itex]\phi(x)[/itex], holds

    [tex]\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)[/tex]

    that's why the [itex]\pi[/itex] disappears! :smile:
     
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