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Given an atomic transition with associated E-field $E(t) = E_{0}\cos(\omega_{0}t)e^{-t/\tau}$ where $\omega_{0}$ is the natural line frequency and $\tau$ is the decay constant of the simple harmonic oscillator. Find an expression for the line flux, that is $I(\omega)/I(\omega_{0})$.

I'm trying to do the following Fourier transform: $$ f(\omega)=\int^{\infty}_{-\infty}E_{0}\cos(\omega_{0}t)e^{-t/\tau}e^{-i\omega t}dt $$

I'm not really sure how to do this integral. I tried calculating it from $-\tau/2$ to $\tau/2$ but I get an answer with the sine of a complex number in the numerator that will not go away when I square it to find Intensity. Any help would be appreciated. Thanks

Colin McFaul
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Sam
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  • Should this be migrated to [math.SE]? – David Z Oct 31 '13 at 21:56
  • @DavidZ, possibly. I always like taking a more expansive definition of on-topic for the math questions that come up in physics. I agree that editing in some more of the physics context could help make this more on-topic here. – Colin McFaul Oct 31 '13 at 22:13

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You definitely don't want to do the integral from $-\tau/2$ to $\tau/2$; you need to go all the way to $\pm \infty$ for the integral to be meaningful. In this particular case, as a comment notes, the integral will diverge if you take the lower limit to $-\infty$; physically, you want to start the integral at $t=0$ because that's when the field comes into existence.

Here is a hint towards the easiest way to do this integral (and many Fourier integrals, for that matter): $\cos(x) = \frac{1}{2}(e^{ix} + e^{-ix})$. The integral should be easy from there.

You can also use integration by parts, but that's a little more tedious and error-prone.

Community
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Colin McFaul
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  • I did it from $-\infty$ to $\infty$ but I get this term that's basically $e^{t}$ from $-\infty$ to $\infty$ – Sam Oct 31 '13 at 23:15
  • @Sam You need to do the integral on the "half-infinite" interval $[0,\infty)$, not infinite inteval $(-\infty,\infty)$ (the latter of course diverges). The electric field has a definite beginning time, assumed to be $t=0$. Quantum mechanically, it is the probability amplitude for an excited atom to be found still excited at time $t$ that is important here, and this is $\exp(-i\omega t - t/(2\tau))$ (note the factor $2\tau$, not $\tau$) the electric field follows from applying the right observables to the Wigner-Weiskopf model; see my answer to http://physics.stackexchange.com/q/79146/26076; – Selene Routley Oct 31 '13 at 23:34