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For electron absorption calculations (with an electric field perturbation $\Delta H = eE_0x cos(\omega t)$) we end up with an integral like:

$$c_2(t) \propto \int \rho(\omega) \left( \frac{\sin(\frac{1}{2}(\omega-\omega_0)}{\frac{1}{2} (\omega - \omega_0)} \right)^2 d\omega$$

and only for $t \rightarrow \infty$ do we get the second function tending to a delta function $\delta(\omega - \omega_0)$ which ensures that $E_f = E_i +\hbar \omega$ (i.e. energy is conserved).

My question is why is this only true for $t \rightarrow \infty$?

My guess is that it is just an artifact in the calculation and that, in reality, if you decomposed the electric field as the source is 'switched on' you would initially have a superposition of many different electric field cosines that have each existed essentially forever, and when the source is 'switched on' the relative amplitudes of these terms in the superposition change, but since they haven't just appeared out of nowhere, you can still assume $t \rightarrow \infty$ in the above integral. Am I right?

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  • $\begingroup$ Related: physics.stackexchange.com/q/89402/2451 $\endgroup$ – Qmechanic Feb 23 at 16:22
  • $\begingroup$ Are you asking why the sinc function tends to the delta function only for $t\to \infty$? Well, that's the math (en.wikipedia.org/wiki/Sinc_function). If that is not your question, keep in mind that Fermi's Golden Rule is based on perturbation theory, so it is only an approximation. $\endgroup$ – oliver Feb 23 at 17:07
  • $\begingroup$ @oliver My question is why is energy not conserved at short times, unless it is actually because of the explanation I guess in the question. $\endgroup$ – Alex Gower Feb 23 at 23:07
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The meaning of this equation is not that the system suddenly has different eigen energies and the sinc function does not violate energy conservation. It is better to think about it as a statement about the incoming time limited field rather than the system under investigation, in my opinion.

Any real finite electromagnetic radiation is basically a pulse due to its temporal finiteness. Real electromagnetic radiation has thus not a single frequency but is spread out around its central frequency and the lineshape function that describes that spreading is the sinc function in this case, which is nothing else but the fourier transform of the temporal rectangle function that limits the temporal field to the interval [-T,T].

If T is large, you get a field with a narrow bandwith, if T is small you get a field with a large bandwith. And i think it makes sense that the field with a narrow bandwith has to be more accurate in hitting the central frequency in comparison to the field with a large bandwith.

And in the idealized case of an infinite temporal electromagnetic field, you get a field with a single frequency which would have to hit the transition energy exactly to cause transitions.

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  • $\begingroup$ Okay do you think my explanation guess in the question was OK the right lines then? That the explanation is to do with the inevitable uncertainty involved with finite time waves rather than a physical effect itself. Is there anything in my guess which was drastically wrong? $\endgroup$ – Alex Gower Feb 23 at 23:09
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    $\begingroup$ I must admit that i am not exactly sure what point you are making in your last paragraph, so i can't say for sure whether you are right or wrong. What exactly do you mean with "artifact in the calculation" ? The sinc function in the expression is correct as long as T is finite. The sinc function also does not imply that energy conservation is violated. On a side note, I was also quite confused by the equation in the past and its easy to misinterpret or to overthink it. $\endgroup$ – Hans Wurst Feb 24 at 9:07

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