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Can someone explain this formula to me? Given a wave packet whose time evolution is $g(t)$, a partially resolved spectrum is found by Fourier transforming its overlap with the same wave packet at time $t=0$, $g$. $$E_T(\omega) = (1/2\pi)\int_{-T}^T \exp(i\omega t)<g|g(t)> dt \\ = (1/\pi)\sum_n \frac {\sin(E_n/\hbar - \omega)T}{(E_n/\hbar - \omega)}~|\langle g|\psi_n\rangle |^2.$$

I don't understand how the second formula is found.

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This is but a sum of cardinal sine functions with peaks at $\omega = E_n/\hbar$, for all n. Absorb $\hbar$ into the Es out of respect for sanity.

Recall $$ |g\rangle=\sum_n |\psi_n\rangle \langle \psi_n|g\rangle, \quad \Longrightarrow \quad |g(t)\rangle=\sum_n |\psi_n\rangle e^{-iE_nt}\langle \psi_n|g\rangle , $$ so that $$ (1/2\pi)\int_{-T}^T \!\!dt ~ e^{i\omega t}\langle g|g(t)\rangle = (1/2\pi)\int_{-T}^T \!\!dt ~ e^{i\omega t}\sum_n e^{-iE_n t}|\langle g|\psi_n\rangle |^2 \\ = (1/\pi)\sum_n\frac{\sin (E_n - \omega)T}{E_n - \omega} ~|\langle g|\psi_n\rangle |^2 ,$$ upon integration.

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  • $\begingroup$ Thank you very much! So clear! $\endgroup$ – Priuk Aug 28 '19 at 12:44
  • $\begingroup$ So you clicked on the check mark? $\endgroup$ – Cosmas Zachos Sep 3 '19 at 14:04

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