0
$\begingroup$

I'm following the derivation of the Kramer-Heisenberg-Dirac formula from the book "Modern Optical Spectroscopy" and I'm having trouble understanding the wording and the derivation of the transition probability of Raman Scattering. The coefficient of the final state $b$ is given by \begin{equation} \begin{aligned} C_{b}(t)=& \alpha_{b a}\left[\frac{\exp \left[i\left(E_{b}-E_{a}+h v_{s}-h v_{e}\right) t / \hbar\right]-1}{E_{b}-E_{a}+h v_{s}-h v_{e}}\right.\\ &+\frac{\exp \left[i\left(E_{b}-E_{a}-h v_{s}-h v_{s}\right) t / \hbar\right]-1}{E_{b}-E_{a}-h v_{s}-h v_{e}} ] \end{aligned} \end{equation} Where $\nu_s$ and $\nu_e$ is the frequency of the incoming and the emitted electric field respectively.

Now, I understand that the first term corresponds to Rayleigh and Raman Scattering and that the first term goes to $it/\hbar$ in Rayleigh scattering but the book states that the second term is negligible for any positive values of $\nu_s$ and $\nu_e$ and I just don't see it. If $E_a = E_b$ and $\nu_s = \nu_e$ as for Rayleigh scattering, surely the second term also goes to $it/\hbar$. Am I missing something?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.