3
$\begingroup$

I'm trying to understand the amplitudes and the ratio between those of Stokes and anti-Stokes scattering in Raman scattering. I understand that from a Boltzmann distribution point of view, the ratio between Stokes and anti-Stokes scattering should be: $$\frac{I_S}{I_{aS}} = e^\frac{E}{k_BT}$$ where $E$ is the energy of the phonon, since this is the ratio of the population of excited and ground state phonons, and because the Stokes process starts with the ground state, and the anti-Stokes with the excited state.

However, I've seen (for example in 'Raman Spectroscopy in Graphene Related Systems' by Jorio) that $I_S \propto n+1 $ and $I_{aS} \propto n $ where $n$ is the thermal population of the phonon $\left( n = \frac{1}{e^\frac{E}{k_BT}-1}\right)$, such that $\frac{n+1}{n}$ gives the usual Boltzmann ratio. I'm not sure about this approach; why should the Stokes scattering amplitude be proportional to the thermal population of phonons plus 1? I understand that in a Stokes process the system goes from $n$ phonons to $n+1$, but I cannot relate this to the proportionality relation, and similarly for the anti-Stokes process.

$\endgroup$

1 Answer 1

0
+100
$\begingroup$

The intensity of the Stokes and anti-Stokes emission peaks $I_S$ and $I_{aS}$ corresponds respectively to the amount of phonon (boson) creation $a^\dagger$ and annihilation $a$. Hence, as for any bosons, $$I_S ∝ |<n+1|a^\dagger|n>|^2=n+1$$ and $$I_{aS} ∝ |<n-1|a|n>|^2=n.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.