Why is raman effect weaker than Rayleigh

1. May 27, 2014

HomerSimps

Hello!

I apologise, if I put this problem into the wrong topic.

I know that when the light is shined on the molecule it can get polarized. And because of that you can get Rayleigh and Raman scaterring. Also I know that only one photon over million or billion will scater as Raman.

The question that puzzles me and I can't get the answer is: Why is Raman effect so weak? Why does light scater more as Rayleigh than Raman?

Thank you for you answer.

2. May 27, 2014

DrDu

The polarizability p is a function of nuclear coordinates Q and you can expand it into a Taylor series:
$p(Q)=p|_{Q_\mathrm{eq}}+\frac{\partial p}{\partial Q}(Q-Q_\mathrm{eq})\ldots$. The first term gives rise to Rayleigh scattering while the second one to Raman scattering. However, $Q-Q_\mathrm{eq}$ is very small for one quantum of vibrational excitation. This is related to the nuclei being heavier than the electrons and therefore it is smaller by some power of (m/M) where m is electronic mass and M is nuclear mass. This is the basic result of Born and Oppenheimers analysis.

3. May 28, 2014

Robert_G

I am not sure the this is right. but here is what I thought.

The point is what's difference between the Raman and Rayleigh scattering. For Raman scattering, the atom or molecule goes from the state $|a\rangle$ to a virtual level $|b\rangle$ while absorbing a photon, and then goes to level $|c\rangle$, which is a totally different level of $|a\rangle$, and emit anther photon. The frequencies of the absorbed and emitted photon is changed by $(E_c-E_a)/\hbar$. So this is an inelastic scattering.

For Rayleigh scattering, that atom goes form $|a\rangle$ to the level $|b\rangle$, which is still a virtual level and end in level $|a\rangle$, that's the exact level which the atoms used be on. The frequencies of the absorbed and emitted photon are the same. So this is an elastic scattering.

We have two ways to understand why the Raman Scattering is weak. First, Just look at the change of the atoms. For the Raman scattering, the atoms end up on the higher level. Well, the atoms don't like to be on a higher Energy levels, right? This is why Raman is weak. Here we assumed that $E_a < E_c$, and this is the Stokes Raman scattering. For anti-Stokes Raman scattering, $E_a > E_c$, there are not much populations on the level $E_a$, So it should be weak.

The second way is similar to the first one. we know that $\langle a|c\rangle = 0$. This means the wave-functions of this two states are not overlapped. Let the scattering operation to be $U$, so $U$ can change the state to $U|a\rangle$, if the atom is coming to $|c \rangle$, that is Raman scatter. The probability amplitude should be $\langle c|U|a \rangle$. Well, For Rayleigh, it will be $\langle a|U|a \rangle$. So, which one is larger, it must be $\langle a|U|a \rangle$, because $U$ will not be so weird that change the atomic state so dramatically, and $\langle a|a\rangle = 1$.

Last edited: May 28, 2014