I'm just wondering why stoke-Raman doesn't go to the ground state. Where does the rest of energy from exciting laser leave?
On the other hand, why does anti-stoke go back to ground state? I know that the overall energy of anti-stoke is higher but why it has to go back to the ground state?
My question is based on this diagram.
As I understand, the anti-stoke has higher energy as it is excited from v1 due to an enhancement. I'm just confused with the last process of anti-stoke when it comes down to the ground state. Why doesn't it come down to the v2 or higher vibrational state like stoke?
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2 Answers
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There is no law that a transition must go to the ground state. Now, each molecule has a multitude of electronic states, and within each of them a multitude of vibrational states. Thus each transition between two pure electronic states is accompanied by a variety of related transitions between higher-to-lower and lower-to-higher vibrational states. Which of these are the Stokes lines and which are the anti-Stokes ones, you'd better figure out for yourself.
answered Sep 1, 2015 at 20:11
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It is really a matter of definition. Rayleigh scattering does not involve any change to the internal energy of the molecule whereas Raman transitions do. The Raman transitions always involve a change in the vibrational /rotational energy levels.
If the molecule has sufficient vibrational energy in its ground state then anti-stokes scattering can be observed since then the $v=1$ level has some population.The Stokes line can always occur as it starts from $v=0$. The total energy remains the same: initial photon energy = scattered photon energy $\pm$ vibrational energy. In general 'selection rules' have to be considered when analysing which spectroscopic transitions are allowed and which are forbidden.
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