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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.
enter image description here
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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    $\begingroup$ Hello, and welcome to Chemistry SE. Feel free to look around, and take the Tour available under 'help' up on the top menu bar. As it stands, your question does not demonstrate much research (even Wikipedia) on Raman scattering. What is unclear? $\endgroup$ – Jon Custer Sep 1 '15 at 18:39
  • $\begingroup$ I'm sorry for this stupid question. My question is based on this diagram. upload.wikimedia.org/wikipedia/commons/8/87/… 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? $\endgroup$ – user19568 Sep 2 '15 at 2:12
  • $\begingroup$ You can edit your question to make it more precise. || @JonCuster You can directly link to the tour with [tour]. $\endgroup$ – Martin - マーチン Sep 2 '15 at 8:33
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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.

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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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