|Nov2-11, 07:44 AM||#1|
How Raman Spectroscopy works?
Hello, I am currently in an analytical chemistry course where I am required to write a report on Raman Spectroscopy. The typical description of Raman Spectroscopy is the excitation of an electron to a "virtual" state, whereupon it can relax to a non zero vibrational level in its ground transitional state and scatter light of a correspondingly different frequency. The difference in the frequencies of the incident and scattered light account for the "Raman Shifts" of a molecule, which can be organized into a spectrum according to wavenumber.
My instructor has informed that this definition is wrong, and that this "virtual" state does not exist. I am trying to find the real way Raman Spectroscopy works, with little success. I have checked a few journal articles and a couple of books, as well as numerous websites. Most provide the traditional description. Some have referenced the polarization of the light changing and interaction with phonons, but do not explain specifically how such interactions can cause a change in the polarization, or how a change in polarization relates to a change in frequency (aka Raman Shift) to generate a spectrum. At least, I have failed to understand the relationship.
Is anyone here able to provide some insight into how Raman really works, then? Or at least clarify this relationship between phonons, changes in polarization of light, and changes in the frequency of light? References to appropriate sources or direct explanations are equally appreciated!
Thank you very much for your time!
|Nov2-11, 12:39 PM||#2|
A Raman interaction is typically a 2-photon process, where the photons used for excitation and de-excitation respectively, have to occur simultaneously. You can drive Raman transitions directly by applying two laser beams, separated by some frequency δω, corresponding to the Raman levels. By changing the difference frequency you can thus easily probe the whole level structure and perform spectroscopy.
The important significance of the fact that it is a 2-photon interaction (and not two events followed after each other sequentially), is that the atom never really spends any time in the excited state. This means for example that the frequency resolution of the Raman spectroscopy is independent of the line width of the excited state, which may otherwise be limiting.
Note however, that two photons interact by themselves only extremely weakly, and so they need the electronically excited state in a Raman process to act like a mediator, i.e. a catalyst for the process, hence the dependence of the detuning to the excited state.
Regarding the polarization changes, the spin is always a conserved quantity, so if your two incoming laser beams have π- and σ+ -polarization respectively, then a total of 1 spin is deposited to the atoms, meaning the atom have to change it's internal spin by +1. This may only be possible for certain incoming angles with respect to your quantizing B-field axis.
Hope that helped at least somewhat.
|Nov21-11, 12:01 PM||#3|
This is very helpful and has been very interesting for me to read. It's strange that this information is not available in the textbooks we use. Thank you for taking the time to reply!
After reading Zarqon's post, I also looked up self-phase modulation, and the whole thing made a little more sense. The Wikipedia pages I found might provide a nice addendum to Zarqon's post and a starting point for anyone else who is also looking for an explanation of Raman Spectroscopy:
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