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How Raman Spectroscopy works?

by blaisem
Tags: raman, spectroscopy
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blaisem
#1
Nov2-11, 07:44 AM
P: 16
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!
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Zarqon
#2
Nov2-11, 12:39 PM
P: 229
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.
blaisem
#3
Nov21-11, 12:01 PM
P: 16
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:

http://en.wikipedia.org/wiki/Self-phase_modulation

http://en.wikipedia.org/wiki/Optical_Kerr_effect

kye
#4
Dec2-13, 03:58 AM
P: 168
How Raman Spectroscopy works?

Quote Quote by Zarqon View Post
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.
Zargon and others. Just like the original poster, I've been reading up about this too and the source doesn't give details like this. Anyway. Can you give an actual example where you can change the value of the the stroke shift (increasing the wavenumber of decrease in the frequency) when the same molecules are probed but the polarization, etc. were altered prior to each Raman probe and what kind of alteration can be done to change the raman stroke value?
Zarqon
#5
Dec4-13, 08:01 AM
P: 229
I'm not sure I understand exactly what you're asking. The Stokes (no 'r') shift is not chosen directly, it's a consequence of the atomic level structure, which is why raman spectroscopy is very good for certains things like determining vibrational of rotational levels for a medium. You can of course change the stokes shift indirectly by applying an external field that changes the level structure of the atoms, which then in turn changes the Stokes shift, but it didn't seem like you were implying this.

I didn't get what you meant about the polarization.
kye
#6
Dec4-13, 09:18 AM
P: 168
Quote Quote by Zarqon View Post
I'm not sure I understand exactly what you're asking. The Stokes (no 'r') shift is not chosen directly, it's a consequence of the atomic level structure, which is why raman spectroscopy is very good for certains things like determining vibrational of rotational levels for a medium. You can of course change the stokes shift indirectly by applying an external field that changes the level structure of the atoms, which then in turn changes the Stokes shift, but it didn't seem like you were implying this.

I didn't get what you meant about the polarization.
Yes. I'm inquiring about the change in the stokes shift indirectly by "applying an external field that changes the level structure of the atoms, which then in turn changes the Stokes shift" as you described. Let's take the case of h20 or water. It is said there is stretching vibrational peak for OH at 3400 cm^-1 and a deformed and weak vibrational peak for HOH at 1635 cm^-1 on average. So to create new shifts at say 2000 cm^-1, where would this come from? what modes affected? Can you give an example with water. About polarization. I read this at wiki "A change in the molecular polarization potential—or amount of deformation of the electron cloud—with respect to the vibrational coordinate is required for a molecule to exhibit a Raman effect". Thanks.


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