Talk:Raman spectroscopy

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Lots of thoughts/questions[edit]

  • I like the tennis ball + spring system analogy presented below. It could be cleaned up and with a few diagrams could make this article much more approachable to newcomers like myself. (talk) 08:46, 9 April 2012 (UTC)
  • To what extent do you need to understand quantum mechanics to begin interpreting Raman data, specifically SERS? How about electromagnetics? Could these prerequisites be simply motivated and explained in this article? (talk) 08:46, 9 April 2012 (UTC)
  • I keep hearing about selection rules, are these related to quantum mechanics somehow? (talk) 08:46, 9 April 2012 (UTC)
  • I also hear that understanding group theory helps explain Raman and a lot of spectroscopic techniques in general. Can someone help explain why? (talk) 08:46, 9 April 2012 (UTC)
    • I know it has something to do with symmetry and all these methods have something to do with squishing and twisting and bending chemical bonds, but the math gets very boring and uninteresting and it's easy to lose sight of why I'm learning it. (talk) 08:46, 9 April 2012 (UTC)
  • Can someone perhaps fix up the article to include some more contextual information like where near-IR, mid-IR, far-IR, etc. methods fit in, perhaps with some nice animated GIFs showing what goes on in each? (talk) 08:46, 9 April 2012 (UTC)

Virtual energy state[edit]

Can somebody explain what a "virtual energy state" is? How is this distinct from a "true" energy transition (presumably one where the electrons are moved to an excited state)?

Since any wavelength radiation can theoretically cause the Raman effect, it then follows that the excited state most of the time does not correspond to any electronic/vibrational/rotational energy level. That is why the term "virtual energy state" is used. Mill haru 01:54, 8 January 2007 (UTC)
Can you provide a more thorough explanation or direct a link its explanation? I'm having trouble visualizing this virtual effect. Thanks. LostLucidity (talk) 19:44, 27 January 2008 (UTC)
In general the incident photon has far more energy than the difference between the ground state and the excited state. (In something like infrared spectroscopy, the incident photon has an energy equal to this difference and can be absorbed when the transition occurs). Thus the photon cannot be completely absorbed by causing the excitation. In essence therefore what happens is that the incident photon loses some of its energy and comes out at a lower frequency. However, in quantum mechanics we talk about transitions absorbing or emitting a photon with energy equal to the energy difference of the transition. To apply this language here, the ground-state electron must absorb the incident photon and move to a ridiculously high energy state, and then immediately emit a slightly less energetic photon, and move down to an excited state. It seems doubtful that this actually happens, and so they call it a virtual energy state, presumably because it sounds more impressive than `ridiculously high energy state'. It's a didactic model more than anything. (talk) 16:29, 5 June 2008 (UTC)
Do you mean that the essence of the Raman effect still not clearly when using "virtual energy state" model to explain. I still wonder about the nature of the phenomenon. Could you explain it with out "virtual energy state"?. In flourescence, UV photons are absorbed and the molecule has a "ridiculously high energy state". So I think that a high energy state is very common. —Preceding unsigned comment added by (talk) 06:10, 21 October 2008 (UTC)

Okay, so here's an analogy: You throw a tennis ball at a wall with a strange spring device in it (the details aren't important, but imagine the spring can be either compressed or relaxed, and it's got a way of being held in compression which can be toggled by the tennis ball). The tennis ball bounces back. If it comes back faster than it went in, then you know the spring was compressed, and got released by the tennis ball. By looking at the change in the ball's speed you can tell how much energy was in the spring. If it comes back slower than it went in, then you know the spring was relaxed and became compressed by the tennis ball. That's all that's going on, except that the photon's energy is given by its frequency rather than its speed. Also it's probably not the same tennis ball, but an identical one because of quantum mechanical stuff. So in the analogy, the spring needs to absorb a tennis ball with a very particular speed to become compressed - a speed much less than the speed of the tennis ball. But somehow it manages to absorb the tennis ball anyway, and it copes with all the extra speed by immediately releasing another tennis ball going slightly slower than the one that came in.Grj23 (talk) 11:21, 4 December 2008 (UTC)

Can someone provide more information on the following statement: "the lateral and depth resolutions were 250 nm and 1.7 µm, respectively, using a confocal Raman microspectrometer with the 632.8 nm line from a He-Ne laser with a pinhole of 100 µm diameter." The microscopic magnification would be an important piece of information if these resolution numbers are to be reproduced. Mill haru 02:24, 8 January 2007 (UTC)

What does it mean that "The fingerprint region of organic molecules is in the range 500-2000 /cm"? If wavelengths are in units of length then why is the "fingerprint" given in inverse length?

Spectroscopists a weird bunch and as such decide to use wierd units, in this case the Wavenumber. They also tend to use the terms energy, wavelength, wavenumber and frequency interchangebly, since they all are related to energy. In Raman spectroscopy what is measured is the difference in energy between the scattered light and the excitation source, and the amount of energy involved is characteristic of a particular bond. For example, silicon has a primary band (or peak) at a Raman shift of about 520 /cm, and that is irrespective of the wavelength of the laser. A skilled spectroscopist will be able to tell simply from looking at a spectrum the major components of a sample as certain bands and features tend to correspond to specific Functional_groups --Wound 22:50, 4 Apr 2005 (UTC)

This article could use some reorganizing: history should come before applications and applications should be merged with Raman microspectroscopy, other types, and see also. Additionally, at least in the RR article, there is a significant amount of repeated information, which may be more appropriately placed the main Raman article. Finally, the theory section should be expanded or an advance theory section should be added to do this topic justice.--Bjsamelsonjones

I think an experimental set-up section would also be a good idea. Including monochromatic incident light, reflection of the sample, filter to remove the rayleigh scattered light of the same frequency as the incoming light (which constitutes the bulk of the reflected light) a disperser (I think), and a detector. What disperser is used nowadays? Do practitioners use a michelson interferometer and fourier transform like in infrared spectroscopy? (talk) 16:36, 5 June 2008 (UTC)

This is a strange sentence: "Consequently in vivo time- and space-resolved Raman spectroscopy is suitable to measure cells, proteins, organs, and erythrocytes." First of all, what is meant by "measuring" cells or organs? Measuring the quantity or characterizing their chemical composition? Secondly, erythrocytes are a type of cell, so to say that Raman spectroscopy is suitable for measuring cells AND erythrocytes is redundant. —Preceding unsigned comment added by (talkcontribs)

I reworded it to fix these problems. the wub "?!" 10:40, 4 July 2007 (UTC)

Comment about lateral and depth resolution[edit]

Hi every one, I would say that I am not completely agree with the example proposed to describe the lateral and depth resolution for a Raman microspectrometer.
First, about the lateral resolution of 250 nm. I think that a lateral resolution of 250 nm could be reached but not with a confocal microscope only. For your example, with a conventional Raman microspectrometer equipped with a He-Ne (632,81 nm) and a dry microscope objective of x100 (Numerical aperture = 0.9), the best lateral resolution you could have is around 1 micrometer. I think a resolution of 250 nm can be reached on AFM instrument coupled with a Raman spectrometer.
Secondly, the depth resolution depend on the nature on the sample. On a Si Wafer, the depth resolution is closed to 1 µm. On bone, the depth resolution decrease to 5 µm. It depends on the homogeneity of the sample. Generally, in spectroscopist review, everybody agree that the depth resolution is close to 1 µm.
If you want, I have reference concerning the use of Polarized Raman experiments in order to determine apatite crystals orientation in human teeth. Capsulcorp (talk) 15:16, 27 June 2008 (UTC)

Isn't the phrase "very high spatial resolution" in fact dubious? For example, X-Ray nanotomography offers resolution down to 50 nm; SEM (okay, rather destructive as we often need to prepare the sample) – easily down to 20 nm. Or this is maybe for non-destructive methods only?--Esmu Igors (talk) 18:22, 15 January 2013 (UTC)

Hi, I have to add something too. Is the graphic showing the raman energy levels correct that way? Atkins (Physical Chemistry) shows ∆J=±2. And the graphic in that book shows a difference in J before and after (anti-)stokes of (-)2. —Preceding unsigned comment added by (talk) 02:41, 9 December 2008 (UTC)

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A citation for use of Raman spectroscopy with a wide range of excitation wavelengths is currently Reference [14], but this citation seems to be actually about biomedical imaging. This citation numbering issue should be investigated/resolved. — Preceding unsigned comment added by Slepkov (talkcontribs) 16:43, 14 October 2011 (UTC)

The doi of the Lombardi paper is wrong, it should be 10.1021/jp800167v. I don't know how to fix this and I don't have the time to read the manual.

It would be nice if the article included more info on isotope analysis with Raman microspectroscopy. This is a technique that is used in ecology and probably elsewhere. -- (talk) 04:22, 9 May 2009 (UTC)

I'm wondering if Brillouin scattering should be included with Rayleigh in the context of scattering close to the exciter line.Periksson28 (talk) 19:46, 18 July 2009 (UTC)

I believe the description of spatially offset Raman spectroscopy, under variations, has been cut off. — Preceding unsigned comment added by (talk) 10:05, 15 June 2012 (UTC)

Image - Raman Energy Levels[edit]

Isn't there a slight mistake in that image? As far I know the selection rule for IR-spectroscopy is v = ± 1 (not 2)!

(regarding also another comment above)
The selection rule is ±1 only for the harmonic oscillator (when a system vibrates just like a spring, according to the Hook's law). For anharmonic oscillator (usually described by the Morse curve) this is not true, and ±2, ±3, etc. are all allowed. But, of course, as these refer to exciting to higher states, often with different symmetry than that of dipole moment components (in IS) or polarizability [tensor] components (in Raman), transitions should be less probable.--Esmu Igors (talk) 18:07, 15 January 2013 (UTC)

Energy level diagram[edit]

Regarding the several comments about the energy level diagram at the top, the infrared absorption transition should match the Raman transition to make the essential point that the energy difference between the excitation photon and the Raman scattered photon is equal to the energy of the infrared transition. v = 0 to v = 1 is a good choice because it is very typical. The file Raman energy levels.jpg should be edited or replaced, so that the short blue arrow on the left (Infrared absorption) extends from 0 to 1, rather than from 0 to 2.

(For example, see Fig. 1.8 of “Introduction to Raman Spectroscopy, 2nd Edition”, Ferraro, Nakamoto and Brown, Academic Press, 2003.)

02:51, 15 April 2016 (UTC) — Preceding unsigned comment added by Propanscience (talkcontribs)

Thank you for explaining the problem more clearly: yes, the arrow for IR absorption should extend only from v=0 to v=1. However in looking through the article revision and image file histories, I realized that the error is actually in the file Raman energy levels.svg (NOT jpg). There was a file Raman energy levels.jpg in the article prior to 19 Sept. 2009, and it showed the IR transition correctly. It was replaced on that date by the svg file, presumably because the svg image looks much nicer. However the jpg file is still available on Wikipedia, and is still used in some language versions of this article: Spanish, Russian, etc.
So since truth is more important than beauty, I will now replace the incorrect svg file by the original jpg file. If someone knows how to correct the svg file by shortening the line for IR absorption, that would be even better. Dirac66 (talk) 00:17, 16 April 2016 (UTC)
Corrected the svg. Do let me know if there are any further problems. :) --Officer781 (talk) 12:27, 26 April 2016 (UTC)
Thanks, that is better. I have inserted your new svg into the French and Spanish articles also. Dirac66 (talk) 13:59, 26 April 2016 (UTC)


I'm just reading the article and encountered the term "rovibronic", which is explained after used. That term isn't in and a lot of other word dictionaries. It might be a bit obscure for casual readers - it's complete jargon. While technically correct and the correct term, consider rewording the passage using different terms to be better understood by a less-technical reader. — Preceding unsigned comment added by (talk) 23:16, 13 December 2016 (UTC)

This is the usual term, but it can of always be better explained. I have now added a wikilink pointing to the article on Rovibrational coupling. This may help clarify the term a little for the reader. Dirac66 (talk) 00:05, 14 December 2016 (UTC)