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I P & Q branches confusion in rovibrational transitions + rant

  1. Nov 11, 2017 #1
    In rovibrational transitions we have following selection rules

    $$ \Delta v = \pm 1 $$

    $$ \Delta J = \pm 1 $$

    where ##v## is the vibrational quantum number and ##J## is rotational quantum number.
    Now based on whether ##\Delta J## changes to +1 or -1 we have two branches of spectroscopic lines.

    They are defined as following:

    - if transition happens such that ##\Delta J = +1## that goes in R-branch

    - if transition happens such that ##\Delta J = -1## that goes in P-branch

    It is also said (eg. https://en.wikipedia.org/wiki/Rotational–vibrational_spectroscopy) that R branch is on the higher frequency side of the Q branch and P branch is on the lower frequency side. This means that the lines in R branch have more photon energy (eg. less wavelength) than lines in the P branch.
    So far so good.

    But notice how I emphasized transition word above? That's because in every site/book I looked at I see they use word transition without specifying whether it's absorption or emission. It seems as if it doesn't matter but looks like it does.
    If we have absorption and ##\Delta J = +1## (say J goes from 0 to 1) then that is R branch. But if we have emission (the same transition in opposite direction where J goes from 1 to 0) we will have ##\Delta J = -1##. Now both of these transitions must be of the same energy but which should go to the R branch? I conclude it must be that the transition specified above in the definition is absorption and not emission.

    r_p_branch.png

    So why I couldn't find in any resource this? Why did I have to lose 3-4 hours on this trying to figure out all this rather than having a precise definition in the first place? When I see transition I usually first think of emission so I started working examples on my own to understand this R and P branches concepts but I couldn't match it with the definition.
    I am disappointed by the physics literature and a loose way of defining things. Or could it be me? Am I missing something from my discussion here? It could be because of lack of sleep I am having these days but I am not seeing it.
     
  2. jcsd
  3. Nov 11, 2017 #2

    Charles Link

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    Usually an IR spectral measurement will measure either an emission or absorption spectrum, but not both at the the same time. And usually I think it is an absorption spectrum. (A broadband source like a high temperature blackbody or an incadescent lamp is used, and the absorption spectrum of the gas is measured). I don't think an emission spectrum (arc discharge) gives much quality lines with an IR spectrum. ## \\ ## Meanwhile, I think there are also Raman measurements, where a visible monochromatic source (laser) is used and IR absorptions and emissions can cause shifts in the measured laser wavelength. With the Raman spectrum, both emission and absorption are measured simultaneously. If a diffraction grating spectrometer is used, the absolute wavelength of the shifted laser wavelengths are measured along with the unshifted laser wavelength. (Here I think is where double spectrometers sometimes become very useful. If the shifted wavelengths are weak compared to the unshifted signal, the first monochromator will eliminate most of the very intense unshifted signal, and the shifted signals can be more readily seen, measured at the exit slit of the second spectrometer).## \\ ## Additional note: If heterodyneing is employed by beating the shifted wavelength against the reference onto a photodiode, (works at r-f frequencies but not IR), then only the absolute value of the frequency shift is known, (detected as an r-f electrical signal at the shift frequency), but not the sign of the frequency shift.
     
    Last edited: Nov 11, 2017
  4. Nov 11, 2017 #3

    Charles Link

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    @misko See edited additions to the above post.
     
  5. Nov 11, 2017 #4
    Sorry I think you missed the point (and pain) of my post.
    I am student, trying to understand rovibrational transitions from books and university lessons, not experimentalist working in a lab.
     
  6. Nov 12, 2017 #5

    blue_leaf77

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    For a given spectrogram it must indeed be specified whether it's for absorption or emission, otherwise it may lead to confusion as you have noticed. In the book "Physics of Atoms and Molecules" by Bransden and Joachain figure 10.4 explicitly mentions that the associated picture is for absorption. Alternatively you can also see this link http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibrot.html where the absorption spectrum of HCl is shown, the R branch is the right wing while the P branch is the left one.
     
    Last edited: Nov 13, 2017
  7. Nov 12, 2017 #6

    Charles Link

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  8. Nov 13, 2017 #7
    Thank you guys.
    I learned another way to think about this in the following video:
    where they specify that "transitions where in the excited state J value is one unit greater than that of the lower energy state gives R lines, and in if they are one unit of J less than in the lower energy state they are P lines".

    So basically if I take take the lower energy level as the reference point for calculating difference in J then I don't have to worry whether transition is absorption or emission.
     
  9. Nov 13, 2017 #8
    Generally, at room temperature, many rotational states are thermally populated, but usually only the ground vibrational state is significantly thermally populated. Thus although Δv = ±1 are both allowed, only Δv = +1 (v = 0→1) really matters practically. However, accompanying this change in vibrational state, you can get rotational state changes of ΔJ = ±1, giving you the R and P branches.
     
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