LIF Spectrum: Understand 12 Branches & Subscripts

[Favry2021]

TL;DR Summary

where does the R21, Q21, etc. branches come from?

I have a bit basic understanding of spectrum, but quite confused with LIF spectrum. It seems existing 12 main branches of Laser induced fluorescence, such P1, P2, Q1, Q2, P21, R21，and so on. I know there should be O P Q R S branches of the rotational transitions, but where these so many subscripts, 1, 2, 21, 12 com from？ thanks for any help.

 

[Twigg]

The single digit subscripts refer to the value of J in the initial state, so Q1 means a transition from J=1 to J=1, P2 means J=2 to J=1, etc.

I'm not sure about the double digits.

 

[EigenState137]

Greetings,

Its sounds to me as if you are interested in the rotationally state resolved rovibronic spectrum of a diatomic molecule having hyperfine structure utilizing 2-photon excitation.

In that case, the O,P,Q, R, and S define the normal change in the value of J. The numerical subscripts (21, 12 ,22, 11) define the the change in the value of the spin component F. By convention the first numerical subscript refers to the upper excited state, the second numerical subscript refers to the lower state in the rovibronic transition.

This situation is of course not limited to LIF detection. It is the same with photoionization detection for example.ES

 

[Twigg]

@EigenState137 That makes sense. Based on that, the example "R21" that the OP quoted is a mistake, yes? If it was a true R branch, it should go up, so R23 or R12 are allowed but R21 is bad, right? Just checking!

Summary:: where does the R21, Q21, etc. branches come from?

R21

Also, @Favry2021 if you're not familiar with the spin component F, just read it as spin component J. Unless you're working on very cold gases (think cryogenics or laser cooling), the hyperfine structure that distinguishes F from J will be washed out by thermal Doppler broadening, so you'll just see a single fat line for J. (Not trying to be dismissive. If you know what F means, nice. Just mentioning this because a lot of chemistry-minded textbooks on molecular spectroscopy don't even bother discussing hyperfine splitting.)

Edit: there are also sub-doppler techniques that get can make hyperfine splitting visible in room temperature gases, but that's way beyond the scope of this thread.

 

[Favry2021]

Thank you ES for your answer, but still not quite sure.

1. Does the hyperfine rovibronic spectrum necessarily mean 2-photon excitation? Can using a single photon excite a molecule and arrive the same results?

2. I was looking at LIF-base CH C-X (0,0) transitions, which shows all the 12 transition lines as shown in the figure below. If the two digits subscripts means upper and lower F, how about the single digit ones?

 

[Favry2021]

@EigenState137 That makes sense. Based on that, the example "R21" that the OP quoted is a mistake, yes? If it was a true R branch, it should go up, so R23 or R12 are allowed but R21 is bad, right? Just checking!Also, @Favry2021 if you're not familiar with the spin component F, just read it as spin component J. Unless you're working on very cold gases (think cryogenics or laser cooling), the hyperfine structure that distinguishes F from J will be washed out by thermal Doppler broadening, so you'll just see a single fat line for J. (Not trying to be dismissive. If you know what F means, nice. Just mentioning this because a lot of chemistry-minded textbooks on molecular spectroscopy don't even bother discussing hyperfine splitting.)

Edit: there are also sub-doppler techniques that get can make hyperfine splitting visible in room temperature gases, but that's way beyond the scope of this thread.

Actually it was not a mistake, but does exist as you can find from my previous post

 

[Twigg]

Does the hyperfine rovibronic spectrum necessarily mean 2-photon excitation?

No. You can easily have single photon hyperfine transitions. The transition rules are the same as for J.

I'm a little confused by these results. If the double subscripts refer to F states, I'd expect to see multiple Q's and P's (say Q11 and Q12 for $\Delta F = 0$ and $\Delta F = 1$). I have a feeling this might be a different convention than what @EigenState137 has in mind. I could be dead wrong though. That being said, the precision is there to see hyperfine splitting, so I'm not really sure.

 

[Twigg]

Any chance you could share a reference that uses this notation? Might help us to track it down.

 

[Favry2021]

I found this kind of notation (P21, R21 and so on) in many papers, such as this one: https://www.sciencedirect.com/science/article/abs/pii/S000926149700239X

I agree with you, Twigg, that it might be a different convention rather than a hyper fine structure. As a paper I read mentioned that there are 12 main branches, as the picture of CH C-X transition shown.

 

[EigenState137]

Thank you ES for your answer, but still not quite sure.

1. Does the hyperfine rovibronic spectrum necessarily mean 2-photon excitation? Can using a single photon excite a molecule and arrive the same results?

2. I was looking at LIF-base CH C-X (0,0) transitions, which shows all the 12 transition lines as shown in the figure below. If the two digits subscripts means upper and lower F, how about the single digit ones?

View attachment 284733

Greetings,

A few points need to be clarified.

You are not resolving hyperfine structure. You are taking advantage of selection rules, and the selection rules for J and F are not the same.

2-photon excitation is not required, but it is the way to obtain O and S rotational branches since it provides for electric dipole allowed transitions with a change of J = +/-2. My apologies, but my latex is rusty in the extreme.

The single subscript transitions most probably refer to 11 and 22 notation.

The molecule CH is a perfect representative of a diatomic with both electron and nuclear spin angular momenta, and therefore hyperfine structure via S dot I coupling. I urge you to scrutinize an energy level diagram of the states of interest that includes rotational and hyperfine structure. Then study the selection rules for either 1-photon or 2-photon excitation. I suspect that will help clarify these points in great detail.

As a side note, the notation is actually more complex in that often a transition also carries two superscripts indicating the symmetries of the upper and lower states. No need to complicate this discussion with that however.ES

 

[Twigg]

You are taking advantage of selection rules, and the selection rules for J and F are not the same.

Can you explain what the difference is? I might've forgotten something. I was under the impression that $\Delta J =0, \pm 1$ & $J=0 \not\to J' = 0$ and that $\Delta F = 0, \pm 1$ & $F = 0 \not\to F'=0$ were the selection rules in question (for single photon processes). Do I have something wrong? I appreciate the correction!

 

[EigenState137]

Greetings,

Those selection rules are indeed correct for one photon electric dipole allowed transitions.

What I was attempting to point out was that the rotational spectrum under discussion does not resolve the hyperfine structure. The transitions are normal rotational state transitions but one needs to account for hyperfine structure and for lambda doubling (hence my prior comment regarding symmetries) to understand exactly what energy levels are involved. Thus my suggestion to @Favry2021 to scrutinize the actual energy level diagram to see exactly which states constitute a transition under the experimental conditions.

My apologies for the confusion.ES

 

[Twigg]

Gotcha, I'm in complete agreement there.

@Favry2021 , I recommend perusing the example spectra in a book like Brown and Carrington. I would point you to specific chapters and sections, but I'm away from home where my copy is. They're definitely deep in the second half of that monster book.

If you're doing research on diatomics, someone in your group probably has a copy. Lefevbre-Brion is also highly reputed.

This might not help with notation for these two photon processes since its a bit niche.

Another thing stands out to me. If these were two photon processes, I would expect them to fall into a different electronic level of the molecular structure as compared to the one photon lines, but the authors seem to include them in the same electronic line. Curious.

 

[EigenState137]

Gotcha, I'm in complete agreement there.

@Favry2021 , I recommend perusing the example spectra in a book like Brown and Carrington. I would point you to specific chapters and sections, but I'm away from home where my copy is. They're definitely deep in the second half of that monster book.

If you're doing research on diatomics, someone in your group probably has a copy. Lefevbre-Brion is also highly reputed.

This might not help with notation for these two photon processes since its a bit niche.

Another thing stands out to me. If these were two photon processes, I would expect them to fall into a different electronic level of the molecular structure as compared to the one photon lines, but the authors seem to include them in the same electronic line. Curious.

Greetings,

Even going back to Herzberg, and Townes and Schalow should provide insights regarding the full picture of the energy levels.

I fail to understand your expectations regarding two-photon excitation to a different electronic state. Such two-photon excitation does not involve some intermediate resonance. Think of it as utilizing a virtual state if you like, but it is just using two-photons each with one-half the energy of the corresponding one-photon excitation. Counter-propagate the beams (via retro-reflection) and presto it is sub-Doppler.ES

 

[Twigg]

What bothers me is that the authors of the ZrN paper linked above say that for example P1 and Q21 are between the same electronic levels. That could make sense if P1=P11, so all the transitions they measure are two-photon. But I don't see any indication in the text that these are two photon transitions. If it was two photon, wouldn't we see a O and S branch in their spectrum?

I wonder if the double subscript notation has something to do with the excited state being Hund's case (b) in the ZrN paper.

 

[EigenState137]

What bothers me is that the authors of the ZrN paper linked above say that for example P1 and Q21 are between the same electronic levels. That could make sense if P1=P11, so all the transitions they measure are two-photon. But I don't see any indication in the text that these are two photon transitions. If it was two photon, wouldn't we see a O and S branch in their spectrum?

I wonder if the double subscript notation has something to do with the excited state being Hund's case (b) in the ZrN paper.

Greetings,

I only have access to the abstract of the paper cited and that is not particularly helpful.

The excited state being Hund's case b is not unusual. Much research has been done on the A-X band of NO in which the A state is case b. If the notation is case dependent, it so only because the energy level structure is case dependent. But for something as heavy as ZrN, I would question the applicability of any Hund's case approximation.

I just found the following paper on CH which utilizes the notation in question for the analysis of UV and VUV absorption spectra and is thus clearly one-photon. The authors are reasonably trustworthy

G. Herzberg and J. W. C. Johns, "New spectra of the CH molecule" Astrophysical Journal, vol. 158, p. 399-418 (1969).ES

 

[Twigg]

Sorry, I forgot about the paywall on the ZrN paper.

Counter-propagate the beams (via retro-reflection) and presto it is sub-Doppler.

Thanks! I didn't know it was that easy for two-photon LIF.

^taken from the OP

I've been reading over this table and it really bugs me. Since there is no row for N''=0, I want to believe that the excited state is a $\Pi$ state ($\Lambda = 1$). However, if that were the case, there would be no P1 line. However, instead there is no P2 line?? So take what I suggested earlier about the single digit subscripts and throw it out the window.

Also, I feel like CH would have nuclear spin I = 1/2 since carbon 12 is an even isotope and hydrogen has I=1/2. Since there's an odd number of electrons, J should have half-integer value and so the subscripts really must be related to F like @EigenState137 was saying or to N. Why the P2 line is missing in the above table is still a mystery to me. Also, if the subscripts are F, isn't O12 also nonsensical? I would expect O31. (Not saying anyone is wrong, just trying to state what I find contradictory here.)

The authors are reasonably trustworthy

You just brought in the heavy artillery

The ZrN paper cited Herzburg's book when they said they expected 12 lines in the body of the paper (but they had no O or S lines). I have a funny feeling he may have established the trend with this notation. I don't own a copy but maybe I should change that.

 

[EigenState137]

Greetings,

The following is what I hope will be a definitive treatment of the rotational branch notation discussed above, and that it will provide clarity.

Two spectral studies have been mentioned explicitly, and one has been on my mind. Namely:

$_{}^{12}\textrm{C}_{}^{1}\textrm{H} \;\; C\: {}^2\, \!\Sigma^+\: -\: X\: {}^2\, \!\Pi_r$

$_{}^{90}\textrm{Zr}_{}^{14}\textrm{N} \;\; A\:{}^2\, \!\Pi_r\: -\: X\: {}^2\, \!\Sigma$

$_{}^{14}\textrm{N}_{}^{16}\textrm{O} \;\; A\: {}^2\, \!\Sigma^+\: -\: X\: {}^2\, \!\Pi$

Obviously, all of these spectral systems are $^2\, \!\Sigma\: -\: {}^2\, \!\Pi$.

Herzberg [1], as one would anticipate, explicitly discusses the rotational structure of such spectral bands. To summarize briefly, $^2\, \!\Sigma$ is always Hund's case (b), while ${}^2\, \!\Pi$ can be either case (a) or case (b).

Now to define the $F_{1}$ and $F_{2}$ notations that have been so problematical to date.

For Hund's case (a), $F_{1}$ is associated with the ${}^2\, \!\Pi_{\frac{1}{2}}$ spin-orbit manifold of states while $F_{2}$ is associated with the ${}^2\, \!\Pi_{\frac{3}{2}}$ spin-orbit manifold of states. This is clearly illustrated in a general energy level diagram on page 259 of reference [1].

For Hund's case (b), and thus a $^2\, \!\Sigma$ state, $F_{1}(K)$ defines the components with $J\, = K+\frac{1}{2}$ while $F_{2}(K)$ defines those components with $J\, = K-\frac{1}{2}$. $K$ is defined as the total angular momentum excluding spin, and $J$ is defined as the total angular momentum including electron spin.

This also serves to explain why there are only $F_{1}$ and $F_{2}$ designations.

[1]. G. Herzberg, "Spectra of Diatomic Molecules" Van Nostrand Reinhold Chapter V, section 2, page 218ff (1950).ES

 

[Twigg]

Ok, I finally think I get it. The F quantum numbers you refer to do not include rotational angular momentum, N, right? That's why P12 makes sense for N=3, because the transition is between the F=1,N=3 to F=2,N=2 state. I believe this is where I got hung up, because in Brown and Carrington (where I learned this stuff from), J and F include the rotational angular momentum.

I think I understand the table better now. N" means N of the excited state. So P1 goes from F=1,N=2 to F"=1,N"=1, so both P1 and P2 are allowed. I don't know how to interpret the exclusion of the P2 line. Perhaps it just didn't show up in the data for whatever reason.

With these clarifications of nomenclature, I think the issue is resolved.

 

[EigenState137]

Greetings,

$F_{1}$ and $F_{2}$ are not quantum numbers. They are state labels, akin perhaps to the symmetry labels, $e$ and $f$, mentioned previously. There is no associated selection rule. They should not be confused with the quantum number $F$ that characterizes the total angular momentum including nuclear spin.

The quantum number $K$ is formally defined within Hund's case (b) as the total angular momentum exclusive of spin such that
$$\vec{K}\equiv \vec{\Lambda} + \vec{N}$$
where $\vec{\Lambda}$ is the total electronic orbital angular momentum along the internuclear axis. For a $\Sigma$ state, $\Lambda \equiv 0$ such that $\vec{K}\equiv \vec{N}$ where $\vec{N}$ is the angular momentum of nuclear rotation.

$\vec{J}$ of course is the total angular momentum including electron spin such that
$$\vec{J}\equiv \vec{K} + \vec{S}$$
and again for a $\Sigma$ state
$$\vec{J}\equiv \vec{N} + \vec{S}$$

By convention, the double prime label is typically used to signify the lower state of a transition, not the upper state. By that convention, the $N{}''$ in the table of data presented above would appear to be the angular momentum of nuclear rotation in the $X\: ^{2}\Pi _{r}$ of $\textup{CH}$ . Quite frankly, that makes no sense to me. I would think that the table would have listed $J{}''$ instead.

It appears that we have bored the OP to tears.ES

 

[Twigg]

It appears that we have bored the OP to tears.

Guilty as charged. @Favry2021 The best advice I can give you in this case is to get ahold of whoever wrote that chart and get them to walk you through it. Twist their arm if you have to If they're in your group or your institute, a box of donuts usually does the trick

Still thinking through your reply, ES. I see I made a mistake confusing $F_1(K)$ and $F_2(K)$ with the total angular momentum $\vec{F}$, but now I'm not sure what these two labels mean. You've already given a lot of time in this thread, and I thank you for that. I'm going to try and hunt this down on my own time.

 

[EigenState137]

Greetings,

For the sake of completeness should any interested party read this in the future:

The rotational branches are defined by
$$\Delta J_{jk}\equiv J_{j}{}'-J_{k}{}''$$
where the prime label ${}'$ defines the upper state of the transition and the double prime label ${}''$ defines the lower state of the transition. Similarly, the $j$ subscript defines the spin component, $F_{1}$ or $F_{2}$, of the upper state and the $k$ subscript defines the spin component of the lower state.

Recalling that in the case of two-photon excitation $\Delta J= -2$ defines an $O$-branch, $\Delta J= -1$ defines a $P$-branch, $\Delta J= 0$ defines a $Q$-branch, $\Delta J= +1$ defines an $R$-branch, and $\Delta J= +2$ defines an $S$-branch, we have for example $R_{12}$ having $\Delta J= +1$ with the upper state having $F_{1}$ and the lower state having $F_{2}$.

I might add, that a reference to the quoted data on $\textup{CH}$ would have been exceedingly helpful.

@Twigg, it was a pleasure to discuss this with youES

 

[Favry2021]

Hi @EigenState137. I would like to express my sincere thanks to you for all your help and discussions on this thread. Particularly, now I realize that you were 100% correct about almost every aspect. Recently I was reading a laser diagnostic book, and now my understanding is much deeper.

Also many thanks to @Twigg. Your discussions and advices really helped to extend the depth of this thread.

Many thanks and best wishes.