Effective molecular Hamiltonian and Hund cases

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  • Thread starter BillKet
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  • #76
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I guess I don't understand what is the mathematical approximation used that allows you to assume that the electronic integral, which is a function of ##R## can be approximated to be constant. In the BO approximation you would use the adiabatic approximation, but I am not sure here, formally, what allows you to do that. Intuitively, if you have, say, the function ##cos^2##, but your response time to this oscillation is too slow, what you see is the average over many periods which is ##1/2##. Given that that electronic integral sees just the average internuclear distance, I assumed it is something similar to this i.e. the electrons see just an average of the internuclear distance.
I'd say it's more of a physical approximation than a mathematical one. For low vibrational states (shorter internuclear distances), the region of the dipole moment function is relatively flat. So just picking the equilibrium distance actually approximates it pretty well. At high vibrational states, where you'd sample large internuclear distances, the curve starts to get wobbly on the outskirts and the approximation breaks down. This makes sense because you'd expect BO breakdown at higher vibrational energies.
 
  • #77
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Oh I see, that makes sense. Thanks a lot! I still have a quick question about the electronic integral. In order to have transitions between different electronic states, as you mentioned, terms of the form $$<\eta'|T_{\pm 1}^1(r)|\eta>$$ should not be zero. But this is equivalent to $$<\eta'|x|\eta>$$ not being zero (and same for ##y##). However, the electronic wavefunctions have cylindrical symmetry, so they should be even functions of x and y (here all the coordinates are in the intrinsic molecular (rotating) frame). Wouldn't in this case $$<\eta'|x|\eta>$$ be zero?

Time to look at some molecular orbitals! Only ##\Sigma## states have cylindrical symmetry, which as you've pointed out, means ##\Sigma## to ##\Sigma## transitions are not allowed, unless you go from ##\Sigma^+## to ##\Sigma^-##, the latter of which is not symmetric along the internuclear axis, but still only ##q = 0## transitions allowed.

Take a look at some ##\Pi## or ##\Delta## orbitals. They are absolutely not cylindrically symmetric, because they look like linear combinations of ##p## and ##d## orbitals.
 
  • #78
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Time to look at some molecular orbitals! Only ##\Sigma## states have cylindrical symmetry, which as you've pointed out, means ##\Sigma## to ##\Sigma## transitions are not allowed, unless you go from ##\Sigma^+## to ##\Sigma^-##, the latter of which is not symmetric along the internuclear axis, but still only ##q = 0## transitions allowed.

Take a look at some ##\Pi## or ##\Delta## orbitals. They are absolutely not cylindrically symmetric, because they look like linear combinations of ##p## and ##d## orbitals.
Thanks for the vibrational explanation! I understand what you mean now.

I should check molecular orbitals indeed, I kinda looked at the rotational part only. But if that is the case, then it makes sense. Thank you for that, too!
 
  • #79
Twigg
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I have a feeling that mathematically the approximation of taking ##\frac{\partial \mu_e}{\partial R} \approx 0## could be obtained from the BO approximation with the adiabatic theorem, taking the dynamical and Berry's phases evolved to be negligibly small since the nuclei barely move over a transition lifetime. I could just be crazy though. I never put a lot of thought into it before.
 
  • #80
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Time to look at some molecular orbitals! Only ##\Sigma## states have cylindrical symmetry, which as you've pointed out, means ##\Sigma## to ##\Sigma## transitions are not allowed, unless you go from ##\Sigma^+## to ##\Sigma^-##, the latter of which is not symmetric along the internuclear axis, but still only ##q = 0## transitions allowed.

Take a look at some ##\Pi## or ##\Delta## orbitals. They are absolutely not cylindrically symmetric, because they look like linear combinations of ##p## and ##d## orbitals.
I came across this reading, which I found very useful in understanding the actual form of the Hund cases (not sure if this is derived in B&C, too), mainly equations 6.7 and 6.12. I was wondering how this would be expanded to the case of nuclear spin (call it ##I##). Given that in most cases the hyperfine interaction is very weak, we can assume that the basis we build including ##I## would be something similar to the Hund case b) coupling of ##N## and ##S## in 6.7 i.e. we would need to use Clebsch–Gordan coefficients.

So in a Hund case a, the total basis wavefunction after adding the nuclear spin, with ##F## the total angular momentum we would have:

$$|\Sigma, \Lambda, \Omega, S, J, I, F, M_F>=\sum_{M_J=-J}^{J}\sum_{M_I=-I}^I <J,M_J;I,M_I|F,M_F> |\Sigma, \Lambda, \Omega, S, J, M_J>|I, M_I>$$

where ##|\Sigma, \Lambda, \Omega, S, J, M_J>## is a Hund case a function in the absence in nuclear spin. For Hund case b we would have something similar, but with different quantum numbers:

$$|\Lambda, N, S, J, I, F, M_F>=\sum_{M_J=-J}^{J}\sum_{M_I=-I}^I <J,M_J;I,M_I|F,M_F> |\Lambda, N, S, J, M_J>|I, M_I>$$

with ## |\Lambda, N, S, J, M_J>## being a Hund case b basis function in the absence of nuclear spin. Is this right? Thank you!
 
  • #81
Twigg
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Yep, that's right. Nuclear angular momentum is just tacked on at the end of the hierarchy (though it need not be the smallest spectral splitting) with another addition of angular momenta.
 
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  • #82
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I came across this reading, which I found very useful in understanding the actual form of the Hund cases (not sure if this is derived in B&C, too), mainly equations 6.7 and 6.12. I was wondering how this would be expanded to the case of nuclear spin (call it ##I##). Given that in most cases the hyperfine interaction is very weak, we can assume that the basis we build including ##I## would be something similar to the Hund case b) coupling of ##N## and ##S## in 6.7 i.e. we would need to use Clebsch–Gordan coefficients.

So in a Hund case a, the total basis wavefunction after adding the nuclear spin, with ##F## the total angular momentum we would have:

$$|\Sigma, \Lambda, \Omega, S, J, I, F, M_F>=\sum_{M_J=-J}^{J}\sum_{M_I=-I}^I <J,M_J;I,M_I|F,M_F> |\Sigma, \Lambda, \Omega, S, J, M_J>|I, M_I>$$

where ##|\Sigma, \Lambda, \Omega, S, J, M_J>## is a Hund case a function in the absence in nuclear spin. For Hund case b we would have something similar, but with different quantum numbers:

$$|\Lambda, N, S, J, I, F, M_F>=\sum_{M_J=-J}^{J}\sum_{M_I=-I}^I <J,M_J;I,M_I|F,M_F> |\Lambda, N, S, J, M_J>|I, M_I>$$

with ## |\Lambda, N, S, J, M_J>## being a Hund case b basis function in the absence of nuclear spin. Is this right? Thank you!

There's also a nice discussion in B&C Section 6.7.8 about the different ways ##I## couples in Hund's cases (a) and (b). For example, if ##I## couples to ##J## in Hund's case (b), that's actually called Hund's case (b##_{\beta J}##), which is one of the different ways it can couple in.
 
  • #83
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Thanks for the clarification, @amoforum! And @BillKet, I'd actually be curious to see what you come up with for the stark shift, if you find the time. I tried to spend some time learning this once but my coworkers weren't having it and sent me back to mixing laser dye :doh: No pressure, of course!

As far as the BO approximation, when we did spectroscopy we didn't really keep a detailed effective Hamiltonian, we would just re-measure the lifetimes and rotational constants in other vibrational states if there was a need to do so. I think in molecules where the BO violation is weak, you can take this kind of pragmatic approach. Then again, we only thought about molecules with highly diagonal Franck-Condon factors so we never really ventured above ##\nu = 2## or so.
@Twigg, here is my take at deriving the Stark shift for a Hund case a. In principle it is in the case of 2 very close ##\Lambda##-doubled levels (e.g. in a ##\Delta## state as in the ACME experiment) in a field pointing in the z-direction, ##E_z##. Please let me know if there is something wrong with my derivation.

$$H_{eff} = <n\nu J M \Omega \Sigma \Lambda S|-dE|n\nu J' M' \Omega' \Sigma' \Lambda' S'>=$$
$$<n\nu J M \Omega \Sigma \Lambda S|-E_z\Sigma_q\mathcal{D}_{0q}^1T_q^1(d)|n\nu J' M' \Omega' \Sigma' \Lambda' S'>=$$
$$-E_z\Sigma_q<n\nu J M \Omega \Sigma \Lambda S|\mathcal{D}_{0q}^1T_q^1(d)|n\nu J' M' \Omega' \Sigma' \Lambda' S'>=$$
$$-E_z\Sigma_q<n\nu|T_q^1(d)|n\nu><J M \Omega \Sigma \Lambda S|\mathcal{D}_{0q}^1| J' M' \Omega' \Sigma' \Lambda' S'>$$

For the ##<n\nu|T_q^1(d)|n\nu>##, given that we are in a given electronic state, the difference between ##\Lambda## and ##-\Lambda## can only be 0, 2, 4, 6..., (for a ##\Delta## state it would be 4) so the terms with ##q=\pm 1## will give zero. So we are left with

$$-E_z<n\nu|T_0^1(d)|n\nu><J M \Omega \Sigma \Lambda S|\mathcal{D}_{00}^1| J' M' \Omega' \Sigma' \Lambda' S'>$$

If we use the variable ##D## for ##<n\nu|T_0^1(d)|n\nu>##, which is usually measured experimentally as the intrinsic electric dipole moment of the molecule (I might have missed a complex conjugate in the Wigner matrix, as it is easier to type without it :D) we have:

$$-E_zD<\Sigma S||\Sigma' S'><\Lambda||\Lambda'><J M \Omega |\mathcal{D}_{00}^1| J' M' \Omega' >$$

From here we get that ##S=S'##, ##\Sigma=\Sigma'## and ##\Lambda = \Lambda'##, which also implies that ##\Omega = \Omega'##. By calculating that Wigner matrix expectation value we get:

$$-E_zD(-1)^{M-\Omega}
\begin{pmatrix}
J & 1 & J' \\
\Omega & 0 & \Omega' \\
\end{pmatrix}
\begin{pmatrix}
J & 1 & J' \\
M & 0 & M' \\
\end{pmatrix}
$$

This gives us that ##M=M'## and ##\Delta J = 0, \pm 1##. If we are in the ##\Delta J = \pm 1## case, we connect different rotational levels, which are much further away from each other relative to ##\Lambda##-doubling levels, so I assume ##\Delta J = 0##. The expression above becomes:

$$-E_zD(-1)^{J-\Omega}\frac{M\Omega}{J(J+1)}$$

Now, the parity eigenstates are linear combinations of hund a cases:

$$|\pm>\frac{|J M S \Sigma \Lambda \Omega>\pm|J M S -\Sigma -\Lambda -\Omega>}{\sqrt{2}}$$

If we build the 2x2 Hamiltonian in the space spanned by ##|\pm>## with the Stark shift included it will then look like this (I will assume the ACME case, with ##J=1## and ##\Omega = 1##):

$$
\begin{pmatrix}
\Delta & -E_zD M\\
-E_zD M & -\Delta \\
\end{pmatrix}
$$

Assuming the 2 levels are very close we have ##\Delta << E_zD## and by diagonalizing the matrix we get for the energies and eigenstates (with a very good approximation): ##E_{\pm} = \pm E_zD M## and ##\frac{|+>\pm|->}{\sqrt{2}}##. Hence the different parities are fully mixed so the system is fully polarized.
 
  • #84
Twigg
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Thank you!! I really appreciate it! Your derivation helped put a lot of puzzle pieces together for me.

I was able to get the polarizability out of your 2x2 Hamiltonian. It has eigenvalues $$E_{\Lambda,M} = \frac{\Lambda}{|\Lambda|} \sqrt{\Delta^2 + (E_z DM)^2} \approx \frac{\Lambda}{|\Lambda|} (\Delta + \frac{1}{2} \frac{E_z ^2 D^2 M^2}{\Delta} +O((\frac{E_z DM}{\Delta})^2))$$
From this, polarizability is $$\alpha = \frac{\Lambda}{|\Lambda|}\frac{D^2 M^2}{2\Delta}$$, since the polarizability is associated with the energy shift that is quadratic in electric field. This seems to be in full agreement with what that review paper was saying (one of these days, I'll find that paper again).


I might have missed a complex conjugate in the Wigner matrix, as it is easier to type without it :D
I can never reproduce something I derived using Wigner matrices because of all the little mistakes here and there. They're just cursed. I'd sell my soul for a simpler formalism :oldbiggrin:

By the way, I found a thesis from the HfF+ eEDM group that derives the Stark shift, and it exactly agrees with your expression for no hyperfine coupling (##F=J## and ##I=0##). Nice work!
 
  • #85
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13
Thank you!! I really appreciate it! Your derivation helped put a lot of puzzle pieces together for me.

I was able to get the polarizability out of your 2x2 Hamiltonian. It has eigenvalues $$E_{\Lambda,M} = \frac{\Lambda}{|\Lambda|} \sqrt{\Delta^2 + (E_z DM)^2} \approx \frac{\Lambda}{|\Lambda|} (\Delta + \frac{1}{2} \frac{E_z ^2 D^2 M^2}{\Delta} +O((\frac{E_z DM}{\Delta})^2))$$
From this, polarizability is $$\alpha = \frac{\Lambda}{|\Lambda|}\frac{D^2 M^2}{2\Delta}$$, since the polarizability is associated with the energy shift that is quadratic in electric field. This seems to be in full agreement with what that review paper was saying (one of these days, I'll find that paper again).



I can never reproduce something I derived using Wigner matrices because of all the little mistakes here and there. They're just cursed. I'd sell my soul for a simpler formalism :oldbiggrin:

By the way, I found a thesis from the HfF+ eEDM group that derives the Stark shift, and it exactly agrees with your expression for no hyperfine coupling (##F=J## and ##I=0##). Nice work!
I am glad it's right! :D Please send me the link to that paper when you have some time. About the polarization, I am a bit confused. Based on that expression it looks like it can go to infinity, shouldn't it be between 0 and 1 (I assumed that if you bring the 2 levels to degeneracy you would get a polarization of 1)?

Side note, unrelated to EDM calculations: I am trying to derive different expressions in my free time just to make sure I understood well all the details of diatomic molecules formalism. It's this term for the Hamiltonian due to parity violation. For example in this paper equation 1 (I just chose this one because I read it recently, but it is the same formula in basically all papers about parity violation) gets turned into equation 3 after doing the effective H formalism. I didn't get a chance to look closely into it, but if you have any suggestions about going from 1 to 3 or any paper that derives it (their references don't help much) please let me know. I guess that cross product comes from the Dirac spinors somehow but it doesn't look obvious to me.
 
  • #86
Twigg
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Here's that thesis. I was looking at equation 6.11 on page 103. Also, I used ##\Lambda## instead of ##\Omega## in my last post, just a careless error.

I don't have APS access right now, so I can't see the Victor Flambaum paper that is cited for that Hamiltonian. Just looking at the form of that Hamiltonian, the derivation might have little to do with the content of Brown and Carrington because it's talking about spin perpendicular to the molecular axis.

If you're reading papers on parity violation, this one on Schiff's theorem is excellent if you can get access. I used to have a copy but lost it. Also, talk about a crazy author list :oldlaugh: What is this, a crossover episode?
 
  • #87
Twigg
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Just noticed I missed your question about polarizability. I'm not sure why it would be limited between 0 and 1. Are you thinking of spin polarization? What I mean here is electrostatic polarizability ##\vec{d}_{induced} = \alpha \vec{E}##. It only appears to go to infinity as ##\Delta \rightarrow 0## because the series expansion I did assumed ##\Delta \gg E_z DM##. The reason for this inequality is that polarizability is usually quoted for ##E_z \rightarrow 0## by convention.
 
  • #89
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13
So I tried to derive the Zeeman effect for a Hund case b, with the nuclear spin included in the wavefunction. The final result seems a bit too simple, tho. I will look only at the ##S\cdot B## term and ignore the ##g\mu_B## prefactor. For a Hund case b, the wavefunction with nuclear spin is:

$$|NS\Lambda J I F M_F> = \Sigma_{M_J}\Sigma_{M_I}<JM_JIM_I|FM_F>|NS\Lambda J M_J>|IM_I>$$

And we also have:

$$|NS\Lambda J M_J> = \Sigma_{M_N}\Sigma_{M_S}<NM_NSM_S|JM_J>|NM_N\Lambda>|SM_S>$$

where ##<JM_JIM_I|FM_F>## and ##<JM_SIM_N|JM_J>## are Clebsch-Gordan coefficients. Now, calculating the matrix element we have:

##<NS\Lambda J I F M_F|S\cdot B|N'S'\Lambda' J' I' F' M_F'>##

I will assume that the magnetic field is in the z direction. Also, given that we are in Hund case b we can look at the spin quantized in the lab frame, so we don't need Wigner rotation matrices, so we get ##S\cdot B = T_{p=0}^1(S)T_{p=0}^1(B) = B_zS_z##, where both ##B_z## and ##S_z## are defined in the lab frame, with ##S_z## being an operator, such that ##S_z|SM_S> = M_S|SM_S>##. So we have:

$$<NS\Lambda J I F M_F|B_zS_z|N'S'\Lambda' J' I' F' M_F'>=$$

$$B_z (\Sigma_{M_J}\Sigma_{M_I}<JM_JIM_I|FM_F><NS\Lambda J M_J|<IM_I|)S_z(\Sigma_{M_J'}\Sigma_{M_I'}<J'M_J'I'M_I'|F'M_F'>|N'S'\Lambda' J' M_J'>|I'M_I'>)$$

As ##S_z## doesn't act on the nuclear spin we get:

$$B_z \Sigma_{M_J}\Sigma_{M_I}\Sigma_{M_J'}\Sigma_{M_I'}<JM_JIM_I|FM_F><J'M_J'I'M_I'|F'M_F'><IM_I||I'M_I'><NS\Lambda J M_J| S_z |N'S'\Lambda' J' M_J'> = $$

$$B_z \Sigma_{M_J}\Sigma_{M_J'}\Sigma_{M_I}<JM_JIM_I|FM_F><J'M_J'IM_I|F'M_F'><NS\Lambda J M_J| S_z |N'S'\Lambda' J' M_J'> = $$

(basically we got ##I=I'## and ##M_I = M_I'##). For the term ##<NS\Lambda J M_J| S_z |N'S'\Lambda' J' M_J'>## we get:

$$(\Sigma_{M_N}\Sigma_{M_S}<NM_NSM_S|JM_J><NM_N\Lambda|<SM_S|)S_z(\Sigma_{M_N'}\Sigma_{M_S'}<N'M_N'S'M_S'|J'M_J'>|N'M_N'\Lambda'>|S'M_S'>)$$

As ##S_z## doesn't act on the ##|NM_N\Lambda>## part we have:

$$\Sigma_{M_N}\Sigma_{M_S}\Sigma_{M_N'}\Sigma_{M_S'}<NM_NSM_S|JM_J><N'M_N'S'M_S'|J'M_J'><NM_N\Lambda||N'M_N'\Lambda'><SM_S|S_z|S'M_S'>$$

From which we get ##N=N'##, ##M_N=M_N'## and ##\Lambda=\Lambda'##. So we have:

$$\Sigma_{M_N}\Sigma_{M_S}\Sigma_{M_S'}<NM_NSM_S|JM_J><NM_NS'M_S'|J'M_J'><SM_S|S_z|S'M_S'> = $$

$$\Sigma_{M_N}\Sigma_{M_S}\Sigma_{M_S'}<NM_NSM_S|JM_J><NM_NS'M_S'|J'M_J'>M_S'<SM_S||S'M_S'> = $$

And now we get that ##S=S'## and ##M_S=M_S'## so we have:

$$\Sigma_{M_N}\Sigma_{M_S}<NM_NSM_S|JM_J><NM_NSM_S|J'M_J'>M_S = $$

$$\delta_{JJ'}\delta_{M_JM_J'}M_S$$

So we also have ##J=J'## and ##M_J=M_J'##. Plugging in in the original equation, which was left at:

$$B_z \Sigma_{M_J}\Sigma_{M_J'}\Sigma_{M_I}<JM_JIM_I|FM_F><J'M_J'IM_I|F'M_F'><NS\Lambda J M_J| S_z |N'S'\Lambda' J' M_J'> = $$

$$B_z \Sigma_{M_J}\Sigma_{M_I}<JM_JIM_I|FM_F><JM_JIM_I|F'M_F'>M_S = $$

$$B_z M_S \delta_{FF'}\delta_{M_FM_F'}$$

So in the end we get ##F=F'## and ##M_F=M_F'##, so basically all quantum numbers need to be equal and the matrix element is ##B_zM_S##. It looks a bit too simple and too intuitive. I've seen mentioned in B&C and many other readings that hund case b calculations are more complicated than Hund case a. This was indeed quite tedious, but the result looks like what I would expect without doing these calculations (for example for the EDM calculations before I wouldn't see that ##\frac{1}{J(J+1)}## scaling as obvious). Also, is there a way to get to this result easier than what I did i.e. figure out that ##B_zM_S## should be the answer without doing all the math? Thank you!
 
  • #90
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I haven't gone through your derivation yet, but yes, there's a way easier method, which is how B&C derive all their matrix elements.

Look at equation 11.3. Its derivation is literally three steps, by invoking only two equations (5.123 first and 5.136 twice, once for ##F## and once for ##J##). The whole point of using Wigner symbols is to avoid the Clebsch-Gordan coefficient suffering.

By the way, almost every known case is in the B&C later chapters for you to look up. Every once in a while it's not. It happened to me actually, but I was able to derive what I needed using the process above.
 
  • #91
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I haven't gone through your derivation yet, but yes, there's a way easier method, which is how B&C derive all their matrix elements.

Look at equation 11.3. Its derivation is literally three steps, by invoking only two equations (5.123 first and 5.136 twice, once for ##F## and once for ##J##). The whole point of using Wigner symbols is to avoid the Clebsch-Gordan coefficient suffering.

By the way, almost every known case is in the B&C later chapters for you to look up. Every once in a while it's not. It happened to me actually, but I was able to derive what I needed using the process above.
Thanks a lot! This really makes things a lot easier!

I have a few questions about electronic and vibrational energy upon isotopic substitution. For now I am interested in the changes in mass, as I understand that there can also be changes in the size of the nucleus, too, that add to the isotope effects.

We obtain the electronic energy (here I am referring mainly to equation 7.183 in B&C) by solving the electrostatic SE with fixed nuclei. Once we obtain these energies, their value doesn't change anymore, regardless of the order of perturbation theory we go to in the effective Hamiltonian. The energy of the vibrational and spin-rotational will change, but this baseline energy of the electronic state is the same. When getting this energy, as far as I can tell, all we care about is the distance between the electrons and nuclei, as well as their charges. We also care about the electron mass, but not the nuclear one. This means that the electronic energy shouldn't change when doing an isotopic substitution. This is reflected in equation 7.199. However in equation 7.207 we have a dependence on the mass of the nuclei. From the paragraphs before, the main reason for this is the breaking of BO approximation. However, this breaking of BO approximation, and hence the mixing of electronic levels is reflected only in the effective Hamiltonian. As I mentioned above, the electronic energy should always be the same as its zero-th order value. Where does this mass dependence of the electronic energy ##Y_{00}## from equation 7.207 come from?

For vibrational energy, we have equation 7.184. I assume that the ##G^{(0)}_{\eta\nu}## term has the isotopic dependence given by 7.199. Do the corrections in 7.207 come from the other 2 terms: ##V^{ad}_{\eta\nu}## and ##V^{spin}_{\eta\nu}##? And if so, is this because these terms can also be expanded as in equation 7.180? For example, from ##V^{ad}_{\eta\nu}## we might get a term of the form ##x_{ad}(\nu+1/2)## so overall the first term in the vibrational expansion becomes ##(\omega_{\nu e}+x_{ad})(\nu+1/2)## which doesn't have the nice expansion in 7.199 anymore but the more complicated one in 7.207? Is this right? Also do you have any recommendations for readings that go into a bit more details about this isotopic substitution effects? Thank you!
 
  • #92
35
15
I'm much less familiar with vibrational corrections. And as you've probably noticed, it's not the main focus of B&C either. A couple places to start would be:

1. Dunham's original paper: http://jupiter.chem.uoa.gr/thanost/papers/papers4/PR_41(1932)721.pdf
It shows the higher order corrections that are typically ignored in all those ##Y_{ij}## coefficients.

2. In that section B&C refer to Watson's paper: https://doi.org/10.1016/0022-2852(80)90152-6
I don't have access to it, but it seems highly relevant to this discussion.
 
  • #93
184
13
I'm much less familiar with vibrational corrections. And as you've probably noticed, it's not the main focus of B&C either. A couple places to start would be:

1. Dunham's original paper: http://jupiter.chem.uoa.gr/thanost/papers/papers4/PR_41(1932)721.pdf
It shows the higher order corrections that are typically ignored in all those ##Y_{ij}## coefficients.

2. In that section B&C refer to Watson's paper: https://doi.org/10.1016/0022-2852(80)90152-6
I don't have access to it, but it seems highly relevant to this discussion.
Thanks for the references, they helped a lot. I was wondering if you know of any papers that extended this isotope shift analysis to molecules that are not closed shell. For example the isotope dependence of spin-orbit, spin-rotation or lambda doubling parameters. I see in B&C that they mention that this hasn't been done, but the book was written in 2003 and perhaps someone did the calculations meanwhile.
 
  • #94
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13
I looked a bit at some actual molecular systems and I have some questions.

1. In some cases, a given electronic state, say a ##^2\Pi## state is far from other electronic states except for one, which is very close (sometimes even in between the 2 spin-orbit states i.e. ##^2\Pi_{1/2}## and ##^2\Pi_{3/2}##) and the rotational energy is very small. Would that be more of a Hund case a or c?

2. I noticed that for some ##^2\Pi## states, some molecules have the electronic energy difference between this state and the other state bigger than the spin-orbit coupling and the rotational energy, which would make them quite confidently a Hund case a. However, the spin orbit coupling is bigger than the vibrational energy splitting of both ##^2\Pi_{1/2}## and ##^2\Pi_{3/2}##. How would I do the vibrational averaging in this case? Wouldn't the higher order perturbative corrections to the spin-orbit coupling diverge? Would I need to add the SO Hamiltonian to the zeroth order hamiltonian, together with the electronic energy?

3. In the Hund case c, will my zeroth order Hamiltonian (and I mean how it is usually done in literature) be ##H_{SO}##, instead of the electronic one, ##H_e## or do I include both of them ##H_e+H_{SO}##? And in this case, if the spin orbit coupling would be hidden in the new effective ##V(R)##, how can I extract the spin-orbit constant, won't it be mixed with the electronic energy?
 

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