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...
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##)...
For more Stark shift/polarization stuff, also take a look at the Gurevich thesis if you haven't already: http://www.doylegroup.harvard.edu/files/bufferpubs/theses/yvg_thesis.pdf
Equation 3.22 shows the bit about polarization going from 0 to 1.
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.
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...
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...
Okay, here's my stab at the first question:
The derivation is completely written out in Lefebvre-Brion/Field in Section 6.1.2.1, and it looks like yours is consistent.
Now as to why there's only cos##\theta## in B&C's version. I suspect this is all because of Eqn. 6.330 in B&C. Notice that the...
Can you elaborate on how you arrived at this interpretation? Why does it imply that the electrons can't catch up? The "crude" BO approximation gives you a dipole moment result at a specific ##R##. If you on average only observe a specific ##R_{eq}## (equilibrium distance), then the electronic...
I think I can answer the second question for now. Eqn. 6.333 I believe has some sloppy notation. The second integral should maybe have a different symbol for ##R_\alpha## for the electronic part. It's meant to be at a single internuclear distance, usually the equilibrium distance. So you don't...
Unfortunately, I can't look at this until later tonight, but I need revisit your derivation above more carefully. Because, I see now that the they way you have it derived, the rotational part forces the electronic part to only ##q = 0## terms, which has to be wrong because inter-electronic...
As to your first question:
For an intra-electronic transition, you're coupling two electronic states that have exactly the same electron spatial distribution. Electron population is distributed symmetrically about the molecular axis, so there is no permanent dipole moment perpendicular to the...
@BillKet Yes, you assume the BO approximation first, then handle the non-adiabatic terms with perturbation theory. i.e. those parameters (or "constants") in the effective Hamiltonian.
As for your second question, that looks right, except I want to clarify: a spherical harmonic is actually a...
Billket, I'm assuming you went through Section 6.11.6 in B&C? In that case, yes, you'll end up keeping the first term in equation 6.333 (for an intra-electronic transition. They mention the first terms goes to zero for inter-electronic transitions.). And then your matrix element is equation...
Oh sorry, it looks like I did misunderstand (##g_l## is one order higher, starting with ##g_L##).
My understanding is that ##g_L## already is an effective parameter. In Eqn. 7.217 it's introduced as a perturbation to the full Hamiltonian. But he mentions right after that it can deviate from 1.0...
Haha okay the answer to this might be pretty funny, assuming I understand the question correctly:
Look at Eq. 7.217, which has a ##g_L## in it. Eqns 7.221 and 7.222 show the result of putting the off-diagonal terms into an effective parameter: ##g_l##. (Notice the capitalized vs. not.) In the...