Subject: non-adiabatic, spin-orbital coupling....

From: <[email-masked]>
Date: Wed, 6 Dec 1995 12:03:01 -0500 (EST)

Hi, CCL members, I've got several basic questions about the Non-adiabatic 
coupling and spin-orbital coupling elements.

1. "selection rule" and Magnitude of the non-adiabatic coupling...
	In principle, non-adiabatic coupling elements are defined as the 
coupling elements of two adiabatic wave function through nuclear momentum 
operator. i.e. <F1(r;R)|P|F2(r;R)>. If we forget about spin-orbital 
coupling, i.e. Fi diagonalize the spin-free Hamiltonian, then the 
"selection rule" of the coupling is fully determined by spatial symmetry. 
If we consider P as the generator of translation, then presumably if 
the direct product of F1 and the displaced F2 (say, along certain 
internal coordinate) has total symmetric component, then the non-adiabatic
coupling is non-zero.( Am I right? ). But how do I get the feeling of 
the magnitude of it? Can I estimate where is the coupling-elements gonna 
be large from the electronic structure before I do the calculation? 
I guess some people will say, it's gonna be large where we have avoided 
crossing. Actually this brings up my another question...=>

2. Avoider vs "perfect" (actually) crossing...
	Again, in principle, the non-crossing rule only holds for diatomic 
systems where one has only one-D. The number of conditions that the system
has to satisfy to acheive a true degeneracy really depends on the behavior
of ur system under time revesal. If one doesn't care about spin-orbital 
coupling, then usually the number of condition is 2 (state of same 
symmetry). Obviously, in poly-atomic systems, the degree-of-freedom is much
larger than 2, namely, one can have actual crossing in polyatomic case, even
between states of same symmetry. Yet, they don't have to, which gives 
avoided crossing. So, after my endless babbling, my question is, in the 
avoided case, the non-adiabtic coupling can be thought to be related to 
the splitting between the two states, which equals to 2 times of the 
diabatic coupling between them (am i right?). What about in the actual 
crossing case? 

3. Rate expression...
	My last question of the day is related to the rate expression of 
radiationless transitions. I've read in many places, where they claim 
that non-adiabatic coupling causes internal conversion, while 
spin-orbital causes intersystem crossing. Of course, this make sense. 
Next what we(they) do is, using Fermi's Golden rule to express the 
transition rate. Obviously, this is the first-order approximation. Let's 
look from another way. To solve spin-forbidden transitions, in principle 
one can go through two ways to solve the Schrodiger's Eq. The first one 
would be using wavefunctions of spin-free Hamiltonian, and treat 
spin-orbital as a perturbation. In this case, the non-adiabatic coupling 
is obviously zero. ( I mean coupling between adiabatic wf. of spin-free 
H). Another way would be, as in many ICN, CH3I studies, solve wave 
function of a Hamiltonian including spin-orbital coupling. In this case,
I guess it's the non-adiabatic coupling between the two states including 
spin-orbital effects that causes the transition. Or, in the dynamic 
calculations, one can diabatrize the potential (wf.) such that the 
off-diagonal elements determines the transition rate. So, which is a 
better, more rigorous approach????

Sorry for such a long message. Thanks VERY MUCH for ur attention...

Please send message directly to: [email-masked]

I'll summerize.

Have a nice day, to all of u!!!

______________________________________________________________

Qiang Cui
Dept. of Chem. Emory Univ.         508 Webster Dr. Apt.#2
Atlanta, GA 30322.                 Decatur, GA 30033.
(404)-727-2381                     (404)-636-6149

http://tswww.cc.emory.edu/~qcui
______________________________________________________________