Calculation of vibronic couplings

[slawa]

Hi all,

I'm referring to the paper Chem. Phys. v.26 pp.169-177 y.1977 from Cederbaum, Domcke and Köppel where the expressions for linear vibronic coupling parameters are derived in second quantization notation. I'm stuck on the point, where the dependence of the fermionic ladder operators on the nuclear coordinates is eliminated. To be more specific:

H = ∑_i ∊_i(0)a_i^†a_i +∑_i (0.5^0.5∂∊_i/∂Q)[a_i^†a_i-n_i]Q

is the part of the Hamiltonian I'm interested in. Here the fermionic ladder operators are still dependend on the nuclear coordinates Q (for simplicity let there be just one of it at this time). Now by projecting the ladder operator on an Q-independend basis

a_i(Q) = ∑_j <ϕ_i(Q)|ϕ_j(0)>a_j(0)

these parts of the hamiltonian should look like:

∑_i ∊_i(0)a_i^†a_i +∑_i 0.5^0.5 ∂_Q∊_i[a_i^†a_i-n_i]Q + ∑_i,j λ^i,j[a_ia_j^†+a_ja_i^†]Q

where the lambdas are given by

λ^i,j = lim _Q→0[∊_i(Q)-∊_j(Q)]<ϕ_j(Q)|∂_Q|ϕ_i(Q)>

Up to now I'm pretty shure that I understand how to get the first two summands of the new hamiltonian but the off diagonal coupling term with the lambdas...My last point was, that from the last term in the old hamiltonian leftover terms like this one

-∂_Q∊₁(0)[<ϕ₁(0)|ϕ₁(Q)>a₁^†<ϕ₁(Q)|ϕ₂(0)>a₂+<ϕ₂(0)|ϕ₁(Q)>a₂^†<ϕ₁(Q)|ϕ₁(0)>a₁]Q-∂_Q∊₂(0)[<ϕ₁(0)|ϕ₂(Q)>a₁^†<ϕ₂(Q)|ϕ₂(0)>a₂+<ϕ₂(0)|ϕ₂(Q)>a₂^†<ϕ₂(Q)|ϕ₁(0)>a₁]Q

have to become the third summand in the new hamiltonian...Here some link is missing.

Further they state, that from going from the general expression for the lambdas to a two state system, the lambdas can be written as:

λ1,2=0.75*(∂Q22[∊1(Q)-∊2(Q)]2)00.5

So here I'm obviously missing some math ;)

I'd appreciate some help and sorry for the load of equations :|

Greetings
Slawa

 

[DrDu]

I think the step you are missing is a simple taylor expansion of their eq 6: $\langle \phi_i(Q)|\phi_j(0)\rangle=\langle \phi_i(0)| \phi_j(0)\rangle +\langle \partial \phi_i(Q)/\partial Q|_{Q=0} |\phi_j(0) \rangle Q$.