I Effective molecular Hamiltonian and Hund cases

[DrDu]

Sorry about that! Here is a screenshot of that table (not sure how to attach the paper here). The table is cut on the right side a bit from the paper (it's quite an old paper).

View attachment 286277

What is $\xi$ and $\eta$?

 

[BillKet]

Does it? As Sigma and Pi have different symmetry, $<\Sigma(R)|\Pi(R')>=0$ for all R and R', hence also
$<\Sigma(R)|T^N|\Pi(R)>=0$.

Actually, for the case of $\Sigma$ and $\Pi$ won't the off diagonal be zero whether I use adiabatic or diabatic? What I mean is, if I use adiabatic, I get the eigenfunctions of $H_{el}$, but the off-diagonal in $T^N$ is zero, as you mentioned. If I use diabatic, I get the eigenfunctions of $T^N$, but the off-diagonal in $H_{el}$ would also be zero, as they have different symmetries again.

 

[BillKet]

What is $\xi$ and $\eta$?

Sorry! $\xi$ is the SO hamiltonian term and $\eta$ the rotational one. So for example $\xi = <\nu_\Sigma|<\Sigma|H_{SO}|\Pi>|\nu_\Pi>$ and $\eta = <\nu_\Sigma|<\Sigma|H_{rot}|\Pi>|\nu_\Pi>$

 

[DrDu]

Actually, for the case of $\Sigma$ and $\Pi$ won't the off diagonal be zero whether I use adiabatic or diabatic? What I mean is, if I use adiabatic, I get the eigenfunctions of $H_{el}$, but the off-diagonal in $T^N$ is zero, as you mentioned. If I use diabatic, I get the eigenfunctions of $T^N$, but the off-diagonal in $H_{el}$ would also be zero, as they have different symmetries again.

As we discussed already in a previous thread
https://www.physicsforums.com/threa...oppenheimer-approximation-doesnt-work.999950/
this is not the case in the adiabatic representation. As the sigma and pi levels intersect, the (non-diagonal) non-adiabatic coupling becomes even singular at the intersection point.
In the diabatic approximation, only small non-diagonal terms, like the SO-coupling remain.

 

[DrDu]

Sorry! $\xi$ is the SO hamiltonian term and $\eta$ the rotational one. So for example $\xi = <\nu_\Sigma|<\Sigma|H_{SO}|\Pi>|\nu_\Pi>$ and $\eta = <\nu_\Sigma|<\Sigma|H_{rot}|\Pi>|\nu_\Pi>$

Ok, so they actually use the vibrational averaging approach in the diabatic representation of the Pi and Sigma states.

 

[BillKet]

Ok, so they actually use the vibrational averaging approach in the diabatic representation of the Pi and Sigma states.

I am pretty sure they are adiabatic. Here is a paper giving a theoretical description of the experimental results in the paper I previously mentioned and they explicitly say that the electronic curves are adiabatic.

But whether they are adiabatic or diabatic, I am not sure how that answers by question about the difference between the 2 approaches.

 

[DrDu]

Concerning the difference between first diagonalizing the hamiltonian and then do vibrational averaging and doing first vibrational averaging: The first possibility is quite hard and can also only be performed perturbationally. Analytical solutions are only available in favourable situations.
For example assuming that the vibrational Pi and Sigma states are all harmonic and only shifted relative to each other and that also the SO coupling depends at most linearly on R.
So if $H=H_0 + \lambda H'$, where $H'$ results from the Spin-Orbit coupling,

$H_0 =\omega a^+ a \sigma_0 + (\alpha+\beta (a+a^+))\sigma_z,$$H' =(\gamma+ \delta(a+a^+))\sigma_x.$

Where $a$ and $a^+$ are the usual HO anihilation and creation operators which can be expressed in terms of R and d/dR. $\sigma_0$ is the 2x2 unit matrix while the other sigmas are the usual Pauli matrices.

We now try to diagonalize $H$ via a unitary transformation
$\exp(-iS)H\exp(iS)$ with

$S=\lambda S_1+ \lambda^2 S_2 \ldots$.
In first order

$i[H_0,S_1]=H'.$

This equation is trivial to solve in an eigenbasis of $H_0$, i.e. using vibrational averaging.
To solve it in terms of R and d/dR, or equivalently in terms of a and $a^+$,
we make for $S_1$ the ansatz

$S_1=(p+qa+ra^+) \sigma_x +(s+ta+ua^+) \sigma_y$

Evidently, the six coefficients p,q,r,s,t and u can be determined comparing terms on the left and right hand side of

$i[H_0,S_1]=H'.$

Now if $\psi_0$ is a zeroth order eigenstate, then

$E= \frac{ <\psi_0| \exp(-i\lambda S_1)H\exp(i\lambda S_1)| \psi_0>}{<\psi_0| \psi_0>}$

is the correct energy eigenvalue up to and including order $\lambda^3$.

What I really want to say is that the non-commutability of the a and $a^+$ or of R and d/dR complicates the diagonalization of the hamiltonian considerably as compared to the case, were the terms depend only on R alone.