0
$\begingroup$

Lets say i have an initial state $|i\rangle$ and a final state $|f\rangle$. A transition from $|i\rangle \rightarrow |f\rangle$ is coupled by an operator $\hat O$. Is the relevant coupling matrix element for the transition starting in state $|i\rangle$ going to $|f\rangle$ the matrix element $$ \langle i|\hat O |f\rangle =O_{if} $$ or is the relevant matrix element $$\langle f | \hat O | i \rangle = O_{fi} $$ ?

I thought that the relevant matrix element is $$ O_{fi} $$

but a paper discussing nonadiabatic coupling is confusing me, since the operator always appears with indices switched in comparison to what i would expect.

For Context:

My source of confusion is this paper, https://doi.org/10.1016/j.chemphys.2008.01.044, which is about Nonadiabatic Surface Hopping. The section "2.2 Velocity adjustment" equation (21) uses the nonadiabatic coupling element of first order $\vec d_{ij}$ to rescale velocities after a hop between two adiabatic electronic states. The hop i.e. coupling is from state $i$ to state $j$, which is why was expecting the coupling element $ \vec d_{ji}$. The indices on this coupling element are reversed in comparison to what i would have expected.

The nonadiabatic coupling element of first order is defined as $$ \vec d_{ij}(R) = \langle \phi_i(r,R)| \nabla_R \phi_j(r,R)\rangle_r $$ where $\phi_{i/j}(r,R)$ are adiabatic electronic states, i.e. eigenstates to the electronic Hamilton operator within the Born Oppenheimer Approximation.

$\endgroup$

1 Answer 1

1
$\begingroup$

You could use either matrix element, since the physical results would necessarily contain combinations of $O_{fi}$ and $O_{fi}$ - either magnitude squared (like in the Fermi golden rule, where the probability of transition is proportional to $|O_{fi}|^2$) or the real/imaginary part of the matrix element (as, e.g., in higher order Fermi golden rules, expressions for the density-of-states, etc.)

This is a nice flashback to the basics of quantum mechanics: the difference between a probability and the probability amplitude, the diagonal and non-diagonal matrix elements of operators, etc.

$\endgroup$
1
  • 1
    $\begingroup$ That is a good point. I guess the my problem is rather a full understanding of the semi-classical methods used in the paper. I think it would make a difference there, if you switched the matrix elements, since the nonadiabatic coupling element is anti hermitian and the momentum of a trajectory is rescaled in the direction of the coupling element. $\endgroup$
    – Hans Wurst
    Commented Dec 15, 2020 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.