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.


1 Answer 1


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.

  • 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

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