How to derive the expression for derivative couplings on using the Born-Oppenheimer expansion?

Question

I am trying to understand the derivation for the non-adiabatic coupling (NAC) elements as presented in Conical Intersections Electronic Structure, Dynamics and Spectroscopy by Wolfgang Domcke and Horst Koppel. Where it is shown that the NAC can be expressed as $$\boldsymbol{\Lambda}_{ji}=\delta_{ji}T_{n}-\left\langle j(\boldsymbol{R})\left|T_{n}\right|i(\boldsymbol{R})\right\rangle $$

Where $R$ are the collective nuclear coordinates. Obviously the above elements are operators in $R$ - space and $T_n$ being the nuclear kinetic energy operator. Now the author states that if we take $$T_n = -\frac{1}{2M}\nabla^2$$ where the gradient $\nabla$ vector acts on the nuclear space. The expression for $\Lambda_{ij}$ becomes

$$\boldsymbol{\Lambda}_{ji}=\frac{1}{2M}[2\textbf{F}_{ji}(\boldsymbol R)\cdot\boldsymbol\nabla+G_{ji}]$$

where

$$\textbf{F}_{ji}=\langle j(\boldsymbol R) | \boldsymbol\nabla i(\boldsymbol R)\rangle$$ and

$$G_{ji}=\langle j(\boldsymbol R) | \nabla^2 i(\boldsymbol R)\rangle$$

These terms are called the derivative and scalar couplings respectively, the boldface symbols denote that they are vectors in the nuclear space. However, I do not reach the same expression on plugging in the expression for $T_n$ into the expression for $\boldsymbol\Lambda_{ji}$. Instead simply plugging it in results in the expression

$$\boldsymbol\Lambda_{ji}=-\frac{\delta_{ji}}{2M}\nabla^2+\frac{\langle j(\boldsymbol R) | \nabla^2 i(\boldsymbol R)\rangle}{2M}$$

The rightmost term seems to be same as the $G_{ji}$ as defined by the author. But I have no idea how to manipulate the first term into the $\textbf{F}_{ji}$ term that the author claims would appear. The same treatment can also be found in here and here (unfortunately also without any further explanation). Can someone please help me figure out how the expression for $\boldsymbol\Lambda_{ji}$ that I attained is equivalent to the one that the author has stated in all of the above references?

Answer 1

Mind the subtle difference between the notations of the forms for $\Lambda_{ji}$: in the first form, there is $\nabla^2|i\rangle$ meaning that $\nabla^2$ acts on everything to the right, while in the second form, we have $|\nabla i\rangle$ and $|\nabla^2i\rangle$, in which the differentiations act only on $|i\rangle$ (admittedly, the authors chose a somewhat non-standard way to notate this). In this notation, the product rule of differentiation looks like $$ \nabla|i\rangle=|\nabla i\rangle+|i\rangle\nabla \ , $$ $$ \nabla^2|i\rangle=|\nabla^2 i\rangle+2|\nabla i\rangle\nabla+|i\rangle\nabla^2 \ . $$ Plugging the latter identity into the first form of $\Lambda_{ji}$ and using $\langle j|i\rangle=\delta_{ij}$ will immediately yield the more well-known second form (remember that although $\nabla$ acts on the nuclear coordinates, the bra-ket refers to integration over electronic coordinates).