# Deriving the equation of motion for adiabatic approximation

In the textbook, Theories of Molecular Reaction Dyanmics by Flemming Y. Hansen and Niels E. Henriksen, there is a derivation of the equation of motion for the adiabatic approximation (pgs. 5-7). The author starts by showing that the wavefunction can be written as a product state by separating the slow and fast components (i.e. nuclear and electronic) as shown below,

$$\Psi(r,R, t) = \chi(R,t)\psi(r;R)$$

where $$\chi(R,t)$$ is the nuclear component and $$\psi(r;R)$$ is the electronic component which depends parametrically on the nuclear position (R).

We take the following time-dependent Schrodinger equation, written as

$$i\hbar \frac{\partial \Psi(r,R,t)}{\partial t} = (\hat{T_{Nuc}} + \hat{H_{e}}) \Psi(r,R,t)$$

and plug in the product state to get the following equation

$$i\hbar \frac{\partial \chi(R,t)}{\partial t} = [\hat{T_{Nuc}} + E_{i}(R) + \left< \psi \right|\hat{T_{Nuc}}\left| \psi \right>_0] \chi(R,t)$$.

It is said that this term is equal to zero, leaving us with the equation of motion of the adiabatic approximation

$$i\hbar \frac{\partial \chi(R,t)}{\partial t} = [\hat{T_{Nuc}} + E_{i}(R) ] \chi(R,t)$$.

Questions

1. How do we get the $$\left< \psi \right|\hat{T_{Nuc}}\left| \psi \right>_0$$ term during the derivation? I am confused on how that arises.

2. What does this term mean? Is it just the coupling between the nuclear and electronic states?

The term is the diagonal element of the nuclear kinetic energy operator with respect to the adiabatic electronic state basis. The term is obtained when the equation is projected on the adiabatic electronic state $$\langle \psi|$$ and integrated over electronic coordinates,
$$i\hbar \partial_t \chi(R,t)\psi(r;R) =(\hat T_N+ \hat H_e) \chi(R,t)\psi(r;R) \stackrel{\langle \psi|_r}{\Rightarrow} \\ \int dr\psi^*(r;R)i\hbar \partial_t \chi(R,t)\psi(r;R)=\int dr\psi^*(r;R)(\hat T_N+ \hat H_e) \chi(R,t)\psi(r;R)\\ \langle \psi|\psi\rangle_ri\hbar \partial_t \chi(R,t) = \big (\langle\psi|\hat T_N|\psi\rangle_r + E(R) \big)\chi(R,t)\\ i\hbar \partial_t\chi(R,t)=\big(\langle\psi|\hat T_N|\psi\rangle_r + \hat T_N + E(R) \big)\chi(R,t)$$ The expansion of $$\hat T_N$$ applied to the product of $$\chi\psi$$, according to the product rule, also leads to a mixed term with first order derivatives $$\sum_{Nuclei}\frac{\hbar ^2}{2M}\langle\psi| \hat P_N|\psi\rangle_r \cdot (\hat P_N \chi(R,t))$$, which I have already omitted since this first order term is exactly zero when using a purely real $$\psi$$. This is also mentioned in the textbook.
But it should be noted that this mixed first order term only vanishes for the diagonal elements of $$\langle \psi |T_N|\psi\rangle$$, i.e. when $$\psi$$ is the same adiabatic electronic state. In the general case, this term leads to nonadiabatic effects, where different adiabatic electronic states become coupled by the motion of the nuclei, inducing nonadiabatic transitions.
The nuclear kinetic energy term $$\langle \psi| \hat T_N|\psi\rangle_r$$ does not vanish when using a real $$\psi$$, but is typically small in comparison to $$E(R)$$, in particular around nuclear equilibrium coordinates. But in principle it is an additional energy term that could/should be included in the Hamiltonian for the nuclear wavefunction, and it is due to the coupling of electronic and nuclear motion.
The term can be calculated if $$|\psi\rangle$$ is known as function of $$R$$, but is typically very small due to the inverse dependency on nuclear masses and the rather slow change of $$\psi$$ with $$R$$. Sometimes it is included at zero-order I.e. the zero-order Taylor expansion term at some $$R_0$$ is added as correction to the potential energy $$E(R)$$, for that reason it is also known as diagonal Born-Oppenheimer correction term. But in most cases it is assumed to be negligible.