What changes when the electronic basis moves

Expand the full state in a local electronic basis

At each nuclear position $R$, choose orthonormal electronic eigenstates $\phi_n(r;R)$. Let $M$ be the nuclear-coordinate mass. Nuclear kinetic energy is $-\hbar^2\partial_R^2/(2M)$. Expand the molecular state with nuclear amplitudes $\chi_n(R,t)$.

The nuclear derivative acts on both factors. Applying the product rule twice produces three terms. Multiplying by a second electronic state and integrating electron coordinates projects the molecular equation into coupled nuclear channels.

Reading: Curchod · John C. Tully

Nuclear motion meets the changing electronic basis

Define $d_{mn}=\langle\phi_m|\partial_R\phi_n\rangle$ and $\tau_{mn}=\langle\phi_m|\partial_R^2\phi_n\rangle$. These inner products integrate electron coordinates only. Potential energy is diagonal in an adiabatic basis; nuclear kinetic energy mixes its channels.

Along a classical nuclear path $R(t)$, the basis changes at a rate $\dot R\,d_{mn}$. Faster motion through the same spatial variation increases the rate at which the electronic state must adjust.

A rotation shows where the coupling goes

Write a real symmetric two-state potential with common energy $s(R)$, diagonal difference $2z(R)$ and off-diagonal coupling $c(R)$. We call this smooth fixed basis diabatic. Nonadiabatic dynamics describes a process; diabatic describes a representation.

A position-dependent rotation through angle $\theta(R)$ diagonalizes this matrix. It also transforms nuclear kinetic energy. The new derivative coupling contains the spatial derivative of the mixing angle. Dropping those terms after diagonalizing the potential changes the approximation.

Worked example of a narrow coupling peak

Near the crossing, set $z=aR$ and $c=\Delta>0$. Here $a$ has units of energy per length. Differentiation gives $\theta'=-a\Delta/[2(a^2R^2+\Delta^2)]$. The central magnitude is $|a|/(2\Delta)$.

Halving the coupling doubles the derivative peak and halves its characteristic spatial width. The nucleus crosses a sharper change of electronic character. At zero coupling the eigenbasis is undefined at the degeneracy, so the divergent expression requires a different numerical treatment.

A constant rotation supplies a small check

At one position let the old coefficients be $(1,0)^{\mathsf T}$. Rotate the electronic basis through half a right angle. The new coefficients are $(1,1)^{\mathsf T}/\sqrt2$, with signs set by the rotation convention. Each new channel has probability one half and total probability remains one.

To measure the original first state, rotate its projector too. Reading only the first new component instead would mistake a new coordinate axis for population transfer. This is why a channel probability needs a representation label.

Test that representation has not changed the prediction

For a unitary rotation $U(R)$, nuclear coefficients obey $\boldsymbol\chi_{\mathrm d}=U\boldsymbol\chi_{\mathrm a}$. The total density is unchanged point by point. Channel probabilities must be projected onto the same physical electronic states before comparison.

A constant unitary rotation provides a simple solver check. A spatially changing rotation also transforms kinetic energy and requires derivative terms. Arbitrary sign flips of local eigenvectors must cancel in correctly transformed overlaps and probabilities.