Phy5646/Born-Oppenheimer Approximation

Consider the problem of two protons and one electron.

As for the two protons, we consider the two bodies problem as one body problem.

$H={\frac {P^{2}}{2M}}+{\frac {p^{2}}{2m}}+V\left({\overset {\rightharpoonup }{R}},{\overset {\rightharpoonup }{r}}\right)$

The wave function is:

$\Psi _{total}=\left|n_{electron}\left({\overset {\rightharpoonup }{R}}\right)>\right|\psi _{proton}\left({\overset {\rightharpoonup }{R}}\right)>$

First step:

Consider R is fixed, to solve the schrodinger equation of electron.

Second step:

Seek an solution of H as: $\left|\Psi >=\int \Psi \left({\overset {\rightharpoonup }{R}}\right)\right|{\overset {\rightharpoonup }{R}}>|n\left({\overset {\rightharpoonup }{R}}\right)>d^{3}R$

To solve: $<{\overset {\rightharpoonup }{R'}}\left|<n\left({\overset {\rightharpoonup }{R}}\right)\right|H|\Psi >$

You will find:

$\left[{\frac {1}{2M}}\left({\overset {\rightharpoonup }{P}}-{\overset {\rightharpoonup }{A}}^{(n)}\right)^{2}+\Phi ^{(n)}+E_{n}\left({\overset {\rightharpoonup }{R}}\right)\right]\Psi \left({\overset {\rightharpoonup }{R}}\right)=E\Psi \left({\overset {\rightharpoonup }{R}}\right)$

The Berry Vector Potential: ${\overrightarrow {A}}^{(n)}=i\hbar \langle n(R)|{\overrightarrow {\nabla }}_{R}|n(R)\rangle$

The Berry Scalar Potential: $\Phi ^{(n)}={\frac {\hbar ^{2}}{2m}}[\langle {\overrightarrow {\nabla }}_{R}n(R)|{\overrightarrow {\nabla }}_{R}n(R)\rangle -\langle {\overrightarrow {\nabla }}_{R}n(R)|n(R)\rangle \langle n(R)|{\overrightarrow {\nabla }}_{R}n(R)\rangle ]$

It's a method to get the energy of the system. Indispensable method in Quantum Chemistry.