Question

Derivation of the Born–Oppenheimer approximation.



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Answer 1

Born–Oppenheimer approximation is the assumption that the motion of atomic nuclei and electrons in a molecule can be separated.

It allows the wavefunction of a molecule to be broken into its electronic and nuclear components.

$\Psi_\mathrm{total} = \psi_\mathrm{electronic} \times \psi_\mathrm{nuclear}$

$E_0(\mathbf{R}) \ll E_1(\mathbf{R}) \ll E_2(\mathbf{R}) \ll \cdots \text{ for all }\mathbf{R}$

We start from the exact non-relativistic, time-independent molecular Hamiltonian:

${\displaystyle H=H_{\text{e}}+T_{\text{n}}}$

with

${\displaystyle H_{\text{e}}=-\sum _{i}{{\frac {1}{2}}\nabla _{i}^{2}}-\sum _{i,A}{\frac {Z_{A}}{r_{iA}}}+\sum _{i>j}{\frac {1}{r_{ij}}}+\sum _{B>A}{\frac {Z_{A}Z_{B}}{R_{AB}}}\quad {\text{and}}\quad T_{\text{n}}=-\sum _{A}{{\frac {1}{2M_{A}}}\nabla _{A}^{2}}.}$

It is useful to introduce the total nuclear momentum and to rewrite the nuclear kinetic energy operator as follows:

${\displaystyle T_{\text{n}}=\sum _{A}\sum _{\alpha =x,y,z}{\frac {P_{A\alpha }P_{A\alpha }}{2M_{A}}}\quad {\text{with}}\quad P_{A\alpha }=-i{\frac {\partial }{\partial R_{A\alpha }}}.}$

Suppose we have K electronic eigenfunctions $\chi_k (\mathbf{r}; \mathbf{R}) of H_{{\text{e}}},$ that is, we have solved

${\displaystyle H_{\text{e}}\chi _{k}(\mathbf {r} ;\mathbf {R} )=E_{k}(\mathbf {R} )\chi _{k}(\mathbf {r} ;\mathbf {R} )\quad {\text{for}}\quad k=1,\ldots ,K.}$

We will assume that the parametric dependence is continuous and differentiable, so that it is meaningful to consider

${\displaystyle P_{A\alpha }\chi _{k}(\mathbf {r} ;\mathbf {R} )=-i{\frac {\partial \chi _{k}(\mathbf {r} ;\mathbf {R} )}{\partial R_{A\alpha }}}\quad {\text{for}}\quad \alpha =x,y,z,}$

which in general will not be zero.

The total wave function ${\displaystyle \Psi (\mathbf {R} ,\mathbf {r} )} is expanded in terms of {\displaystyle \chi _{k}(\mathbf {r} ;\mathbf {R} )}:$

${\displaystyle \Psi (\mathbf {R} ,\mathbf {r} )=\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} ),}$

with

${\displaystyle \langle \chi _{k'}(\mathbf {r} ;\mathbf {R} )|\chi _{k}(\mathbf {r} ;\mathbf {R} )\rangle _{(\mathbf {r} )}=\delta _{k'k},}$

Answer 2

Answer:

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