# Born-Oppenheimer approximation and perturbation theory

In the book Molecular Physics by Demtroder there is an explanation of the Born-Oppenheimer approximation and the adiabatic approximation in terms of a perturbative series. The Hamiltonian is $$H_0 + T_\text{nuc} = H_0 + \lambda W$$, where $$T_\text{nuc}$$ is the nuclear kinetic energy term. The solutions of $$H_0$$ are $$\phi_m(r, R)$$ and a general wavefunction is written as a linear combination $$\psi(r, R) = \sum_m \chi_m(R) \phi_m(r,R)$$. The Schroedinger equation $$(H_0 + \lambda W)\psi = E\psi$$ is then solved by writing $$E$$ and $$\chi_m$$ as a series expansion of $$\lambda$$. According to the book, the energy to second order is $$E_n = E_n^{(0)} + W_{nn} + \sum_{k\neq n} \frac{W_{nk}W_{kn}}{E_n^{(0)} - E_k^{(0)}}$$ where $$W_{nk} = \int \phi_n^{(0)*} T_\text{nuc} \phi_k^{(0)} dr.$$

Now I am lost on how this was obtained. They then relate the second term $$W_{nn}$$ to the adiabatic correction.

Since the energies are labelled, I would assume that then $$\psi$$ would also have to be labelled by $$n$$.

I would appreciate any suggestions or explanations for this.

Edit: My attempt.

So if I use the $$\chi$$ and $$E$$ expansion as well as their form of $$\psi$$ then I obtain $$(H_0 + \lambda W) \psi = \sum_m \chi_m(R) E_m(R) \phi_m(r,R) + \lambda \sum_m \phi_m(r,R) W \chi_m(R) + \chi_m(R) W \phi(r,R) = \sum_m E \chi_m(R) \phi_m(r,R)$$

Now I multiply $$\phi_k^*(r,R)$$ and integrate over $$r$$ to obtain $$E_k(R) \chi_k(R) + \lambda W \chi_k(R) + \lambda \sum_m \chi_m(R) W_{mk} = E \chi_k(R).$$ I should have made clear that in the book the energy was expanded as $$E_n = E_n^{(0)} + \lambda E_n^{(1)}...$$ Here the energy is labelled so the $$\psi$$ need to be labelled but I will ignore this. If $$\psi$$ is to be normalised then $$\chi_m(R)^*\chi_m(R)$$ over $$m$$ must sum to 1. The zeroth order is then $$E^{(0)} = E_k(R)$$ which is not possible as it depends on $$R$$. I must be reading something wrong here.

Let's apply the fundamentals of perturbation theory. We want to calculate the eigenspectrum of a Hamiltonian. $$H = H_0 + H'$$

We know the eigenspectrum of the unperturbed Hamiltonian, $H_0 \phi_n^0=E_n^0\phi_n^0$

The first-order correction to the energy is the expectation value of the of the perturbing operator: $$E_n^1 = <\phi_n^0|\lambda W|\phi_n^0>$$

See Griffiths for a derivation of first and second order perturbation theory. Griffiths describes the above equation as the most important equation in quantum mechanics. You should see that this is precisely your $W_{nn}$.

Since the energies are labelled, I would assume that then ψ would also have to be labelled by n.

In the adiabatic approximation this is true. An adiabatic perturbation would keep a particle in state n within state n so that $E_n^0->E_n'$ and $\phi_n^0->\psi_n$. However, it is not true in general. This is why each $\psi$ is written as a sum over $\phi$

• Thanks for answering. I do understand the derivation for perturbation theory in general. When I try to apply it in this case, I am unable to obtain the answer. What I mean by $\psi$ being labelled is that each $\chi$ should have two labels as different linear combinations of $\phi$ are needed for different $\psi$. Commented Aug 3, 2018 at 17:07
• To be more specific, $H\psi = (H_0 + \lambda T) \sum_m \chi_m \phi_m = \sum_m E_m(R) \chi_m \phi_m + \lambda \sum_m ( \phi_m T \chi_m + \chi_m T \phi_m) = E \sum_m \chi_m \phi_m$. Commented Aug 3, 2018 at 17:15
• It looks like it should be $\chi_k(R)$ not $E_k(R)$. They only wrote a single wavefunction, not the entire basis of the system, and so the expression they have is general. To finish your derivation you additonally need to multiply on the left by $\chi_j(R)$ and integrate, and apply some orthonormalization condition to $\chi(R)$. E.g. $<\chi_m|\chi_n>=\delta_{mn}$. To summarize, the ground state energy is $<\psi|H|\psi>$ not $<\phi|H|\psi>$ Commented Aug 3, 2018 at 21:10
• Can $<\chi_m | chi_n > = \delta_{mn}$? Since $\psi$ are presumably orthogonal, after integrating over r you would obtain $\int \psi^* \psi dr dR = \sum_m \int \chi_m^*\chi_m dR$ I think. Commented Aug 5, 2018 at 19:36
• $\chi_m(R)$ can contain weights similar to the variational method for a single particle (the weights would probably largest for the lowest energy states of $\phi_m(R)$ anyway) or it could contain occupation functions (although this is a temperature-dependent formalism). They did not give a precise description for $\chi_m(R)$, and it is complicated by the fact that we will have to deal with multiple electrons. This thought experiment makes a little sense if you apply it to a hydrogen atom, but for most other things it makes more sense to treat the nucleus classically. Commented Aug 5, 2018 at 23:28