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 Schrondinger 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 $\phi$

  • $\begingroup$ 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 $\psi$. $\endgroup$ – user110503 Aug 3 '18 at 17:07
  • $\begingroup$ To be more specific, $H\psi = (H_0 + \lambda T) \sum_m \chi_m \phi_m = \sum_m E_m(R) \chi_m \phi_m + \lambda \sum_m ( \phi_m T \chi_m + \chi_m T \phi_m) = E \sum_m \chi_m \phi_m$. $\endgroup$ – user110503 Aug 3 '18 at 17:15
  • $\begingroup$ So then multiplying by $\phi^*_k$ and integrating, I obtain $ E_k(R) \chi_k + \lambda T\chi_k + \lambda \sum_m \chi_m W_{km} = E \chi_k.$ The zeroth order energy is then supposedly $E_k(R)$ which can't be right as it depends on R. $\endgroup$ – user110503 Aug 3 '18 at 17:17
  • $\begingroup$ 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 $<\phi|H|\psi>$ $\endgroup$ – Tony Ruth Aug 3 '18 at 21:10
  • $\begingroup$ 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. $\endgroup$ – user110503 Aug 5 '18 at 19:36

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