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I am trying to follow the derivation of the Born-Oppenheimer approximation proposed on Wikipedia.

I know this is not the sexiest source, but i thought it could give a nice first glimpse. But actually, I have troubles understanding their point.

My first problem is at the very beginning: why do they include the $$\sum_{A>B}\frac{Z_AZ_B}{R_{AB}}$$ term in the electronic hamiltonian $H_e$? I thought the whole point of BOE was to separate electronic and nuclear equations, but here they seem to integrate the nuclei-nuclei interaction in the electronic term (they only leave the nuclear kinetic energy out). It doesn't sound like the thing to do... What did I get wrong ?

Another problem comes later down the line. I really Don't understand how they get the relation:

$$(H_n(\boldsymbol{R}))_{k'k} = \delta_{k,k'}T_n-\sum_{A,\alpha}\frac{1}{M_A}< \chi_k'|P_{A\alpha}|\chi_k>_{(\boldsymbol{r})}P_{A\alpha}+< \chi_k'|T_n|\chi_k>_{(\boldsymbol{r})}$$

In my understanding of what is above in the derivation, I thought that the very definition of $(H_n(\boldsymbol{R}))_{k'k}$ was: $(H_n(\boldsymbol{R}))_{k'k} = < \chi_k'|T_n|\chi_k>_{(\boldsymbol{r})}$ But clearly I'm wrong because it looks lije this is only a part of the term. What is the definition of the matrix elements of $H_n$ then ?

Sorry in advance if those misunderstandings are basic mistakes !

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I thought the whole point of BOE was to separate electronic and nuclear equations, but here they seem to integrate the nuclei-nuclei interaction in the electronic term (they only leave the nuclear kinetic energy out).

We're working with 2 sets of coordinates: $\mathbf{r}$, for the electrons, and $\mathbf R$ for the nuclei. In the electronic term, we've gone and the "nailed" down the nuclei, in some predetermined positions, letting the electrons move freely. The term itself doesn't interact with the electrons, and is meant to represent the potential energy contribution of the static configuration of the nuclei. It doesn't really matter if you include it or not: it's only worth a constant at the end of the day.

In my understanding of what is above in the derivation, I thought that the very definition of $(H_n(\boldsymbol{R}))_{k'k}$ was: $(H_n(\boldsymbol{R}))_{k'k} = < \chi_k'|T_n|\chi_k>_{(\boldsymbol{r})}.$

Wikipedia is being very confusing with it's notation. To see why $(H_n(\boldsymbol{R}))_{k'k} = < \chi_k'|T_n|\chi_k>_{(\boldsymbol{r})}$ is wrong, lets just apply $\hat H$ to $\Psi(\mathbf r, \mathbf R)$: $${\displaystyle H\Psi (\mathbf {R} ,\mathbf {r} )=E\Psi (\mathbf {R} ,\mathbf {r} )}, $$hence, $$\left[H_{\text{e}}+T_{\text{n}}\right]\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} ) = E\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} ).$$ Hitting the right side with $\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )$, and integrating over the electronic wavefunctions (notice the $*$ and $k'$ I've attached!), $$\int d\mathbf r \Big(\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )\left[H_{\text{e}}+T_{\text{n}}\right]\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} )\Big) = \int d\mathbf r \Big(\chi _{k'}^*(\mathbf {r} ;\mathbf {R} ) E\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} )\Big).$$ On the right side, we may invoke the orthonormality of $\chi_k$ since $E$ is a constant, but on the left side, $H_e$ and $T_n$ are operators, and we can't. Fortunately, we can still separate left side and simplify it considerably, like this: $$\begin{align*}&\int d\mathbf r \Big(\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )\left[H_{\text{e}}+T_{\text{n}}\right]\sum _{k=1}^{K}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} )\Big) \\ = &\int d\mathbf r \Big(\sum _{k=1}^{K}\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )H_{\text{e}}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} ) + \sum _{k=1}^{K}\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )T_{\text{n}}\chi _{k}(\mathbf {r} ;\mathbf {R} )\phi _{k}(\mathbf {R} )\Big) \\ = & E_e(\mathbf{R})\phi _{k'}(\mathbf {R} ) + \sum _{k=1}^{K}\int d\mathbf r \Big(\chi _{k'}^*(\mathbf {r} ;\mathbf {R} )T_{\text{n}}\chi _{k}(\mathbf {r} ;\mathbf {R} ) \phi _{k}(\mathbf {R} )\Big)\end{align*}.$$

These represent a total of $K$ equations. The first term is the same as the Wikipedia term, just in component form. The second term looks somewhat familiar if we choose to define $$\int d\mathbf r \chi _{k'}^*(\mathbf {r} ;\mathbf {R} )T_{\text{n}}\chi _{k}(\mathbf {r} ;\mathbf {R} ) = \big (\mathbb {H} _{\text{n}}(\mathbf {R} ){\big )}_{k'k},$$ but it's clear that this isn't the case, because $T_{\text{n}}$ is really acting on $\chi _{k}(\mathbf {r} ;\mathbf {R} ) \phi _{k}(\mathbf {R} )$ inside the integral, not just $\chi _{k}(\mathbf {r} ;\mathbf {R} )$. It certainty helps to use the matrix notation to clean up the clutter, but it's easy to forget what is really going on if you're not famailiar with it. If you use the product rule to expand out $T_{\text{n}}\chi _{k}(\mathbf {r} ;\mathbf {R} ) \phi _{k}(\mathbf {R} )$, it will lead to the relation you want. If you find anything confusing, feel free to ask.

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  • $\begingroup$ Thank you very much ! This is a very clear answer that solves my problems. For the second part of my question I was using the bra-ket notation when trying to work it out, and I stupidly dropped the coordinates dependance out of my notations too soon. So I got phik(R) out of the bra-ket product as if it was some kind of constant unaffected by Tn. I should pay more attention to what is acting on what (because I was losing pretty much all the physical meaning of this term in the context of BOE by doing so). Thanks again for your help ! $\endgroup$ – Barbaud Julien Dec 22 '18 at 13:05

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