Given a molecular system, the time-independent Schrodinger Equation (SE) that we want to solve is $$\hat H \psi(r_e, r_n) = E \psi(r_e, r_n)$$ $\hat H$ is hermitian, so the set of solutions $\psi(r_e, r_n)$ are orthonormal.

Under Born-Oppenheimer approximation, $\{r_e \}$ decouples with $\{ r_n \}$, thus each $\psi(r_e, r_n)$ can be represented as $$\psi(r_e, r_n) = \psi_e(r_e|r_n) \chi(r_n)$$ in which $\hat H_e(r_n) \psi_e(r_e|r_n) = E_e(r_n) \psi_e(r_e|r_n)$.

My confusions:

In every reference I read, the eigenfunction of $\hat H$ is represented as

$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty}\psi_e^i (r_e|r_n) \chi^i(r_n)$$ in which {$\psi_e^i(r_e|r_n)$} is the set of eigenfunctions of $\hat H_e$

Why an eigenfunction of $\hat H$ can be represented in such a way? For me, this expression is just $$\psi(r_e, r_n) = \sum_{i = 1}^{\infty}\psi^i (r_e, r_n)$$ it makes no sense that an eigenfunction can be expanded in some bases.

  • 1
    $\begingroup$ Your last equation is just some basis set expansion, something you can always do and not an approximation. But it does not decouple electrons and nuclei, so it is not very useful on its own. As an approximation you can decouple them in each basis function, as is done in your second last equation. This is called Born-Huang-Expansion, which is closely related to the BO Approximation. $\endgroup$
    – Feodoran
    Oct 2, 2018 at 7:13
  • $\begingroup$ @ Feodoran Thank you for the clarification. But I still don't understand why the eigenvectors of $\hat H$ can be expanded by eigenvectors of $\hat H_e$. My understanding of basis expansion is limited to group representation theory, in which we can find the basis of the irreducible subspaces, then any functions can be expanded by the basis. I cannot see the relation between this and the expansion used in BO approximation. $\endgroup$ Oct 2, 2018 at 7:26
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    $\begingroup$ As $\hat{H}_e$ is an hermitian operator it's eigenfunctions represent a complete basis. This in turns means that any arbitrary other function can be expressed as a linear combination of those eigenfunctions. $\endgroup$
    – Raven
    Oct 2, 2018 at 8:30
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    $\begingroup$ The eigenvectors $\psi$ of $\hat H$ are not expanded in the eigenvectors $\psi_e^i$ of $\hat H_e$, but in a product containing $\psi_e^i$ (and $\chi^i$). Also as @Raven already pointed out: From a mathematical point of view it does not matter in what basis we expand something, as long as it is a valid basis. We can choose a physically motivated basis, which makes interpretation of the results easier, or we can choose something more abstract. In this case, the choice of $\psi_e^i$ and $\chi^i$ is physically motivated as we need to include both $r_e$ and $r_n$ in $\hat H$. $\endgroup$
    – Feodoran
    Oct 2, 2018 at 8:43
  • $\begingroup$ We prefer to not use MathJax in the title field, see here for details. $\endgroup$ Oct 22, 2018 at 14:57

1 Answer 1


After reading this lecture note I have a better understanding.

The reason I'm so confused is that the form $$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \psi_e^i(r_e|r_n) \chi^i(r_N)$$ seems to be a linear combination with coefficients that are all $1$. And this is very importand for the derivation of the set of entangled nucleus time-independent SEs since there is no arbitrary coefficients in the set of equations.

After I read the lecture note, here is my understanding:

For a fixed atomic configuration $R^{\alpha}$, the electronic state can be represented by a linear combination of the eigenstates of $\hat H_e(R^{\alpha})$ $$\psi(r_e| R^{\alpha}) = \sum_{i = 1}^{\infty} c_i(R^{\alpha})\psi^{i}_e(r_e|R^{\alpha})$$ although $\psi(r_e| R^{\alpha})$ is an electronic state, this is essentially a solution to $\hat H \psi = E \psi$ at a fixed atomic configuration.

Since by varying $R^{\alpha}$, the set of linear coefficients {$c_i$} will change. So the general $\psi(r_e, r_n)$ can be represented as $$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \chi^i(r_N) \psi_e^i(r_e|r_n)$$ in which the set of {$\chi^i(r_n)$} is just the set {$c_i({R^{\alpha}})$}.

Essentially, the $\psi(r_e, r_n)$ that we plug back into $\hat H \psi = E \psi$ is still an electron wavefunction.


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