# Born Oppenheimer Approximation: Why can any molecular state be represented as a linear combination of electronic states?

in the Born Oppenheimer Approximation, one expands the molecular wavefunction $\Psi(x,X)$ in terms of the electronic wavefunctions $\phi(x;X)$: $$\Psi(x,X)= \sum_k(c(X)_k\phi(x;X)_k)$$ ($x$ are the electronic coordinates and $X$ are the nucleonic coordinates)

Now, since the electronic wavefunctions are eigenstates of the electronic Hamiltonian, the constitute a complete basis of the electronic space. Thus any electronic wavefunction can be expanded in terms of the eigenfunctions. But, how can we be sure that any molecular wavefunction can be expanded in terms of the electronic wavefunctions? How can we be sure that the molecular Hilbert space is not larger than the space which is spanned by the eigenstates of the electronic Hamiltonian?

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I thought the Born Approximation was the latest Matt Damon movie. Thanks for the clarification! – twistor59 Feb 4 '13 at 7:24

Even though this problem confused me a lot, it is actually quite simple, and I understood it best using some mathematical language. I call the space of electronic coordinates $A$ (so $x \in A$) and $\Psi(x, X)$ is the exact solution of the complete (electronic and nuclear) Schrödinger equation. Then for each set of nuclear coordinates $X$, we can define a function $$\Psi_X: A \rightarrow \mathbb{C}, \quad \Psi_X(x) := \Psi(x, X).$$ We know that for all $X$, the set of eigenfunctions of the electronic Hamiltonian, {$\phi(x;X)_k$}, is a complete basis for electronic wave functions. We now expand each electronic wave function $\Psi_X$ in a different set of basis functions, namely $$\Psi_X(x) = \sum_k c(X)_k \phi(x;X)_k,$$ where $c(X)_k$ is the expansion coefficient belonging to the $k$th basis function associated with the electronic wave function parametrized by $X$. But since $\Psi_X(x) = \Psi(x, X)$ we already obtained what OP asked for.