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in the derivation of the Born-Oppenheimer-Approximation you have the Hamiltonoperator H=T_n + H_e, where T_n is the kinetic energy of the nuclei and H_e the electronic Hamiltonian.

The Schroedinger equation to solve is:

[itex]H \Psi = E \Psi[/itex]

Now, what people do, is the following ansatz:

[itex]\Psi = \sum_k \Psi^n_k \Psi^e_k[/itex]

where the [itex] \Psi^e_k[/itex] are soultions of the electronic problem:

[itex]H_e \Psi^e_k = E_e \Psi^e_k[/itex]

and [itex]\Psi^n_k is a coefficient that depends on the nucleonic coordinates.[/itex]

My question:

Using the ansatz:

[itex]\Psi = \sum_k \Psi^n_k \Psi^e_k[/itex]

how can you tell, that you don't loose some possible solutions of [itex]H \Psi = E \Psi[/itex]?

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# Born Oppenheimer Approximation and Product Asnatz

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