- #1
Derivator
- 149
- 0
Hi,
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]?
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]?