Given a trial function like this one:
$$\lvert\hat{\Psi}\rangle = \sum_i c_i\lvert\psi_i\rangle$$
where the trial function is expanded using exact solutions $\psi_i$ to the Time Independent Schroedinger Equation (as it is done in the variations principle.)
Then, according to the variations principle, the expected energy will be an upper bound to the systems true energy ($E_0$).
$$\langle\hat{\Psi}\lvert H \lvert\hat{\Psi}\rangle \geq E_0$$
Assuming it normalised for simplicity.
Hence, those $\psi_i$, are obviously eigenfunctions of H, but the trial function seems not an eigenfunction of $H$.
When we do it in the real world, using atomic orbitals, or just any suitable set of basis functions,
- Do the basis functions that we use (especially in linear variations) need to be eigenfunctions of the Hamiltonian of the system ?
Why this question ? I am not sure how could be use the theorem otherwise, since it uses eigenfunctions to derive its result.
And if they need not, why? And how do we know that this combinations won't actually (for some unknown reason) be actually below the system's ground energy ?