Can ground states computed through the variational method not be eigenvectors of a Hamiltonian?

Question

Suppose you have to apply the variational principle to compute the ground state of a system whose Hamiltionian is $H_s$ and your ansatz is a linear combination of ground states of a simpler system whose Hamiltionian is $H_a$, like you might for an $H^+$ ion by assuming that it's some linear combination of $1s$ orbitals around each of two protons. Let $\psi_a$ be the ground state eigenvector of $H_a$, $\psi_{a1}$ and $\psi_{a2}$ be symmetric versions of $\psi_a$, like a $1s$ orbital around a left proton and a $1s$ orbital around the right proton, and the linear combination $\psi_{guess} = a\psi_{a1} + b\psi_{a2}$ be the guess ground state of $H_s$.

Generally, $\psi_{guess}$ will not be an eigenvector of $H_s$, even though it may have the same general shape and boundary conditions as $\psi_{true}$, the true ground state of $H_s$. It may not even be a linear combination of the eigenvectors of $H_s$. I'm not certain of this, so this is my first question. But assuming it's true, then $\psi_{guess}$ could be expressed as a projection onto the eigenbasis of $H_s$ plus some error term like so: $\psi_{guess} = \sum_{i} a_i\psi_{si} + \psi_{error}$. Now, if the ansatz is a good one, then the projection should be close to $\psi_{true}$ and the $a_{true}$ coefficient should be close to one. In other words, $\psi_{guess} = a_{true}\psi_{true} + \sum_{i{\ne}true} a_i\psi_{si} + \psi_{error}$, where $\left\lvert a_{true} \right\rvert$ is close to 1.

Are the above statements true? And would the energy of the guess be $E_{guess} = a_{true}\lambda_{true} + \sum_{i{\ne}true} a_i\lambda_{si}$, with the error term not having any contribution?

Answer 1

Can ground states computed through the variational method not be eigenvectors of a Hamiltonian?

Given your description of the computation the answer is yes, the variational approximation is not a true eigenvector of the full Hamiltonian.

Are the above statements true?

Which of the above statements? Some are, some aren't. It is too much to ask in one question to go through each statement and discuss it's truth or falsity (or some intermediate position).

$\psi_{guess} = a_{true}\psi_{true} + \sum_{i{\ne}true} a_i\psi_{si} + \psi_{error}$...

And would the energy of the guess be...

You don't need the "error" state if your true Hamiltonian basis is complete. $$ \psi_{guess} = a_0\psi_{true,0} + \sum_{n\neq 0}a_n \psi_{true, n} $$

Generally your guess will have a larger energy. This is because, if the eigenstates of the true Hamiltonian are a complete basis (which we assume they are) then the guess can be decomposed into a combination of the true states, and since the guess is not the true ground state it will have contributions from true higher energy states.

This type of thing is discussed in many intermediate quantum mechanics books. For example, I think Bethe's QM book is literally titled "Intermediate Quantum Mechanics" and I'm pretty sure he discusses this sort of thing.