Existence of effective Hamiltonian for variational ansatz

I recently saw on Wikipedia that in perturbation theory one can define in a systematic way effective Hamiltonian $\hat{H}_{\rm eff}$ that produces the same kind of states and energies as if one applied perturbative expansion to the quantum states and energies (I don't how exactly it works, but maybe someone can explain). I wonder if similar thing could be possible when we have some variational ansatz for the wavefunction. Let's say that this is the state $|\tilde{\psi} \rangle$ of some specific form that is not the solution of stationary Schrodinger equation $$\hat{H}|\tilde{\psi}\rangle \neq E |\tilde{\psi}\rangle$$ but there might exist some different Hamiltonian $\hat{H}_{\rm eff}$ that will actually generate state of the assumed form $$\hat{H}_{\rm eff}|\tilde{\psi}\rangle = \tilde{E}|\tilde{\psi}\rangle$$

Maybe this is true in mean-field bosonic theories, where for a number-conserving Hamiltonian a variational ansatz that does not conserve particle number is suggested. What do you think?

• Given any orthogonal basis and desired set of energies you can define a Hamiltonian by requiring the states in your basis to have the required eigenvalues and extending to the full Hilbert space by linearity. With only a single state your $H_{eff}$ is underdefined. – By Symmetry Jul 27 '16 at 13:52