Finding the Hamiltonian for which a certain wavefunction is the ground state

Question

Consider a 1D system that has the Hamiltonian $H=\frac{P^2}{2m}+V(x)$. At a certain moment, its wavefunction is $\psi$ and has energy $E$. Can we find another Hamiltonian $H'=\frac{P^2}{2m}+V_{eff}(x)$ for which $\psi$ is its ground state with ground state energy $E$?

Note: I know that the $\psi$ is an eigenstate of the operator $\rho H$ with an eigenvalue $E$ where $\rho$ is the density matrix. It is not clear to me though whether this matrix can be represented in the form $\frac{P^2}{2m}+V_{eff}(x)$ or whether $\psi$ is its ground state.

EDIT: Since the ground state of a Hermitian $H'$ is a real function while $\psi$ is complex function, then I release the requirement of the ground state of $H'$ to be the best approximation to $\psi$

It sounds to me as if you're effectively asking for the Hohenberg-Kohn theorem. — ACuriousMind, Apr 27 '17 at 10:49
Related: Is it possible to reconstruct the Hamiltonian from knowledge of its ground state wave function? and In quantum mechanics, given certain energy spectrum can one generate the corresponding potential? — Emilio Pisanty, Apr 27 '17 at 14:39

score 1 · Answer 1 · answered Apr 27 '17 at 11:11

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The general answer is no. For example, Ground state wavefunctions in 1d have no nodes. So even if the wavefunction $\psi$ has a node anywhere, there is no $V_{eff}$ for which it is a ground state.

answered Apr 27 '17 at 11:11

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Finding the Hamiltonian for which a certain wavefunction is the ground state

1 Answers1