Does One-to-One Correspondence of Hohenberg Kohn Theorem Mean Bijective or Injective and How to Prove it?

Question

I have asked a similar question but after thinking about it I have a more specific question.

According to Ullrich, Carsten A.. Time-Dependent Density-Functional Theory : Concepts and Applications, the Hohenberg–Kohn theorem states

In a finite, interacting N-electron system with a given particle–particle interaction there exists a one-to-one correspondence between the external potential $V(r)$ and the ground-state density $n_0(r)$. In other words, the external potential is a unique functional of the ground-state density, $V[n_0](r)$, up to an arbitrary additive constant.

The way I understand it, assuming V differs by more than a constant and psi differs by more than a phase, the logic is: one potential (V) yields one hamiltonian (H) which yields a wave function (Ψ) which yields a density (n). V -> Ψ -> n.

V -> Ψ (ignoring constant) This is proven in HK theorem via proof by contradiction

Ψ -> n (ignoring phase factor) This is proven in HK theorem via proof by contradiction.

Then they conclude that: We have thus shown that $Ψ_0$ and $Ψ′_0$ give different densities $n_0$ and $n′_0$; but in the first step we showed that $Ψ_0$ and $Ψ′_0$ also come from different potentials $V$ and $V′$. Therefore, a unique one-to-one correspondence exists between potentials and ground-state densities, which can be formally expressed by writing $V[n_0](r)$, and thus $V[n_0]$.

This confuses me because they have only proven "one direction." They have proven that two V's cannot give the same Ψ but they haven't proven that one V cannot yield more than one Ψ. Likewise they have proven that two Ψ's cannot give the same n but haven't proven that one Ψ cannot yield more than one n. Perhaps I'm missing something obvious but any insight would be appreciated.

Answer 1

I'm not familiar with the Carsten Ullrich text you mention. However, one possibility is that he followed Hohenberg and Kohn's example of assuming a non-degenerate ground state. If the ground state $\Psi$ is non-degenerate, $V$ can by definition only produce one $\Psi$.

You're correct that the proofs have to be modified when there is a degenerate ground state manifold. There's a nice discussion in K. Capelle, C. A. Ullrich, G. Vignale, Degenerate ground states and nonunique potentials: breakdown and restoration of density functionals, Physical Review A 76, p. 012508 (2007) (arXiv link) that may answer some of your questions.

I agree with Kevin's comment that the density $n$ is fixed by the inner product $|\psi|^2$.