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

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.

• Isn't one $\Psi$ yielding multiple $n$ trivially dealt with by the fact that $n\left( \textbf{r} \right) = {q\left|\psi\left( \textbf{r} \right)\right|^{2}}$? The other part ($V$ yielding multiple $\Psi$) seems to go back to your other question about uniqueness of solutions to the Schrodinger equation. Oct 21, 2020 at 1:20

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