Does the Schrodinger Equation yield a unique wave function and density?

Question

I am learning DFT and the Hohenberg Kohn Theorem of Existence. And it says that there is a one-to-one correspondence between the external potential and the density. However the proofs that I have seen only show that potential gives a unique density. But how do we know that a density gives a unique potential? This would require that the Schrodinger equation gives a unique solution. Is this true and is there a proof for this?

Answer 1

A 1D case is with $\hbar/2m$ set to 1 :

$$L\psi(x)=-\partial_{xx}\psi(x)+V(x)\psi(x)=E\psi(x)$$

Dividing by $\psi(x)$ and deriving gives the derivative of the potential as a function of the wavefunction $$V'(x)=\partial_x(\frac{\psi''(x)}{\psi(x)})$$

The unknown energy is :$$E=\int L\psi(x)\psi(x)=\int-\psi''(x)\psi(x)+\underbrace{V(s)\int^s\psi(u)^2du|^\infty_{s=-\infty}}_{\rightarrow 0 if V\rightarrow 0}-(\frac{\psi''(x)}{\psi(x)})'\int^x\psi(s)^2ds+C*V'(x)dx$$

$C$ is a constant of integration after integrating by parts, it yields $C(V(\infty)-V(-\infty))$ that should be 0 in most cases (at least harmonic oscillator and Coulomb potential) so the result is independent of $V$ and $C$.

This indicates the energy of a wavefunction solution of the Schrödinger equation (stationary) can be expressed solely in terms of the wavefunction itself. This corresponds to the principle that the wavefunction "completely" describes the system : once you know it is an eigenstate of the SE, the potential and energy is "encoded" in this eigen-wf.

However I don't know how it is possible to go from the one electron density to the many particles wavefunction, since I think the 2 particles density could be $$n_2(x_1,x_2)=\frac{n(f_1(x_1,x_2))n(f_2(x_1,x_2))}{2|D\vec{f}|}$$ with the $f$'s any invertible symmetric functions ?