I am learning DFT and the Hohenberg Kohn Theorem of Existence. 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. How do we know that a density gives a unique potential? This would require that the Schrodinger equation has a unique solution. Is this true and is there a proof for this? Does the Schrödinger equation yield a unique wavefunction and density?

I asked this at Physics SE and got no answers.

  • 2
    $\begingroup$ +1. Welcome to our community, and thank you for bringing your question here !!! We hope to see much more of you! $\endgroup$ – Nike Dattani Oct 6 at 21:20
  • 2
    $\begingroup$ The electronic wavefunction can also have a phase, so in that sense the wavefunction is not unique also pubs.acs.org/doi/10.1021/acs.accounts.7b00220# $\endgroup$ – Cody Aldaz Oct 7 at 3:59
  • 2
    $\begingroup$ This is a bit of a meta comment, but the fact that this question got no answers on Physics SE is a great example of why the matter modeling SE is so useful. $\endgroup$ – taciteloquence Oct 7 at 10:44

The solutions to the Schrödinger equation are not unique in general, and uniqueness depends on several things such as the form of the potential and boundary conditions. Many papers have discussed uniqueness of solutions to the Schrödinger equation for specific classes of potentials and boundary conditions, but in general it is possible to come up with cases where the solution is not unique.

In fact for most potentials and boundary conditions, the Schrödinger equation has several solutions (a ground state solution, and several excited state solutions).

Is at least the ground state solution unique?

For many "physically realistic" potentials and boundary conditions that we normally use (for example, Morse potentials and generalizations of it) we do have a unique ground state energy, but keep in mind that multiple wavefunction solutions can have the same ground state energy: These are called degenerate solutions.

What about DFT and the Hohenberg-Kohn Theorem of Existence?

The theorem states that "the external potential (and hence the total energy), is a unique functional of the electron density." This means that for a given density $\rho$, there is a unique energy functional $E[\rho(r)]$.

Let's say there's a degenerate energy level that has two degenerate solutions to the Schrödinger equation, with densities $\rho_1(r)$ and $\rho_2(r)$. There is a unique energy functional of $\rho_1(r)$, which is $E_1[\rho_1(r)]$, and a unique functional of $\rho_2(r)$, which is $E_2[\rho_2(r)]$. When these two functionals are applied, the resulting energy is the same either way (because the energy is degenerate), but that doesn't mean the functionals are the same or the densities are the same.

An energy can correspond to two different solutions of the Schrödinger equation, each with their own unique functional that gives that energy. Since these two solutions of the Schrödinger equation are not the same, the solutions are not unique.

| cite | improve this answer | |
  • 2
    $\begingroup$ Thank you for your answer. I appreciate it. $\endgroup$ – GalliumBeryllium Oct 7 at 2:31
  • 2
    $\begingroup$ So is the Hohenberg Kohn Theorem of Existence not holding a consequence when there are degenerate ground state wave functions? Also I understand that solving the Schrodinger equation can be very difficult in general. Is there a proof showing that once we have a solution, that it is unique? $\endgroup$ – GalliumBeryllium Oct 7 at 2:39
  • 3
    $\begingroup$ I've updated my answer: the H-K theorem still holds even for degenerate states, and no there is not a proof showing that once you have a solution to the Schrödinger equation it is unique. Schrödinger equation's are unique only in very specific cases (for very specific types of potentials and boundary conditions), and you can see proofs in those cases by searching "unique solution schrodinger equation" in any search engine of the world-wide-web (here's one example, but not the only example: link.springer.com/article/10.1007/s10231-019-00896-z). $\endgroup$ – Nike Dattani Oct 7 at 3:43
  • 1
    $\begingroup$ Excellent and thorough answer! $\endgroup$ – taciteloquence Oct 7 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.