How can I tell how many solutions I will have for an electronic Schrödinger equation ? For example, solving it for the hydrogen atom we get infinitely many solutions \begin{equation} H_e(\mathbf{R})\Psi_i(\mathbf{R},\mathbf{r}) = E_i(\mathbf{R}) \Psi_i(\mathbf{R},\mathbf{r}), \qquad i = 1, 2, ..., \infty \end{equation} They all are bound in the potential.

But for a different potential, e.g. Morse potential it gives a finite number.Wikipedia claims that "This failure [to match the real anharmonicity] is due to the finite number of bound levels in the Morse potential".

I am looking at molecules and wondered if there would be an infinite number of molecular orbitals in general.

  • 2
    $\begingroup$ In general, the best way to find out the number of solutions to an equation is to solve it. Do you have any reason to believe there's another way? $\endgroup$
    – ACuriousMind
    May 13, 2020 at 16:22
  • $\begingroup$ Martin, I think your question is how many bound solutions there are, correct? $\endgroup$
    – zonksoft
    May 13, 2020 at 17:50
  • $\begingroup$ @ ACuriousMind and zonksoft : Well, that of course. But I have molecules in my mind, so solutions will only be found computationally and/or with approximations. And yes, I guess bound solutions, since if the potential does not go to infinity, there will be a continuous space of solutions (thinking of free electrons). $\endgroup$
    – Martin
    May 14, 2020 at 9:51
  • 5
    $\begingroup$ This is generally a hard problem. Atomic anions, for example, are known to have only a finite number of bound states (see ref. 7 here and other references here and here), but this is not territory for the faint of heart. For neutral molecules, one would expect a Rydberg series that's increasingly hydrogenic as it approaches the ionization threshold -- but I'm unsure how much of this has been rigorously proved. $\endgroup$ May 20, 2020 at 16:40
  • $\begingroup$ The Morse potential is a potential for the atoms, to find vibrational states of a molecule. It is not used in the electronic Schrödinger equation. $\endgroup$
    – user137289
    Feb 21, 2021 at 9:47

2 Answers 2


One brief remark on the first part of the question:

How can I tell how many solutions I will have for an electronic Schrödinger equation ?

For spherically symmetric potentials (such as the Coulomb or Morse potential given as examples in the OP), Levinson's theorem provides a way to compute the number of bound states. In particular, it allows to obtain this number without computing all the solutions of a given potential.


Sticking my head out with guesswork, if the potential has infinite range, or is infinitely deep but with finite range, the number of bound states is infinite. For an atom or a molecule the number of bound states is definitely infinite. The energy separation of the bound states goes to zero when the onset of unbound states is approached.

  • $\begingroup$ Even in a simple electric field (as in the Stark effect), there are no bound states for the hydrogen atom: there is a probability that the electron will get ionized (tunnel away). $\endgroup$
    – user137289
    Feb 21, 2021 at 9:50
  • $\begingroup$ You can have an negative ion/anion (atom or molecule), and then for sure there is a maximum number of bounded states. For atoms with a number of electrons $N \le Z$ Hunzinker-van Winter-Zhislin theorem establishes that the number of bound states is inifinite. $\endgroup$
    – Davius
    Jul 23, 2022 at 20:29

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