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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.

Emilio Pisanty
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Martin
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2 Answers2

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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.

Wolpertinger
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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.

my2cts
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