Slater type orbitals (STO) are considered to be more accurate than gaussian type orbitals (GTO) for atomic and molecular QM calculations because - among other reasons - they decay with $e^{-\alpha r}$ as $r \to \infty$. But GTOs are more popular because they are easier to calculate with. GTOs decay with $e^{-\alpha r^2}$, so its adequate to sometimes add diffuse functions to the GTO basis set to compensate for gaussian decay behaviour.

Also, exact hydrogen wavefunctions decay exponentially, so the motivation for STOs.

I understand that the only boundary requirement for solving the Schrödinger equation for atoms and molecules in free space is that the wavefunction goes zero as $r \to \infty$, but there are no a priori requirements for the way it decays as it does so.

My question is: do we have theoretical (ab initio) and/or experimental reasons to believe that all atomic and molecular wavefunctions decay like $e^{-\alpha r}$ as $r \to \infty$.

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    $\begingroup$ The title change was just to make the post more easily searchable and accessible across the SE network, see here for some details. $\endgroup$
    – Tyberius
    Jan 24, 2022 at 18:10
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    $\begingroup$ @Tyberius, cool, will keep that in mind in the next ones, thanks (bad habits from Math.SE die hard :). $\endgroup$
    – Arc
    Jan 25, 2022 at 5:53

1 Answer 1


I'll answer this question from the theoretical side. The exponential behavior follows simply from the Schrödinger equation. Consider the one-electron Schrödinger equation: $$ (-\frac{1}{2}\nabla^2 + V(\mathbf{r}))\psi(\mathbf{r}) = \epsilon\psi(\mathbf{r}), \epsilon < 0 $$ At spatial points that are very far away from the nucleus, $V(\mathbf{r})\approx 0$, so that the asymptotic solution is given by $$ -\frac{1}{2}\nabla^2\psi(\mathbf{r}) = \epsilon\psi(\mathbf{r}), \epsilon < 0 $$ This differential equation has basic solutions of the form $$ \psi(\mathbf{r}) = Ce^{-\sqrt{-2\epsilon}\mathbf{k}\cdot\mathbf{r}} $$ for some unit vector $\mathbf{k}$. The real asymptotic behavior of $\psi(\mathbf{r})$ is thus a linear combination of these basic solutions. The linear combination may bring a polynomial prefactor to the exponential, but will never alter the exponent. Thus we have not only proved the exponential behavior, but also derived the correct exponent $\alpha = \sqrt{-2\epsilon}$. For a multi-electronic, non-interacting system, the overall decay rate is governed by the slowest decaying orbital, i.e. the HOMO.

Of course, the real wavefunction can only be described by a multi-electron Schrödinger equation. But we can work on the equivalent Kohn-Sham system and show that the Kohn-Sham wavefunction decays at a rate given by the Kohn-Sham HOMO energy. By Janak's theorem, the Kohn-Sham HOMO energy is just the negative of the ionization potential of the exact system. To see this, consider a huge ensemble of $N$ identical, non-interacting molecules. If we remove one electron from the ensemble and let the hole delocalize evenly between all the molecules, then as $N\to +\infty$, the electron removal has a negligible impact on the electron density of any molecule (and therefore the Kohn-Sham potential of each molecule). Therefore under the Kohn-Sham framework we see that removing such an electron costs an energy of $-\epsilon_{\mathrm{HOMO}}$ (it does not matter whether the HOMO refers to that of the ensemble or that of a molecule, since their orbital energies are equal), since the electron is taken from an energy level whose energy is $\epsilon_{\mathrm{HOMO}}$ and the Hamiltonian is not changed in this process. On the other hand, from the perspective of the real system it is clear that the energy cost is equal to the first ionization energy of one of the molecules, $I$. Therefore we have $\epsilon_{\mathrm{HOMO}} = -I$, which means that the Kohn-Sham wavefunction decays like (again up to a possible polynomial prefactor; the precise determination of this polynomial prefactor is a much more difficult question) $$ \psi(\mathbf{r}) = Ce^{-\sqrt{2I}\mathbf{k}\cdot\mathbf{r}} $$ Although the Kohn-Sham wavefunction is fictional, its density is equal to the true multielectronic density, and in order for the true density to have the same asymptotic behavior as the Kohn-Sham density, the true wavefunction must have the same asymptotic behavior as the Kohn-Sham wavefunction. Q.E.D.

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    $\begingroup$ A very sound and well developed demonstration. $\endgroup$
    – Arc
    Jan 24, 2022 at 12:37
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    $\begingroup$ We already know the behaviour for hydrogen, so only the final paragraph really gets to the interesting part. But I don't follow your argument there, I think you should expand – in particular on Janak's theorem. $\endgroup$ Jan 25, 2022 at 10:21
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    $\begingroup$ I agree that the real-world field(s) must continue to decrease as otherwise there'd be a stable point at some distance from the particle. This means the field either is asymptotic to zero or actually hits zero at some finite distance. The latter case doesn't really make sense unless you quantize the field strength, and the theory (if any) about that is way beyond me. $\endgroup$ Jan 25, 2022 at 12:59
  • $\begingroup$ @leftaroundabout Sorry for the delay - the part about Janak's theorem has been expanded $\endgroup$
    – wzkchem5
    Feb 4, 2022 at 10:33

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