For hydrogen, other than in the 1s state, the electron wavefunctions have radial and/or angular nodes where the electron probability density is zero.

In helium or further atoms with more than one electron, considering correlation of electrons, what happens to these nodes? Are there still radial and angular nodes where electron probability density is zero?

For example, if helium is in a 2s2s state, is there anywhere that an electron cannot be?

  • $\begingroup$ If you haven't already seen it, this answer of mine might be of interest to you. Maybe I'm overlooking something but this question might be rather difficult to answer. For one thing it will depend on which "kind" of wavefunction you mean. If you are talking about the many-particle wavefunctions then I think it is very difficult to say something definite about the nodal structure. But if you are talking about one-particle states, like Hartree-Fock or Kohn-Sham orbitals, then much more can be said. $\endgroup$
    – Philipp
    Mar 3, 2015 at 20:20
  • $\begingroup$ For Kohn-Sham orbitals the sperical symmetry of an atom should always permit us to seperate the orbital into a spherical harmonic (whose nodes we are familiar with) and a radial component, which should have $n-l-1$ nodes. $\endgroup$
    – Philipp
    Mar 3, 2015 at 20:26
  • $\begingroup$ Technically, wavefunction nodes for atomic orbitals only exist in solutions for the non-relativistic Schrödinger equation. Addition of relativistic terms means every region has a non-zero wavefunction value for every orbital; what used to be nodes then becomes "quasi-nodes". $\endgroup$ Mar 4, 2015 at 3:05
  • 2
    $\begingroup$ @NicolauSakerNeto "according to Dirac’s relativistic treatment of an electron, which has now been extended to many-electron atoms, atomic orbitals possess no nodes whatsoever." chem.ucla.edu/dept/Faculty/scerri/pdf/Atkins_critique.pdf Foundations of Chemistry 1: 297–305, 1999 $\endgroup$
    – DavePhD
    Mar 5, 2015 at 16:52
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    $\begingroup$ @NicolauSakerNeto Thanks for pointing out the relativistic effect. I'm trying to read more about it starting with "Relativistic quantum chemistry: The electrons and the nodes" pubs.acs.org/doi/abs/10.1021/ed045p558 and "Contour diagrams for relativistic orbitals" pubs.acs.org/doi/abs/10.1021/ed046p678 $\endgroup$
    – DavePhD
    Mar 6, 2015 at 16:06

2 Answers 2


My answer to this question is the following. Nodes can happen for highly excited states, whose average distances are typically much larger than the Bohr radius. Consider Na, Cs etc, which have one outermost shell electron. When the outermost shell electron is highly excited, the case is not much different from that of a hydrogen atom (i.e. the indistinguishability of electrons may be ignored when electrons are well separated.) Thus there can be nodes in these highly excited states.

A more non-trivial question is, whether the total electron density can have nodes for the ground state. I tend to think that it's not possible. My following naive, sketchy explanation may be seriously wrong. For a second-order eigenvalue problem, we have the sturm theorem (?; in doubt), which guarantees that the roots of successive states occur in an interlacing manner. That means, where state $i$ has a node at $r_i$, state $i+1$ cannot have a node at $r_i$. Thus the density, which is equal to the sum of square of wave function of all the occupied states, cannot be zero at one point. We have a similar second-order eigenvalue problem for each angular momentum $l$, and so the argument goes.

The above too sketchy, so let's formalise it a bit based on the one-particle approximation. The total ground-state electron density is proportional to $ \sum_{\{n,l, m, s\}; \textrm{occupied}}|\psi_{nlm}(r)|^2 |Y_{lm}(\theta, \phi)|^2$. Here $s$ is for the spin, $n$ labels the energy state (1 for ground, 2 for first excited, etc), and the sum is over all occupied states. Note that in order for the electronic density to be zero at $r_0$, each term has to vanish at $r_0$. But remember that for $l=0$, $m=0$, the $n=1$ radial wave function $\psi_{100}(r)$ can have no node, and $Y_{00}$ is a constant, which means the term $\psi_{100}(r)Y_{00}$ can vanish nowhere in space. This state has to be occupied in the ground state.

  • $\begingroup$ Indeed as I stated in one of my comments to the original question: You are right in thinking that the ground state of an atom should not show any nodes. There is even a strict proof for this: According to Courant's nodal line theorem the eigenfunction corresponding to the lowest eigenvalue cannot have any nodes (except the trivial one at infinity). $\endgroup$
    – Philipp
    Mar 5, 2015 at 11:04

There is a proof of node for the triplet-S state of helium in What Do We Know About Wavefunction Nodes? Recent Advances in Computational Chemistry vol. 2. (alternative link to preprint version):

The wavefunction can be written as a function of only 3 coordinates: the distance ($r_1$) of the first electron from the nucleus, the distance ($r_2$) of the second electron from the nucleus, and the distance ($r_{12}$) between the electrons.

$\Psi = \Psi(r_1, r_2, r_{12})$

by the Pauli exclusion principle:

$\Psi(r_1, r_2, r_{12}) = -\Psi(r_2, r_1, r_{12})$

Then, suppose both electrons are the same distance ($r$) from the nucleus, but not necessarily at the same point:

$\Psi(r, r, r_{12}) = -\Psi(r, r, r_{12})$


$\Psi(r, r, r_{12}) = 0$

In other words, there is a node defined by $r_1 = r_2$.

The two electrons can never be the same distance from the nucleus.

The reference goes on to say that in 6-dimensional space, the node is a 5-dimensional hypersurface.


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