# Typical application of Pauli exclusion principle in atoms

Typically when we talk about the electron orbitals around atoms we talk about them getting "filled up" starting with s1, s2, and so on (with spin taken into account as well). This relies on the Pauli exclusion principle, since electrons are fermions, so they cannot be in the same quantum state.

Something that has been bothering me for a while now is that in the usual interpretation of quantum mechanics we cannot say that the electrons are in any state until we measure them. So really when we talk about the Pauli exclusion principle what we mean is that we will never measure two identical fermions as being in the same quantum state. (Correct me if I am wrong in this)

So I do not understand how we talk about "filling up orbitals" when the only way we can see that the electrons are in these states is to measure the state of every electron at once. Is this whole "filling up orbitals" approach actually correct, or just a nice way to explain the periodic table at first? I know this interpretation is useful in explaining things such as why certain elements form ions in the way that they do.

One possible answer I have been thinking of is using the fact that quantum states referring to energy levels in the atom are stationary states. So maybe we can say that the electrons are in definitive states rather than being in a superposition of states until we make a measurement. I am not sure in this answer though.

• While we may not know (without measurement), in which specific states the electrons are, they still can be only in the states that exist in the system (or their superposition). For example, if there are two states on the s1 orbit in helium, only two electrons can be there regardless of your knowledge of their specific state. The states of these electrons can be uncertain, but only within of what is allowed. I am sure the experts here can easily show this in math. Jun 7, 2018 at 17:47
• @safesphere If the electrons are in a superposition of states, then they are not in a definitive state until we make a measurement. But everything else you said makes sense. Jun 7, 2018 at 18:04
• I didn't say that electrons in a superposition of states are in a definitive state. My point is that, if there are two possible states, then there are only two superpositions of these states possible at a time, so the total number of electrons is the same. Uncertainty does not affect the exclusion. Jun 7, 2018 at 19:06
• @safesphere I am sorry I didn't mean to imply that you said electron's in a superposition are in a definitive state. Jun 7, 2018 at 20:46

You are correct that the standard explanation of "filling up single-electron orbitals" is confusing. That's because it makes two key simplifying assumptions which are rarely stated explicitly:

First, it neglects the Coulomb interaction between the elections, so that the Hamiltonian can be decomposed as $$H_\text{full} = \sum_{i=1}^n H^{(1)}_i,$$ where $$H^{(1)}$$ represents a single-electron Hamiltonian (e.g. the hydrogen atom Hamiltonian). In this very special case, it can be shown that the eigenfunctions can all be represented as Slater determinants of single-electron eigenfunctions, so that we really can meaningfully talk about the wave functions of the individual electrons without having to measure all of them at once. In the general entangled case where we fully incorporate the Coulomb interaction, we can't do this, and the "orbital" picture breaks down, and as you say, the physical consequences of the Pauli exclusion principle become very hard to intuit.

In practice, we very often use a hybrid approach called the "Hartree-Fock" approximation (which works surprisingly well and is ubiquitous in quantum chemistry). It's a variational approximation in which we try to minimize the energy of the exact interacting Hamiltonian, but only over the space of Slater determinants of single-particle wavefunctions. In this case it turns out that the best energies come from giving different electrons effective hydrogen-like orbitals, but with different effective nuclear charges that are less than the true nuclear charge $$Ze$$. Physically, this represents the fact that the interelectron repulsion is being approximately incorporated into a "screening" effect that the inner electrons have on the outer ones, by partially cancelling out the nuclear charge. (Moreover, the best effective nuclear charge turns out to depend on the angular momentum quantum number $$l$$ (although not on $$m$$). This breaks the energy degeneracy between orbitals with different values of $$l$$ that one finds in the hydrogen atom.) But it is inherently just an approximation; in the exact ground-state wave function, you can't talk about individual electron wave functions.

Within the HF approximation, we can assume that each electron has a well-defined orbital, but any arbitrary set of orbitals (that respects Pauli exclusion) is a valid eigenstate. Why do we always assume that they get filled up from lowest to highest energy? Because of the second implicit assumption, which is that the electrons are in thermal equilibrium at zero temperature, so that they are in the ground state of the full multi-electron Hamiltonian. This is an excellent approximation: except in exotic high-temperature systems like plasmas, the electrons are almost always found to be in their ground state. (This is lucky, because it turns out that for the exact exited states, the Hartree-Fock approximation works much less well than for the exact ground state.)

The Pauli exclusion principle says the electrons have to be in an antisymmetric (under particle interchange). This of course rules out them every being in the same state, which makes for a better informal statement of the principle.

Any single electron is never in a definitive state. Even in a ground state helium atom, neither electron has a definite spin state--only their total spin is known $S=0$. In this case that are in the same spatial state.

When you get to lithium, the 3 electron ground state is a Slater determinant:

$$\psi =\left|\begin{array}{+++}^1S_1\uparrow_1&^1S_1\downarrow_1&^2S_1\downarrow_1\\^1S_2\uparrow_2&^1S_2\uparrow_2&^2S_2\uparrow_2 \\ ^1S_3\uparrow_3&^1S_3\downarrow_3&^2S_3\uparrow_3\end{array}\right|$$

which can't even be factored into space and spin. No single electron is in a definite state.

• Your answer clarifies that electrons are in a superposition of states. However, the question is on the possible number of electrons. Perhaps you could add a clarification that the number of electrons in superposition of states cannot exceed the number of states? (Perhaps this is implied in your first paragraph, but is not clear enough for a non-specialist). Jun 8, 2018 at 0:16
• Yes, I already knew all of this. I am interested in how we use this to talk about filling up orbitals. It seems like I am on the right track in that it seems to be just a convenient way to explain orbitals for introductory chemistry students. Jun 8, 2018 at 16:24
• @safesphere I think the Slater determinant is the only answer. If I have an He atom in the ground state, how do the 2 electrons differ? Their spatial wave function is the same, and the spin states are the same (they're not in a $S_3$ eigenstate). They only differ in their collective behavior: you interchange them and unmeasurable phase changes.
– JEB
Jun 8, 2018 at 20:34