Why identical particles change matter
Helium, with two electrons, holds on to them so tightly that it takes 24.6 eV to remove one, and it reacts with nothing. Lithium, with three, gives up its outermost electron for only 5.4 eV and is a soft, reactive metal. If every electron could settle into the lowest state, lithium's third electron would sit alongside the other two, close to a nucleus with three times hydrogen's charge, and lithium would be as unreactive as helium. Why don't all electrons sit in the lowest orbital?
The answer is a rule with no classical counterpart, about what it means for two particles to be exactly alike. It says that no two electrons can occupy the same state, and it follows from a single symmetry of the wavefunction. Combined with spin, it builds the shells of the periodic table. This essay derives the rule, shows what it does to where two particles are found, and fills the first shells of atoms.
Two particles at once
Until now every wavefunction has described one particle. A state of two particles is a single wavefunction of both positions, $\psi(x_1, x_2)$, and $|\psi(x_1, x_2)|^2$ is the probability density for finding one particle at $x_1$ and the other at $x_2$. When the two are independent, one in state $\varphi_a$ and the other in $\varphi_b$, it is the product $\varphi_a(x_1)\,\varphi_b(x_2)$, just as the probability of two independent events is the product of their probabilities.
Particles with no labels
Two billiard balls can be told apart even if they look alike: follow each along its path, and you always know which is which. Two electrons cannot be followed, because they have no paths. Every electron is exactly like every other, and quantum mechanics has to treat them as having no labels at all.
That has a precise consequence. If we swap the two positions in the wavefunction, nothing observable may change, since "electron 1 here and electron 2 there" is the same situation as the reverse: $|\psi(x_2, x_1)|^2 = |\psi(x_1, x_2)|^2$. So swapping can at most multiply $\psi$ by a phase, $\psi(x_2, x_1) = e^{i\delta}\,\psi(x_1, x_2)$. Swap again and you are back where you started, so $e^{2i\delta} = 1$, and the phase can only be $+1$ or $-1$. Wavefunctions of identical particles are either symmetric or antisymmetric under exchange.
Nature uses both. Particles with symmetric wavefunctions are called bosons: photons, and helium-4 atoms, for instance. Particles with antisymmetric wavefunctions are fermions: electrons, protons and neutrons. Which kind a particle is follows its spin: whole-number spin means boson, half-number spin means fermion. That link, the spin–statistics connection, is established only in relativistic quantum field theory; here it is taken as a fact of observation, not proved.
What antisymmetry forbids
Take two identical particles, one in state $\varphi_a$ and one in state $\varphi_b$. The plain product $\varphi_a(x_1)\varphi_b(x_2)$ says which is which, so it will not do. The combinations that treat them alike are
$$\begin{aligned} \psi_\pm(x_1, x_2) = \tfrac{1}{\sqrt2}\bigl[\,&\varphi_a(x_1)\,\varphi_b(x_2)\\[2pt] \pm\;&\varphi_a(x_2)\,\varphi_b(x_1)\bigr], \end{aligned}$$symmetric with the plus sign and antisymmetric with the minus. Now put both fermions in the same state, $a = b$. The antisymmetric combination becomes $\varphi_a(x_1)\varphi_a(x_2) - \varphi_a(x_2)\varphi_a(x_1) = 0$: no wavefunction at all. Two electrons cannot be in the same state. That is the Pauli exclusion principle, proposed by Wolfgang Pauli in 1925 to explain atomic spectra, and here it falls out of the minus sign.
The minus sign does more than forbid identical states. Even with $a \neq b$, set $x_1 = x_2$: the antisymmetric wavefunction vanishes, so two fermions are never found at the same place. The symmetric one does the opposite and makes coincidences more likely. The figure shows the effect for two particles in a box.
Spin, and two electrons per orbital
An electron's state has two parts: its orbital, where it is likely to be, and its spin. The exclusion principle applies to the whole state. Two electrons can therefore share an orbital if their spins differ, one up and one down, and a third is excluded. Each orbital holds at most two electrons, and level $n$ of hydrogen, with $n^2$ orbitals, holds at most $2n^2$: 2, 8, 18, 32.
Worked example: filling the first shells
Treat each electron as moving independently in the attraction of the nucleus, shielded by the others, and fill the lowest available states, two per orbital.
| Atom | Electrons | Filling | Energy to remove one |
|---|---|---|---|
| Hydrogen | 1 | 1s¹ | 13.6 eV |
| Helium | 2 | 1s², shell $n = 1$ full | 24.6 eV |
| Lithium | 3 | 1s² 2s¹ | 5.4 eV |
| Neon | 10 | 1s² 2s² 2p⁶, shell $n = 2$ full | 21.6 eV |
| Sodium | 11 | 1s² 2s² 2p⁶ 3s¹ | 5.1 eV |
Helium puts both electrons in 1s, spins opposite. That fills the $n = 1$ shell's two places.
Lithium's third electron is excluded from 1s and must go into $n = 2$, where the orbital is about four times larger and the two inner electrons screen most of the nucleus's charge. It is loosely held: 5.4 eV against helium's 24.6 eV. That is the answer to the opening.
Neon fills all eight $n = 2$ places, $2\times2^2 = 8$, and is inert again. Sodium's eleventh electron starts the next shell and, like lithium's, is easily removed. The rhythm of the periodic table, a reactive metal just after each inert gas, is the exclusion principle counting to $2n^2$.
Where the simple count fails. The independent-electron picture is an approximation. Screening makes 2s fill before 2p, which hydrogen alone would not explain, and from potassium on the 4s orbital fills before 3d. So the table's rows hold 2, 8, 8, 18, 18, 32 elements rather than $2n^2$ in order. The exclusion principle still decides how many electrons fit in each orbital; which orbital comes next depends on the interactions the simple picture leaves out.
Optional: many electrons and determinants
For $N$ fermions the antisymmetric wavefunction is built as a determinant, whose rows are the occupied states and whose columns are the particles:
$$\psi(x_1, \dots, x_N) = \frac{1}{\sqrt{N!}}\det\bigl[\varphi_i(x_j)\bigr].$$Swapping two particles swaps two columns, which flips the sign of a determinant: antisymmetry is automatic. Putting two electrons in the same state makes two rows equal, and a determinant with two equal rows is zero: exclusion is automatic too. This form, introduced by John Slater in 1929, is the starting point for calculating the electronic structure of atoms and molecules.
One consequence is visible in helium's spectrum. With one electron excited, the two electrons can have their spins parallel or opposite. When the spins are parallel, the spatial part of the wavefunction must be antisymmetric, so the electrons keep apart, as in the figure, and feel less of each other's repulsion. Those states lie lower in energy, and the two families of helium lines were once thought to come from two different gases.
Two questions
1. The third shell. How many electrons can the $n = 3$ level hold? The third row of the periodic table, sodium to argon, has only eight elements. Is that a failure of the exclusion principle?
Worked answer
Level $n = 3$ has $3^2 = 9$ orbitals (one 3s, three 3p, five 3d), each holding two electrons: 18 in all. The third row has eight elements because only 3s and 3p fill there. After argon, the 4s orbital lies lower than 3d in a many-electron atom, so potassium and calcium fill 4s first, and 3d fills in the next row, among the transition metals.
The exclusion principle is not at fault: no orbital ever holds more than two electrons. The slip is reading $2n^2$ as the length of row $n$. It counts the places available in a hydrogen level; the order in which those places fill depends on the other electrons.
2. An electron on the Moon. Strictly, the wavefunction of every electron in the universe should be antisymmetric under exchange. Does that mean an electron in the 1s state of a hydrogen atom in your laboratory prevents another, in the 1s state of a hydrogen atom on the Moon, from having the same spin?
Worked answer
No. "1s of the atom here" and "1s of the atom on the Moon" are different states, since their orbitals are centred in different places, so exclusion does not forbid anything. And when two particles' wavefunctions do not overlap, the exchange term in $|\psi_\pm|^2$ is zero wherever either particle could be found: every prediction is the same as for distinguishable particles.
The tempting wrong turn is to picture exclusion as a force or a signal between electrons. It is a property of the joint wavefunction, and it has observable effects only where the particles' states overlap: inside an atom, a metal or a star, not across the distance to the Moon.
Further reading
- Feynman, Lectures on Physics, Vol. III, Ch. 4 "Identical Particles", §4-1 "Bose particles and Fermi particles" and §4-7 "The exclusion principle", with the periodic table in §19-6.
- Born, Atomic Physics (Dover, 8th ed.), Ch. VI §4 "The Helium Atom", §5 "Pauli's Exclusion Principle" and §6 "The Periodic System. Closed Shells".
- Landau & Lifshitz, Quantum Mechanics (2nd ed.), §61 "The principle of indistinguishability of similar particles".
Where this leaves us
Electrons do not all sit in the lowest orbital because identical fermions have antisymmetric wavefunctions, and an antisymmetric wavefunction for two electrons in the same state is zero. With two spin states, each orbital holds two electrons, the shells fill in turn, and a lone electron outside a closed shell is loosely held. That is lithium, sodium and the whole rhythm of the periodic table.
But the shell picture rested on an approximation: electrons moving independently, their repulsion folded into a vague "screening". Put the repulsion in properly and even helium, with just two electrons, has no exact solution. Nearly every real problem in quantum mechanics is like this. The next essay is about how to get reliable numbers anyway, and how to tell when an approximation can be trusted.