The Pauli Exclusion Principle

The previous lesson assigned fermions an antisymmetric multiparticle state. That single property, applied to two fermions in the same single-particle state, produces one of the most consequential results in all of physics.

Antisymmetry forbids double occupancy

Take two identical fermions and try to place both in the same single-particle state $\psi_1$. The antisymmetric two-particle wavefunction is

The two terms are now identical, so they cancel exactly:

A wavefunction that is identically zero describes no state at all. Therefore two identical fermions cannot occupy the same single-particle state. This is the Pauli exclusion principle, and you have just derived it from antisymmetry alone — it is not an extra postulate but a direct consequence of the minus sign under exchange.

The same conclusion from the exchange operator

There is an even shorter argument. If two fermions were in the identical state $\psi_1$, then swapping them changes nothing physically and even relabels nothing in the formula, so $\hat{P}_{12}\Psi = +\Psi$. But fermions require $\hat{P}_{12}\Psi = -\Psi$. The only vector satisfying both $\Psi = +\Psi$ and $\Psi = -\Psi$ is $\Psi = 0$. Again the state is excluded.

What "the same state" includes

A single-particle state is specified by all of its quantum numbers, spin included. For an electron in an atom that means the spatial orbital quantum numbers $(n, \ell, m_\ell)$ and the spin projection $m_s = \pm\tfrac12$. Two electrons may share the same spatial orbital only if their spins differ — one up, one down. A third electron has no remaining distinct spin label for that orbital and must go elsewhere. This is exactly why atomic orbitals hold at most two electrons and why electrons stack into successive shells.

Why matter has volume

Exclusion is what keeps electrons from collapsing into the lowest orbital of every atom. Because each state holds at most one electron of each spin, electrons fill up to a high energy — the Fermi energy — even at zero temperature. The resulting degeneracy pressure holds up white dwarf and neutron stars against gravity and gives ordinary solids their incompressibility. The stability and size of the matter around you trace back to the cancellation you computed below.

Try it

Compute the antisymmetric combination $\tfrac{1}{\sqrt{2}}(ab - ba)$ when both fermions are placed in the same single-particle state, so $a = b$. The result must be exactly zero — the quantitative statement of exclusion.

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