Pauli Exclusion Revisited

Pauli’s exclusion rule began with atomic spectra and shell filling. Fermi’s and Dirac’s 1926 work turned that rule into a statistics of many-particle states. The useful question is now: which occupations are possible, and what probabilities does equilibrium assign to them?

The earlier historical account explains the atomic motivation; the formal derivation proves exclusion from antisymmetry. Here a three-mode example separates the counting rule from the thermal distribution.

Two Fermions in Three Modes

Choose three orthonormal complete one-particle modes $a,b,c$, including spin, with energies $\epsilon_a,\epsilon_b,\epsilon_c$. For two identical fermions the allowed occupation basis consists of

$$\lvert1,1,0\rangle,\qquad \lvert1,0,1\rangle,\qquad \lvert0,1,1\rangle.$$

There are $\binom32=3$ states. An occupation such as $(2,0,0)$ is absent, regardless of its hypothetical energy. With two bosons in these same three modes, double occupancy is allowed and the number instead is $\binom{2+3-1}{2}=6$.

For noninteracting fermions the three energies are

$$\epsilon_a+\epsilon_b,\qquad \epsilon_a+\epsilon_c,\qquad \epsilon_b+\epsilon_c.$$

The exclusion rule determines the allowed basis; the Hamiltonian determines the energies. Changing an interaction changes those energies and mixes allowed states, but does not create a forbidden double occupation of one fermionic mode.

From Counting to Thermal Probabilities

At fixed particle number $N=2$, canonical equilibrium assigns weights $e^{-\beta E}$, with $\beta=1/(k_{\mathrm B}T)$. Define $w_{ij}=e^{-\beta(\epsilon_i+\epsilon_j)}$. Then

$$Z_2=w_{ab}+w_{ac}+w_{bc},\qquad \langle n_a\rangle=\frac{w_{ab}+w_{ac}}{Z_2}.$$

This exact finite-system answer satisfies $0\le\langle n_a\rangle\le1$. The occupations of the three modes are correlated by the fixed total: filling two modes leaves the third empty.

For independent modes in the grand-canonical ensemble, particle number fluctuates and a chemical potential fixes its mean. A single fermionic mode has weights $1$ and $e^{-\beta(\epsilon_i-\mu)}$, giving

$$\langle n_i\rangle=\frac{1}{e^{\beta(\epsilon_i-\mu)}+1}.$$

This is the Fermi–Dirac distribution. The bound of one is a state-space property; the displayed thermal function also assumes independent modes and that ensemble. It is not the exact bare-orbital occupation law for every interacting fermion system.

Antisymmetry Without a New Force

The occupation $\lvert1,1,0\rangle$ corresponds, with a fixed ordering convention, to

$$\frac{\phi_a(q_1)\phi_b(q_2)-\phi_b(q_1)\phi_a(q_2)}{\sqrt2},$$

where $q$ includes position and spin. Setting $\phi_b=\phi_a$ makes the numerator zero. This is a forbidden state, not a state made expensive by a new repulsive force.

A determinant describes one occupied set of orbitals. A correlated many-electron state can be a superposition of determinants and need not assign each electron a definite orbital. Nevertheless, in any chosen orthonormal mode basis the occupation operator has eigenvalues zero and one. Its expectation can be fractional; that fraction describes quantum or statistical uncertainty, not fractional fermions detected in a mode.

Opposite-spin electrons can share a spatial orbital because their complete spin-orbitals differ. Their combined state must still be antisymmetric. Spin and Spatial Wavefunctions explains the singlet and triplet constructions.

Exercises

1. For $\epsilon_a=0$, $\epsilon_b=\Delta$, and $\epsilon_c=2\Delta$ with $\Delta>0$, calculate the mean occupations of the fixed-two-particle example at zero temperature and in the high-temperature limit.

Solution

At zero temperature the lowest pair is $ab$, so $(\langle n_a\rangle,\langle n_b\rangle,\langle n_c\rangle)=(1,1,0)$. At high temperature all three many-particle states have equal weight. Each mode appears in two of the three states, giving $(2/3,2/3,2/3)$. The sum remains exactly two.

2. For that high-temperature fixed-$N$ ensemble, calculate $\langle n_a n_b\rangle$ and compare it with $\langle n_a\rangle\langle n_b\rangle$.

Solution

Both $a$ and $b$ are occupied only in the $ab$ state, so $\langle n_a n_b\rangle=1/3$. The product of means is $4/9$. Their covariance is $-1/9$: fixed total number makes the occupations correlated even without interactions.

3. Can three electrons occupy one spatial orbital by choosing three different spin directions?

Solution

No. Electron spin space has dimension two. Three spinors in that same spatial orbital are linearly dependent, so their three-particle Slater determinant vanishes. Changing the spin basis does not increase the number of available spin states.

References

• P. A. M. Dirac, “On the Theory of Quantum Mechanics,” Proceedings of the Royal Society A 112, 661–677, 1926, DOI: 10.1098/rspa.1926.0133.
• E. Fermi, “Sulla quantizzazione del gas perfetto monoatomico,” Rendiconti Lincei 3, 145–149, 1926.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover, 2003.
• W. Pauli, “Exclusion Principle and Quantum Mechanics,” Nobel Lecture, 1946, NobelPrize.org.