Pauli Exclusion Principle

The Pauli exclusion principle says that no two identical fermions can occupy the same complete one-particle quantum state.

The word complete matters. For an electron, a one-particle state includes spin as well as spatial quantum numbers. Two electrons may occupy the same spatial orbital only if their spin states differ so that the total two-electron state remains antisymmetric. In occupation-number language, each fermionic mode has occupation

$$n_i=0\quad\text{or}\quad n_i=1.$$

Pauli exclusion is not an extra force between fermions. It is a direct consequence of the antisymmetry required by the symmetrization postulate.

In operator language, the same content is carried by fermionic anticommutation relations, especially $(c_i^\dagger)^2=0$.

Helpful background. Symmetrization Postulate supplies fermionic antisymmetry. Two Spin-1/2 Particles introduces the singlet used in the atomic examples.

Occupation restriction for fermions

Let $\mathcal h$ be the one-particle Hilbert space for a species of identical fermions. If $\lvert\alpha\rangle\in\mathcal h$ is a normalized one-particle state, then the two-fermion state with both slots assigned to $\lvert\alpha\rangle$ is zero:

$$\lvert\alpha,\alpha\rangle_A=0.$$

Equivalently, an antisymmetric $N$-fermion state vanishes whenever two fermions are assigned the same complete one-particle state.

For electrons in atoms, the “complete one-particle state” is often called a spin-orbital: a spatial orbital together with a spin state. Two electrons can share a spatial orbital only because spin provides two different spin-orbitals.

Derivation from Two-Fermion Antisymmetry

For two identical fermions in orthonormal one-particle states $\lvert\alpha\rangle$ and $\lvert\beta\rangle$, the antisymmetric state is

$$\lvert\alpha,\beta\rangle_A = \frac{1}{\sqrt2} \bigl( \lvert\alpha\rangle_1\lvert\beta\rangle_2 - \lvert\beta\rangle_1\lvert\alpha\rangle_2 \bigr).$$

If the two one-particle states are the same, set $\lvert\beta\rangle=\lvert\alpha\rangle$. Then

$$\lvert\alpha,\alpha\rangle_A = \frac{1}{\sqrt2} \bigl( \lvert\alpha\rangle_1\lvert\alpha\rangle_2 - \lvert\alpha\rangle_1\lvert\alpha\rangle_2 \bigr) = 0.$$

There is no normalized two-fermion state of this form. The state is not forbidden because it has high energy; it is absent from the fermionic Hilbert space.

Coefficient Form

Choose a one-particle basis $\{\lvert i\rangle\}$ for $\mathcal h$. A two-slot state may be expanded as

$$\lvert\Psi\rangle = \sum_{ij} C_{ij} \lvert i\rangle_1\lvert j\rangle_2.$$

Fermionic antisymmetry requires

$$C_{ij}=-C_{ji}.$$

Setting $i=j$ gives

$$C_{ii}=-C_{ii},$$

so

$$C_{ii}=0.$$

Thus the amplitude for both fermions occupying the same basis state vanishes. This is the exclusion principle in coefficient form.

Arbitrary modes and correlated states

The restriction holds in every orthonormal one-particle basis, not only in energy eigenmodes. For a normalized mode $\alpha$, the number operator is $n_\alpha=c_\alpha^\dagger c_\alpha$. The anticommutation relation and $(c_\alpha^\dagger)^2=0$ imply

$$n_\alpha^2=n_\alpha,\qquad 0\le\langle n_\alpha\rangle\le1.$$

Thus measurements of one-mode occupation yield only zero or one, while an expectation value can be fractional. For instance, one fermion in $(\lvert a\rangle+\lvert b\rangle)/\sqrt2$ has mean occupation $1/2$ in each of two orthogonal modes. This is a superposition, not half a particle in each measurement.

A Slater determinant implements exclusion by vanishing when occupied orbitals become linearly dependent. Correlated states can be superpositions of determinants and obey the same occupation bound.

Atomic Shells Preview

In a central-potential approximation, an electron state is labeled by quantum numbers such as

$$n,\quad \ell,\quad m_\ell,\quad m_s.$$

For a fixed spatial orbital specified by $n,\ell,m_\ell$, an electron has two possible spin projections:

$$m_s=\frac12, \qquad m_s=-\frac12.$$

Therefore a single spatial orbital can hold at most two electrons, one in each spin state. A subshell with orbital angular momentum $\ell$ has $2\ell+1$ values of $m_\ell$, so its electron capacity is

$$2(2\ell+1).$$

In the ideal hydrogenic counting, a shell with principal quantum number $n$ has total capacity

$$2n^2.$$

Real atoms require electron-electron interactions, spin-orbit effects, screening, and many-body approximations, but the exclusion principle is the structural reason electrons fill shells rather than all collapsing into the same one-particle state.

The helium ground state gives the first useful example. Both electrons occupy the same spatial $1s$ orbital in a simple approximation, but they have opposite spin projections and combine into an antisymmetric spin singlet. The spatial part is symmetric, the spin part is antisymmetric, and the total electronic state is antisymmetric.

Pauli Principle in Atoms develops the atomic consequences in full, including equivalent-electron term restrictions, closed-subshell structure, particle–hole counting, and the qualified connection to periodicity.

Stability of Matter Preview

Pauli exclusion is a major ingredient in the stability and bulk structure of ordinary matter. Since electrons are fermions, many electrons cannot all occupy the same low-energy one-particle state. Filling many distinct states forces a kinetic-energy cost that grows with density.

This effect is often called degeneracy pressure in many-body settings. It is not a classical pressure caused by collisions. It is the energy cost of antisymmetric fermionic state filling. In atoms, molecules, solids, white dwarfs, and neutron-rich matter, this filling structure combines with Coulomb forces, gravity, interactions, and boundary conditions.

The exclusion principle alone is not a complete proof of matter stability. Rigorous stability results also use kinetic-energy estimates, electrostatic inequalities, and the detailed form of the Hamiltonian. Still, without fermionic antisymmetry, ordinary matter would have a radically different structure.

Fermi Gas Preview

A free spin-$1/2$ Fermi gas illustrates exclusion in momentum space. Put particles in a box and label one-particle states by momentum and spin. At zero temperature, the lowest-energy many-fermion state fills momentum modes up to a Fermi wave number $k_F$ (momentum $p_F=\hbar k_F$). The Degenerate Fermi Gas page develops the resulting thermal shell, Pauli-blocked final states, pressure, and low-temperature heat capacity.

In three dimensions, with spin degeneracy $g$, the number density is

$$n = \frac{g\,k_F^3}{6\pi^2}.$$

The corresponding Fermi energy for nonrelativistic particles of mass $m$ is

$$E_F = \frac{\hbar^2 k_F^2}{2m}.$$

This simple model is the entry point to electron gases in metals, neutron matter, and many condensed-matter estimates. The full many-body theory is much richer, but the first step is just one fermion per mode.

Exercises

1. Show directly that two identical fermions cannot both occupy the same one-particle state $\lvert\alpha\rangle$.

Solution

The antisymmetric two-fermion construction gives

$$\frac{1}{\sqrt2} \bigl( \lvert\alpha\rangle_1\lvert\alpha\rangle_2 - \lvert\alpha\rangle_1\lvert\alpha\rangle_2 \bigr) = 0.$$

The zero vector is not a physical state, so two identical fermions cannot occupy the same complete one-particle state.

2. Why can the two electrons in the helium ground-state approximation share a spatial $1s$ orbital?

Solution

They do not occupy the same complete one-particle state. The spatial orbital is the same, but the spin states differ. In the usual ground-state approximation, the spatial wavefunction is symmetric and the spin state is the antisymmetric singlet, making the total electronic state antisymmetric as required for fermions.

3. How many electrons can fit in a $p$ subshell in the central-potential counting?

Solution

A $p$ subshell has $\ell=1$, so $m_\ell=-1,0,1$: three spatial orbital states. Each can be paired with two spin projections. The capacity is therefore

$$2(2\ell+1) = 2(3) = 6.$$

4. A spin-$1/2$ Fermi gas has spin degeneracy $g=2$. Express $k_F$ in terms of the number density $n$.

Solution

Use

$$n = \frac{g\,k_F^3}{6\pi^2}.$$

With $g=2$,

$$n = \frac{k_F^3}{3\pi^2}.$$

Therefore

$$k_F = (3\pi^2 n)^{1/3}.$$

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• E. H. Lieb and R. Seiringer, The Stability of Matter in Quantum Mechanics, Cambridge University Press, 2010.
• A. M. L. Messiah and O. W. Greenberg, “Symmetrization Postulate and Its Experimental Foundation”, Physical Review 136, B248–B267 (1964).
• W. Pauli, “Über den Zusammenhang des Abschlusses der Elektronengruppen im Atom mit der Komplexstruktur der Spektren”, Zeitschrift für Physik 31, 765–783 (1925).
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.