Pauli Exclusion Principle

Equivalent forms

\langle \alpha | \beta \rangle = 0 \quad \text{for distinct fermions}

n_i \in \{0, 1\}

A minus sign under particle exchange explains chemistry, neutron stars, and electron degeneracy pressure.

Symbol	Name	SI unit	Dimension	Range
$n_i$	Occupation numberoutput Number of fermions in single-particle state i	1	1	0 – 1
$N$	Particle number Total number of identical fermions	1	1	1 – 1000
$E_F$	Fermi energy Energy of the highest occupied state at T = 0	J	ML^2T^-2	1e-19 – 1e-16
$spin$	Spin label Internal spin state (+1/2 or -1/2 for electrons)	1	1	-0.5 – 0.5

Unit systems

Symbol	SI	CGS	Imperial
$n_i$	1	1	1
$N$	1	1	1
$E_F$	J	erg	ft*lbf
$spin$	1	1	1

Where it holds

Holds for all spin-1/2 particles (and any half-integer spin): electrons, protons, neutrons, quarks. Bosons obey the opposite (symmetric) rule.

Dimensional analysis

\Psi \to [\psi ]

-

\Psi \to [\psi ]

Both sides are wavefunctions — relation is purely symmetry-based, dimensionless.

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Discovery

Wolfgang Pauli · 1925

Pauli postulated that no two electrons share the same set of four quantum numbers to explain the structure of the periodic table. The deeper antisymmetry was clarified by Dirac and Heisenberg the next year. Pauli received the 1945 Nobel Prize for it.

Try this

Why doesn't every electron in an atom fall into the lowest energy state?

Identical fermions refuse to share quantum states — and this single rule builds the entire periodic table, holds up white dwarfs against gravity, and explains why solids are solid.

Research status: stable

Real-world applications

Periodic table and chemical bonding
Electron degeneracy pressure in white dwarfs and neutron stars
Semiconductor band structure (conduction vs valence)
Quantum simulation of fermionic systems

Common misconceptions

Pauli is not a 'force' — it is a kinematic constraint from wavefunction symmetry
Applies to identical fermions only; an electron and a proton can share orbitals
Bosons follow the opposite rule (symmetric wavefunction) — they can pile up (BEC, laser light)

Experimental verification

Periodic table structure (shell filling 2, 8, 18, 32). White dwarf mass-radius relation (Chandrasekhar 1931). Electron degeneracy pressure in metals. Two-electron interference asymmetry in beam experiments. Hong-Ou-Mandel anti-bunching for fermions.

Derivation

For identical particles, |Psi(r1,

r_{2})|^2 = |\Psi (r_{2}

,r1)|^2.

So Psi(r1,

r_{2}) = e^(i*\varphi )*\Psi (r_{2}

,r1); doing the swap twice gives

e^(2i*\varphi ) = 1

, so

\varphi = 0

(bosons) or pi (fermions).

Setting

r_{1} = r_{2}

in the antisymmetric case gives

\Psi = -\Psi

, so

\Psi = 0

— two fermions cannot occupy the same state.

The Slater determinant

\Psi = (1/\sqrt{N!})*det[\varphi _i(r_j)]

enforces antisymmetry automatically.

Limiting cases

T = 0

⟶

n_i = 1

if E_i < E_F, else 0All states below Fermi level filled; above empty — step function.

k_B*T >> E_F⟶ Recovers Maxwell-BoltzmannHot, dilute fermion gas behaves classically — quantum statistics negligible.

Two electrons, same orbital, same spin⟶

\Psi \equiv 0

— forbiddenAntisymmetry forces vanishing wavefunction.

What if…

What if electrons were bosons?

All electrons would collapse into the 1s orbital — no chemistry, no periodic table, no life.

What if Pauli pressure vanished in a white dwarf?

It collapses under gravity to a neutron star — Chandrasekhar limit $\approx 1.44$ solar masses.

What if two electrons are made distinguishable (different spin states)?

They can share spatial orbital — explains why s-orbitals hold 2 and not 1 electron.

1

Maximum electrons in n=3 shell

Given ·

n:: 3

Find · max_electrons

Steps

l ∈ {0, 1, $2} \to 3$ subshells
$Sum (2l+1) = 1 + 3 + 5 = 9$ orbitals
Each holds 2 electrons (opposite spins $) \to 18$ total

Result ·

2*n^2 = 18

electrons

(l = 0

,1,2;

m_l = -l

..l;

m_s = \pm 1/2)

2

Fermi energy of copper

Given ·

n electron density:: 8.45e+28

Find · E_F

Steps

$(3*pi^2*n)^(1/3) \approx (3*9.87*8.45e28)^(1/3) \approx 1.36e10 /m$
$k_F^2 \approx 1.85e20 /m^2$
$E_F \approx (1.054571817e-34)^2*1.85e20/(2*9.1093837015e-31) \approx 1.12e-18\,\mathrm{J} \approx 7.0\,\mathrm{eV}$

Result ·

E_F =

(hbar^

2/(2*m_e))*(3*pi^2*n)^(2/3) \approx 1.12e-18\,\mathrm{J} \approx 7.0\,\mathrm{eV}

Spin-1/2 Operators→Time-Independent Schrödinger Equation→bohr hydrogen energy levels