Fermions take the same bookkeeping with one change: their operators anticommute. Define $\hat{c}^{\dagger}_{i}$ and $\hat{c}_{i}$ with $\{\hat{c}_{i},\hat{c}^{\dagger}_{j}\}=\delta_{ij}$ and all other anticommutators zero. Two consequences follow at once. Putting both operators equal in $\{\hat{c}^{\dagger}_{i},\hat{c}^{\dagger}_{i}\}=0$ gives $(\hat{c}^{\dagger}_{i})^{2}=0$: a fermion cannot be added twice to one state, and the exclusion principle becomes an identity between operators. And $\hat{c}^{\dagger}_{a}\hat{c}^{\dagger}_{b}\ket{0}=-\hat{c}^{\dagger}_{b}\hat{c}^{\dagger}_{a}\ket{0}$: changing the order of adding two fermions changes the sign, which is the antisymmetry of a Slater determinant. The number operator $\hat{c}^{\dagger}_{i}\hat{c}_{i}$ equals its own square, so its eigenvalues are 0 and 1. Because order matters, occupation-number states are written with the operators in a fixed order, and an operator acting on a state picks up a minus sign for each occupied state it passes. The lowest state of $N$ free fermions fills the $N$ lowest levels — the Fermi sea — and its simplest excitation lifts one fermion from a filled level to an empty one, leaving a hole behind.

A car park of numbered single spaces: each is empty or holds one car, and no second car fits in a taken space. The analogy stops at the sign — swapping the order in which two cars arrive changes nothing in a car park and changes the sign of a fermion state.

Four fermions in seven levels. In the lowest state they fill the four lowest levels, the Fermi sea; here $\hat{c}^{\dagger}_{6}\hat{c}_{4}$ has lifted the top one to the sixth level, leaving a hole in the sea and a particle above it.

For fermions, $\hat{c}^{\dagger}_{i}$ adds a particle to state $i$ and $\hat{c}_{i}$ removes one, with anticommutators :

• $(\hat{c}^{\dagger}_{i})^{2}=0$: no state holds two fermions — the exclusion principle as an identity between operators.

Rewrite the middle pair of $\hat{c}^{\dagger}\hat{c}\hat{c}^{\dagger}\hat{c}$ with the anticommutator. The leftover term contains , which is zero, so the number operator equals its own square; an eigenvalue of such an operator has .

The anticommutator of $\hat{c}^{\dagger}$ with itself is $2(\hat{c}^{\dagger})^{2}$.

From $\{\hat{c},\hat{c}^{\dagger}\}=1$.

The second term contains $(\hat{c}^{\dagger})^{2}=0$.

Apply $\hat{n}^{2}=\hat{n}$ to an eigenvector.

Fermionic creation operators anticommute; with both equal the anticommutator gives $(\hat{c}^{\dagger})^{2}=0$: the exclusion principle.

Swapping two fermionic creation operators changes the sign of the state they make: the antisymmetry of fermions.

The fermionic number operator equals its own square, so every occupation is 0 or 1.

With the states in a fixed order, the creation operator of state $i$ picks up a factor $-1$ for each occupied state before $i$.

The lowest state of $N$ free fermions fills the $N$ lowest levels. The operator $\hat{c}^{\dagger}_{p}\hat{c}_{h}$ moves one fermion from a filled level to an empty level , leaving a hole.

Great work! You now know how to:

• use anticommuting creation and annihilation operators
• derive the exclusion principle and the signs of fermion states
• describe the Fermi sea and its particle–hole excitations