Fermionic Anticommutation Relations

Fermionic anticommutation relations are the operator algebra that makes antisymmetric Fock space usable in occupation-number language. They encode Pauli exclusion, the minus sign from exchanging identical fermions, and the sign bookkeeping required when modes are ordered.

For orthonormal fermionic modes labeled by $i,j$, the canonical anticommutation relations are

$$\{c_i,c_j^\dagger\} = \delta_{ij}I, \qquad \{c_i,c_j\} =0, \qquad \{c_i^\dagger,c_j^\dagger\} =0.$$

Here

$$\{A,B\} = AB+BA$$

is the anticommutator. The operators $c_i^\dagger$ and $c_i$ create and annihilate a fermion in a complete one-particle mode $i$, such as a spin-orbital or a momentum-spin state.

Helpful background. Creation and Annihilation Operators fixes the occupation ordering; Fermionic Fock Space explains antisymmetric sectors.

Canonical fermionic anticommutators

For one fermionic mode, the algebra is

$$\{c,c^\dagger\} = I, \qquad \{c,c\} =0, \qquad \{c^\dagger,c^\dagger\} =0.$$

Equivalently,

$$cc^\dagger = I-c^\dagger c.$$

The sign is the important difference from bosonic creation and annihilation operators. Bosons obey commutation relations; fermions obey anticommutation relations.

For distinct modes $i\ne j$,

$$c_i^\dagger c_j^\dagger = -c_j^\dagger c_i^\dagger, \qquad c_i c_j = -c_j c_i, \qquad c_i c_j^\dagger = -c_j^\dagger c_i.$$

Thus reordering fermionic operators changes signs. This is the most common source of mistakes in hand derivations and many-body code.

Ordered occupation states

Use the convention $|n_1,\ldots,n_M\rangle_F=(c_1^\dagger)^{n_1}\cdots(c_M^\dagger)^{n_M}|0\rangle$. Inserting or removing mode $i$ requires moving its operator past the $S_i=\sum_{k<i}n_k$ occupied modes to its left. The resulting factor is $(-1)^{S_i}$, as derived with examples in Creation and Annihilation Operators. Creation gives zero on an occupied mode; annihilation gives zero on an empty mode.

Pauli Exclusion from the Algebra

The anticommutator of a fermionic creation operator with itself gives

$$\{c_i^\dagger,c_i^\dagger\} = 2(c_i^\dagger)^2 =0.$$

Therefore

$$(c_i^\dagger)^2 =0.$$

Applying the same fermionic creation operator twice gives zero:

$$(c_i^\dagger)^2\lvert\Psi\rangle =0$$

for every Fock-space vector $\lvert\Psi\rangle$. In particular,

$$(c_i^\dagger)^2\lvert0\rangle =0.$$

This is the Pauli exclusion principle in operator form: a complete fermionic mode can be either empty or occupied, but not doubly occupied.

Similarly,

$$c_i^2=0.$$

Trying to annihilate the same occupied mode twice also gives zero, because after one annihilation the mode is empty.

Number Operators

The fermionic number operator for mode $i$ is

$$N_i = c_i^\dagger c_i.$$

It has eigenvalues $0$ and $1$:

$$N_i \lvert n_1,n_2,\ldots\rangle_F = n_i \lvert n_1,n_2,\ldots\rangle_F.$$

Using the anticommutation relations,

$$N_i^2 = N_i.$$

Thus $N_i$ is a projector onto states with mode $i$ occupied. The total number operator is

$$N = \sum_i N_i.$$

The creation and annihilation commutators with $N_i$ have the same raising-and-lowering meaning as in the bosonic case:

$$[N_i,c_j^\dagger] = \delta_{ij}c_j^\dagger, \qquad [N_i,c_j] = -\delta_{ij}c_j.$$

The algebra used to prove them is different, but the physical interpretation is the same: creation raises a mode occupation by one when possible, and annihilation lowers it by one when possible.

Mode Ordering and Signs

For two modes, define the canonical two-occupied state by

$$\lvert1,1\rangle_F = c_1^\dagger c_2^\dagger\lvert0\rangle.$$

Reversing the creation order gives

$$c_2^\dagger c_1^\dagger\lvert0\rangle = -c_1^\dagger c_2^\dagger\lvert0\rangle = -\lvert1,1\rangle_F.$$

The minus sign is not optional. It is the occupation-number version of antisymmetry under fermion exchange.

For a three-mode example,

$$c_3^\dagger\lvert1,0,0\rangle_F = -\lvert1,0,1\rangle_F,$$

because one occupied mode lies before mode $3$. By contrast,

$$c_1^\dagger\lvert0,0,1\rangle_F = \lvert1,0,1\rangle_F,$$

because no occupied mode lies before mode $1$.

In practical calculations, signs should be derived from the chosen operator order rather than guessed from particle labels. Fermionic occupation basis states label modes, not distinguishable particles.

Antisymmetry and even local observables

An ordered product of creation operators for distinct orthonormal modes corresponds to the normalized antisymmetric product of their one-particle states. Reversing two creation operators reverses its sign; repeating one makes it vanish. This is the exterior-power construction in operator form.

The CAR do not mean that every fermionic observable must be combined using anticommutators. Time evolution uses the ordinary commutator, and

$$[c_i,c_i^\dagger]=I-2N_i$$

is a valid, state-dependent operator identity. Even observables with disjoint mode support commute: moving an even product past another disjoint even product produces an even number of minus signs. For example, $[N_i,N_j]=0$. This is how local even observables remain compatible despite the anticommutation of the odd single-fermion operators.

The field anticommutator follows by expanding a complete one-particle basis; Mode Expansions gives that derivation.

Exercises

1. Verify $\{c,c^\dagger\}\lvert0\rangle=\lvert0\rangle$ for a single fermionic mode.

Solution

Since $c\lvert0\rangle=0$ and $c^\dagger\lvert0\rangle=\lvert1\rangle$,

$$cc^\dagger\lvert0\rangle = c\lvert1\rangle = \lvert0\rangle.$$

The other term is

$$c^\dagger c\lvert0\rangle =0.$$

Therefore

$$\{c,c^\dagger\}\lvert0\rangle = \lvert0\rangle.$$

2. With canonical mode order $1,2,3$, compute $c_2\lvert1,1,0\rangle_F$.

Solution

For mode $2$, the number of occupied modes before it is

$$S_2=n_1=1.$$

Therefore

$$c_2\lvert1,1,0\rangle_F = (-1)^{1}\lvert1,0,0\rangle_F = -\lvert1,0,0\rangle_F.$$

3. Even bilinears on disjoint modes. Show that $[c_1^\dagger c_2,c_3^\dagger c_4]=0$ when all four modes differ. Why does the same argument fail if a creation and annihilation index coincide?

Solution

Each of the two operators in one bilinear crosses both operators in the other, giving four minus signs and hence a plus sign. If indices coincide, a swap can use $c_i c_i^\dagger=1-c_i^\dagger c_i$ and produces an additional contraction. In general, $[c_i^\dagger c_j,c_k^\dagger c_l]=\delta_{jk}c_i^\dagger c_l-\delta_{il}c_k^\dagger c_j$.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• 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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.