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 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.

Algebra

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.

Action on Occupation States

Choose and keep a fixed ordering of modes:

$$1,2,3,\ldots .$$

The canonical occupation basis is defined by applying occupied creation operators in that order:

$$\lvert n_1,n_2,\ldots,n_M\rangle_F = (c_1^\dagger)^{n_1} (c_2^\dagger)^{n_2} \cdots (c_M^\dagger)^{n_M} \lvert0\rangle,$$

where each $n_i$ is $0$ or $1$.

Let

$$S_i = \sum_{k<i}n_k$$

be the number of occupied modes before mode $i$ in the chosen ordering. Then creation acts as

$$c_i^\dagger \lvert n_1,\ldots,0_i,\ldots\rangle_F = (-1)^{S_i} \lvert n_1,\ldots,1_i,\ldots\rangle_F,$$

and

$$c_i^\dagger \lvert n_1,\ldots,1_i,\ldots\rangle_F = 0.$$

Annihilation acts as

$$c_i \lvert n_1,\ldots,1_i,\ldots\rangle_F = (-1)^{S_i} \lvert n_1,\ldots,0_i,\ldots\rangle_F,$$

and

$$c_i \lvert n_1,\ldots,0_i,\ldots\rangle_F = 0.$$

The phase $(-1)^{S_i}$ counts how many occupied modes the operator must pass through to reach the canonical position. Different authors may choose different ordering conventions, but a calculation must use one convention consistently.

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.

Relation to Antisymmetric Fock Space

The fermionic Fock space over a one-particle Hilbert space $\mathcal h$ is

$$\mathcal F_F(\mathcal h) = \bigoplus_{N=0}^{\infty} \wedge^N\mathcal h.$$

Given ordered one-particle modes $\lvert\varphi_i\rangle$, the state

$$c_{i_1}^\dagger c_{i_2}^\dagger\cdots c_{i_N}^\dagger\lvert0\rangle, \qquad i_1<i_2<\cdots<i_N,$$

corresponds to the antisymmetric slot state

$$\frac{1}{\sqrt{N!}} \sum_{\pi\in S_N} \operatorname{sgn}(\pi) \lvert\varphi_{i_{\pi(1)}}\rangle_1 \cdots \lvert\varphi_{i_{\pi(N)}}\rangle_N.$$

Swapping two creation operators changes the sign exactly as swapping two columns in a Slater determinant changes the sign. Repeating a creation operator gives zero exactly as a determinant with two identical columns vanishes.

The anticommutation relations are therefore not merely a formal trick. They are the compact operator expression of the exterior-power structure of identical-fermion Hilbert spaces.

After a one-particle basis is chosen, the same relations imply the anticommutator of the position-space field operators through the mode expansion.

Common Mistakes

• Using commutators instead of anticommutators for fermionic creation and annihilation operators.
• Forgetting that $c_i^\dagger c_j^\dagger=-c_j^\dagger c_i^\dagger$ for $i\ne j$.
• Applying $c_i^\dagger$ twice and expecting a nonzero doubly occupied state.
• Dropping the sign $(-1)^{S_i}$ when acting on occupation bitstrings.
• Changing the mode ordering halfway through a calculation.
• Treating fermionic occupation bitstrings as if they label distinguishable particles.

Cross-Links

• Creation and Annihilation Operators
• Fermions
• Number Operators
• Mode Expansions
• Field Operators
• Normal Ordering
• Fermionic Fock Space
• XXZ Spin Chain
• Occupation-Number Basis
• Pauli Exclusion Principle
• Bosonic Commutation Relations
• Fock Space Exercises
• Reference Bridge: Second Quantization

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

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. Show that $(c_i^\dagger)^2=0$ follows from the anticommutation relations.

Solution

Set $i=j$ in

$$\{c_i^\dagger,c_j^\dagger\}=0.$$

Then

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

Over the complex numbers, this implies

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

3. 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

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

4. Let $\lvert1,1\rangle_F=c_1^\dagger c_2^\dagger\lvert0\rangle$. What is $c_2^\dagger c_1^\dagger\lvert0\rangle$?

Solution

The anticommutation relation gives

$$c_2^\dagger c_1^\dagger = -c_1^\dagger c_2^\dagger.$$

Thus

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