Fermionic Fock Space

Fermionic Fock space is the Hilbert space that collects all possible particle-number sectors for identical fermions. If $\mathcal h$ is the one-particle Hilbert space, the fermionic Fock space is

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

Here $\wedge^N\mathcal h$ is the antisymmetric $N$-particle subspace of $\mathcal h^{\otimes N}$. It is also called the $N$th exterior power of $\mathcal h$.

The main difference from bosonic Fock space is that each mode can be occupied at most once. Fermionic occupation-number basis states are therefore bitstrings of zeros and ones, with signs controlled by a fixed ordering convention for the modes.

Helpful background. Occupation-Number Basis introduces the labels; Fermions explains antisymmetry and complete one-particle modes.

Vacuum Sector

The $N=0$ sector is again one-dimensional:

$$\wedge^0\mathcal h \cong \mathbb C.$$

Its normalized basis vector is the vacuum,

$$\lvert0\rangle.$$

The vacuum is a no-particle state, not the zero vector. It belongs to both bosonic and fermionic Fock-space constructions because every Fock space includes a zero-particle sector.

Two-Particle Antisymmetric Sector

The two-fermion sector is

$$\wedge^2\mathcal h \subset \mathcal h\otimes\mathcal h.$$

For two distinct orthonormal modes $\lvert a\rangle$ and $\lvert b\rangle$, the occupation state with both modes occupied corresponds to

$$\lvert 1_a,1_b\rangle_F \longleftrightarrow \frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 - \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

There is no state $\lvert2_a\rangle_F$. If two identical fermions are assigned to the same complete one-particle state, the antisymmetric combination is zero.

The two-particle antisymmetrizer is

$$\Pi_A = \frac12(I-P_{12}).$$

It projects a two-slot vector onto the antisymmetric sector.

Antisymmetric sectors and determinants

The fixed-number sector $\wedge^N\mathcal h$ consists of vectors with $U(\pi)\lvert\Psi\rangle=\operatorname{sgn}(\pi)\lvert\Psi\rangle$. At $N=1$ this is just $\mathcal h$.

An occupation vector selecting $N$ orthonormal spin-orbitals corresponds to the normalized Slater determinant of those orbitals. The general state is a linear combination of determinants; antisymmetry alone does not make it a single determinant. The antisymmetrizer gives the explicit construction without assuming independent particles.

Direct Sum Over Particle Number

A vector in fermionic Fock space is a sequence

$$\lvert\Psi\rangle = \psi_0\oplus\psi_1\oplus\psi_2\oplus\cdots, \qquad \psi_N\in\wedge^N\mathcal h.$$

The norm is

$$\lVert\Psi\rVert^2 = \sum_{N=0}^{\infty} \lVert\psi_N\rVert^2.$$

A definite-$N$ fermionic state has only one nonzero component. A mathematical Fock vector may have several particle-number components. With the usual fermion-parity superselection rule, physical density operators have no coherence between even and odd total number. Superpositions within one parity, such as vacuum plus a pair, are compatible with that rule even when exact number is not conserved. See Particle-Number Superselection for the difference.

Finite-Dimensional Limit

If $\dim\mathcal h=M$, then

$$\wedge^N\mathcal h=0 \qquad \text{for }N>M.$$

There are only $M$ available one-particle modes, and each can be occupied at most once. The full fermionic Fock space is then finite-dimensional:

$$\dim\mathcal F_F(\mathcal h) = \sum_{N=0}^{M} \binom{M}{N} = 2^M.$$

Each basis vector is therefore labeled by an $M$-bit string. After choosing an order, the vector space can be identified with $M$ qubits, but the physical operator representation must retain fermionic signs: operators on different modes anticommute. A Jordan–Wigner transformation implements those signs with strings of parity operators.

Occupation-Number Basis

Choose an ordered orthonormal mode basis

$$\lvert\varphi_1\rangle, \lvert\varphi_2\rangle, \ldots .$$

A fermionic occupation basis vector is

$$\lvert n_1,n_2,n_3,\ldots\rangle_F,$$

with

$$n_i\in\{0,1\}, \qquad N=\sum_i n_i.$$

The ordering of the modes is part of the convention. Once fermionic creation operators are introduced, a finite-mode occupation state is written schematically as

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

Changing the order of fermionic creation operators can introduce minus signs. This is not a nuisance added by notation; it is how antisymmetry is carried in occupation language.

The normalized occupation basis vectors are developed in Number States.

Exercises

1. For four fermionic modes, how many basis states are in the full Fock space?

Solution

Each of the four modes may be empty or occupied. Therefore the full fermionic Fock space has

$$2^4=16$$

basis states.

2. For four fermionic modes, how many states are in the $N=2$ sector?

Solution

Choose which two of the four modes are occupied:

$$\binom{4}{2}=6.$$

Thus $\dim\wedge^2\mathbb C^4=6$.

3. Parity sectors. For $M\ge1$, show that the even and odd sectors of the full $M$-mode Fock space both have dimension $2^{M-1}$. Does this mean only one of them exists physically?

Solution

The sums of even and odd binomial coefficients have sum $(1+1)^M=2^M$ and difference $(1-1)^M=0$. Both sectors exist; the parity rule restricts coherent superpositions between them. It permits either sector and classical mixtures of the sectors.

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.