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.

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.

One-Particle Sector

The $N=1$ sector is the one-particle Hilbert space:

$$\wedge^1\mathcal h = \mathcal h.$$

If $\{\lvert\varphi_i\rangle\}$ is a mode basis, then

$$\lvert 1_i\rangle_F$$

means one fermion in mode $i$ and no fermions in the other displayed modes.

For one particle, antisymmetry is not visible. It becomes visible when at least two fermions are present.

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.

N-Particle Sector

For $N$ identical fermions, the fixed-particle-number Hilbert space is

$$\wedge^N\mathcal h.$$

Vectors in this sector satisfy

$$U(\pi)\lvert\Psi\rangle = \operatorname{sgn}(\pi)\lvert\Psi\rangle \qquad \text{for every }\pi\in S_N.$$

The full antisymmetrizer is

$$\Pi_A^{(N)} = \frac{1}{N!} \sum_{\pi\in S_N} \operatorname{sgn}(\pi)U(\pi).$$

In wavefunction language, an $N$-fermion state made from orbitals $\varphi_1,\ldots,\varphi_N$ is represented by a Slater determinant:

$$\Psi(q_1,\ldots,q_N) = \frac{1}{\sqrt{N!}} \det \begin{pmatrix} \varphi_1(q_1) & \cdots & \varphi_N(q_1)\\ \vdots & \ddots & \vdots\\ \varphi_1(q_N) & \cdots & \varphi_N(q_N) \end{pmatrix}.$$

Exchanging two particles exchanges two rows and changes the sign. Repeating an orbital makes two columns equal, so the determinant vanishes. That is Pauli exclusion in determinant form.

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 general Fock-space state may involve several particle-number sectors if the physical setting permits such superpositions.

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:

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

This is why fermionic Fock states for a finite mode set can be represented as bitstrings. Each of the $M$ modes is either empty or occupied.

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.

Pauli Exclusion in Fock Space

In fermionic occupation notation, Pauli exclusion becomes the simple rule

$$n_i=0\quad\text{or}\quad n_i=1.$$

The state

$$\lvert 1,0,1,1\rangle_F$$

is allowed. The state

$$\lvert 2,0,0,0\rangle_F$$

is not a fermionic basis state.

In operator language, the same rule appears as

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

Applying the same fermionic creation operator twice gives zero, so a mode cannot be doubly occupied.

Examples

For two fermionic modes $a$ and $b$, the full Fock space has four basis states:

$$\lvert0_a,0_b\rangle_F, \qquad \lvert1_a,0_b\rangle_F, \qquad \lvert0_a,1_b\rangle_F, \qquad \lvert1_a,1_b\rangle_F.$$

There are no states with occupations $2$ in either mode.

For three modes, the $N=2$ sector has

$$\lvert1,1,0\rangle_F, \qquad \lvert1,0,1\rangle_F, \qquad \lvert0,1,1\rangle_F.$$

Those three states correspond to choosing which two of the three modes are occupied.

Common Mistakes

• Allowing occupations $n_i>1$ for a fermionic mode.
• Forgetting that the mode order matters for fermionic signs.
• Treating the bitstring itself as a list of labeled particles.
• Confusing the finite-dimensional size $2^M$ with a tensor product of $M$ distinguishable particles.
• Forgetting that a “mode” may be a spin-orbital, not just a spatial orbital.
• Assuming the vacuum has different meaning in bosonic and fermionic Fock spaces.

Cross-Links

• Occupation-Number Basis
• Direct Sums versus Tensor Products
• Fermions
• Vacuum State
• Number States
• Mode Occupations
• Particle-Number Superselection Preview
• Fock Space Examples
• Bosonic Fock Space
• Creation and Annihilation Operators
• Fermionic Anticommutation Relations
• Fock Space Exercises
• Pauli Exclusion Principle
• Slater Determinants
• Symmetrization Postulate
• Indistinguishability
• 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. 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. Write the antisymmetric slot state corresponding to $\lvert1_a,1_b\rangle_F$ for orthonormal modes $a$ and $b$.

Solution

The corresponding state is

$$\frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 - \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

4. Why is $\lvert2_a\rangle_F$ not a fermionic basis state?

Solution

Fermionic modes have occupation $0$ or $1$ only. If two identical fermions are assigned to the same complete one-particle state $a$, the antisymmetric two-slot state cancels to zero. In operator notation this is expressed by $(c_a^\dagger)^2=0$.