CAR and CCR Algebras in Many-Body Theory

The CAR algebra makes fermionic field operators into bounded algebra elements before a vacuum or thermal state is chosen. Its finite-mode pieces are full matrix algebras; infinitely many modes are handled by a norm completion. This page constructs those pieces, explains fermionic locality, and gives the positive-contraction parametrization of gauge-invariant quasi-free states. The separate CCR algebra treatment supplies the universal Weyl construction for bosons.

Required background. Creation and annihilation as operators proves the fermionic norm bound. Operator algebra minimum explains states and C*-completion.

Finite-mode CAR generate a full matrix algebra

For a complex one-particle Hilbert space $\mathfrak h$, use annihilation symbols $a(f)$ antilinear in $f$ and relations

$$a(f)a(g)+a(g)a(f)=0,$$ $$a(f)a^*(g)+a^*(g)a(f)=\langle f,g\rangle I.$$

Take $\mathfrak h=\mathbb C^d$ with orthonormal basis $e_j$, and write $a_j=a(e_j)$. On $(\mathbb C^2)^{\otimes d}$ a concrete realization is

$$a_j=Z^{\otimes(j-1)}\otimes |0\rangle\langle1|\otimes I^{\otimes(d-j)}, \qquad Z=|0\rangle\langle0|-|1\rangle\langle1|.$$

The preceding parity strings make distinct modes anticommute. Omitting them would give commuting qubit lowering operators.

Every word can be reduced by the CAR to a linear combination of normally ordered monomials with each creator and annihilator appearing at most once. There are $2^{2d}$ such monomials, so the abstract algebra has dimension at most $4^d$. Conversely the displayed representation generates all matrix units: the commuting occupations $n_j=a_j^*a_j$ isolate basis configurations, while creators and annihilators move between them, with controllable signs. Its image is $M_{2^d}(\mathbb C)$, of dimension $4^d$. The dimension bounds prove that the abstract finite CAR algebra is exactly this full matrix algebra.

This is an algebra statement. A mixed state can have a reducible GNS representation of the same finite matrix algebra.

Infinitely many modes require completion

For finite-dimensional subspaces $E\subset F\subset\mathfrak h$, identify the generators labeled by $E$ with the same generators in $\operatorname{CAR}(F)$. The embedding is injective: the finite matrix algebra is simple, and a unital representation cannot have its entire algebra as kernel.

The union over finite-dimensional subspaces is therefore a consistent star algebra of bounded elements, with compatible C*-norms. Its norm completion is $\operatorname{CAR}(\mathfrak h)$. Equivalently it has the universal property for bounded families satisfying the CAR. In a separable space one may take an increasing sequence of finite spans of an orthonormal basis.

The Fock vacuum defines one state, not the algebra itself. Other states yield their own representations. Passing to infinitely many modes can produce inequivalent representations even though every finite subalgebra remains a matrix algebra.

Parity replaces naive tensor-product locality

The automorphism $\vartheta(a(f))=-a(f)$ defines even and odd elements. Monomials in disjoint orthogonal mode subspaces obey

$$AB=(-1)^{pq}BA$$

when their parities are $p,q\in\{0,1\}$. Indeed, moving one monomial through the other uses one sign per interchange, and all mixed contractions vanish by orthogonality.

Hence disjoint even observables commute, but disjoint odd fields anticommute. The even algebra is often the physical observable algebra when fermion parity is superselected. A Jordan–Wigner identification with qubit matrices preserves the abstract algebra while making some fermionic fields into strings across the chosen ordering. It does not make odd fields spatially local spin observables.

For lattice applications, support and parity must therefore be stated together. A locality theorem for commuting disjoint observable algebras cannot be transferred to arbitrary odd fields without a graded version.

Quasi-free states are fixed by a positive contraction

A gauge-invariant quasi-free state has vanishing expectations for monomials with unequal numbers of creators and annihilators, and satisfies

$$\omega_Q(a^*(f_1)\cdots a^*(f_n) a(g_n)\cdots a(g_1)) =\det[\langle g_i,Qf_j\rangle]_{i,j=1}^n.$$

Here $Q$ is a bounded operator with $0\leq Q\leq I$. Necessity of these bounds follows from positivity applied to $a^*(f)a(f)$ and from the CAR applied to $a(f)a^*(f)$.

For existence in finite dimension, diagonalize $Q$ with eigenvalues $p_j\in[0,1]$. On the occupation basis take the product density matrix with one-mode probabilities $1-p_j,p_j$. Direct expansion gives the determinant formula; the basis-independent expression also handles a non-diagonal $Q$.

In infinite dimension compress $Q$ to each finite-dimensional subspace. The finite quasi-free states agree on intersections because all their polynomial expectations use the same two-point function. They extend by continuity to a state of the CAR algebra. This establishes positivity, which would not follow merely by declaring a formal Wick rule.

At thermal equilibrium for a free one-particle Hamiltonian, $Q=(I+e^{\beta(h-\mu)})^{-1}$ is such a contraction. This formula can define an algebraic state even when no trace-class many-body Gibbs density exists in the vacuum Fock representation.

The bosonic comparison needs Weyl operators

For bosons, nonzero $a(f)$ is unbounded and cannot be an element of a C*-algebra. The Weyl formulation replaces fields by bounded unitaries and treats regularity as a representation condition. In particular, the abstract Weyl generators are not norm continuous in their labels. Fermionic $a(f)$, by contrast, satisfies $\|a(f)-a(g)\|=\|f-g\|$.

Quasi-free bosonic states also exist, but their covariance obeys a symplectic uncertainty inequality rather than the fermionic occupation bound $Q\leq I$. The two positivity tests should not be interchanged.

Exercises

Two modes. Evaluate $\omega_Q(n_1n_2)$ in an arbitrary orthonormal two-mode basis.

Solution

Reordering gives $n_1n_2=a_1^*a_2^*a_2a_1$. The determinant formula yields $Q_{11}Q_{22}-|Q_{12}|^2$, with $Q_{ij}=\langle e_i,Qe_j\rangle$. This is nonnegative since $Q\geq0$. Occupations factorize only when the off-diagonal correlation vanishes in that basis.

An impossible two-point function. Why cannot $Q=2I$ define a fermionic state?

Solution

For a unit $f$, positivity would give $\omega(a(f)a^*(f))\geq0$, whereas the CAR give $1-\omega(a^*(f)a(f))=1-2=-1$. Positivity of $Q$ alone is insufficient; the upper bound is essential.

References

• J. Dereziński, “Introduction to representations of the canonical commutation and anticommutation relations,” in Large Coulomb Systems, Lecture Notes in Physics 695, Springer, 2006, pp. 63–143, doi:10.1007/3-540-32579-4_3, author text.