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Countable-Mode CAR Has Inequivalent Irreducible Representations

Statement

Let πF be the countable-mode occupation representation on F∞ and let α be the particle-hole automorphism of Particle-Hole Exchange Is an Automorphism of Countable-Mode CAR. Then πF and πh:=πF∘α are irreducible representations of CARN, and they are not unitarily equivalent. Here two representations are unitarily equivalent if a unitary operator between their Hilbert spaces intertwines their actions. This proves existence of inequivalent irreducible representations; it does not classify all representations.

Facts & Assumptions

Given: The countable-mode CAR algebra, the occupation representation πF, and the particle-hole automorphism α.

[F1]

The occupation representation is irreducible, and no nonzero vector is killed by every creator cj† (The Countable-Mode CAR Fock Representation Is Irreducible).

[F2]

The particle-hole map α is a surjective ∗-automorphism with α(aj)=aj† (Particle-Hole Exchange Is an Automorphism of Countable-Mode CAR).

Proof

1.1givenF1F2algebra

Since α is a ∗-automorphism, πh=πF∘α is a representation. If a closed subspace were invariant under πh(CARN), surjectivity of α would make it invariant under πF(CARN); [F1] therefore makes it trivial or all of F∞. Thus πh is irreducible.

2.1givenF1F2algebra∎

Suppose a unitary U:F∞→F∞ intertwines πF with πh. Then for every j, Ucj=πh(aj)U=cj†U. Applying this to the vacuum Ω=e∅ gives cj†UΩ=UcjΩ=0 for every j. Since UΩ≠0, this contradicts the common-creator-kernel statement in [F1]. No such unitary exists.

Remarks

The explicit obstruction is the absence of a nonzero vector in the infinite occupation space annihilated by every creator. In finite mode number such a fully occupied vector exists, so this argument is specifically about countably infinitely many modes.

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