Fermionic creation operator

This record is written in two registers, and the division is the point.

What no source establishes. No paper's subject is the fermionic creation operator. It was searched for — including in the mathematical-physics literature, where a paper about the operator algebra itself would most plausibly sit — and none was found. It is a textbook primitive: universally used, entirely correct, and not the result of anybody's paper. That sentence is the honest core of this record and is not replaced by the citation below.

What a source does state. Bravyi and Kitaev's Fermionic quantum computation is about a model of computation with local fermionic modes, not about this operator — but §2 of it states the definition, and a claim may be taken at the level a paper states it. Eq. (1) gives the action on occupation-number basis vectors,

$a_j\,|n_0,\dots,n_{j-1},1,n_{j+1},\dots\rangle = (-1)^{\sum_{s<j} n_s}\,|n_0,\dots,n_{j-1},0,n_{j+1},\dots\rangle,$

with $a_j$ annihilating any vector where mode $j$ is already empty, and $a^\dagger_j$ the Hermitian conjugate. Eq. (2) gives the algebra: $a_j a_k + a_k a_j = 0$, $a^\dagger_j a^\dagger_k + a^\dagger_k a^\dagger_j = 0$, and $a_j a^\dagger_k + a^\dagger_k a_j = \delta_{jk}$.

The sign is the content. The paper draws attention to it in its own words — the definition depends on the order of the modes — and that dependence is not bookkeeping. It is antisymmetry, it is why fermionic operators on disjoint sites fail to commute where qubit operators on disjoint sites succeed, and it is the reason every fermion-to-qubit encoding in this catalog exists at all. The operator alone is a definition; the sign it carries is what the rest of the shelf is about.

Why the record stops here. Real primary results exist one level up — about products of these operators, and about when a transformation of them is implementable — but none of them was read, so none is cited. A partial deepening that blurs which sentence rests on which source is worse than the stub it replaces, because a stub claims nothing.