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Computational basis states have an extended stabilizer tableau

Statement

For n≥0 and a bit string x=(x0,…,xn−1)∈F2n, the basis state ∣x⟩ has rows indexed by a=1,…,n given by sj+1=(−1)xjZj+1 and dj+1=Xj+1 for each 0≤j<n. Their labels satisfy the extended tableau pairing conditions.

Facts & Assumptions

Given: The ordered computational basis of an n-qubit register and x=(x0,…,xn−1)∈F2n.

[F1]

Finite registers and their ordered computational bases: the ordered tensor register has the lexicographically ordered computational basis; for n=0 it is C with basis {1}.

[F2]

Big-endian encoding of ordered bit strings: the first wire is the most significant bit and coordinate xj occupies factor position j, which is Pauli wire j+1 by [F9].

[F3]

Pauli group on the empty register: when n=0, the ambient Pauli group is scalar and the empty generator list generates {I}.

[F4]

The computational basis is orthonormal: the computational basis of each factor Q2 is orthonormal.

[A1]

Phase-aware stabilizer tableau: a row (k,u,v) represents P(k,u,v)=ikXuZv and row pairings are computed by ω.

[F5]

Binary symplectic form: ω((u,v),(u′,v′))=u⋅v′+v⋅u′.

[F6]

Binary Pauli quotient isomorphism: a nonzero binary label cannot represent a scalar Pauli.

[F7]

Pauli matrix algebra and trace identities and The Pauli matrices: on the one-qubit basis, X∣b⟩=∣b⊕1⟩ and Z∣b⟩=(−1)b∣b⟩, and both are Hermitian involutions.

[F8]

Pauli commutator is symplectic pairing: Pauli rows commute exactly when their label pairing is zero.

[F9]

Phase aware n qubit pauli group: Pauli wire a∈{1,…,n} is tensor factor a−1; its Xa and Za operators act on that factor.

[F10]

Ordered tensor registers, their inner product, and wire permutations: elementary tensors of factor basis vectors form a basis, and their inner product is the product of factor inner products. For n=0 the space is C, with ⟨1,1⟩=1.

Proof

Given: The data in the Statement.

1.1F1F2F4F7F9F10given

By [F4] and [F10], ⟨∣x⟩,∣y⟩⟩=∏j<nδxjyj=δxy for computational words x,y, including the empty product 1 when n=0. Thus this basis is orthonormal and ∣x⟩ is normalized. For each 0≤j<n, [F7] and [F9] give Zj+1∣x⟩=(−1)xj∣x⟩, so sj+1=(−1)xjZj+1 fixes ∣x⟩. In the same basis each sj+1 is a real diagonal involution and each dj+1=Xj+1 flips bit j, an involutive permutation; hence all these rows are Hermitian involutions.

2.1A1F3F5F6F8F9step 1.1algebra∎

For 0≤i,j<n, the labels of sj+1 are (0,ej) and of di+1=Xi+1 are (ei,0). Therefore ω(di+1,sj+1)=ei⋅ej=δij, which is the required pairing δi+1,j+1; pairings within either family vanish by [F5]. By [F8], the stabilizer rows commute. Since they are involutions, every generated element is a product over a subset of the rows. A nonempty subset has nonzero label by independence and the label homomorphism [F6], so is not scalar; the empty subset gives I. Thus the generated group excludes −I. For n=0, the empty row families generate {I} by [F3], and all pairing and independence conditions hold vacuously.

Sources

Aaronson–Gottesman §III, PDF pp. 3–4, gives the identity tableau for computational-basis input. De Wolf §20.1, PDF p. 185, notes that every basis state is a stabilizer state: https://arxiv.org/pdf/quant-ph/0406196; https://arxiv.org/pdf/1907.09415.

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