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Stabilizer Circuits and Gottesman–Knill Simulation

1 · Prerequisites

2 · Summary

Objective

Develop the finite phase-aware stabilizer formalism for ordered qubit registers. Prove the stabilizer average and code-dimension formula, characterize deterministic and fair Pauli measurement branches, and show that exact signed-tableau updates reproduce the projective instrument, including impossible outcomes. The final simulation result gives explicit bit and workspace bounds for adaptive circuits generated by H, S, S†, and CNOT from computational-basis inputs. The current QC09 syntax, unitary-denotation, and resource bodies provide the circuit interfaces; this page states the Pauli-specific finite-history tree used by the theorem. The proof uses instrument branch laws directly and does not require uniqueness of a POVM induced by a general instrument.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Pauli measurement branch semantics

Definition

Let Hn=(C2)⊗n be an ordered n-qubit register, let ρ∈M2n(C) be a density operator, and let M∈M2n(C) be a Hermitian Pauli observable with M2=I. For b∈{0,1} define

Pb:=I+(−1)bM2,pb:=Tr⁡(ρPb),Ib(ρ):=PbρPb.

The outcome b has probability pb. Its conditional density operator is ρb:=Ib(ρ)/pb when pb>0; it is ⊥ (no assigned conditional density operator) when pb=0. The branch map Ib is the projective measurement instrument for the (−1)b eigenspace of M. A non-Hermitian phase-aware Pauli is outside this definition. For n≥1, Hermiticity of P(k,x,z)=ikXxZz means k≡x⋅z(mod2); in particular Y=iXZ is admitted. For n=0, the admitted observables are IH0 and −IH0.

The hypotheses and meanings of density operator, Pauli observable, projective measurement, POVM outcome law, and instrument are the cited suppliers. The local projector calculation is Pauli measurement projectors and total probability ↗.

Sources

Ronald de Wolf, Quantum Computing: Lecture Notes, §20.2, PDF pp. 186–188, defines the two Pauli eigenspace projectors and their outcome probabilities: https://arxiv.org/pdf/1907.09415.

John Watrous, Understanding Quantum Information and Computation, Lesson 14, §14.1, printed pp. 405–407 (PDF pp. 411–413), treats a Hermitian Pauli tensor as a two-outcome observable: https://jhwatrous.github.io/UQIC.pdf.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Pauli measurement projectors and total probability

Statement

Let H be a nonzero finite-dimensional complex inner-product space, let ρ be a density operator on H, and let M be a Hermitian Pauli observable on H with M2=IH. For b∈{0,1} set

Pb:=IH+(−1)bM2,pb:=Tr⁡(ρPb).

Then P0,P1 are complementary orthogonal projections and p0,p1≥0, with p0+p1=1.

Facts & Assumptions

Given: A Hermitian Pauli observable M with M2=I and a density operator ρ on the same finite-dimensional register.

[F1]

The setup defines Pb=(IH+(−1)bM)/2 and pb=Tr⁡(ρPb).

[A1]

Density operator: ρ is positive semidefinite and Tr⁡(ρ)=1.

[F2]

Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes: matrix products have entries (AB)ij=∑kaikbkj, and I has entries δij.

[F3]

For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA): for compatible finite matrices, Tr⁡(AB)=Tr⁡(BA), including square matrices of the same size.

[F5]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: trace is basis-independent and equals the matrix trace in any ordered basis.

Proof

Given: The data in the Statement.

1.1givenF1F2algebra

For s=(−1)b, Pb†=Pb because s is real and M†=M. In any ordered basis write M=(mij). By [F2], (Pb2)ij=14∑k(δik+smik)(δkj+smkj)=14(δij+2smij+(M2)ij). Since M2=IH, this equals (Pb)ij. Thus both Pb are self-adjoint idempotents.

2.1F1step 1.1algebra

Direct multiplication gives P0+P1=I and P0P1=(I−M2)/4=0=P1P0. Hence their ranges are orthogonal and sum to the full register.

2.2A1step 1.1algebra

Put Qb=PbρPb. Self-adjointness of Pb and ρ gives ⟨Qbu,v⟩=⟨ρPbu,Pbv⟩=⟨u,Qbv⟩, so Qb is self-adjoint. For every vector v, ⟨Qbv,v⟩=⟨ρPbv,Pbv⟩≥0, using the root-linear-first inner-product convention. Therefore Qb is positive semidefinite.

3.1A1F1F3F4F5step 2.1step 2.2∎

By cyclicity [F3] and Pb2=Pb, pb=Tr⁡(ρPb)=Tr⁡(PbρPb)=Tr⁡(Qb). In the orthonormal eigenbasis supplied by [F4], the diagonal entries of Qb are nonnegative; [F3] and [F5] therefore give pb=Tr⁡(Qb)≥0. Linearity of the finite diagonal sum and P0+P1=I give p0+p1=Tr⁡(ρ)=1.

Sources

The two projectors and the two-outcome probabilities are treated in de Wolf, §20.2, PDF pp. 186–188, and Watrous, Lesson 14 §14.1, printed pp. 405–407 (PDF pp. 411–413). The finite-dimensional trace manipulations above are proved locally from the cited matrix trace suppliers.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A maximal stabilizer label space is self-dual

Statement

Let V=F22n have the standard symplectic form ω((x,z),(x′,z′))=x⋅z′+z⋅x′. If L≤V is isotropic and r=dim⁡L, then

dim⁡L⊥=2n−r.

Consequently r≤n, and if r=n then

L⊥:={v∈V:ω(v,ℓ)=0 for every ℓ∈L}=L.

Facts & Assumptions

Given: A finite-dimensional space V=F22n, the stated form, and an isotropic subspace L≤V of dimension r.

[F1]

For n≥1, Binary symplectic form supplies the displayed coordinate formula and bilinearity, alternation, and nondegeneracy. For n=0, V={0} and the displayed form is the unique zero form.

[F2]

Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T: for a linear map on finite-dimensional spaces, dimension of the domain is nullity plus rank.

[F3]

If dim⁡FV=n and U is a linear subspace of V, then U is finite-dimensional, dim⁡FU≤n, and dim⁡FU=n if and only if U=V: every finite independent list extends to a basis in finite dimension without choice, and a subspace with the same dimension as its ambient space equals the ambient space.

Proof

Given: The data in the Statement.

1.1givenF1algebra

If n=0, then V=L=L⊥={0} and r=0, so dim⁡L⊥=0=2n−r, r≤n, and L⊥=L. Hence the result holds.

1.2F1givenalgebra

Assume now n≥1. If v=(x,z)∈V pairs to zero with every vector, pairing with (ej,0) and (0,ej) gives zj=xj=0 for every j. More generally, for any linear functional ϕ on V, set aj=ϕ(ej,0) and bj=ϕ(0,ej). Then v=(b,a) satisfies ω(v,(x′,z′))=b⋅z′+a⋅x′=ϕ(x′,z′) by linearity. Thus v↦ω(v,⋅) is both injective and onto.

2.1F2F3step 1.2choose

Put r=dim⁡L and choose a basis ℓ1,…,ℓr of L. Extend it to a basis of V using [F3]. For any y∈F2r, define a linear functional on this basis by ϕ(ℓj)=yj and by zero on the added basis vectors. Step 1.2 supplies v∈V with ω(v,⋅)=ϕ, so T:V→F2r, T(v)=(ω(v,ℓj))j=1r, is onto. Its kernel is L⊥, and [F2] gives dim⁡L⊥=2n−r. This includes r=0, where both the basis and F2r are empty or zero-dimensional.

3.1givenF3step 2.1algebra

For n≥1, isotropy gives L⊆L⊥. Comparing dimensions with step 2.1 gives r≤2n−r, hence r≤n. If r=n, both subspaces have dimension n, so [F3] gives L=L⊥. This argument uses only finite basis extension and no choice principle.

4.1step 1.1step 1.2step 3.1∎

Step 1.1 proves the result for n=0, and step 1.2 and step 2.1 and step 3.1 prove it for n≥1. These cases exhaust all dimensions in the Statement.

Sources

The symplectic label space and its self-orthogonality are treated in de Wolf §20.1 and Aaronson–Gottesman §II. The dimension argument above is supplied locally from the current batch-15 symplectic-form definition; that supplier is present as a draft body and is not independently accepted.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Stabilizer subgroup

Definition

For an ordered register of n qubits, a stabilizer subgroup is an abelian subgroup S of the Hermitian n-qubit Pauli operators such that −I∉S. The binary label of a phase-aware Pauli P(k,x,z) is (x,z)∈F22n. Define the rank of S by

rank⁡(S):=dim⁡F2span⁡{(x,z):P(k,x,z)∈S}.

The binary label convention and commutation form are supplied by the cited Pauli items. Stabilizer labels form an isotropic space of the same rank ↗ proves that this label span is isotropic, its dimension is independent of any generating set, and ∣S∣=2rank⁡(S). Matrix multiplication gives the phase-aware Pauli group operation by Pauli product and adjoint. For n=0, the Pauli group is the empty-register convention of Pauli group on the empty register.

Sources

De Wolf §20.1, PDF pp. 185–186, and Watrous Lesson 14 §§14.1 and 14.3, printed pp. 403–404 and 415–416 (PDF pp. 409–410 and 421–422), describe commuting Pauli stabilizer groups, independence, and exclusion of −I: https://arxiv.org/pdf/1907.09415; https://jhwatrous.github.io/UQIC.pdf.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Pauli projectors define a finite measurement instrument

Statement

Let H be a nonzero finite-dimensional complex inner-product space and let M be a Hermitian Pauli observable on H with M2=IH. Put Pb=(IH+(−1)bM)/2 and define Ib(X)=PbXPb for b∈{0,1}. Then the maps (I0,I1) form a finite instrument from H to H. For every density operator ρ on H, the outcome weight of branch b is Tr⁡(ρPb).

Facts & Assumptions

Given: The data in the Statement.

[F1]

The setup gives Ib(X)=PbXPb and the projectors Pb.

[F2]

Pauli measurement projectors and total probability: Pb†=Pb, Pb2=Pb, P0+P1=I, and P0P1=P1P0=0.

[F3]

Completely positive maps: complete positivity means that for every finite ancilla dimension r≥1, the block amplification Ib(r) preserves nonnegative operators.

[F4]

Finite ancilla amplification is a well-defined block map: if X=[Xij] is the block matrix of an operator on H⊗Cr, then Ib(r)(X)=[Ib(Xij)].

[F5]

Ordered tensor registers, their inner product, and wire permutations: the ordered product inner product on H⊗Cr makes its ancilla basis orthonormal.

[F6]

Non-negative and positive operators: a nonnegative operator is self-adjoint and has nonnegative quadratic form on every vector.

[F7]

For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA): trace is cyclic for compatible finite-dimensional endomorphisms.

[F8]

Finite instruments and conditional states: a finite family of completely positive branch maps is an instrument when the sum of their output traces equals the input trace for every operator; its branch weight on a density operator ρ is Tr⁡(Ib(ρ)).

Proof

Given: The data in the Statement.

1.1F1algebra

Each Ib is complex-linear because matrix multiplication distributes over finite sums and scalars: Pb(aX+cY)Pb=aPbXPb+cPbYPb.

1.2F2F4F5F6givenalgebra

Fix r≥1, take an orthonormal basis (ej)j<r of the ancilla, and write X=[Xij] for a nonnegative operator on H⊗Cr. Self-adjointness of X by [F6] gives Xij†=Xji: for ξ,η∈H, ⟨Xijξ,η⟩=⟨X(ξ⊗ej),η⊗ei⟩=⟨ξ⊗ej,X(η⊗ei)⟩=⟨ξ,Xjiη⟩ by the product pairing [F5]. Consequently the blocks PbXijPb satisfy the same adjoint relation, so Ib(r)(X) is self-adjoint.

1.3F1F2F7algebra

For every operator X, cyclicity [F7] and idempotence [F2] give Tr⁡(I0(X))+Tr⁡(I1(X))=Tr⁡(XP0)+Tr⁡(XP1)=Tr⁡(X(P0+P1))=Tr⁡(X). For a density operator ρ, the same calculation gives Tr⁡(Ib(ρ))=Tr⁡(ρPb), the branch trace.

2.1F2F3F4F5F6step 1.2algebra

Write v=∑j<rvj⊗ej and w=∑j<rPbvj⊗ej. By [F4] and the product pairing [F5], ⟨Ib(r)(X)v,v⟩=∑i,j<r⟨PbXijPbvj,vi⟩=∑i,j<r⟨XijPbvj,Pbvi⟩=⟨Xw,w⟩≥0. Thus the ampliation preserves nonnegative operators. Since r was arbitrary, Ib is completely positive by [F3].

3.1F8step 2.1step 1.3

The two maps are completely positive by step 2.1 and their total output trace equals the input trace by step 1.3. They therefore form a finite instrument by [F8].

4.1F8step 1.3step 3.1∎

By the branch-weight clause of [F8], the outcome weight on ρ is Tr⁡(Ib(ρ)). Step 1.3 identifies this with Tr⁡(ρPb).

Sources

Watrous Lesson 14 §14.1, PDF pp. 405–407, treats the two Pauli branches as a projective measurement; de Wolf §20.2, PDF pp. 186–188, gives their branch maps and probabilities. Complete positivity and trace preservation are verified locally from the block definition and the two projector identities.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Positive-probability Pauli branches are density operators

Statement

Let H be a nonzero finite-dimensional complex inner-product space, let ρ be a density operator on H, and let M be a Hermitian Pauli observable on H with M2=IH. For b∈{0,1} put Pb=(IH+(−1)bM)/2 and pb=Tr⁡(ρPb). If pb>0, then the normalized branch

ρb:=PbρPbpb

is a density operator.

Facts & Assumptions

Given: The data in the Statement.

[F1]

The setup defines Pb=(IH+(−1)bM)/2, pb=Tr⁡(ρPb), and ρb=PbρPb/pb when pb>0.

[F2]

Pauli measurement projectors and total probability: Qb=PbρPb is nonnegative and Tr⁡(Qb)=pb.

[F3]

Trace is a linear functional on Mn(F): trace respects multiplication by a scalar.

[F4]

Density operator: a density operator is a nonnegative endomorphism of trace one on a nonzero finite-dimensional complex-inner-product space.

Proof

Given: The data in the Statement.

1.1F2givenalgebra

Set Qb=PbρPb. By [F2], Qb is nonnegative and has trace pb. Since pb>0, the scalar 1/pb is positive. Hence, for every v, ⟨(Qb/pb)v,v⟩=(1/pb)⟨Qbv,v⟩≥0.

2.1F2F3F4step 1.1algebra

By [F3] and [F2], Tr⁡(Qb/pb)=pb−1Tr⁡(Qb)=1. Thus Qb/pb is a density operator by [F4].

3.1F1step 2.1∎

The conditional state in [F1] is exactly Qb/pb, so it is the density operator established in step 2.1.

Sources

Watrous Lesson 14 §14.1, printed pp. 405–407 (PDF pp. 411–413), defines the projective conditional state; de Wolf §20.2, PDF pp. 186–188, gives the Pauli branch update. The positivity and trace-one check is proved locally.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Zero-probability Pauli measurement branch

Statement

Let H be a nonzero finite-dimensional complex inner-product space, let ρ be a density operator on H, and let M be a Hermitian Pauli observable on H with M2=IH. For b∈{0,1} put Pb=(IH+(−1)bM)/2 and pb=Tr⁡(ρPb). If pb=0, then PbρPb=0. The outcome has probability zero and, under the conditional-state convention, has no normalized conditional state.

Facts & Assumptions

Given: The H,ρ,M,b,Pb,pb data of the Statement, with pb=0.

[F1]

The setup defines Pb=(IH+(−1)bM)/2 and pb=Tr⁡(ρPb).

[F2]

Pauli measurement projectors and total probability: Pb∗=Pb and Pb2=Pb.

[A1]

Density operator: ρ is a nonnegative endomorphism on a nonzero finite-dimensional complex inner-product space.

[F3]

For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA): trace is cyclic for compatible finite-dimensional endomorphisms.

[F5]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: trace is basis-independent and equals the matrix trace in any ordered basis.

Proof

Given: The data in the Statement.

1.1givenF2A1algebra

Set Qb=PbρPb. For every v, self-adjointness of Pb in [F2] gives ⟨Qbv,v⟩=⟨ρPbv,Pbv⟩≥0 by [A1]. Thus Qb is nonnegative.

1.2givenF1F2F3F4F5algebra

By [F1], [F2], and cyclicity [F3], Tr⁡(Qb)=Tr⁡(PbρPb)=Tr⁡(ρPb2)=pb=0. By [F4], choose an orthonormal eigenbasis of Qb whose eigenvalues are nonnegative; [F5] expresses its trace as their finite sum. A finite sum of nonnegative reals is zero only when each term is zero, so Qb=0.

2.1givenstep 1.2algebra∎

The branch output is therefore zero. Normalization would require dividing it by pb=0, which is undefined; the conditional-state convention assigns no state to this zero-probability outcome.

Sources

De Wolf §20.3, PDF pp. 187–188, separates deterministic outcomes from random outcomes; Watrous Lesson 14 §14.1, printed pp. 405–407 (PDF pp. 411–413), specifies the two projective branches. The zero-trace implication is proved locally using the finite spectral theorem.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Finite adaptive Pauli-measurement circuit tree

Definition

Fix an ordered n-qubit register and a finite gate alphabet as in Finite circuit syntax and ordered wire labels. A finite adaptive Pauli-measurement circuit tree is a finite rooted tree whose leaves are terminal and whose nonterminal nodes are of one of these two types:

  1. A gate node is labelled by a gate occurrence on the ordered register and has one successor. Its state update is conjugation by the global unitary assigned to that occurrence by Denotational semantics of a finite layered circuit.
  2. A measurement node is labelled by a Hermitian Pauli observable M and has two outgoing edges labelled b=0,1. On input density operator ρ, edge b has weight pb=Tr⁡(ρPb) and unnormalized output PbρPb, where Pb=(I+(−1)bM)/2 as in Pauli measurement branch semantics. Its child state is PbρPb/pb only if pb>0; an edge of weight zero is impossible and has no conditional state.

The observed measurement transcript is the ordered list of measurement-edge labels on a root-to-leaf path. The tree itself specifies every later instruction from that observed history, so its classical control is deterministic. A path whose edges all have positive weights has probability equal to their product. A path passing through a zero-weight edge has probability zero, and its descendants are not evaluated because no conditional state is assigned. A leaf outputs its terminal state; a ray is assigned only on a positive-probability path whose input and gate/measurement updates remain pure.

The complementary-projector and total-probability facts for each measurement node are supplied by Pauli measurement projectors and total probability. This definition does not identify an instrument with a POVM or require uniqueness of such a POVM.

Sources

De Wolf §20.3, PDF pp. 187–188, defines measurement-controlled Clifford circuits with feed-forward from earlier outcomes. Watrous Lesson 14 §14.1, printed pp. 405–407 (PDF pp. 411–413), defines the two Pauli measurement branches. The local tree fixes the finite branch and zero-probability conventions used here.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Stabilizer state

Definition

Let S be a rank-n stabilizer subgroup on an ordered n-qubit register Hn=(C2)⊗n. A pure stabilizer state is the unique ray in the common +1 fixed space

C(S):={ψ∈Hn:gψ=ψ for every g∈S}.

The cited projector-rank theorem proves that this common space is nonzero and one-dimensional, so the ray is well defined. A normalized representative may be used for probabilities, but the state itself is the ray, not a chosen vector.

Sources

De Wolf §20.1, PDF p. 185, and Watrous Lesson 14 §14.3, printed pp. 415–416 (PDF pp. 421–422), define a stabilizer state from a full-rank commuting Pauli subgroup and its common +1 eigenspace: https://arxiv.org/pdf/1907.09415; https://jhwatrous.github.io/UQIC.pdf.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Stabilizer labels form an isotropic space of the same rank

Statement

Let S be a stabilizer subgroup on n≥0 qubits and let λ(P(k,x,z))=(x,z). Then λ∣S is injective, its image L is an isotropic F2-subspace, the rank dim⁡F2L depends only on S and not on a generating list, and ∣S∣=2dim⁡F2L.

Facts & Assumptions

Given: A stabilizer subgroup S and the phase-aware Pauli convention P(k,x,z)=ikXxZz.

[F1]

Stabilizer subgroup: S is an abelian subgroup of Hermitian Paulis and −I∉S.

[F2]

Binary Pauli label and quotient: the binary label is λ(P(k,x,z))=(x,z) and forgets only scalar phase.

[F3]

Binary Pauli quotient isomorphism: the label map is a group homomorphism, and its kernel on phase-aware Paulis is the scalar center.

[F4]

Pauli commutator is symplectic pairing: two Pauli operators commute exactly when their labels have symplectic pairing zero.

[F5]

Hermitian Pauli phase criterion: P(k,x,z) is Hermitian exactly when k≡x⋅z(mod2); every Hermitian Pauli squares to I.

[F7]

A d-dimensional vector space over a field with q elements has exactly qd elements: an r-dimensional vector space over a field of order q has qr elements.

[F8]

Phase aware n qubit pauli group and Pauli group on the empty register: the ambient Pauli set is finite, so every subgroup S is finite.

[F9]

Pauli group on the empty register: for n=0, the Hermitian scalar Paulis are I and −I, and the binary-label space has one element.

Proof

Given: The data in the Statement.

1.1F1F9given

If n=0, then S={I}: a subgroup contains I and its only other possible Hermitian scalar Pauli is −I, which is excluded. Thus λ∣S is injective, its image is the zero subspace of dimension 0, this dimension is independent of any generating list, and ∣S∣=1=20.

1.2F1F2F3F6given

Assume now n≥1. The identity of S has zero label. If P,Q∈S, then PQ∈S and λ(PQ)=λ(P)+λ(Q) by [F3]. Thus L=λ(S) contains zero and is closed under addition; over F2, closure under the only scalars 0,1 follows, so L is a subspace.

1.3F1F3F5given

Suppose P,Q∈S have the same label. By the homomorphism in [F3], λ(P−1Q)=0, so P−1Q is a scalar Pauli. Since Hermitian Paulis square to I by [F5], P−1=P; as P,Q commute and are Hermitian, P−1Q=PQ is Hermitian. The phase criterion [F5] then makes this scalar equal to I or −I. The latter is excluded by [F1], hence P=Q and λ∣S is injective.

2.1F1F4step 1.2algebra

For n≥1, any two labels in L come from commuting elements of the abelian group S, so [F4] gives symplectic pairing zero. Bilinearity then makes every pair in their span pair to zero; hence L is isotropic. The image L is defined from S alone, so its dimension is unchanged by replacing one generating list of S by another.

2.2F6F7F8step 1.3

For n≥1, the map λ∣S:S→L is onto by the definition of L and injective by step 1.3, so it is a bijection of finite sets by [F8]. If r=dim⁡F2L, [F6] and [F7] give ∣L∣=2r. Therefore ∣S∣=∣L∣=2r, as claimed.

3.1step 1.1step 1.2step 2.2∎

Step 1.1 proves the claim for n=0, and step 1.2 and step 1.3 and step 2.1 and step 2.2 prove it for n≥1. These cases exhaust all n≥0.

Sources

De Wolf §20.1, PDF pp. 185–186, describes the finite Pauli group, commuting independent stabilizers, and the rank-n state condition. Aaronson–Gottesman §II, PDF pp. 2–4, gives the binary phase-aware Pauli row representation. The current batch-15 label-quotient and commutator supplier bodies are listed in the Facts & Assumptions above; this draft does not claim their independent acceptance.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Stabilizer average is the fixed-space projector

Statement

For a stabilizer subgroup S on n qubits, the finite operator average

ΠS:=1∣S∣∑g∈Sg

is self-adjoint and idempotent, and its image is exactly the common +1 fixed space C(S). If this image is the span of a unit vector ψ, then ΠS=∣ψ⟩⟨ψ∣ under the root-linear-first Dirac convention.

Facts & Assumptions

Given: A finite abelian stabilizer subgroup S of Hermitian Pauli operators.

[F1]

Stabilizer subgroup: each g∈S is Hermitian and the group is closed under multiplication and inverses.

[F2]

Stabilizer labels form an isotropic space of the same rank: S is finite, ∣S∣=2rank⁡(S), and its identity occurs once in the sum.

[F3]

Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes: multiplication is associative and the identity acts on both sides.

[A1]

Dirac Notation Conversion: in the root-linear-first convention, ∣x⟩⟨y∣ acts by z↦⟨z,y⟩rootx.

Proof

Given: The data in the Statement.

1.1F1F2F3given

Since every g is Hermitian, ΠS†=ΠS. In the product ΠS2=∣S∣−2∑g,h∈Sgh, each k∈S occurs exactly ∣S∣ times: for each g there is exactly one h=g−1k∈S. Therefore ΠS2=∣S∣−1∑k∈Sk=ΠS.

2.1F1F2step 1.1

For any s∈S, left multiplication g↦sg permutes S, so sΠS=ΠS. Hence every vector in im⁡ΠS is fixed by every s∈S. Conversely if sv=v for all s∈S, then ΠSv=∣S∣−1∑s∈Ssv=v. Thus im⁡ΠS=C(S).

3.1A1step 1.1step 2.1algebra∎

If C(S)=Cψ with ∥ψ∥=1, then ΠS is a self-adjoint idempotent onto that line. Its kernel is orthogonal to its image: for u=ΠSw and ΠSv=0, ⟨u,v⟩=⟨w,ΠSv⟩=0. Therefore it is the orthogonal projection onto Cψ. By [A1] that projection is ∣ψ⟩⟨ψ∣.

Sources

Watrous Lesson 14 §14.3, printed pp. 420–421 (PDF pp. 426–427), derives the stabilizer average as the code-space projector. De Wolf §20.3, PDF pp. 187–188, treats the rank-n stabilizer projector. The group-average proof above supplies the full multiplication count locally.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Phase-aware stabilizer tableau

Definition

For n≥0, x,z∈F2n, and k∈Z/4, write P(k,x,z)=ikXxZz, with the fixed tensor order and Xx=⨂j<nXxj, Zz=⨂j<nZzj. A row (k,x,z) is Hermitian when k≡x⋅z(mod2); it stores the exact Pauli phase as well as its binary label (x,z).

For n=0, use the scalar Pauli convention of Pauli group on the empty register; the generator list is empty.

A generator tableau is an ordered list of n Hermitian rows whose Pauli operators commute, have linearly independent labels, and generate a subgroup not containing −I. It represents the unique common +1 ray when the rows stabilize a pure state. An extended tableau also stores Hermitian destabilizer rows d1,…,dn and stabilizer rows s1,…,sn with

ω(di,sj)=δij,ω(di,dj)=0,ω(si,sj)=0.

The exact phase and symplectic conventions are those of the cited Pauli suppliers. The generator-only simulator uses the sj rows; the extended pairing is retained as a second representation invariant.

Sources

Aaronson–Gottesman §III, PDF pp. 3–5, defines the phase-bit tableau and its destabilizer/stabilizer pairing. De Wolf §20.3, PDF pp. 187–188, gives the generator-only stabilizer representation used by the simulator: https://arxiv.org/pdf/quant-ph/0406196; https://arxiv.org/pdf/1907.09415.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

An anticommuting stabilizer measurement is fair

Statement

Let ρ=∣ψ⟩⟨ψ∣ be a pure stabilizer state density operator, and let M be a Hermitian Pauli observable that anticommutes with one of its stabilizer generators. Then Tr⁡(P0ρ)=Tr⁡(P1ρ)=1/2.

Facts & Assumptions

Given: A rank-n stabilizer subgroup S fixing the ray of ψ, a normalized representative ψ, and a Hermitian Pauli M with gM=−Mg for some g∈S.

[F1]

Stabilizer state: every g∈S fixes ψ.

[F2]

Hermitian Pauli phase criterion: a Hermitian Pauli is self-adjoint and squares to I.

[F3]

Pauli measurement branch semantics: Pb=(I+(−1)bM)/2, and its probability on a pure state is pb=Tr⁡(∣ψ⟩⟨ψ∣Pb).

[F4]

Pure-state projective branch probability is a squared norm: for a unit vector ψ and orthogonal projection P, Tr⁡(∣ψ⟩⟨ψ∣P)=∥Pψ∥2.

[F5]

Ray-to-projector is well-defined and bijective: the projector of a unit representative of a pure-state ray is a positive trace-one density operator.

[F6]

Pauli measurement projectors and total probability: P0,P1 are orthogonal complementary projections.

Proof

Given: The data in the Statement.

1.1givenF1F2F3algebra

Since g=g† and g2=I by [F2], it is unitary. From gψ=ψ and gMg=−M, we get gPbg=P1−b and gPbψ=P1−bψ. Hence ∥Pbψ∥=∥P1−bψ∥.

2.1F3F4F5F6step 1.1algebra

By [F5], the unit ray projector is a density operator, so its branch law [F3] and the norm identity [F4] give pb=∥Pbψ∥2=∥P1−bψ∥2=p1−b. The complementary orthogonal projections in [F6] satisfy ∥P0ψ∥2+∥P1ψ∥2=∥ψ∥2=1, so equal probabilities must both be 1/2.

3.1step 2.1∎

This proves that each outcome bit has probability 1/2.

Sources

De Wolf §20.3, PDF p. 188, proves equality of the two outcome probabilities by conjugating with an anticommuting stabilizer. Aaronson–Gottesman §III, PDF pp. 4–5, gives the corresponding tableau measurement case.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

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.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Phase-aware tableau updates for Clifford generators

Statement

Write a Hermitian row as P(k,x,z)=ikXxZz, with k∈Z/4. In this lemma's formulas, label coordinates are numbered by Pauli wires a=1,…,n in tensor order: the coordinate denoted xa (respectively za) here is the coordinate denoted xa−1 (respectively za−1) in Phase-aware stabilizer tableau. This reindexing changes no operator or phase. For wire a with 1≤a≤n, conjugation by Ha, Sa, and Sa† updates the row by, respectively,

Ha:(k,xa,za)↦(k+2xaza,za,xa), Sa:(k,xa,za)↦(k+xa,xa,za⊕xa), Sa†:(k,xa,za)↦(k−xa,xa,za⊕xa).

For distinct wires c,t, CNOT⁡c→t updates xt↦xt⊕xc, zc↦zc⊕zt, with all other label coordinates and k unchanged. Applying these conjugations to every row preserves the generator tableau and, when present, the extended pairing invariant.

Facts & Assumptions

Given: A valid phase-aware tableau on n≥1 wires, an index 1≤a≤n for a one-qubit gate, or distinct indices c,t for CNOT⁡c→t.

[F1]

Phase-aware stabilizer tableau: a row represents ikXxZz and commutation is recorded by ω. Its zero-based label coordinates are reindexed in tensor order by the one-based wire names used in this lemma.

[A1]

Hadamard pauli action and Phase gate pauli action: the exact H and S conjugation updates are the corresponding formulas in the Statement.

[F2]

Exact one-qubit gates and rotations: H,S,S† are unitary.

[F3]

CNOT Pauli action: CNOT is unitary and conjugates Xc,Xt,Zc,Zt to XcXt,Xt,Zc,ZcZt, respectively.

[F4]

Binary symplectic form: ω((x,z),(x′,z′))=x⋅z′+z⋅x′ over F2.

Proof

Given: The data in the Statement.

1.1A1F1F2algebra

For H, [A1] gives the exact coordinate swap and phase 2xaza. For S, SXS†=iXZ and SZS†=Z give the stated shear and phase increment. The second identity and unitarity give S†ZS=S†(SZS†)S=Z. Then X=S†(SXS†)S=i(S†XS)Z, so it follows that S†XS=−iXZ; hence the inverse shear decrements the phase by xa. All other tensor factors are unchanged.

1.2F1F3algebra

For CNOT, conjugation sends Xc↦XcXt, Xt↦Xt, Zc↦Zc, and Zt↦ZcZt by [F3]. The image of the ordered Xx part is a product of X factors and the image of Zz is a product of Z factors; the order XxZz is retained, so no X crosses a Z and k is unchanged. This gives the stated toggles.

2.1F4step 1.1step 1.2algebra

For H, swapping xa,za in both labels leaves their contribution xaza′+zaxa′ unchanged. For S, replacing za,za′ by za+xa,za′+xa′ adds xaxa′+xaxa′=0. For CNOT, the control and target contributions change by, respectively, xczt′+ztxc′ and xczt′+ztxc′, whose sum is zero in F2. Thus each update preserves the displayed form [F4] and all row pairings.

3.1F1F2F3F4step 2.1algebra∎

Each label update is invertible (the H swap is its own inverse and the S and CNOT shears are invertible over F2); conjugation also preserves Hermiticity and fixes −I. Consequently row independence, commutation, exclusion of −I, and every extended pairing remain valid.

Sources

The one- and two-qubit Pauli conjugation tables appear in de Wolf §20.2, PDF pp. 186–187, and Aaronson–Gottesman §III, PDF pp. 4–5. The actual H, S, CNOT, and binary symplectic-form supplier bodies are present in batch 15 and are cited above; they remain draft, not accepted.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Tableau row reduction returns the deterministic Pauli sign

Statement

Let n≥0 and let g1,…,gn be the independent signed rows of a rank-n stabilizer tableau with unit stabilizer representative ψ, and let a Hermitian Pauli M commute with every row. Binary Gaussian elimination finds the unique c∈F2n whose row-label sum is λ(M). The phase-aware product G=∏j=1ngjcj, taken in increasing row order, is exactly (−1)bM for the deterministic measurement bit b.

Facts & Assumptions

Given: A valid generator tableau and a Hermitian Pauli M commuting with its stabilizer subgroup.

[F1]

Phase-aware stabilizer tableau: tableau rows store exact phase-aware Paulis and independent binary labels.

[F2]

Stabilizer labels form an isotropic space of the same rank: the row labels span an isotropic space L, the label map is injective on the stabilizer group, and the row labels form a basis of L in a rank-n tableau.

[F3]

A maximal stabilizer label space is self-dual: for this rank-n stabilizer, L⊥=L.

[F4]

Pauli commutator is symplectic pairing: commutation is equivalent to zero label pairing.

[F5]

Hermitian Pauli phase criterion: P(k,x,z) is Hermitian exactly when k≡x⋅z(mod2), and a Hermitian Pauli squares to I.

[F6]

Stabilizer state: every stabilizer row fixes the represented ray.

[F7]

Pauli product and adjoint: phase-aware row multiplication uses the exact formula P(k,x,z)P(k′,x′,z′)=P(k+k′+2(z⋅x′),x+x′,z+z′), with the binary dot product evaluated as 0 or 1 before doubling in Z/4Z.

[F8]

Pauli measurement branch semantics: outcome b has projector Pb onto the (−1)b eigenspace and probability Tr⁡(∣ψ⟩⟨ψ∣Pb).

[F9]

Pure-state projective branch probability is a squared norm: for normalized ψ, Tr⁡(∣ψ⟩⟨ψ∣Pb)=∥Pbψ∥2.

[F10]

Binary Pauli quotient isomorphism: equal binary labels mean the Pauli operators differ by a scalar phase.

[F11]

Ray-to-projector is well-defined and bijective: a unit representative of the stabilizer ray defines a density operator.

[F12]

Pauli group on the empty register: on the empty register, the Hermitian Paulis are I and −I, and the binary-label space is the zero space.

Proof

Given: The data in the Statement.

1.1F8F9F11F12givenalgebra

If n=0, the row list and its label list are empty, so the unique coefficient vector is c=() and the row product is the empty product G=I. By [F12], M=±I, so there is a unique bit b with G=(−1)bM. The stabilizer ray has a unit representative ψ and Mψ=(−1)bψ. Therefore Pbψ=ψ and P1−bψ=0. By [F11], [F8], and [F9], the branch probabilities are one and zero, so the deterministic measurement bit is b.

1.2F1F2F3F4givenalgebra

Assume now n≥1. Since M commutes with every stabilizer row, [F4] gives λ(M)∈L⊥=L by [F3]. The row labels form a basis of L by [F2], so the linear system with those rows and target λ(M) is consistent and has a unique coefficient vector c. Gaussian elimination over F2 returns it by pivoting on the independent rows.

2.1F1F2F5F7F10step 1.2

Multiplying the selected commuting rows in increasing order using [F7] produces the group element G with label ∑jcjλ(gj)=λ(M). By [F10], G and M differ by a scalar; as both are Hermitian, [F5] makes that scalar +1 or −1. The exact product phase therefore determines the unique bit b with G=(−1)bM.

3.1F6F8F9F11F12step 1.1step 2.1algebra∎

Since G is a product of stabilizer rows (the empty product when n=0), [F6] gives Gψ=ψ. Thus Mψ=(−1)bψ. By [F11], the unit ray projector is a density operator, and [F8] and [F9] give that the corresponding branch has probability one and the opposite branch has probability zero. Hence the phase returned by the row reduction is exactly the deterministic measurement bit.

Sources

Aaronson–Gottesman §III, PDF pp. 4–5, explains phase accumulation and the deterministic measurement sign; de Wolf §20.3, PDF p. 188, identifies the sign as the measured eigenvalue.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Stabilizer projector rank and code dimension

Statement

Let S be a stabilizer subgroup on n qubits of rank r, and let ΠS be its fixed-space projector. Then

Tr⁡(ΠS)=dim⁡Cim⁡ΠS=2n∣S∣=2n−r.

In particular, if r=n, then C(S) is one-dimensional; for arbitrary r it is the 2n−r-dimensional stabilizer code space.

Facts & Assumptions

Given: An n-qubit stabilizer subgroup S of rank r.

[F1]

Stabilizer average is the fixed-space projector: ΠS is an idempotent whose image is C(S).

[F2]

Stabilizer labels form an isotropic space of the same rank: ∣S∣=2r, and the identity is the unique element with zero binary label.

[F3]

The Pauli matrices: in the ordered computational basis, X=(0110) and Z=(100−1).

[F4]

Phase aware n qubit pauli group: for n≥1, P(k,x,z)=ikXxZz in the fixed tensor order.

[F5]

Tensor products of finitely many linear maps: tensor products of operators act factorwise on elementary tensor vectors.

[F6]

Ordered tensor registers, their inner product, and wire permutations: the ordered computational basis is the lexicographically ordered product basis of the tensor factors, and the zero-qubit register is C.

[F7]

Pauli group on the empty register and Stabilizer subgroup: when n=0, the only stabilizer subgroup element is IC.

[F9]

The trace of an idempotent is its rank as a field scalar: for an idempotent on a finite-dimensional complex space, its trace is its image dimension viewed as a scalar.

[F10]

A maximal stabilizer label space is self-dual: an isotropic stabilizer label space of dimension r satisfies r≤n.

[F11]

Proof

Given: The data in the Statement.

1.1F1F2F3F4F5F6F7F8F10F11givenalgebra

By [F1], [F2], and [F10], 0≤r≤n and Tr⁡(ΠS)=∣S∣−1∑g∈STr⁡(g). For n>0, the identity is I⊗n. In the ordered product basis [F5]–[F6], its trace is a sum over bit words; [F11] factors this finite sum into the product of its one-qubit diagonal sums, each equal to 2. Thus its trace is 2n. When n=0, [F6]–[F8] give trace 1=20. For n>0, every other g∈S has nonzero binary label by [F2], hence at least one nonidentity tensor factor. By [F4] that factor is a scalar multiple of X, Z, or XZ. For any tensor Pauli, its trace is the sum over bit words of products of one-qubit diagonal coefficients; [F11] factors this finite sum into the product of the one-qubit traces. By [F3], X and XZ are off-diagonal, while Z has diagonal entries 1,−1, so the nonidentity factor has trace zero. Hence every such g has trace zero. Therefore Tr⁡(ΠS)=2n/∣S∣=2n−r.

2.1F1F9step 1.1

As ΠS is idempotent, [F9] gives Tr⁡(ΠS)=dim⁡Cim⁡ΠS. Combining with step 1.1 proves the dimension formula, including r=0 and r=n.

3.1step 2.1algebra∎

When r=n, step 2.1 gives a one-dimensional common fixed space, so the space is nonzero and all of its nonzero vectors determine the same ray. This is the existence and uniqueness assertion used in the definition of a pure stabilizer state.

Sources

Watrous Lesson 14 §14.3, printed pp. 420–422 (PDF pp. 426–428), expands commuting stabilizer projectors and computes code-space dimension. De Wolf §20.3, PDF pp. 187–188, gives the rank-n stabilizer-state and projector calculation. The local proof keeps the binary-label injectivity, finite tensor trace factorization, and trace-zero cases explicit.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Deterministic Pauli measurement on a pure stabilizer state

Statement

Let S be a rank-n stabilizer subgroup on n≥0 qubits and let M be a Hermitian Pauli observable commuting with every element of S. There is a unique g∈S with the same binary label as M, and g=(−1)bM for a unique b∈{0,1}. On the stabilizer state of S, outcome b has probability one and the other outcome has probability zero. Moreover Pbψ=ψ, so the conditional state in the nonzero branch is the original ray.

Facts & Assumptions

Given: A rank-n stabilizer subgroup S, its common +1 state, and a Hermitian Pauli M commuting with S.

[F1]

Stabilizer labels form an isotropic space of the same rank: L=λ(S) is isotropic, λ∣S is injective, and dim⁡L=n.

[F2]

Pauli commutator is symplectic pairing: commutation of M with S means its label lies in L⊥.

[F3]

A maximal stabilizer label space is self-dual: L⊥=L for this rank-n isotropic space.

[F4]

Binary Pauli quotient isomorphism: equal binary labels mean the Pauli operators differ by a scalar phase.

[F5]

Hermitian Pauli phase criterion: if both Paulis are Hermitian, their scalar ratio is +1 or −1.

[F6]

Stabilizer state: every element of S fixes the state.

[F7]

Pure state and ray: ψ is a normalized representative of the stabilizer ray.

[F8]

Ray-to-projector is well-defined and bijective: the projector of a unit representative of a pure-state ray is a positive trace-one density operator.

[F9]

Pauli measurement branch semantics: the probability of bit b is Tr⁡(∣ψ⟩⟨ψ∣Pb).

[F10]

Pure-state projective branch probability is a squared norm: this branch probability equals ∥Pbψ∥2 for normalized ψ.

[F11]

Pauli group on the empty register: on the empty register, the Hermitian Paulis are I and −I, both with the unique zero binary label.

Proof

Given: The data in the Statement.

1.1F8F9F10F11givenalgebra

If n=0, then S={I} by [F11] and the exclusion of −I. Write M=(−1)bI for its unique sign bit b. The empty row product is g=I=(−1)bM, so the matching group element and bit are unique. On H0=C, the given stabilizer ray has a unit representative ψ. The projector formula gives Pbψ=ψ and P1−bψ=0. By [F8]–[F10] the probabilities are 1 and 0, respectively; the nonzero branch is the original ray and the other branch has no normalized state by Zero-probability Pauli measurement branch.

1.2F1F2F3given

Assume now n≥1. Since M commutes with S, [F2] gives λ(M)∈L⊥. By [F3], this is L, so there is g∈S with λ(g)=λ(M). Injectivity in [F1] makes this g unique.

2.1F4F5F6step 1.2algebra

Both g and M are Hermitian and have the same binary label, so [F4] and [F5] give g=(−1)bM for exactly one bit b. Since gψ=ψ by [F6], Mψ=(−1)bψ.

3.1F6F7F8F9F10step 2.1algebra

For n≥1, the relation Mψ=(−1)bψ gives Pbψ=ψ and P1−bψ=0. By [F8], the unit ray has its rank-one density projector; [F9] and [F10] then give branch probabilities 1 and 0. The nonzero branch is the original ray, while the zero-probability branch has no normalized state by Zero-probability Pauli measurement branch.

4.1step 1.1step 1.2step 3.1∎

Step 1.1 proves the full claim for n=0, and step 1.2 and step 2.1 and step 3.1 prove it for n≥1. Thus the statement holds for every n≥0.

Sources

De Wolf §20.3, PDF p. 188, identifies the deterministic case from maximal commuting stabilizers and the measured Pauli's signed product. Aaronson– Gottesman §III, PDF pp. 4–5, describes phase recovery for deterministic tableau measurements.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Random Pauli measurement updates stabilizer generators

Statement

Let n≥1 and let S=⟨g1,…,gn⟩ be a stabilizer subgroup whose displayed generators have linearly independent binary labels and fix a unit vector ψ on n qubits. Let M be a Hermitian Pauli observable anticommuting with g1. For either outcome b∈{0,1}, define g1′=(−1)bM and, for j>1, set gj′=gjg1 if gj anticommutes with M, and gj′=gj otherwise. Then the updated rows generate a rank-n stabilizer subgroup for the normalized post-measurement state Pb∣ψ⟩/∥Pb∣ψ⟩∥.

Facts & Assumptions

Given: Independent commuting Hermitian stabilizer generators g1,…,gn, a Hermitian Pauli M anticommuting with g1, and an outcome b with positive probability.

[F1]

An anticommuting stabilizer measurement is fair: both outcomes have probability 1/2.

[A1]

Pauli product and adjoint: exact multiplication of signed Pauli rows uses P(k,x,z)P(k′,x′,z′)=P(k+k′+2⟨z,x′⟩,x+x′,z+z′), where the binary dot product is evaluated as 0 or 1 before doubling in Z/4Z.

[F2]

Stabilizer subgroup: the old rows commute and their group excludes −I.

[F3]

Pauli measurement branch semantics: Pb projects onto the (−1)b eigenspace and the unnormalized branch is Pb∣ψ⟩.

[F4]

Pure-state projective branch probability is a squared norm: the branch probability on a unit vector ψ is ∥Pbψ∥2.

[F5]

The displayed generator labels are independent by hypothesis. By Stabilizer labels form an isotropic space of the same rank, they span the isotropic label space L of S, which therefore has dimension n.

[F6]

Binary Pauli quotient isomorphism: the label map adds under Pauli multiplication, and zero label means a scalar Pauli.

[F7]

Binary symplectic form: ω((x,z),(x′,z′))=x⋅z′+z⋅x′ over F2.

[F8]

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

[F9]

Hermitian Pauli phase criterion: a Hermitian Pauli squares to I.

Proof

Given: The data in the Statement.

1.1A1F2F9givenalgebra

Each unchanged row commutes with M. If gj anticommutes with M, then gjg1 commutes with M because both factors anticommute with M. The changed and unchanged rows commute with each other because they are products of elements of the old abelian group. The new pivot (−1)bM commutes with each other row by the first two calculations and is Hermitian. Every changed row is Hermitian because its two factors commute and are Hermitian; its exact phase-aware product is the one computed by [A1].

2.1F5F6F7F8F9givenstep 1.1algebra

Write vj=λ(gj) and m=λ(M). By [F6] and [F8], the updated rows for j>1 have labels vj′=vj+ω(vj,m)v1. They remain independent: in any zero linear combination, the coefficients of the independent vectors v2,…,vn are the coefficients of the updated rows, so each is zero. Also ω(vj′,m)=ω(vj,m)+ω(vj,m)ω(v1,m)=0; unchanged rows already pair to zero with m. Every vj′ lies in L, but m∉L because it pairs nontrivially with v1 while isotropy in [F5] makes every vector in L pair trivially with v1. Thus adjoining m gives n independent labels. Any nonempty product of the new commuting generators has one of these nonzero labels, so it cannot equal the zero-label scalar −I. The generated subgroup therefore excludes −I.

2.2F1F3F4step 1.1algebra

Since g1 anticommutes with M, [F1] gives probability 1/2 to each bit; [F4] therefore gives ∥Pb∣ψ⟩∥2=1/2, so the branch vector is nonzero. The new pivot stabilizes it because MPb=(−1)bPb. Each unchanged row commutes with Pb and fixes ∣ψ⟩. For an updated row gjg1, commutation with Pb and gjg1∣ψ⟩=∣ψ⟩ give gjg1Pb∣ψ⟩=Pb∣ψ⟩. Thus all updated generators fix the normalized branch.

3.1step 2.1step 2.2algebra∎

The new generators have rank n, so their common fixed space is one-dimensional by Stabilizer projector rank and code dimension. It contains the normalized post-measurement vector from step 2.2, hence it is exactly that post-measurement ray.

Sources

De Wolf §20.3, PDF p. 188, gives the pivot replacement and row multiplication rule. Aaronson–Gottesman §III, PDF pp. 4–5, verifies the commuting-row update and signed tableau representation. The product, quotient-label, symplectic, commutator, and Hermitian-phase supplier bodies are present in batch 15 and are quoted in the Facts & Assumptions; they remain draft, not accepted.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Exact generator-tableau simulator for stabilizer circuits

Definition

Inputs and admissible conditions. An integer n≥0, a basis string x=(x0,…,xn−1)∈F2n, and a finite adaptive Pauli-measurement circuit tree on Hn=(C2)⊗n. Circuit wire labels are j∈[n]={0,…,n−1}. Tableau/Pauli wire labels are a∈{1,…,n}, with circuit wire j represented by Pauli wire a=j+1. Unitary instructions are Hj,Sj,Sj† and CNOT⁡c→t with distinct c,t∈[n]. Here S†:=S∗, the conjugate transpose of the matrix S=diag⁡(1,i), so S∗=diag⁡(1,−i). Measurement instructions are Hermitian phase-aware Paulis M=P(k,u,v) satisfying the Hermitian phase criterion. The tree uses the branch semantics of Finite adaptive Pauli-measurement circuit tree. Its classical bit operations are charged separately under Finite circuit resource accounting. A malformed input, unsupported gate, out-of-range wire, or non-Hermitian measurement operand returns INVALID.

Output. For a valid input, the sampled measurement transcript and the final signed generator rows representing the final pure stabilizer ray. Fair randomness is used exactly when the measured Pauli anticommutes with a stabilizer row.

Initialization. For each circuit wire j∈[n], store the signed stabilizer row sj+1=(−1)xjZj+1. The simulator stores only these generator rows; if n=0, the row list is empty.

Execution. Set the current tree node to the root and the transcript to the empty list. While the current node is not a leaf, perform the instruction at that node and move to its indicated child:

  1. Gate node. For Hj, Sj, or Sj†, apply Phase-aware tableau updates for Clifford generators to every generator row using Pauli wire j+1. For CNOT⁡c→t, apply the same supplier using control wire c+1 and target wire t+1. Then move to the gate node's unique successor.
  2. Measurement node with a pivot. Compute ω(λ(sa),λ(M)) for each row a=1,…,n. If any value is 1, choose the least such a and retain the old row g=sa. Sample a fair bit b. For every other row sℓ that anticommutes with M, replace it by the exact phase-aware product sℓg. Replace the pivot row by (−1)bM, keep the other rows, append b to the transcript, and move to the child edge labelled b.
  3. Measurement node without a pivot. If every row commutes with M, form the augmented matrix over F2 whose columns are λ(s1),…,λ(sn) in the fixed (x,z) coordinate order and whose right-hand side is λ(M). Scan the generator columns in increasing order. For each column, choose the least row not already used as a pivot that has entry 1 in that column, swap it into the next pivot row, and add that pivot row to every other row having entry 1 in the column. If a column has no pivot, or a row with zero coefficient part has right-hand side 1, return INVALID. Otherwise read the unique coefficient vector c=(c1,…,cn) from the right-hand sides of the pivot rows. For n=0 this is the empty system with the empty solution. Multiply the selected signed rows in increasing row order using the exact phase-aware product rule, obtaining G. Set b=0 if G=M and b=1 if G=−M; the unique matching sign is guaranteed by Tableau row reduction returns the deterministic Pauli sign. Leave the tableau unchanged, append b to the transcript, and move to the child edge labelled b.

For a valid rank-n pure stabilizer tableau, each deterministic system is consistent and has full column rank. Each iteration moves to a child, and the finite tree ensures that execution reaches a leaf. Return the transcript and current stabilizer rows there. Correctness and resource bounds are separate results.

Sources

De Wolf §20.3, PDF pp. 187–188, gives the generator-only random and deterministic measurement updates. Aaronson–Gottesman §III, PDF pp. 3–5, gives the phase-aware tableau algorithm: https://arxiv.org/pdf/1907.09415; https://arxiv.org/pdf/quant-ph/0406196.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Phase-aware tableau measurement agrees with the projective instrument

Statement

For a rank-n stabilizer tableau and a Hermitian Pauli measurement M, the tableau measurement rule returns exactly the projective instrument branches: if some stabilizer row anticommutes with M, both bits occur with probability 1/2 and the pivot update represents each conditional state; if every row commutes with M, the signed row product returns the unique probability-one bit, leaves the state unchanged, and the opposite zero-probability branch has no normalized state.

Facts & Assumptions

Given: A valid rank-n stabilizer tableau, a Hermitian Pauli M, and the measurement projectors and instrument from the Statement.

[F1]

An anticommuting stabilizer measurement is fair: an anticommuting row makes both branch probabilities 1/2.

[F2]

Random Pauli measurement updates stabilizer generators: the pivot and row replacement represent the normalized conditional state in either branch.

[F3]

Deterministic Pauli measurement on a pure stabilizer state: when all rows commute, exactly one outcome has probability one.

[F4]

Tableau row reduction returns the deterministic Pauli sign: the phase-aware row product computes that deterministic bit.

[F5]

Zero-probability Pauli measurement branch: the opposite branch map is zero and has no normalized state.

[F6]

Positive-probability Pauli branches are density operators: every positive-probability Pauli branch has a normalized density operator.

[F7]

Pure-state projective branch probability is a squared norm: for a pure input, the normalized branch density is the rank-one projector of the normalized branch vector.

[F8]

Ray-to-projector is well-defined and bijective: a pure-state ray determines a unique rank-one projector.

Proof

Given: The data in the Statement.

1.1F1F2F6F7F8given

If some row anticommutes with M, [F1] gives probability 1/2 to each bit. For either sampled bit, [F2] proves the updated rows are a rank-n stabilizer for the normalized state obtained by applying Pb. Thus the tableau state is the ray of that branch vector. [F6] defines the instrument's normalized branch density, and [F7] identifies it with the vector's rank-one projector; [F8] makes that projector independent of the chosen unit representative. Therefore the tableau and instrument have the same two branch states and probabilities.

1.2F3F4F5F6F7F8given

If every row commutes with M, [F3] gives one deterministic outcome and [F4] computes its bit from the signed row product. The post-measurement state for that bit is unchanged because it is already in the eigenspace. Facts F6, F7, and F8 identify its probability-one branch density with the rank-one projector of the same ray; [F5] says the other branch is impossible and has no normalized state. Hence the tableau output agrees with the instrument in this case as well.

2.1step 1.1step 1.2algebra∎

Every Hermitian Pauli either commutes or anticommutes with each stabilizer row, so the two cases above exhaust all measurements. Therefore the tableau rule agrees with every defined branch of the projective instrument, including its zero-probability boundary.

Sources

The two cases and their state updates are given in de Wolf §20.3, PDF pp. 187–188. Aaronson–Gottesman §III, PDF pp. 4–5, provides the extended tableau version and deterministic sign calculation.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Exact Gottesman–Knill simulation with measurement

Statement

Let C be a finite adaptive circuit on n≥0 ordered qubits, starting in a computational-basis state ∣x⟩. Its unitary instructions are from the finite Clifford generating alphabet H,S,S†,CNOT⁡ and its measurements are Hermitian Pauli observables. Its finite deterministic controller is represented by a circuit tree as in Finite adaptive Pauli-measurement circuit tree. Let T be a worst-case bound on quantum instructions per branch and Ccl a worst-case bound on controller bit operations per branch. Then Exact generator-tableau simulator for stabilizer circuits outputs exactly the circuit's distribution on measurement transcripts and final stabilizer rays.

With dense binary rows and elementary bit operations, tableau and elimination workspace is O((n+1)2) bits. Each gate costs O((n+1)2); each random measurement costs O((n+1)2); each deterministic measurement costs O((n+1)3). Initializing and outputting the dense signed rows costs O((n+1)2) bit operations. Thus the simulator uses O((n+1)2+T(n+1)3+Ccl) bit operations and stores an O(T)-bit transcript, in addition to the tableau and controller workspace. For n≥1 the time bound is O(n2+Tn3+Ccl). No amplitude vector is formed.

Facts & Assumptions

Given: A finite adaptive circuit, ordered basis input, and its declared resource accounting.

[F1]

Finite circuit syntax and ordered wire labels: a circuit has a finite ordered wire set and finite gate occurrences with distinct ordered operands.

[F2]

Denotational semantics of a finite layered circuit: global gate embeddings are unitary and chronological layers act in the declared order; the empty circuit acts as the identity.

[F3]

Finite adaptive Pauli-measurement circuit tree: each Pauli measurement node has two outcome edges with the corresponding instrument weights, and its tree selects later instructions from observed outcomes; zero-weight edges have no conditional state.

[F4]

Finite circuit resource accounting: gate occurrences, wire width, and operation-model costs are separately counted; the model may charge classical bit operations explicitly.

[F5]

Computational basis states have an extended stabilizer tableau: the initial signed rows represent ∣x⟩ and satisfy the tableau invariants.

[F6]

Phase-aware tableau updates for Clifford generators: gate conjugation preserves the row representation and tableau invariant.

[F7]

Phase-aware tableau measurement agrees with the projective instrument: every Pauli measurement branch has exactly the projective probability and conditional tableau prescribed by the instrument.

[F8]

Exact generator-tableau simulator for stabilizer circuits: the simulator processes the same finite instructions and samples a fair bit exactly for a random Pauli measurement branch.

[F9]

Pauli group on the empty register: on the empty register, a Hermitian Pauli is I or −I.

Proof

Given: A circuit and input as in the Statement.

1.1F5given

Before the first instruction, [F5] gives a tableau for the exact input state and transcript. This establishes the simulation invariant at prefix length zero.

1.2F4algebra

A gate updates at most 2n rows, each with O(n) bits. A measurement tests at most n symplectic pairings, each in O(n) bit operations. In its random case it updates at most n rows by an O(n) Pauli product each. In its deterministic case Gaussian elimination on a 2n-by-n binary system uses O(n3) bit operations, and multiplying at most n selected rows uses O(n2). Thus every quantum instruction costs O((n+1)3). Initializing and writing the n signed rows each takes O((n+1)2) bit operations because each row stores 2n+O(1) bits; writing the transcript takes O(T). Summing instruction costs over T steps, then adding initialization, output, and Ccl gives O((n+1)2+T(n+1)3+Ccl). The tableau has O((n+1)2) bits, elimination scratch has O((n+1)2) bits, and the output transcript has at most T bits. The register width and controller charge use the accounting in [F4]; gate and measurement instructions are counted in T and charged by the bit-operation bounds just proved.

2.1F2F3F6F7F8step 1.1

Suppose the invariant holds after a prefix of k instructions. If the next instruction is a Clifford generator, [F6] conjugates every stabilizer row by the same unitary as the circuit, whose action is ordered by [F2]. If it is a Pauli measurement, [F3] supplies the finite instrument-history branch and [F7] gives its probability and updated tableau; [F8] samples a fair bit exactly in the random case and computes the signed bit in the deterministic case. Equal transcripts therefore select the same next child by [F3]. Thus the conditional transition laws agree after instruction k+1. A terminated branch remains in the same final state. Each simulator iteration advances to a child of the finite tree, so it reaches a leaf after at most T quantum instructions. Induction through that path proves equality of the complete transcript and final ray distributions.

3.1F1F8F9givenalgebra∎

If n=0, no wire gate is admissible and [F9] makes every valid Pauli observable I or −I. Its label is the unique zero vector, so the empty row product determines the sign and each measurement is deterministic; the tableau uses constant space. The uniform (n+1) bounds therefore include this boundary.

Sources

De Wolf, Theorem 6 and proof in §20.3, PDF pp. 187–188, proves exact polynomial-time simulation for measurement-controlled Clifford circuits. Aaronson–Gottesman §III, PDF pp. 3–6, gives the signed tableau and the O(n2) random and O(n3) deterministic generator-only update bounds.

5 · Examples, counterexamples and false statements

None yet.

Sources