Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 324 / 2

An $n$-qubit stabilizer state is the unique simultaneous $+1$ eigenstate of an abelian group $G$ of Pauli operators having $2^n$ elements and not containing $-I$. The group $G$ is the state's stabilizer group; equivalently, it is generated by $n$ independent commuting Hermitian Pauli operators.

The state $|001⟩$ has independent stabilizer generators $ZII$, $IZI$, and $-IIZ$. Its complete stabilizer group isFor $|{\Phi }^{+}⟩\otimes |+⟩$, take the generators $XXI$, $ZZI$, and $IIX$. Their products giveIn particular $(XX)(ZZ)=-YY$, which accounts for the two minus signs.

The Gottesman--Knill theorem states that stabilizer-state preparation, Clifford circuits, and adaptive measurements of Pauli observables can be simulated in classical polynomial time. For $|{\Psi }_{\mathrm{in}}⟩=|+{⟩}^{\otimes 3}$, choose generators $XII$, $IXI$, and $IIX$. With each row written as $[x_1{x}_2{x}_3\mid z_1{z}_2{z}_3]$, its sign-free stabilizer tableau isConjugating these generators successively by ${\mathrm{CNOT}}_{21}$, $H_2\otimes H_3$, and ${\mathrm{CNOT}}_{13}$ gives $XIX$, $XZX$, and $ZIZ$. Hence the output tableau, again ignoring signs, is

At each qubit, two matrices from $\{I,X,Y,Z\}$ either commute or anticommute. Moving every $Q_i$ past the corresponding $P_i$ therefore giveswhere $r$ is the number of positions containing distinct nonidentity Pauli matrices. Thus two Pauli strings always commute or anticommute. If they anticommute and a vector $|\psi ⟩$ were stabilized by both, then $PQ|\psi ⟩=|\psi ⟩$ and $QP|\psi ⟩=|\psi ⟩$, contradicting $PQ=-QP$. Their common stabilizer subspace is consequently the zero subspace $\{0\}$.

The group averageis Hermitian. In its square, every $R\in G$ occurs exactly $|G|$ times among products $PQ$, and therefore ${\Pi }_G^2={\Pi }_G$. Moreover $g{\Pi }_G={\Pi }_G$ for every $g\in G$, so its image lies in the stabilizer subspace $V_G$, while ${\Pi }_G|\psi ⟩=|\psi ⟩$ for every $|\psi ⟩\in V_G$. Thus ${\Pi }_G$ is the orthogonal projector onto $V_G$. If $g_1,\dots ,g_l$ are independent generators, expanding the product chooses each element of $G$ exactly once and gives the stabilizer-projector formula