The Exercise

Let $S$ be a subgroup of the Pauli group $G_n$ on $n$ qubits, the group of operators $i^s$ times a tensor product of $I, X, Y, Z$ with $s$ an integer modulo $4$, defined below, that does not contain $-I$. The exercise asks for two facts about every element $g$ of $S$: it squares to the identity, and it is Hermitian.

The Pauli Group

The Pauli matrices are the four complex $2 \times 2$ matrices $\sigma_0, \sigma_1, \sigma_2, \sigma_3$, with $\sigma_0 = I$ the identity and $\sigma_1 = X$, $\sigma_2 = Y$, $\sigma_3 = Z$, where $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(1)

For a choice $a = (a_1, \dots, a_n)$ of one index in $\{0,1,2,3\}$ for each of $n$ qubits, the Pauli string $\sigma_{a_1} \otimes \cdots \otimes \sigma_{a_n}$ is the operator on the $n$ qubits whose matrix, with rows and columns labelled by bit strings $x, y \in \{0,1\}^n$, has the entries $$(\sigma_{a_1} \otimes \cdots \otimes \sigma_{a_n})_{xy} = \prod_{k=1}^{n} (\sigma_{a_k})_{x_k y_k}.$$(2)^✦

The phase $i^s$ of an integer $s$ modulo $4$ is the imaginary unit raised to the representative of $s$ in $\{0,1,2,3\}$, one of the four fourth roots of unity $1, i, -1, -i$.^✦

The Pauli group $G_n$ on $n$ qubits consists of the operators $i^s \sigma_{a_1} \otimes \cdots \otimes \sigma_{a_n}$. An element $g$ of $G_n$ is given by its phase power $s \in \mathbb{Z}_4$ and its Pauli string $a \in \{0,1,2,3\}^n$, which puts one of $I, X, Y, Z$ on each qubit.^✦ The matrix $M_g$ of $g$ is its Pauli string multiplied by its phase, $$M_g = i^s \, \sigma_{a_1} \otimes \cdots \otimes \sigma_{a_n},$$(3)^✦ a complex $2^n \times 2^n$ matrix with rows and columns labelled by $\{0,1\}^n$.

The element $-I$ of $G_n$ is the identity string $I \otimes \cdots \otimes I$ with phase power $s = 2$, whose matrix is $i^2 I = -I$.^✦

The Claim

What Is Checked

The claim is stated over the Pauli group $G_n$ for any number $n$ of qubits and over any subgroup $S$ of it that leaves out $-I$; its second half is stated over the matrix $M_g$ of (3). The library proves of these constructions that the one hypothesis $-I \notin S$ suffices: the square in (4) is the square in the group $G_n$ itself, and the adjoint in (5) is the conjugate transpose of the complex $2^n \times 2^n$ matrix $M_g$.