The Exercise

Take two qubits, with the Pauli operators $X_1 = X \otimes I$ and $X_2 = I \otimes X$ acting on the first and on the second, and likewise $Z_1 = Z \otimes I$ and $Z_2 = I \otimes Z$, where $X$ and $Z$ are the Pauli matrices and $I$ is the $2 \times 2$ identity. Two unitaries $U$ and $V$ on the pair transform these four operators by conjugation in the same way when $$U X_k U^\dagger = V X_k V^\dagger, \qquad U Z_k U^\dagger = V Z_k V^\dagger, \qquad k = 1, 2.$$(1) The exercise asks whether $U$ and $V$ must then be equal.

The library answers with two theorems, stated for unitary matrices on two qubits and for the operators $X_k$ and $Z_k$ built from the Pauli matrices.

Pauli Strings

The Pauli matrices $\sigma_0, \sigma_1, \sigma_2, \sigma_3$ are the identity $I$ and the three complex $2 \times 2$ matrices $X$, $Y$ and $Z$: $$\sigma_0 = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \sigma_1 = X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$(2)^✦

An operator $A$ on $n$ qubits is a complex $2^n \times 2^n$ matrix whose rows and columns are labelled, like the computational basis states $|i\rangle$, by the strings $i = i_1 \cdots i_n$ of $n$ bits. A unitary $U$ on $n$ qubits is an operator with $U^\dagger U = I$.

For a choice $g = (g_1, \dots, g_n)$ of an index $g_k \in \{0, 1, 2, 3\}$ for each qubit $k$ of $n$ qubits, the Pauli string $\sigma_g$ is the operator on the $n$ qubits with the entries $$(\sigma_g)_{ij} = \prod_{k=1}^{n} (\sigma_{g_k})_{i_k j_k},$$(3)^✦ which is the tensor product $\sigma_{g_1} \otimes \cdots \otimes \sigma_{g_n}$.

The $X$ generator $X_k$ on qubit $k$ of $n$ qubits is the Pauli string with $X$ on qubit $k$ and the identity on every other qubit^✦, and the $Z$ generator $Z_k$ is the Pauli string with $Z$ on qubit $k$ and the identity on every other.^✦ For $n = 2$ they are the four operators of (1).

The Claim

What Is Checked

For these generators, two unitaries on two qubits that conjugate all four of them in the same way agree up to a global phase of modulus one.^✦ There are also two different unitaries on two qubits that conjugate all four generators in the same way.^✦