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$U = |0\rangle\langle 0|\otimes U_1 + |1\rangle\langle 1|\otimes U_2$ is a block-diagonal unitary matrix. For this question we will assume $U$ acts on qubits. Then for some integer $N\ge 1$, $U$ is a $2^N \times 2^N$ unitary matrix and $U_1,U_2$ are $2^{N-1} \times 2^{N-1}$ unitary matrices.

My question is: For what $U_1,U_2$ does $U$ equal a tensor product of Pauli matrices?

More specifically, I'd like to prove (or disprove) that $U$ equals a tensor product of Pauli matrices iff $U_1,U_2$ are tensor products of Pauli matrices and $U_1 = \pm U_2$.

$(\implies)$ If $U_1, U_2$ are tensor products of Pauli matrices and $U_1 = U_2 = P$ or $P = U_1 = -U_2$ then $U = I\otimes P$ or $U=Z\otimes P$, respectively.

For the other direction we know that $U^2 = I$ (up to a global phase $\pm 1$) and therefore $U_1^2 = U_2^2 = I$, but I can't quite make the proof go through. Any help is much appreciated.

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Assume $U$ equals a Pauli matrix and has the form $|0\rangle \langle 0| \otimes U_1 + |1 \rangle \langle 1| \otimes U_2$. The set of Pauli matrices $\mathcal{P}_N := \{I, X, Y, Z\}^{\otimes N}$ is an orthogonal basis for $2^N \times 2^N$ complex matrices, so there's exactly one $P \in \mathcal{P}_N$ for which $\text{Tr}(PU) \neq 0$.

Pick any $P$ of the form $P = X \otimes P_{N-1}$ (with $P_{N-1} \in \mathcal{P}_{N-1}$) and you'll compute \begin{align} \text{Tr}(PU) &= \text{Tr}\left( (X \otimes P_{N-1})(|0\rangle \langle 0| \otimes U_1 + |1 \rangle \langle 1| \otimes U_2) \right) \\&= \text{Tr}(X |0\rangle \langle 0|) \text{Tr}(P_{N-1} U_1) + \text{Tr}(X|1 \rangle \langle 1|)\text{Tr}( P_{N-1} U_2) \\&=0, \end{align} so $U$ cannot start with $X$ (or $Y$, for the same reason). There will be exactly one $P_{N-1}$ such that $U \in \{I_2 \otimes P_{N-1}, Z \otimes P_{N-1}\}$, i.e. $U_1 = P_{N-1}$ and $U_2 = \pm P_{N-1}$.

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  • $\begingroup$ Isn't Z the only block diagonal Pauli matrix? $\endgroup$ Oct 8, 2023 at 0:39
  • $\begingroup$ Above when I refer to Pauli matrices I mean $\{I, X, Y, Z\}$ $\endgroup$
    – forky40
    Oct 8, 2023 at 1:05
  • $\begingroup$ So then why does your answer have so many words? Aren't the only block diagonal matrices in that set "I" and "Z"? By the way Pauli matrices are usually just X, Y and Z. I've edited the question so your answer makes more sense. $\endgroup$ Oct 8, 2023 at 3:08
  • $\begingroup$ Thanks! Looks good. Should be "...one $P\in \mathcal{P}_N$ for which Tr$(PU)\ne 0$.", right? $\endgroup$ Oct 8, 2023 at 20:51
  • $\begingroup$ yep - I've edited that typo $\endgroup$
    – forky40
    Oct 9, 2023 at 16:16

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