$U = |0\rangle\langle 0|\otimes U_1 + |1\rangle\langle 1|\otimes U_2$ 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.


1 Answer 1


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 $U_2 = \pm P_{N-1}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.