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Given a non-Clifford circuit $U$, say $U = \prod_{i=1}^k e^{i \theta_i P_i} $ for $P_i \in \{ I,X,Y,Z\}^{\otimes n} $ and $\theta_i \in \mathbb{R}$.

Is it possible to construct a non-trivial Clifford circuit $C$ such that it commutes with $U$? .

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    $\begingroup$ I would guess not since it is possible to decompose U as {Clifford+T}. If there is C that commutes with U while being Clifford shouldn't we have Cliffords commuting with T, which is not possible I guess? Interesting question (+1) $\endgroup$
    – R.W
    Apr 17 at 6:42
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    $\begingroup$ @R.W Any diagonal Clifford commutes with $T$ ... $\endgroup$ Apr 18 at 11:42

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