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Let $C$ be a Clifford circuit, is there necessarily a Clifford circuit $C'$ such that $CT=TC'$ (where $T$ is taken as applying the $T$ gate to the same qubit on both sides)?

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No.

There always exists a unique unitary $U$ such that $CT=TU$. Namely, $U = T^\dagger C T$. The question is whether $U$ is Clifford. It turns out that this is not guaranteed. For a simple counterexample take Hadamard for $C$. Then

$$ U=T^\dagger H T = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & e^{i\pi/4} \\ e^{-i\pi/4} & -1 \end{bmatrix} $$

is not Clifford, because $U|0\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle+e^{-i\pi/4}|1\rangle\right)$ is not one of the stabilizer states $|0\rangle$, $|1\rangle$, $|+\rangle$, $|-\rangle$, $|{+i}\rangle$ and $|{-i}\rangle$.

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    $\begingroup$ Caution: this is correct, but sensitive to details of the question that may not be intentional. In particular, if $C^\prime$ were allowed to include an ancilla qubit initialized to $|+\rangle$ and were allowed to move the T gate to the ancilla and were allowed to do measure+feedback then it becomes possible because you can inject a T state and do T gate teleportation. $\endgroup$ May 3 at 16:25
  • $\begingroup$ @CraigGidney Can you please elaborate? Although this is not what I intended in the question, I'd really love to see more detail. $\endgroup$
    – Haim
    May 7 at 0:48
  • $\begingroup$ I think @CraigGidney is referring to the construction shown in circuit $(14)$ on page 6 in this paper and also in figure 10.25 on page 486 in Nielsen & Chuang. $\endgroup$ May 7 at 2:52
  • $\begingroup$ @AdamZalcman Thanks! $\endgroup$
    – Haim
    May 10 at 22:24
  • $\begingroup$ @CraigGidney Just to see if I got this right, does your comment imply that for every Clifford circuit $C$ on $n$ qubits and $i\in \{0,...,n-1\}$, $CT$ (with $T$ applied on the $i$th wire) equals $TC'$ where $C'$ is Clifford on $n+1$ qubits and $T$ is applied to the $i$th wire as well? $\endgroup$
    – Haim
    May 10 at 22:52

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