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)?


1 Answer 1



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 $|{-i}\rangle$.

  • 1
    $\begingroup$ 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, 2021 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, 2021 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, 2021 at 2:52
  • $\begingroup$ @AdamZalcman Thanks! $\endgroup$
    – Haim
    May 10, 2021 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, 2021 at 22:52

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.