# If $C$ is a Clifford circuit, is there necessarily a Clifford circuit $C'$ such that $CT=TC'$?

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)?

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 $$|{-i}\rangle$$.

• 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. May 3 at 16:25
• @CraigGidney Can you please elaborate? Although this is not what I intended in the question, I'd really love to see more detail.
– Haim
May 7 at 0:48
• 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. May 7 at 2:52
• @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?