It's well known how to implement a controlled-U gate when $U$ is a single qubit gate via the decomposition of $U$ into the product $U=e^{i\phi}AXBXC$ where $ABC=I$.

My question is: is it possible to implement the controlled-S gate in such a way that $B$ belongs to the Clifford group?

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1 Answer 1


TL;DR: After sending $D$ across the equals sign, the right hand side is similar to a controlled Pauli operator and the left hand side is a diagonal operator. Their spectra turn out to be incompatible, so they cannot be equal.

If $B$ is Clifford, then $BXB^\dagger=P$ is a Pauli operator with eigenvalues $\pm 1$ and $Q:=XP$ is either $\pm I$ or a Pauli operator with eigenvalues $\pm i$. The former renders the circuit on the right hand side non-entangling, so assume the latter. Note that $$ \begin{align} AXBXC&=AXPBC=AQA^\dagger\tag1 \end{align} $$ so shifting the $B$ gate left to the position between $C$ and the first CNOT on the right hand side of the diagram in the question yields controlled-$AQA^\dagger$ followed by $D\otimes I$.

Moreover, by considering the action of the two sides of the diagram on the computational basis we see that $D=\mathrm{diag}(d_1,d_2)$ for some $d_1,d_2\in\mathbb{C}$. Now, hit both sides of the diagram with $D^\dagger\otimes I$. On the left hand side we obtain $$ (D^\dagger\otimes I)\circ\text{CS}=\mathrm{diag}(\overline{d}_1, \overline{d}_1, \overline{d}_2, i\overline{d}_2).\tag2 $$ On the right hand side we obtain an operator, similar to a controlled Pauli, with eigenvalues $+1, +1, +i, -i$. We conclude that the two operators cannot be equal if $B$ is Clifford.

  • 1
    $\begingroup$ Great answer; I want to point out that your proof also proves that the impossibility remains even if we consider the implementation of approximation of controlled-S instead of the exact gate. $\endgroup$
    – Dudu Ponar
    Mar 20 at 11:11

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