Trying to prove that all Clifford gates can be constructed with CNOT, H and S gates, I'm following the classical path by induction (Nielsen and Chuang, Quantum Computation and Quantum Information -- Exercice 10.40).
Assuming that
- $U (Z_1 \otimes I_2 \otimes \cdots \otimes I_n)U^* = X \otimes g$ where $g \in Pauli_{n-1}$
- $U (X_1 \otimes I_2 \otimes \cdots \otimes I_n)U^* = Z \otimes g^\prime$ where $g^\prime \in Pauli_{n-1}$
The classical circuit proposed in the exercice for the Clifford gate $U$ is $(Controlled-g) (H \otimes I) (controlled-g^\prime)(I \otimes U^\prime)$, or
$U = \frac{1}{\sqrt{2}} \begin{pmatrix} I & 0 \\\\ 0 & g\end{pmatrix}\begin{pmatrix} I & I \\\\ I & -I\end{pmatrix}\begin{pmatrix} I & 0 \\\\ 0 & g^\prime\end{pmatrix}\begin{pmatrix} U^\prime & 0 \\\\ 0 & U^\prime \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} U^\prime & g^\prime U^\prime \\\\ gU^\prime & gg^\prime U^\prime\end{pmatrix}$
By construction, $U^\prime = \langle 0| U (|0 \rangle \otimes |\phi \rangle)$ is unitary -- and to be proven, also a Clifford gate . By the fact that $g, g^\prime \in Pauli_{n-1}$, $g=g^*$ and $g^\prime = g^{\prime *}$. Thus,the assumption 2 can be verified as follows:
$\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \begin{pmatrix} U^\prime & g^\prime U^\prime \\\\ gU^\prime & gg^\prime U^\prime\end{pmatrix} \begin{pmatrix} 0 & I \\\\ I & 0 \end{pmatrix} \begin{pmatrix} U^{\prime *} & U^{\prime *} g \\\\ U^{\prime *} g^\prime & U^{\prime *} g^\prime g \end{pmatrix} = \begin{pmatrix} g^\prime & 0 \\\\ 0 & -g g^\prime g \end{pmatrix} \neq \begin{pmatrix} g^\prime & 0 \\\\ 0 & g^\prime \end{pmatrix} as \ expected$
At least not always. Since the Pauli gates anticommute, $g g^\prime g = (-1)^k g g g^\prime = (-1)^k g^\prime$, where $k$ is the number of pairs $(g_i, g^\prime_i)$, which are not containing the Identity gates.
I should have made an error but I do not see it.
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Addendum (answering the obvious question when you put all the details)
The Pauli matrices form a non commutative group. In fact their products anti-commute, e.g. $XY = -YX$. On the other hand, $XI = IX = X$. Thus, the two operators $O_X = X_1 \otimes I_2 \otimes \cdots \otimes I_n$ and $O_Z = Z_1 \otimes I_2 \otimes \cdots \otimes I_n$ anti-commute, since $X_1$ and $Z_1$ anti-commute. This gives the following reasoning:
$U(O_X O_Z)U^* = -U(O_Z O_X)U^* \Rightarrow (UO_X U^*) (U O_ZU^*) = -(UO_Z U^*) (UO_XU^*)$
Therefore, the operators $X \otimes g$ and $Z \otimes g^\prime$ must anti-commute. Since $X$ and $Z$ anti-commute, $g$ and $g^\prime$ must commute. This is sufficient to prove that $-gg^\prime g = -gg g^\prime = -g^\prime$. Obvious when you mentioned all the assumptions.
Naturally, there are more details to complete the proof of the problem but that was not the original question.
