a few more lines

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Daniel Knüttel 2020-02-17 18:49:25 +01:00
parent 614796c35a
commit d6df56a51b

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@ -659,4 +659,18 @@ as $o_a$ is a Clifford operator:
\end{equation}
Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. It is therefore enough to study the measurements
of any Pauli operator on the vertex operator free graph states.
of any Pauli operator on the vertex operator free graph states. The commutators of the observable
with the $K_G^{(i)}$ are quite easy to compute, note that Pauli matrices wither commute or anticommute
so it is easier to list the operators that anticommute:
\begin{equation}
\begin{aligned}
A_{\pm X_a} &= \left\{j | \{j, a\} \in E\right\}\\
A_{\pm Y_a} &= \left\{j | \{j, a\} \in E\right\} \cup \{a\} \\
A_{\pm Z_a} &= \{a\}\\
\end{aligned}
\end{equation}
This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a$
is measured the result $+1$ is obtained with probability $1$ and $(V, E, O)$ is unchanged.