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added something about measurements

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Daniel Knüttel 2020-02-17 10:18:04 +01:00
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thesis/chapters/stabilizer.tex
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@ -205,6 +205,9 @@ efficient as only $n$ stabilizers have to be modified, each being just the tenso
product of $n$ Pauli matrices. This has led to the simulation using stabilizer tableaux product of $n$ Pauli matrices. This has led to the simulation using stabilizer tableaux
\cite{gottesman_aaronson2008}. \cite{gottesman_aaronson2008}.
\subsubsection{Measurements on Stabilizer States}
\label{ref:meas_stab}
Interestingly also measurements are dynamics covered by the stabilizers. Interestingly also measurements are dynamics covered by the stabilizers.
When an observable $g_a \in \{\pm X_a, \pm Y_a \pm Z_a\}$ acting on qbit $a$ is measured When an observable $g_a \in \{\pm X_a, \pm Y_a \pm Z_a\}$ acting on qbit $a$ is measured
one has to consider the projector one has to consider the projector
@ -626,3 +629,34 @@ the graphical representation of a stabilizer state is indeed able to represent a
If one wants to do computations using this formalism it is however also necessary to perform measurements. If one wants to do computations using this formalism it is however also necessary to perform measurements.
\subsubsection{Measurements on Graph States} \subsubsection{Measurements on Graph States}
Recalling \ref{ref:meas_stab} it is clear that one has to compute the commutator of
the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes.
This is a quite expensive computation in theory however it is possible to simplify
the problem by pulling the observable behind the vertex operators. For this consider
the projector $P_a = \frac{I + (-1)^sZ_a}{2}$:
\begin{equation}
\begin{aligned}
P_a \ket{\psi} &= P_a \left(\prod\limits_{o_i \in O} o_i \right) \ket{\bar{G}} \\
&= \left(\prod\limits_{o_i \in O \setminus o_a}o_i \right)P_a o_a \ket{\bar{G}} \\
&= \left(\prod\limits_{o_i \in O \setminus o_a}o_i \right) o_a o_a^\dagger P_a o_a \ket{\bar{G}} \\
&= \left(\prod\limits_{o_i \in O} o_i \right) o_a^\dagger P_a o_a \ket{\bar{G}} \\
&= \left(\prod\limits_{o_i \in O} o_i \right) \tilde{P}_a \ket{\bar{G}} \\
\end{aligned}
\end{equation}
This transformed projector has the important property that it still is a Pauli projector
as $o_a$ is a Clifford operator:
\begin{equation}
\begin{aligned}
\tilde{P}_a &= o_a^\dagger P_a o_a \\
&= o_a^\dagger \frac{I + (-1)^sZ_a}{2} o_a \\
&= \frac{I + (-1)^s o_a^\dagger Z_a o_a}{2} \\
&= \frac{I + (-1)^s \tilde{g}_a}{2} \\
\end{aligned}
\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.