From 614796c35a3999f1ac63a4f528795fdb5995fcc1 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= <daniel.knuettel@daknuett.eu>
Date: Mon, 17 Feb 2020 10:18:04 +0100
Subject: [PATCH] added something about measurements

---
 thesis/chapters/stabilizer.tex | 34 ++++++++++++++++++++++++++++++++++
 1 file changed, 34 insertions(+)

diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex
index 4955f7c..c0d20cd 100644
--- a/thesis/chapters/stabilizer.tex
+++ b/thesis/chapters/stabilizer.tex
@@ -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
 \cite{gottesman_aaronson2008}.
 
+\subsubsection{Measurements on Stabilizer States}
+\label{ref:meas_stab}
+
 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
 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.
 
 \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.