From 8c1be4a8851700f81226a6c28d2af27f2faf6fb1 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= <daniel.knuettel@daknuett.eu>
Date: Tue, 3 Mar 2020 11:12:52 +0100
Subject: [PATCH] some more work on the presentation

---
 presentation/main.tex | 36 +++++++++++++++++++++++++++++++++---
 1 file changed, 33 insertions(+), 3 deletions(-)

diff --git a/presentation/main.tex b/presentation/main.tex
index 5b08079..7fab365 100644
--- a/presentation/main.tex
+++ b/presentation/main.tex
@@ -78,7 +78,7 @@
         \pause
         \item Fault tolerant QC needs a way to correct for those errors.
         \pause
-        \item Several strategies exist one important class of quantum error correction codes are \textbf{stabilizer codes}.
+        \item Several strategies exist; one important class of quantum error correction codes are \textbf{stabilizer codes}.
     \end{itemize}
 
 \end{frame}
@@ -368,6 +368,7 @@
         \item{
             The generators are not unique. For instance $C_1$ can be generated using $H, S$ or $\sqrt{-iX}, \sqrt{iZ}$.
             }
+        \pause
         \item{
             The generators of a group have some kind of independence property.
             }
@@ -446,7 +447,7 @@
             \begin{equation}
                 \ket{\psi'} = U\ket{\psi} = US^{(i)}\ket{\psi} = US^{(i)}U^\dagger U\ket{\psi} = US^{(i)}U^\dagger\ket{\psi'}
             \end{equation}
-            this gives that if $U \in C_n$ $\ket{\psi'}$ is a stabilizer state again with the stabilizers
+            using this it is clear that for $U \in C_n$ the state $\ket{\psi'}$ is a stabilizer state again with the stabilizers
             $S' = \langle US^{(i)}U^\dagger\rangle_{i=1,...,n}$
             }
         \item{
@@ -454,7 +455,7 @@
             stabilizers.
             }
         \item{Because the stabilizers are given by $n$ matrices which are the tensor product of $n$ Pauli matrices
-            this can be simulated in $n^2$ time instead of $2^n$.}
+            this can be simulated in $\mathcal{O}\left(n^2\right)$ time instead of $\mathcal{O}\left(2^n\right)$.}
     \end{itemize}
 
 \end{frame}
@@ -464,6 +465,35 @@
 \begin{frame}{Measurements on Stabilizer States}
     \begin{itemize}
         \item{
+                Consider a Pauli observable $g_a \in \{(-1)^s X_a, (-1)^s Y_a, (-1)^s Z_a\}$ and the projector
+                onto its eigenspace $\frac{I + g_a}{2}$.
+            }
+        \item{If $g_a$ commutes with all $S^{(i)}$ the observable $g_a$ is diagonal in this basis and 
+            the stabilizer state $\ket{\psi}$ is the $+1$ eigenstate of $g_a$. Therefore the measurement is 
+            deterministic and the stabilizers remain unchanged.}
+    \end{itemize}
+
+\end{frame}
+}
+{
+\begin{frame}{Measurements on Stabilizer States}
+    \begin{itemize}
+        \item{If $g_a$ does not commute with all stabilizers $A := \left\{S^{(i)} \middle| \left[g_a, S^{(i)}\right] \neq 0\right\}$
+            is the set of stabilizers that anticommute with $g_a$.}
+        \item{
+                To compute the probability to measure a result of $s=0$ one can use the trace formula
+                \begin{equation}
+                \begin{aligned}
+                    P(s=0) &= \Tr(\frac{I + g_a}{2} \ket{\psi}\bra{\psi}) \\
+                            &= \Tr(\frac{I + g_a}{2} S^{(j)} \ket{\psi}\bra{\psi}) \\
+                            &= \Tr(S^{(j)}\frac{I - g_a}{2} \ket{\psi}\bra{\psi}) \\
+                            &= \Tr(\frac{I - g_a}{2} \ket{\psi}\bra{\psi}) \\
+                            &= P(s=1)\\
+                \end{aligned}
+                \end{equation}
+
+                where $S^{(j)} \in A$. The stabilizer $S^{(j)}$ pulled to the right using the cyclic property of the
+                trace and absorbed into the $\bra{\psi}$.
             }
     \end{itemize}