some more work on stabilizers

computations/projector_after_commute_C_L.ipynb

@@ -2,7 +2,7 @@
  "cells": [
   {
    "cell_type": "code",
-   "execution_count": 11,
+   "execution_count": 2,
    "metadata": {},
    "outputs": [
     {
@@ -53,7 +53,7 @@
        "    ⎦⎦"
       ]
      },
-     "execution_count": 11,
+     "execution_count": 2,
      "metadata": {},
      "output_type": "execute_result"
     }
@@ -108,7 +108,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 12,
+   "execution_count": 3,
    "metadata": {},
    "outputs": [
     {
@@ -123,7 +123,7 @@
        "⎣0  -1⎦"
       ]
      },
-     "execution_count": 12,
+     "execution_count": 3,
      "metadata": {},
      "output_type": "execute_result"
     }

thesis/chapters/stabilizer.tex

@@ -80,6 +80,66 @@ This is an important insight that is used for simulations\cite{gottesman_aaronso
 updating the $n$ stabilizers that are a tensor product of $n$ Pauli matrices scales with roughly
 $\mathcal{O}(n^2)$ instead of $\mathcal{O}(2^n)$ for the state vector approach.
 
+Every qbit can be measured in the $X, Y$ or  $Z$ basis which is a projection using
+
+\begin{equation}
+    \frac{I + (-1)^s g_a}{2}
+\end{equation}
+
+Where $g_a \in \{Z_a, Y_a, X_a\}$ and $s \in \{0, 1\}$. How the stabilizers change when
+measuring a qbit is given by the following lemma:
+
+\begin{lemma}
+    \label{lemma:stab_measurement}
+    Let $J := \{ S_i | [g_a, S_i] \neq 0\}$. Then
+
+    \begin{enumerate}
+        \item{If $J = \{\}$, one value is measured with probability $1$ and the stabilizers are unchanged.}
+        \item{If $J \neq \{\}$, $1$ and $0$ are measured with probability $\frac{1}{2}$ and the new state
+            $\ket{\psi'}$ is stabilized by 
+            \begin{equation}
+                \langle \{(-1)^s g_a\} \cup \{K_G^{(i)} K_G^{(j)} | j \in J, i \in J \setminus \{j\} \} \cup \{K_G^{(i)} | i \in J^c\}\rangle
+            \end{equation}}
+    \end{enumerate}
+\end{lemma}
+
+\begin{proof}
+    \begin{enumerate}
+        \item{As $g_a$ commutes with all stabilizers, $\ket{\psi}$ is an eigenstate of $g_a$,
+            so the result of the measurement is deterministic and $\ket{\psi}$ is left unchanged.}
+        \item{As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli operators,
+            $S_i$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then
+            
+            \begin{equation}
+            \begin{aligned}
+                P(s=+1) &= \hbox{Tr}\left(\frac{I + g_a}{2}\ket{\psi}\bra{\psi}\right) \\
+                        &= \hbox{Tr}\left(\frac{I + g_a}{2}S^{(j)} \ket{\psi}\bra{\psi}\right)\\
+                        &= \hbox{Tr}\left(S^{(j)}\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
+                        &= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}S^{(j)}\right)\\
+                        &= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
+                        &= P(s=-1)
+            \end{aligned}
+            \notag
+            \end{equation}
+
+            With $P(s=+1) + P(s=-1) = 1$ follows $P(s=+1) = \frac{1}{2} = P(s=-1)$.
+            Further for $S^{(i)},S^{(j)} \in J$
+
+            \begin{equation}
+            \begin{aligned}
+                \frac{I + (-1)^sg_a}{2}\ket{\psi} &= \frac{I + (-1)^sg_a}{2}S^{(j)}S^{(i)} \ket{\psi} \\
+                    &= S^{(j)}\frac{I + (-1)^{s + 1}g_a}{2}S^{(i)} \ket{\psi} \\
+                    &=  S^{(j)}S^{(i)}\frac{I + (-1)^{s + 2}g_a}{2}\ket{\psi} \\
+                    &=  S^{(j)}S^{(i)}\frac{I + (-1)^sg_a}{2}\ket{\psi}
+            \end{aligned}
+            \notag
+            \end{equation}
+
+            the state after measurement is stabilized by $S^{(j)}S^{(i)} i,j \in J$, and by 
+            $S^{(i)} \in J^c\setminus\{a\}$. $g_a$ trivially stabilizes $\ket{\psi'}$.
+            }
+    \end{enumerate}
+\end{proof}
 
 \subsection{The Vertex Operator-Free Graph States}
 
@@ -98,7 +158,7 @@ were derived from the vertex operator-free graph states.
 
     for all $i \in V$ where for some operator $O$ $O_i$ indicates that it acts on the $i$-th qbit.
 
-    A state $\ket{\overline{G}}$ is a $+1$ eigenstate of all $n$ $K^{(i)}_G$.
+    $\ket{\overline{G}}$ is the $+1$ eigenstate of all $n$ $K^{(i)}_G$.
 \end{definition}
 
 \begin{corrolary}
@@ -163,5 +223,61 @@ were derived from the vertex operator-free graph states.
 \end{proof}
 
 These insights can be used to understand how measurement works on the vop-free graph state \cite{nielsen_chuang_2010}:
-Consider a state $\ket{\psi}$ that is stabilized by $g_1, ... g_n$ and a hermitian $g$ that is to be measured.
+Consider a state $\ket{\psi}$ that is stabilized by $g_1, ... g_n$ and a Pauli operator $g \in \{Z_a, Y_a, X_a\}$ 
+that is to be measured.
+Recalling lemma \ref{lemma:stab_measurement} the following relations follow immideately:
+
+\begin{enumerate}
+    \item{For $g = X_a$ if $a$ is an isolated qbit $+1$ is measured with probability $1$
+        and the state $\ket{\bar{G}}$ is unchanged.}
+    \item{For $g = X_a$ and $a$ is non-isolated, choose $b \in n_a$ and the new stabilizers are
+        \begin{equation}
+            \langle \{(-1)^sX_a\} 
+                \cup \{K_G^{(b)}K_G^{(i)} | i \in n_a \setminus \{b\}\} 
+                \cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle
+        \end{equation}}
+    \item{For $g = Z_a$ the new stabilizers are
+        \begin{equation}
+            \langle \{(-1)^sZ_a\} 
+                \cup \{K_G^{(i)} | i \in V \setminus n_a\} \rangle
+        \end{equation}}
+    \item{For $g = Y_a$ the new stabilizers are
+        \begin{equation}
+            \langle \{(-1)^sY_a\} 
+                \cup \{K_G^{(a)}K_G^{(i)} | i \in n_a\} 
+                \cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle
+        \end{equation}}
+\end{enumerate}
+
+The states after the measurement are in general no vop-free graph states anymore,
+the following discussion will allow to construct new vop-free graph states and
+Clifford transformations from the vop-free graph state to the resulting state.
+In the case $g = X_a$ and $n_a = \{\}$ the graph is obviously unchanged.
+
+\begin{lemma}
+    \begin{enumerate}
+        \item{For a result $+Z_a$ the new state is
+                $\ket{+_Z}_a \otimes \ket{\bar{G}'}$
+                with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for
+                $i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$.
+            }
+        \item{For a result $-Z_a$ the new state is $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$
+            with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for $i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$
+            and $U = \prod\limits_{i \in n_a}Z_i$.}
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    \begin{enumerate}
+        \item{
+                It is trivial that $Z_a$ and $K_G^{(i)} i \neq a$ stabilize $\ket{+_Z}_a \otimes \ket{\bar{G}'}$.
+                $Z_aK_G^{(i)}$ for $i \in n_a$ do not act on $a$, so $\ket{\bar{G}'}$ is well-defined.
+            }
+        \item{
+                The state $\ket{-_Z}_a \otimes \ket{\bar{G}'}$ is the $-1$ eigenstate of $K_G^{(i)}$ for all
+                $i \in n_a$. This can be corrected by transforming $K_G^{(i)}$ to $-K_G^{(i)} = Z_i K_G^{(i)} Z_i^\dagger$
+                which stabilizes the state $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$.
+            }
+    \end{enumerate}
+
+\end{proof}