some work on the paper

thesis/chapters/stabilizer.tex

@@ -371,6 +371,7 @@ This transformation toggles the neighbourhood of $a$ which is an operation
 that will be used later.
 
 \begin{lemma}
+    \label{lemma:M_a}
     When applying $M_a$ to a state $\ket{\bar{G}}$ the new state
     $\ket{\bar{G}'}$ is again a VOP-free graph state and the 
     graph is updated according to 
@@ -454,6 +455,7 @@ can be constructed from $CZ$ and $C_L$ and in the following discussion it will b
 clear that both $C_L$ and $CZ$ can be applied to a general graph state.
 
 \subsubsection{Graph States and Vertex Operators}
+\label{ref:g_states_vops}
 
 \begin{definition}
     \label{def:g_state}
@@ -522,19 +524,105 @@ it is clear that just the vertex operators are changed and the new vertex operat
 \end{equation}
 
 The action of a $CZ$ gate on the state $(V, E, O)$ is in most cases less trivial.
-Let $i \neq j$ be two qbits, now consider the action of $CZ_{i,j}$ on $(V, E, O)$.
+Let $i \neq j$ be two qbits, now consider the action of $CZ_{a,b}$ on $(V, E, O)$.
-The following discussion closely follows \cite{andersbriegel2005} and is analogous
+The cases given here follow the implementation of a $CZ$ application in \cite{pyqcs},
-to the implementation.
+the respective paragraphs from \cite{andersbriegel2005} are given in italic.
 
 
-\textbf{Case 1:}
+\textbf{Case 1}(\textit{Case 1})\textbf{:}\\
-Both $o_i$ and $o_j$ commute with $CZ_{i,j}$. This is the case for exactly
+Both $o_a$ and $o_b$ commute with $CZ_{a,b}$. This is the case for exactly
 four vertex operators: $\mathcal{Z} = \left\{I, Z, S, S^\dagger\right\}$.
 In this case the CZ can be pulled past the vertex operators and just the edges
-are changed to $E' = E \Delta \left\{\{i,j\}\right\}$.
+are changed to $E' = E \Delta \left\{\{a,b\}\right\}$.
 
-\textbf{Case 2:}
+\textbf{Case 2}(\textit{Sub-Sub-Case 2.2.1})\textbf{:}\\
-At least one vertex operator does not commute with $CZ_{i,j}$. 
+The two qbits are isolated: From the definition of the graph state it is clear that
+any isolated clique of the graph can be treated independently. Therefore the two isolated qbits
+can be treated as an independent state and the set of two qbit stabilizer states is finite. An
+upper bound to the number of two qbit stabilizer states is given by $2\cdot24^2$: with or without 
+an edge between the qbits and $24$ Clifford operators on each vertex.
 
-\textbf{Sub-Case 2.1}
+All those states and the resulting state after a $CZ$ application can be computed and while doing so one
-Both 
+gets another interesting result that will be useful later: If one vertex has the vertex operator $I$ the
+resulting state can be chosen such that at least one of the vertex operators is $I$ again and in particular
+the identity on the vertex can be preserved under the application of a $CZ$.
+
+\textbf{Case 3}(\textit{Case 2, but one sub-case has been handled})\textbf{:}\\
+At least one vertex operator does not commute with $CZ$ and at least one vertex 
+has non-operand neighbours. In this case one can try to clear the vertex operators
+which will succeed for at least one qbit.
+
+The transformation given in 
+Lemma \ref{lemma:M_a} is used to "clear" the vertex operators. Recalling that
+the transformation $M_j$ toggles the neighbourhood of vertex $j$ gives substance
+to the following theorem:
+
+\begin{theorem}
+    A graph state $\ket{G}$ associated with $(V, E, O)$ is invariant when
+    simultaneously toggling the neighbourhood of a non-isolated qbit $j$
+    and right-multiplying $M_j^\dagger$ to the vertex operators in the sense
+    that $\sqrt{-iX}^\dagger = \sqrt{iX}$ is right-multiplied to $o_j$ and 
+    $\sqrt{iZ}^\dagger = \sqrt{-iZ}$ is right-multiplied to $o_l$ for all 
+    neighbours $l$ of $j$.
+
+    Without proof.
+\end{theorem}
+
+As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$ and $\sqrt{iZ}$.
+This yields an algorithm to reduce the vertex operator of a non-isolated qbit $j$ to the identity.
+The combined operation of toggling the neighbourhood of $j$ and right-multiplying
+$M_j^\dagger$ is called $L_j$ transformation and it transforms $(V, E, O)$ into a so-called
+local Clifford equivalent graphical representation.
+
+\begin{enumerate}
+    \item{Check whether $n_a \setminus \{b\} \neq \{\}$. If this condition does not hold
+            the vertex operator on $a$ cannot be cleared }
+    \item{Express the vertex operator $o_a$ as a product of $\sqrt{-iX}$ and $\sqrt{iZ}$.}
+    \item{Move from right to left through the product
+        \begin{enumerate}
+            \item{If the current matrix is $\sqrt{-iX}$ apply $L_a$}
+            \item{If the current matrix is $\sqrt{iZ}$ select a vertex
+                $j \in n_a \setminus \{b\}$ (which is possible as it has been checked before)
+                and apply $L_j$.}
+        \end{enumerate}
+        }
+\end{enumerate}
+
+This algorithm has the important properties that if the algorithm succeeds 
+$o_a = I$ and $o_b$ has picked up powers of $\sqrt{iZ}^\dagger$.
+If $b$ has non-operand neighbours after clearing the vertex operators on $a$ the vertex operators on $b$
+can be cleared using the same algorithm which gives $o_b = I$ and $o_a = (\sqrt{iZ}^\dagger)^l = S^l \in \mathcal{Z}$
+for some $l \in \{1, ..., 5\}$. Therefore Case 1 can now be applied.
+
+If $o_a$ could not be cleared $o_b$ can be cleared using the same procedure and after
+clearing $o_b$ one can retry to clear $o_a$. 
+
+In any case at least one vertex operator has been cleared. If both vertex operators have been
+cleared Case 1 will be applied, if there is just one cleared vertex operator it is sure that it 
+is the vertex with non-operand neighbours. Using this one can still apply a $CZ$: Without loss of generality
+assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form\cite{andersbriegel2005}
+
+\begin{equation}
+    \begin{aligned}
+        \ket{G} &= \left(\prod\limits_{o_i \in O} o_i\right) \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+}_n \\
+                &= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) o_b (CZ_{a,b})^s \ket{+}_n \\
+                &= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) \ket{+}_{n-2} \otimes \left(o_b (CZ_{a,b})^s \ket{+}_2\right) \\
+    \end{aligned}
+\end{equation}
+
+as $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicates whether there is an edge between $a$ and $b$.
+
+\begin{equation}
+    \begin{aligned}
+        CZ_{a,b}\ket{G} &= CZ_{a,b}\left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) o_b (CZ_{a,b})^s \ket{+} \\
+                        &= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) \ket{+}_{n-2} \otimes \left(CZ_{a,b} o_b (CZ_{a,b})^s \ket{+}_2\right) \\
+    \end{aligned}
+\end{equation}
+
+Using this one can re-use the method used in Case 2 to apply the $CZ$ while keeping the $o_a = I$.
+
+As both $CZ$ and $C_L$ can be applied to a graph state $\ket{G}$ this proofs constructively that
+the graphical representation of a stabilizer state is indeed able to represent any stabilizer state.
+If one wants to do computations using this formalism it is however also necessary to perform measurements.
+
+\subsubsection{Measurements on Graph States}

thesis/main.bib

@@ -129,3 +129,13 @@
     author={Peter Shor},
     note={https://journals.aps.org/pra/pdf/10.1103/PhysRevA.52.R2493}
 }
+
+@online{
+    pyqcs,
+    url={https://github.com/daknuett/pyqcs},
+    urldate={13.01.2020},
+    author={Daniel Knüttel},
+    title={PyQCS},
+    year=2020,
+    note={https://github.com/daknuett/pyqcs},
+}