some more changes by Simon

thesis/chapters/stabilizer.tex

@@ -331,22 +331,38 @@ from the graph.
     \end{equation}
 \end{lemma}
 \begin{proof}
-    Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$.
+    Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before.
+    Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$.
+    In the following discussion the direction $\prod\limits_{\{l,k\} \in E} := \prod\limits_{\{l,k\} \in E, l < k}$
+    is introduced as the graph is undirected and edges must not be handled twice.
     Set $\ket{\tilde{G}} := \left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right)\ket{+}$.
 
     \begin{equation}
         \begin{aligned}
-            K_G^{(i)} \ket{\tilde{G}} & = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right) \ket{+} \\
+            K_G^{(i)} \ket{\tilde{G}} 
-            & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
+            & = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)
-            & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l}}\ket{1}\bra{1}_k \otimes Z_l\right) X_i \ket{+} \\
+                \left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right) \ket{+} \\
-            & = \prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l} + \sum\limits_{\{i,j\} \in E} \delta_{j,k}}\ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
+            & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,k\} \in E}
+                \left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
+            & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,k\} \in E}
+                \left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l}}\ket{1}\bra{1}_k \otimes Z_l\right) X_i \ket{+} \\
+        \end{aligned}
+    \end{equation}
+    As $X,Z$ anticommute. $X_i$ can now be absorbed into $\ket{+}$. The next step is a bit tricky:
+    A $Z_j$ can be absorbed into a $\ket{0}\bra{0}_j$ giving no phase or into a $\ket{1}\bra{1}_j$ yielding
+    a phase of $-1$. If there is no projector on $j$ the $Z_j$ can be commuted to the next projector. 
+    It is guaranteed that a projector on $j$ exists by the definition of $\ket{\tilde{G}}$.
+
+    \begin{equation}
+        \begin{aligned}
+            K_G^{(i)} \ket{\tilde{G}} 
+            & = \prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l} + \delta_{j,k}}\ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
 			& = \prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\	
 		    & = +1 \ket{\tilde{G}}	
         \end{aligned}
     \end{equation}
 
-    as $X, Z$ anticommute and $Z\ket{1} = -1\ket{1}$.
+    The $\delta_{i,l} + \delta_{j,k}$ is either $0$ or $2$ by the definitions of $K_G^{(i)}$ and $\ket{\tilde{G}}$.
-
 \end{proof}
 
 \subsubsection{Dynamics of the VOP-free Graph States}
@@ -357,8 +373,8 @@ resulting in a multiplication of $Z_j$ to $K_G^{(i)}$ and $Z_i$ to $K_G^{(j)}$.
 is done by using the symmetric set difference:
 
 \begin{definition}
-    For to finite sets $A,B$ the symmetric set difference $\Delta$ is 
+    For two finite sets $A,B$ the symmetric set difference $\Delta$ is 
-    defined as
+    defined as:
 
     \begin{equation}
         A \Delta B = (A \cup B) \setminus (A \cap B)
@@ -366,7 +382,7 @@ is done by using the symmetric set difference:
 \end{definition}
 
 Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$.
-Another transformation on the VOP-free graph states is for a vertex $a \in V$
+Another transformation on the VOP-free graph states is for a vertex $a \in V$:
 
 \begin{equation}
     M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}
@@ -379,7 +395,7 @@ that will be used later.
     \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 
+    graph is updated according to:
     \begin{equation}
         \begin{aligned}
             n_a' &= n_a \\
@@ -413,10 +429,10 @@ that will be used later.
         \end{aligned}
     \end{equation}
 
-    One can now construct a new set of $K_{G'}^{(i)}$ s.t. $M_a \ket{\bar{G}}$ is the $+1$ eigenvalue
+    One can now construct a new set of $K_{G'}^{(i)}$ s.t. $M_a \ket{\bar{G}}$ is the $+1$ eigenstate
     of the new $K_{G'}^{(i)}$. It is clear that $\forall j \notin n_a$  $K_{G'}^{(j)} = K_G^{(j)}$.
     To construct the $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a = \{j\} \cup I$ and $n_j = \{a\} \cup J$.
-    Then
+    Then follows:
 
     \begin{equation}
         \begin{aligned}
@@ -431,7 +447,7 @@ that will be used later.
         \end{aligned}
     \end{equation}
 
-    Using this ine can show that $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$:
+    Using this one can show that $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$:
 
     \begin{equation}
         \ket{\bar{G}'} = S^{(j)}\ket{\bar{G}'} = K_{G}^{(a)} K_{G'}^{(j)}\ket{\bar{G}'} 
@@ -443,7 +459,7 @@ that will be used later.
     multi-local Pauli operators where the $S^{(i)}$ can be generated from the $K_{G'}^{(i)}$
     and $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$ 
     $\langle\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}\rangle$
-    are the stabilizers of $\ket{\bar{G}'}$ and the associated graph is changed as given
+    are the stabilizers of $\ket{\bar{G}'}$. Therefore the associated graph is changed as given
     in the third equation.
 \end{proof}
 
@@ -451,12 +467,12 @@ that will be used later.
 
 The definition of a VOP-free graph state above raises an obvious question: 
 Can any stabilizer state be described using just a graph?
-The answer is quite simple: No. The most simple cases are the single qbit
+The answer is straight forward: No. The most simple cases are the single qbit
-stated $\ket{0},\ket{1}$ and $\ket{+_Y}, \ket{-_Y}$. But there is a simple extension
+stated $\ket{0},\ket{1}$ and $\ket{+_Y}, \ket{-_Y}$. But there is an extension
 to the VOP-free graph states that allows the representation of an arbitrary
 stabilizer state. The proof that indeed any state can be represented is
-just constructive. As seen in theorem \ref{thm:clifford_group_approx} any $c \in C_n$
+purely constructive. As seen in theorem \ref{thm:clifford_group_approx} any $c \in C_n$
-can be constructed from $CZ$ and $C_L$ and in the following discussion it will become
+can be constructed from $CZ$ and $C_L$. In the following discussion it will become
 clear that both $C_L$ and $CZ$ can be applied to a general graph state.
 
 \subsubsection{Graph States and Vertex Operators}
@@ -467,7 +483,7 @@ clear that both $C_L$ and $CZ$ can be applied to a general graph state.
     A tuple $(V, E, O)$ is called the graphical representation of a stabilizer state
     if $(V, E)$ is a graph as in Definition \ref{def:graph} and $O = \{o_1, ..., o_n\}$ where $o_i \in C_L$.
 
-    The state $\ket{G}$ is defined by the eigenvalue relation
+    The state $\ket{G}$ is defined by the eigenvalue relation:
 
     \begin{equation}
         +1 \ket{G} = \left(\prod\limits_{j=1}^no_j\right) K_G^{(i)} \left(\prod\limits_{j=1}^no_j\right)^\dagger \ket{G}
@@ -485,8 +501,8 @@ Recalling the dynamics of stabilizer states the following relation follows immed
 The great advantage of this representation of a stabilizer state is its space requirement:
 Instead of storing $n^2$ $P_1$ matrices only some vertices (which often are implicit),
 the edges and some vertex operators ($n$ matrices) have to be stored. The following theorem
-will improve this even further: instead of $n$ matrices it is enough to store $n$ integers
+will improve this even further: instead of $n$ matrices it is sufficient to store $n$ integers
-representing the vertex operators is enough:
+representing the vertex operators:
 
 \begin{theorem}
     $C_L$ has $24$ degrees of freedom.
@@ -498,14 +514,14 @@ representing the vertex operators is enough:
     of $X,Z$ only.
 
     As the transformations are unitary they preserve eigenvalues, so $X$ can be mapped
-    to $\pm X, \pm Y, \pm Z$ which gives $6$ degrees of freedom, the image of $Z$ has to 
+    to $\pm X, \pm Y, \pm Z$ which gives $6$ degrees of freedom. Furthermore the image of $Z$ has to 
     anti-commute with the image of $X$ so $Z$ has four possible images under the transformation.
     This gives another $4$ degrees of freedom and a total of $24$.
 \end{proof}
 
-From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be used,
+From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be used.
-one can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is 
+One can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is 
-$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which will be used in one specific operation on graph states.
+$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in one specific operation on graph states.
 
 \begin{equation}
     S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)
@@ -525,7 +541,7 @@ basically a stabilizer tableaux that might require less memory than the tableaux
 CHP. The true power of this formalism is seen when studying its dynamics. The simplest case
 is a local Clifford operator $c_j$ acting on a qbit $j$: The stabilizers of are changed to 
 $\langle c_j S^{(i)} c_j^\dagger\rangle_i$. Using the definition of the graphical representation
-it is clear that just the vertex operators are changed and the new vertex operators are given by
+it is clear that just the vertex operators are changed and the new vertex operators are given by:
 
 \begin{equation}
     \begin{aligned}
@@ -550,11 +566,11 @@ are changed to $E' = E \Delta \left\{\{a,b\}\right\}$.
 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 
+upper bound to the number of two qbit stabilizer states is given by $2\cdot24^2$: With and without 
 an edge between the qbits and $24$ Clifford operators on each vertex.
 
-All those states and the resulting state after a $CZ$ application can be computed and while doing so one
+All those states and the resulting state after a $CZ$ application can be computed which leads to
-gets another interesting result that will be useful later: If one vertex has the vertex operator $I$ the
+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 $CZ$.