From 606949d6b4a690d65aefe76d1e5ba501f5b89c72 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Tue, 25 Feb 2020 11:19:09 +0100 Subject: [PATCH] more changes from Simon --- thesis/chapters/stabilizer.tex | 47 +++++++++++++++++----------------- 1 file changed, 24 insertions(+), 23 deletions(-) diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex index e6c4315..1213ccd 100644 --- a/thesis/chapters/stabilizer.tex +++ b/thesis/chapters/stabilizer.tex @@ -382,7 +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 for a vertex $a \in V$ is: \begin{equation} M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j} @@ -599,15 +599,15 @@ As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$ and 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. +local Clifford equivalent graphical representation. The algorithm is given by the following steps: \begin{enumerate} \item{Check whether $n_a \setminus \{b\} \neq \{\}$. If this condition does not hold - the vertex operator on $a$ cannot be cleared } + 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 + \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 $\sqrt{-iX}$ 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 $L_j$.} @@ -625,9 +625,9 @@ If $o_a$ could not be cleared $o_b$ can be cleared using the same procedure and 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 +cleared Case 1 will be applied. If there is just one cleared vertex operator 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} +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} @@ -637,7 +637,7 @@ assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket \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 $b$. +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} @@ -646,25 +646,26 @@ as $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicat \end{aligned} \end{equation} -Using this one can re-use the method used in Case 2 to apply the $CZ$ while keeping the $o_a = I$. +This allows to re-use the method 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 +As it has been shown how both $CZ$ and $C_L$ act on a graph state $\ket{G}$ and the resulting state +is a graph state as well 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} 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 observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes +which is a quite expensive computation in theory. It is possible to simplify the problem by pulling the observable behind the vertex operators. For this consider the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$: \begin{equation} \begin{aligned} P_{a,s} \ket{\psi} &= P_{a,s} \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 \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,s} \ket{\bar{G}} \\ \end{aligned} @@ -682,10 +683,10 @@ as $o_a$ is a Clifford operator: \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 +Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. Therefore, it is enough to study the measurements of any Pauli operator on the vertex operator free graph states. The commutators of the observable -with the $K_G^{(i)}$ are quite easy to compute, note that Pauli matrices wither commute or anticommute -so it is easier to list the operators that anticommute: +with the $K_G^{(i)}$ are quite easy to compute. Note that Pauli matrices either commute or anticommute +and it is easier to list the operators that anticommute: \begin{equation} \begin{aligned} @@ -697,8 +698,8 @@ so it is easier to list the operators that anticommute: This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a (-X_a)$ is measured the result $s=0(1)$ is obtained with probability $1$ and $(V, E, O)$ is unchanged. -In any other case the results $s=1$ and $s=0$ both have probability $\frac{1}{2}$ and both -graph and vertex operators have to be updated. Further it is clear that measurement of $-\tilde{g}_a$ +In any other case the results $s=1$ and $s=0$ have probability $\frac{1}{2}$ and both +graph and vertex operators have to be updated. Further it is clear that measurements of $-\tilde{g}_a$ and $\tilde{g}_a$ are related by just inverting the result $s$. The calculations to obtain the transformation on graph and vertex operators are lengthy and follow @@ -712,10 +713,10 @@ the steps required to obtain the following results: \end{aligned} \end{equation} -These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement} -and the second part makes sure that the qbit $a$ is diagonal in the correct state of the measured state. +These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}. +The second part makes sure that the qbit $a$ is diagonal in measured observable and has the correct eigenvalue. When comparing with Lemma \ref{lemma:stab_measurement} in both cases $Y,Z$ the anticommuting stabilizer -$K_G^{(a)}$ is chosen. The graph is changed according to +$K_G^{(a)}$ is chosen. The graph is changed according to: \begin{equation} \begin{aligned} @@ -725,7 +726,7 @@ $K_G^{(a)}$ is chosen. The graph is changed according to \end{equation} -For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are +For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are: \begin{equation} \begin{aligned}