From 650d73cccac084754081642036eec035c13ddbc0 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Thu, 5 Mar 2020 10:48:13 +0100 Subject: [PATCH] some changes proposed by Andreas --- thesis/chapters/quantum_computing.tex | 54 ++++++------ thesis/chapters/stabilizer.tex | 118 ++++++++++++++------------ 2 files changed, 91 insertions(+), 81 deletions(-) diff --git a/thesis/chapters/quantum_computing.tex b/thesis/chapters/quantum_computing.tex index efcddfc..a42458a 100644 --- a/thesis/chapters/quantum_computing.tex +++ b/thesis/chapters/quantum_computing.tex @@ -9,7 +9,7 @@ $ \{\ket{\uparrow} \equiv \ket{1}, \ket{\downarrow} \equiv \ket{0}\} $ with $\braket{\uparrow}{\downarrow} = 0$. In the following this basis will be called the $Z$ basis in analogy to the conventions used in spin systems ($\sigma_Z$). For some computations - it can be useful to have component vectors, $\ket{\uparrow} \equiv \left(\begin{array}{c} 0 \\ 1\end{array} \right)$ and + it can be useful to have component vectors; $\ket{\uparrow} \equiv \left(\begin{array}{c} 0 \\ 1\end{array} \right)$ and $\ket{\downarrow} \equiv \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ are used in these cases. \end{definition} @@ -17,7 +17,7 @@ A gate acting on a qbit is a unitary operator $G \in SU(2)$. One can show that $\forall G \in SU(2)$ $G$ can be approximated arbitrarily good as a product of unitary generator matrices \cite[Chapter 4.3]{kaye_ea2007}\cite[Chapter 2]{marquezino_ea_2019}; -common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ with +common choices for the generators are $ X, H, R_{\phi}$ or $Z, H, R_{\phi}$ with \label{ref:singleqbitgates} \begin{equation} @@ -26,19 +26,19 @@ common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ wit & Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right) \\ & H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right) \\ & R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)\\ - & I := \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right) \\ + & I := \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right). \\ \end{aligned} \end{equation} Note that $X = HZH$ and $Z = R_{\pi}$, so the set of $H, R_\phi$ is sufficient. Further note that the basis vectors are chosen s.t. $Z\ket{0} = +\ket{0}$ and $Z\ket{1} = -\ket{1}$; -transforming to the other Pauli eigenstates is done using $H$ and $SH$: +transforming to the other Pauli eigenstates is done using $H$ and $SH$ \begin{equation} S := R_{\frac{\pi}{2}} = \left(\begin{array}{cc} 1 & 0 \\ 0 & i\end{array}\right) \end{equation} \begin{equation} - S H Z H^\dagger S^\dagger = S X S^\dagger = Y + S H Z H^\dagger S^\dagger = S X S^\dagger = Y. \end{equation} The following states are the $\pm 1$ eigenstates of the $X,Y,Z$ operators @@ -59,12 +59,12 @@ and will be used in some calculations later: \label{ref:many_qbits} \begin{postulate} - A $n$-qbit quantum mechanical state is the tensor product \cite[Definition 14.3]{wuest1995} of the $n$ one-qbit -states. + A $n$-qbit quantum mechanical state is the tensor product of the $n$ one-qbit + states \cite[Postulate 4]{nielsen_chuang_2010}. \end{postulate} Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$ -be the basis of the one-qbit systems. Then two-qbit basis states are: +be the basis of the one-qbit systems. Then two-qbit basis states are \begin{equation} \ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right) @@ -76,21 +76,21 @@ be the basis of the one-qbit systems. Then two-qbit basis states are: \ket{2} := \ket{0b10} := \ket{1}_s \otimes \ket{0}_s := \left(\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right) \end{equation} \begin{equation} - \ket{3} := \ket{0b11} := \ket{1}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right) + \ket{3} := \ket{0b11} := \ket{1}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right). \end{equation} The $n$-qbit basis states can be constructed in a similar manner. -A general $n$-qbit state can then be written as a superposition of the -integer states: +A general $n$-qbit state is now a superposition of the +integer states \begin{equation} \label{eq:ci} - \ket{\psi} = \sum\limits_{i = 0}^{2^n - 1} c_i \ket{i} + \ket{\psi} = \sum\limits_{i = 0}^{2^n - 1} c_i \ket{i} . \end{equation} -With the normation condition: +with the normation condition \begin{equation} - \sum\limits_{i = 0}^{2^n - 1} |c_i|^2 = 1 + \sum\limits_{i = 0}^{2^n - 1} |c_i|^2 = 1. \end{equation} The states $\ket{i}$ for $i = 0, ..., 2^{n}-1$ are called integer states. Note @@ -133,12 +133,12 @@ is acting on qbit $j$. where without loss of generality $j < i$; the other case is analogous. - In particular for $X, Z$: + In particular for $X, Z$ \begin{equation}\label{eq:CX_pr} CX(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes X_i \end{equation} \begin{equation}\label{eq:CZ_pr} - CZ(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes Z_i + CZ(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes Z_i . \end{equation} \end{definition} @@ -148,13 +148,13 @@ $CU$ applies the gate $U$ to the act-qbit if the control-qbit is in its $\ket{1} One can show that the gates in \ref{ref:singleqbitgates} together with an entanglement gate, such as $CX$ or $CZ$ are enough to generate an arbitrary $n$-qbit gate \cite[Chapter 4.3]{kaye_ea2007}\cite{barenco_ea_1995}. -The matrix representation of $CX$ and $CZ$ for two qbits is given by: +The matrix representation of $CX$ and $CZ$ for two qbits is given by \begin{equation} CX_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right) \end{equation} \begin{equation} - CZ_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right) + CZ_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right). \end{equation} \subsubsection{Measurements} @@ -181,13 +181,13 @@ $i$ $Z_i$ is measured. The $+1$ eigenvalue of $Z_i$ is $\ket{0}_i$, $\ket{1}_i$ \end{corrolary} \begin{proof} - The measuerment in not injective: Measuring both + The measuerment is not injective: Measuring both $\ket{0}$ and $\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ (can) map to $\ket{0}$. Any unitary matrix $U$ has the inverse $U^\dagger \equiv U^{-1}$. \end{proof} -Because a measurement is not unitary it is not a gate in the sense the definition above. +Because a measurement is not unitary it is not a gate in the sense of the definition above. In the following discussion the term \textit{measurement gate} will be used from time to time as a measurement can be treated similarely while doing numerics. @@ -196,27 +196,27 @@ to time as a measurement can be treated similarely while doing numerics. As mentioned in \ref{ref:many_qbits} one can approximate an arbitrary $n$-qbit gate $U$ as a product of some single-qbit gates and either $CX$ or $CZ$. -Writing (possibly huge) products of matrices is quite unpractical and very much +Writing (possibly huge) products of matrices is quite unpractical and unreadable. To address this problem quantum circuits have been introduced. These represent the qbits as a horizontal line and a gate acting on a qbit is a box with a name on the respective line. Quantum circuits are read from left to right. This means that a gate $U_i = Z_i X_i H_i$ has the -circuit representation: +circuit representation \[ \Qcircuit @C=1em @R=.7em { & \gate{H} & \gate{X} & \gate{Z} &\qw \\ -} +}. \] The controlled gates (such as $CX$ and $CZ$) have a vertical line from the control-qbit to -the gate, for instance the circuit for $CZ_{2, 1}CX_{2,0}$ is: +the gate, for instance the circuit for $CZ_{2, 1}CX_{2,0}$ is \[ \Qcircuit @C=1em @R=.7em { & \qw & \ctrl{2} & \qw & \qw &\qw \\ & \qw & \qw & \qw & \ctrl{1} &\qw \\ -& \qw & \gate{X} & \qw & \gate{Z} &\qw \\ +& \qw & \gate{X} & \qw & \gate{Z} &\qw &\rstick{.}\\ } \] @@ -241,12 +241,12 @@ algorithm that can be used to analyze the spectrum of the transfer matrix: T = \exp(itH) \end{equation} -The eigenvalue of $T$ can now be estimated by using the phase estimation circuit: +The eigenvalue of $T$ can now be estimated by using the phase estimation circuit \[ \Qcircuit @C=1em @R=.7em { & \lstick{\ket{0}} & {/^N} \qw & \gate{H^{\otimes n}} & \ctrl{1} & \gate{FT^\dagger} & \qw & \meter & \rstick{x}\\ - & \lstick{\ket{\varphi}} & {/} \qw & \qw & \gate{T} & \qw & {/} \qw & \qw& \rstick{\ket{\varphi}} \\ + & \lstick{\ket{\varphi}} & {/} \qw & \qw & \gate{T} & \qw & {/} \qw & \qw& \rstick{\ket{\varphi}.} \\ } \] diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex index 1213ccd..4587a1d 100644 --- a/thesis/chapters/stabilizer.tex +++ b/thesis/chapters/stabilizer.tex @@ -18,10 +18,10 @@ performance has since been improved to $n\log(n)$ time on average \cite{andersbr \begin{definition} \begin{equation} - P := \{\pm 1, \pm i\} \cdot \{I, X, Y, Z\} + P := \{\pm I, \pm X, \pm Y, \pm Z, \pm iI, \pm iX, \pm iY, \pm iZ\} \end{equation} - Is called the Pauli group. + with the matrix product is called the Pauli group \cite{andersbriegel2005}. \end{definition} The group property of $P$ can be verified easily. Note that @@ -34,7 +34,7 @@ the elements of $P$ either commute or anticommute. P_n := \left\{\bigotimes\limits_{i=0}^{n-1} p_i \middle| p_i \in P\right\} \end{equation} - is called the multilocal Pauli group on $n$ qbits. + is called the multilocal Pauli group on $n$ qbits \cite{andersbriegel2005}. \end{definition} The group property of $P_n$ and the (anti-)commutator relationships follow directly from its definition @@ -44,13 +44,15 @@ via the tensor product. \subsubsection{Stabilizers} +The discussion below follows the argumentation given in \cite{nielsen_chuang_2010}. + \begin{definition} \label{def:stabilizer} An abelian subgroup $S = \{S^{(0)}, ..., S^{(N)}\}$ of $P_n$ is called a set of stabilizers iff \begin{enumerate} \item{$\forall i,j = 1, ..., N$: $[S^{(i)}, S^{(j)}] = 0$%: $S^{(i)}$ and $S^{(j)}$ commute } - \item{$-I \notin S$} + \item{$-I \notin S$ } \end{enumerate} \end{definition} @@ -61,7 +63,7 @@ via the tensor product. \begin{enumerate} \item{$\pm iI \notin S$} \item{$(S^{(i)})^2 = I$ for all $i$} - \item{$S^{(i)}$ are hermitian for all $i$} + \item{$S^{(i)}$ are hermitian for all $i$ } \end{enumerate} \end{lemma} \begin{proof} @@ -71,7 +73,7 @@ via the tensor product. \item{From the definition of $S$ ($G_n$ respectively) follows that any $S^{(i)} \in S$ has the form $\pm i^l \left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$ where $\tilde{p}_j \in \{X, Y, Z, I\}$ and $l \in \{0, 1\}$. As $\left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$ - is hermitian $(S^{(i)})^2$ is either $+I$ or $-I$. As $-I \notin S$ $(S^{(i)})^2 = I$ follows directly. + is hermitian and unitary $(S^{(i)})^2$ is either $+I$ or $-I$. As $-I \notin S$ $(S^{(i)})^2 = I$ follows directly. } \item{Following the argumentation above $\left(S^{(i)}\right)^2 = -I \Leftrightarrow l=1$ therefore $\left(S^{(i)}\right)^2 = -I \Leftrightarrow \left(S^{(i)}\right)^\dagger \neq S^{(i)}$.} @@ -113,7 +115,7 @@ can be diagonalized simultaneously. This motivates and justifies the following d \end{equation} is called the space of stabilizer states associated with $S$ and one says - $\ket{\psi}$ is stabilized by $S$. + $\ket{\psi}$ is stabilized by $S$ \cite{nielsen_chuang_2010}. \end{definition} It is clear that to show the stabilization property of @@ -121,13 +123,13 @@ $S$ the proof for the generators is sufficient, as all the generators forming an element in $S$ can be absorbed into $\ket{\psi}$. The dimension of $V_S$ is not immediately clear. One can however show that for a set of stabilizers $\langle S^{(i)} \rangle_{i=1, ..., n-m}$ the dimension -$dim V_S = 2^m$ \cite[Chapter 10.5]{nielsen_chuang_2010}. This yields the following important +$\dim V_S = 2^m$ \cite[Chapter 10.5]{nielsen_chuang_2010}. This yields the following important result: \begin{theorem} \label{thm:unique_s_state} For a $n$ qbit system and stabilizers $S = \langle S^{(i)} \rangle_{i=1, ..., n}$ the stabilizer - space $V_S$ has $dim V_S = 1$, in particular there exists an up to a trivial phase unique + space $V_S$ has $\dim V_S = 1$, in particular there exists an up to a trivial phase unique state $\ket{\psi}$ that is stabilized by $S$. Without proof. @@ -147,7 +149,7 @@ and a unitary transformation $U$ that describes the dynamics of the system, i.e. \end{equation} It is clear that in general $\ket{\psi'}$ will not be stabilized by $S$ anymore. There are -however some statements that can still be made: +however some statements that can still be made \cite{nielsen_chuang_2010}: \begin{equation} \begin{aligned} @@ -169,7 +171,7 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers. C_n := \left\{U \in SU(2^n) \middle| \forall p \in P_n: UpU^\dagger \in P_n \right\} \end{equation} - is called the Clifford group. $C_1 =: C_L$ is called the local Clifford group. + is called the Clifford group. $C_1 =: C_L$ is called the local Clifford group \cite{andersbriegel2005}. \end{definition} \begin{theorem} @@ -212,10 +214,10 @@ product of $n$ Pauli matrices. This has led to the simulation using stabilizer t Interestingly also measurements are dynamics covered by the stabilizers. When an observable $g_a \in \{\pm X_a, \pm Y_a, \pm Z_a\}$ acting on qbit $a$ is measured -one has to consider the projector: +one has to consider the projector \begin{equation} - P_{g_a,s} = \frac{I + (-1)^s g_a}{2} + P_{g_a,s} = \frac{I + (-1)^s g_a}{2}. \end{equation} If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with probability $1$ @@ -230,19 +232,20 @@ and the stabilizers are left unchanged: \end{aligned} \end{equation} -As the state that is stabilized by $S$ is unique $\ket{\psi'} = \ket{\psi}$. +As the state that is stabilized by $S$ is unique $\ket{\psi'} = \ket{\psi}$ \cite{nielsen_chuang_2010}. If $g_a$ does not commute with all stabilizers the following lemma gives the result of the measurement. \begin{lemma} \label{lemma:stab_measurement} - Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$. When measuring + Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$ and + $J^c := \left\{S^{(i)} \middle| S^{(i)} \notin J \right\}$. When measuring $\frac{I + (-1)^s g_a}{2} $ $s=1$ and $s=0$ are obtained with probability $\frac{1}{2}$ and after choosing - a $j \in J$ the new state $\ket{\psi'}$ is stabilized by + a $j \in J$ the new state $\ket{\psi'}$ is stabilized by \cite{nielsen_chuang_2010} \begin{equation} - \langle \{(-1)^s g_a\} \cup \left\{S^{(i)} S^{(j)} \middle| S^{(i)} \in J \setminus \{S^{(j)}\} \right\} \cup J^c\rangle + \langle \{(-1)^s g_a\} \cup \left\{S^{(i)} S^{(j)} \middle| S^{(i)} \in J \setminus \{S^{(j)}\} \right\} \cup J^c \rangle. \end{equation} \end{lemma} @@ -276,7 +279,7 @@ the result of the measurement. \end{equation} The state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$, and by - $S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$. + $S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$ \cite{nielsen_chuang_2010}. \end{proof} \subsection{The VOP-free Graph States} @@ -287,17 +290,17 @@ vertex operator-free will be clear in the following section about graph states. \begin{definition} \label{def:graph} - The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$. + The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$ elements. In the following $V = \{0, ..., n-1\}$ will be used. - $E$ is the set of edges $E = \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$. + $E$ is the set of edges $E \subset \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$. For a vertex $i$ $n_i := \left\{j \in V \middle| \{i, j\} \in E\right\}$ is called the neighbourhood - of $i$. + of $i$ \cite{hein_eisert_briegel2008}. \end{definition} -This definition of a graph is way less general than the definition of a mathematical graph. +This definition of a graph is way less general than the definition of a graph in graph theory. Using this definition will however allow to avoid an extensive list of constraints on the -mathematical graph that are implied in this definition. +graph from graph theory that are implied in this definition. \begin{definition} For a graph $G = (V = \{0, ..., n-1\}, E)$ the associated stabilizers are @@ -305,7 +308,7 @@ mathematical graph that are implied in this definition. K_G^{(i)} := X_i \prod\limits_{\{i,j\} \in E} Z_j \end{equation} for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is the state stabilized by - $\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$. + $\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$ \cite{hein_eisert_briegel2008}. \end{definition} It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute @@ -321,16 +324,19 @@ from the graph. \begin{lemma} For a graph $G = (V, E)$ the associated state $\ket{\bar{G}}$ is - constructed using + constructed using \cite{hein_eisert_briegel2008} \begin{equation} \begin{aligned} \ket{\bar{G}} &= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right)\left(\prod\limits_{i \in V} H_i\right) \ket{0} \\ - &= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+} \\ + &= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+} .\\ \end{aligned} \end{equation} \end{lemma} \begin{proof} + FIXME: This + + 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 $\prod\limits_{\{l,k\} \in E} := \prod\limits_{\{l,k\} \in E, l < k}$ @@ -374,28 +380,28 @@ is done by using the symmetric set difference: \begin{definition} For two finite sets $A,B$ the symmetric set difference $\Delta$ is - defined as: + defined as \cite{hein_eisert_briegel2008} \begin{equation} - A \Delta B = (A \cup B) \setminus (A \cap B) + A \Delta B = (A \cup B) \setminus (A \cap B). \end{equation} \end{definition} Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$. -Another transformation on the VOP-free graph states for a vertex $a \in V$ is: +Another transformation on the VOP-free graph states for a vertex $a \in V$ is \cite{andersbriegel2005} \begin{equation} - M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j} + M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}. \end{equation} This transformation toggles the neighbourhood of $a$ which is an operation -that will be used later. +that will be used later\cite{andersbriegel2005}. \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: + graph is updated according to\cite{andersbriegel2005}: \begin{equation} \begin{aligned} n_a' &= n_a \\ @@ -489,7 +495,7 @@ clear that both $C_L$ and $CZ$ can be applied to a general graph state. +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} \end{equation} - $o_i$ are called the vertex operators of $\ket{G}$. + $o_i$ are called the vertex operators of $\ket{G}$ \cite{andersbriegel2005}. \end{definition} Recalling the dynamics of stabilizer states the following relation follows immediately: @@ -499,16 +505,16 @@ Recalling the dynamics of stabilizer states the following relation follows immed \end{equation} 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), +Instead of storing $n^2$ $P$ 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 sufficient to store $n$ integers representing the vertex operators: \begin{theorem} - $C_L$ has $24$ degrees of freedom. + $C_L$ has $24$ degrees of freedom \cite{andersbriegel2005}. \end{theorem} \begin{proof} - It is clear that $\forall a \in C_L$ a is a group isomorphism $P_1 \circlearrowleft$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$. + It is clear that $\forall a \in C_L$ a is a group isomorphism $P \rightarrow P$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$. Therefore $a$ will preserve the (anti-)commutator relations of $P$. Further note that $Y = iXZ$, so one has to consider the anti-commutator relations of $X,Z$ only. @@ -521,7 +527,7 @@ representing the vertex operators: 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 -$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required 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 \cite{andersbriegel2005}. \begin{equation} S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right) @@ -541,12 +547,12 @@ 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} o_i' &= o_i &\mbox{if } i \neq j\\ - o_i' &= c o_i c^\dagger &\mbox{if } i = j\\ + o_i' &= c o_i c^\dagger &\mbox{if } i = j.\\ \end{aligned} \end{equation} @@ -554,6 +560,7 @@ 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 $CZ_{a,b}$ on $(V, E, O)$. The cases given here follow the implementation of a $CZ$ application in \cite{pyqcs}, the respective paragraphs from \cite{andersbriegel2005} are given in italic. +Most of the discussion follows the one given in \cite{andersbriegel2005} closely. \textbf{Case 1}(\textit{Case 1})\textbf{:}\\ @@ -590,7 +597,7 @@ to the following theorem: 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 $j$. + neighbours $l$ of $j$ \cite{andersbriegel2005}. Without proof. \end{theorem} @@ -627,13 +634,13 @@ 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 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} \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) \\ + &= \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} @@ -655,11 +662,14 @@ If one wants to do computations using this formalism it is however also necessar \subsubsection{Measurements on Graph States} +This is adapted from \cite{andersbriegel2005}; measurement results and updating the graph after +a measurement is described in \cite{hein_eisert_briegel2008}. + 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 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 $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$: +the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$ \begin{equation} \begin{aligned} @@ -667,32 +677,32 @@ the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$: &= \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}} \\ + &= \left(\prod\limits_{o_i \in O} o_i \right) \tilde{P}_{a,s} \ket{\bar{G}} .\\ \end{aligned} \end{equation} This transformed projector has the important property that it still is a Pauli projector -as $o_a$ is a Clifford operator: +as $o_a$ is a Clifford operator \begin{equation} \begin{aligned} \tilde{P}_{a,s} &= o_a^\dagger P_a o_a \\ &= o_a^\dagger \frac{I + (-1)^sZ_a}{2} o_a \\ &= \frac{I + (-1)^s o_a^\dagger Z_a o_a}{2} \\ - &= \frac{I + (-1)^s \tilde{g}_a}{2} \\ + &= \frac{I + (-1)^s \tilde{g}_a}{2} .\\ \end{aligned} \end{equation} 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 either commute or anticommute -and it is easier to list the operators that anticommute: +and it is easier to list the operators that anticommute \begin{equation} \begin{aligned} A_{\pm X_a} &= \left\{j \middle| \{j, a\} \in E\right\}\\ A_{\pm Y_a} &= \left\{j \middle| \{j, a\} \in E\right\} \cup \{a\} \\ - A_{\pm Z_a} &= \{a\}\\ + A_{\pm Z_a} &= \{a\}.\\ \end{aligned} \end{equation} @@ -704,29 +714,29 @@ 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 the scheme of Lemma \ref{lemma:M_a}. \cite[Section IV]{hein_eisert_briegel2008} also contains -the steps required to obtain the following results: +the steps required to obtain the following results \begin{equation} \begin{aligned} U_{Z,s} &= \left(\prod\limits_{b \in n_a} Z_b^s\right) X_a^s H_a \\ - U_{Y,s} &= \prod\limits_{b \in n_a} \sqrt{(-1)^{1-s} iZ_b} \sqrt{(-1)^{1-s} iZ_a}\\ + U_{Y,s} &= \prod\limits_{b \in n_a} \sqrt{(-1)^{1-s} iZ_b} \sqrt{(-1)^{1-s} iZ_a}.\\ \end{aligned} \end{equation} 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} E'_{Z} &= E \setminus \left\{\{i,a\} \middle| i \in V\right\}\\ - E'_{Y} &= E \Delta (n_a \otimes n_a) \setminus \left\{\{i,a\} \middle| i \in V\right\}\\ + E'_{Y} &= E \Delta (n_a \otimes n_a) \setminus \left\{\{i,a\} \middle| i \in V\right\}.\\ \end{aligned} \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} @@ -739,7 +749,7 @@ For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are: E'_{X} = E &\Delta (n_b \otimes n_a) \\ & \Delta ((n_b \cap n_a) \otimes (n_b \cap n_a)) \\ & \Delta (\{b\} \otimes (n_a \setminus \{b\})) \\ - & \setminus \left\{\{i,a\} \middle| i \in V\right\}\\ + & \setminus \left\{\{i,a\} \middle| i \in V\right\}.\\ \end{aligned} \end{equation}