From ee3f52198f48c599310d4ce7ed18bff26d0383a1 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Sat, 14 Dec 2019 10:29:34 +0100 Subject: [PATCH] some work on the paper --- thesis/chapters/graph_simulator.tex | 159 ++++++++++++++++------------ thesis/chapters/introduction_qc.tex | 2 - thesis/chapters/stabilizer.tex | 125 +++++++++++++++++++++- 3 files changed, 214 insertions(+), 72 deletions(-) diff --git a/thesis/chapters/graph_simulator.tex b/thesis/chapters/graph_simulator.tex index f1cbf37..ccdebae 100644 --- a/thesis/chapters/graph_simulator.tex +++ b/thesis/chapters/graph_simulator.tex @@ -30,67 +30,67 @@ generated using $CZ$ and $C_1$ gates. \end{equation} \end{proof} -\begin{lemma} - One cannot measure phases by projecting states. -\end{lemma} -\begin{proof} - Let $\ket{\psi}$ be a state, $\ket{\varphi}\bra{\varphi}$ a projector. $\ket{\psi'} := \exp(i\phi)\ket{\psi}$ for some $\phi \in [0, 2\pi)$. - - \begin{equation} - \begin{aligned} - \bra{\psi'}\ket{\varphi}\bra{\varphi}\ket{\psi'} &= \exp(-i\phi)\bra{\psi}\ket{\varphi}\bra{\varphi}\exp(i\phi)\ket{\psi} \\ - & = \exp(-i\phi)\exp(i\phi)\bra{\psi}\ket{\varphi}\bra{\varphi}\ket{\psi} \\ - & = \bra{\psi}\ket{\varphi}\bra{\varphi}\ket{\psi} - \end{aligned} - \end{equation} -\end{proof} - -\begin{definition} - A phase $\phi \in [0, 2\pi)$ is called qbit-global, if for some qbit states $\ket{\psi}, \ket{\varphi}$ $\ket{\psi} = \exp(i\phi)\ket{\varphi}$. -\end{definition} - -\begin{lemma} - When entangling qbits via projections one can disregard qbit-global phases. - - Two qbits are entangled via projection, if for some single qbit gates $M,N$ - and two orthonormal states $\ket{a}, \ket{b}$ - - \begin{equation} - C^{M,N}(i,j) = \ket{a}\bra{a}_j \otimes M_i + \ket{b}\bra{b}_j \otimes N_i - \end{equation} - - \textbf{Remark.} - In particular when entangling states using $CX$ and $CZ$ one can disregard qbit-global phases. - This is immideatly clear when recalling \eqref{eq:CX_pr} and \eqref{eq:CZ_pr}. -\end{lemma} - -\begin{proof} - Let $\alpha, \beta \in [0, 2\pi)$ be some phases, $\ket{\psi}, \ket{\varphi}, \ket{\psi'} := \exp(i\alpha)\ket{\psi}, \ket{\varphi'} := \exp(i\beta)\ket{\varphi}$ some single qbit states, - $M, N, \ket{a}, \ket{b}, C^{M,N}(i,j)$ as defined above. - -\begin{equation} -\begin{aligned} - C^{M,N}(1, 0) (\ket{\psi'}\otimes\ket{\varphi'}) & = \ket{a}\braket{a}{\varphi'}\otimes M\ket{\psi'} + \ket{b}\braket{b}{\varphi'} \otimes N\ket{\psi'} \\ - & = \exp(i\beta)\ket{a}\braket{a}{\varphi}\otimes\exp(i\alpha)M\ket{\phi} + \exp(i\beta)\ket{b}\braket{b}{\varphi}\otimes\exp(i\alpha)N\ket{\phi}\\ - & = \exp(i(\beta + \alpha))(\ket{a}\braket{a}{\varphi}\otimes M\ket{\psi} + \ket{b}\braket{b}{\varphi} \otimes N\ket{\psi})\\ - & = \exp(i(\beta + \alpha))C^{M,N}(1, 0) (\ket{\psi}\otimes\ket{\varphi}) -\end{aligned} -\end{equation} - - Where $\exp(i(\beta + \alpha))$ is a multi-qbit-global phase which can be (following the above argumentation) disregarded. - -\end{proof} - -\begin{corrolary} - One can disregard global phases of elements of the $C_1$ group. -\end{corrolary} - -\begin{proof} - As it has been shown above a quantum computer cannot measure global phases. Also - the entanglement gates $CX, CZ$ map qbit-global phases to multi-qbit-global phases which cannot - be measured. It has been shown above that one can choose the $C_1$ operators such that they do not yield - a phase. -\end{proof} +%\begin{lemma} +% One cannot measure phases by projecting states. +%\end{lemma} +%\begin{proof} +% Let $\ket{\psi}$ be a state, $\ket{\varphi}\bra{\varphi}$ a projector. $\ket{\psi'} := \exp(i\phi)\ket{\psi}$ for some $\phi \in [0, 2\pi)$. +% +% \begin{equation} +% \begin{aligned} +% \bra{\psi'}\ket{\varphi}\bra{\varphi}\ket{\psi'} &= \exp(-i\phi)\bra{\psi}\ket{\varphi}\bra{\varphi}\exp(i\phi)\ket{\psi} \\ +% & = \exp(-i\phi)\exp(i\phi)\bra{\psi}\ket{\varphi}\bra{\varphi}\ket{\psi} \\ +% & = \bra{\psi}\ket{\varphi}\bra{\varphi}\ket{\psi} +% \end{aligned} +% \end{equation} +%\end{proof} +% +%\begin{definition} +% A phase $\phi \in [0, 2\pi)$ is called qbit-global, if for some qbit states $\ket{\psi}, \ket{\varphi}$ $\ket{\psi} = \exp(i\phi)\ket{\varphi}$. +%\end{definition} +% +%\begin{lemma} +% When entangling qbits via projections one can disregard qbit-global phases. +% +% Two qbits are entangled via projection, if for some single qbit gates $M,N$ +% and two orthonormal states $\ket{a}, \ket{b}$ +% +% \begin{equation} +% C^{M,N}(i,j) = \ket{a}\bra{a}_j \otimes M_i + \ket{b}\bra{b}_j \otimes N_i +% \end{equation} +% +% \textbf{Remark.} +% In particular when entangling states using $CX$ and $CZ$ one can disregard qbit-global phases. +% This is immideatly clear when recalling \eqref{eq:CX_pr} and \eqref{eq:CZ_pr}. +%\end{lemma} +% +%\begin{proof} +% Let $\alpha, \beta \in [0, 2\pi)$ be some phases, $\ket{\psi}, \ket{\varphi}, \ket{\psi'} := \exp(i\alpha)\ket{\psi}, \ket{\varphi'} := \exp(i\beta)\ket{\varphi}$ some single qbit states, +% $M, N, \ket{a}, \ket{b}, C^{M,N}(i,j)$ as defined above. +% +%\begin{equation} +%\begin{aligned} +% C^{M,N}(1, 0) (\ket{\psi'}\otimes\ket{\varphi'}) & = \ket{a}\braket{a}{\varphi'}\otimes M\ket{\psi'} + \ket{b}\braket{b}{\varphi'} \otimes N\ket{\psi'} \\ +% & = \exp(i\beta)\ket{a}\braket{a}{\varphi}\otimes\exp(i\alpha)M\ket{\phi} + \exp(i\beta)\ket{b}\braket{b}{\varphi}\otimes\exp(i\alpha)N\ket{\phi}\\ +% & = \exp(i(\beta + \alpha))(\ket{a}\braket{a}{\varphi}\otimes M\ket{\psi} + \ket{b}\braket{b}{\varphi} \otimes N\ket{\psi})\\ +% & = \exp(i(\beta + \alpha))C^{M,N}(1, 0) (\ket{\psi}\otimes\ket{\varphi}) +%\end{aligned} +%\end{equation} +% +% Where $\exp(i(\beta + \alpha))$ is a multi-qbit-global phase which can be (following the above argumentation) disregarded. +% +%\end{proof} +% +%\begin{corrolary} +% One can disregard global phases of elements of the $C_1$ group. +%\end{corrolary} +% +%\begin{proof} +% As it has been shown above a quantum computer cannot measure global phases. Also +% the entanglement gates $CX, CZ$ map qbit-global phases to multi-qbit-global phases which cannot +% be measured. It has been shown above that one can choose the $C_1$ operators such that they do not yield +% a phase. +%\end{proof} \begin{definition} \begin{equation} @@ -100,9 +100,7 @@ generated using $CZ$ and $C_1$ gates. \end{definition} \textbf{Remark.} When computing the elements of $C_L$ and their products one will realize that $C_L$ is not a group. -If one however disregards a global phase the product of two $C_L$ elements will be in $C_L$ again. Because the global phases -can be disregarded as discussed above $C_L$ will be used from now on instead of $C_1$. - +If one however disregards a global phase the product of two $C_L$ elements will be in $C_L$ again. \begin{theorem} \begin{equation} | C_L | = 24 @@ -111,13 +109,37 @@ can be disregarded as discussed above $C_L$ will be used from now on instead of \begin{proof} It is clear that $\forall a \in C_L$ a is a group isomorphism $P \circlearrowleft$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$. - Therefore $a$ will preserve the (anti-)commutator relations of $P$. Also $P$ is generated by $X,Z$ when disregarding a phase wich - does not matter for anticommutator relations. - This means that $X$ can be mapped to any $p \in P$ which are six elements disregarding + 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. + + 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 + anti-commute with the image of $X$, so there are four degrees of freedom left which gives + a total of $24$ degrees of freedom. - FIXME \end{proof} +\begin{theorem} + One can use $C_L$ instead of $C_1$ when studying stabilizer states and + the choice of $C_L$ is arbitrary. +\end{theorem} +\begin{proof} + Let $\ket{\psi}$ be a stabilizer stabilized by $\langle S_i \rangle_i$. When applying an $a \in C_1$ to + $\ket{\psi}$ $a\ket{\psi}$ is stabilized by $\langle a S_i a^\dagger \rangle_i$, let $b \in C_L$ s.t. + $a = \exp(i\phi)b$ then $\forall j$ $a S_j a^\dagger = \exp(i\phi)b S_j \exp(-i\phi)b^\dagger = b S_j b^\dagger$, + so the dynamics of a state under the local Clifford group is fully described by $C_L$ and the choice of + the phases are arbitrary. +\end{proof} + +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 +$\langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ that will be used in one specific operation on graph states. + +$$ S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)$$ +$$ \sqrt{-iX} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right)$$ +$$ \sqrt{-iZ} = \exp(-i\frac{\pi}{4})\left(\begin{array}{cc} 1 & 0 \\ 0 & -i \end{array}\right)$$ + \subsection{Introduction to the Graph Formalism} The first step towards the simulation in the graph formalism has been @@ -125,7 +147,6 @@ the discovery of the stabilizer states and stabilizer circuits \cite{gottesman20 They led to the faster simulation using stabilizer tableaux\cite{gottesman_aaronson2008} and later to the graph formalism\cite{schlingenmann2001}\cite{andersbriegel2005}\cite{vandennest_ea2004}. The following discussion eludicates the graph formalism and explains how the graph simulator works. -Some parts will be kept short as they can be looked up in \cite{andersbriegel2005}. A naive state is just a vector containing the coefficients $c_i$ as defined in \ref{ref:nqbitsystems}. It is a quite straight forward approach and gates are applied by updating the coefficients according diff --git a/thesis/chapters/introduction_qc.tex b/thesis/chapters/introduction_qc.tex index 583370c..e9686af 100644 --- a/thesis/chapters/introduction_qc.tex +++ b/thesis/chapters/introduction_qc.tex @@ -68,7 +68,6 @@ FIXME: rewrite this. 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}. - The matrix representation of $CX$ and $CZ$ for two qbits is given by \begin{equation} @@ -80,7 +79,6 @@ The matrix representation of $CX$ and $CZ$ for two qbits is given by Where $1$ is the act-qbit and $0$ the control-qbit. In words $CX$ ($CZ$) apply an $X$ ($Z$) gate on the act-qbit, if the control-qbit is set. - The following notation\cite{dahlberg_ea2019} can be more handy when discussing more qbits: \begin{equation}\label{eq:CX_pr} diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex index 1727dc5..a81ed77 100644 --- a/thesis/chapters/stabilizer.tex +++ b/thesis/chapters/stabilizer.tex @@ -149,7 +149,7 @@ were derived from the vertex operator-free graph states. \begin{definition} \label{def:vop_free_g_state} - A $n$ qbit vertex operator-free graph state $\ket{\overline{G}}$ is associated with a graph $(V, E)$ + A $n$ qbit vertex operator-free graph state $\ket{\overline{G}}$ is associated with a graph $\bar{G} = (V, E)$ by the $n$ operators \begin{equation} @@ -222,6 +222,94 @@ were derived from the vertex operator-free graph states. as $X, Z$ anticommute and $Z\ket{1} = -1\ket{1}$. \end{proof} +\begin{definition} + Let $\bar{G} = (V, E)$ be a graph as in definition \ref{def:vop_free_g_state}. + For a vertex $i \in V$ $n_i := \{j \in V | \{i,j\} \in E\}$ is called the neighbourhood + of $i$. +\end{definition} + +\begin{lemma} + Let $\bar{G}$, $\ket{\bar{G}}$, $K_G^{(i)}$ be as in definition \ref{def:vop_free_g_state} and for + a vertex $a \in V$ set + + \begin{equation} + M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j} + \end{equation} + + Then the graph $\bar{G}'$ associated with $\ket{\bar{G}'} = M_a\ket{\bar{G}}$ is changed according to + the following equations: + + \begin{equation} + \begin{aligned} + n_a' &= n_a \\ + n_j' &= n_j, \hbox{ if } j \notin n_a\\ + n_j' &= (n_j \cup n_a) \setminus (n_j \cap n_a), \hbox{ if } j \in n_a + \end{aligned} + \end{equation} + + I.e. the neighbourhood of $a$ is toggled. +\end{lemma} + +\begin{proof} + $\ket{\bar{G}'}$ is stabilized by $\langle M_a K_G^{(i)} M_a^\dagger \rangle_i$, so it is sufficient + to study how the $ K_G^{(i)}$ change under $M_a$. + At first note that $[K_G^{(a)}, M_a] = 0$ and $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$, + so the first two equations follow trivially. For $j \in n_a$ set + + \begin{equation} + \begin{aligned} + S^{(j)} &= M_a K_G^{(j)} M_a^\dagger \\ + &= \sqrt{-iX_a} \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}\right) + \sqrt{iZ_j} K_G^{(j)} \sqrt{iZ_j}^\dagger + \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}^\dagger\right) + \sqrt{-iX_a}^\dagger \\ + &= \sqrt{-iX_a} \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}\right)\sqrt{iZ_j} + X_j \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) Z_a + \sqrt{iZ_j}^\dagger + \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}^\dagger\right) + \sqrt{-iX_a}^\dagger \\ + &= \sqrt{iZ_j} X_j\sqrt{iZ_j}^\dagger\left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) + \sqrt{-iX_a} Z_a \sqrt{-iX_a}^\dagger \\ + &= (-1)^2 Y_j Y_a \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right)\\ + &= (-1)i^2 Z_j X_a X_j Z_a \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) + \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 + 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 + + \begin{equation} + \begin{aligned} + S^{(j)} &= Z_j X_a X_j Z_a \prod\limits_{l \in J} Z_l\\ + &= Z_j X_a X_j Z_a \left(\prod\limits_{l \in J} Z_l\right) + \left(\prod\limits_{l \in I}Z_l\right) + \left(\prod\limits_{l \in I}Z_l\right) \\ + &= Z_j X_a X_j Z_a \left(\prod\limits_{l \in ((I\cup J) \setminus (I\cap J))} Z_L\right) + \left(\prod\limits_{l \in I}Z_l\right) \\ + &= K_{G'}^{(a)} K_{G'}^{(j)} \\ + &= K_{G}^{(a)} K_{G'}^{(j)} + \end{aligned} + \end{equation} + + Using this ine 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}'} + = K_{G'}^{(j)}K_{G}^{(a)}\ket{\bar{G}'} = K_{G'}^{(j)}\ket{\bar{G}'} + \end{equation} + + Because $\{K_G^{(i)} | i \notin n_a\} \cup \{S^{(i)} | i\in n_a\}$ and + $\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}$ are both $n$ commuting + 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 $K_{G'}^{(j)}$ + $\langle\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}\rangle$ + are the stabilizers of $\ket{\bar{G}'}$ and the associated graph is changed as given + in the third equation. + +\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 Pauli operator $g \in \{Z_a, Y_a, X_a\}$ that is to be measured. @@ -255,6 +343,7 @@ 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} + \label{lemma:Z_measurement} \begin{enumerate} \item{For a result $+Z_a$ the new state is $\ket{+_Z}_a \otimes \ket{\bar{G}'}$ @@ -281,3 +370,37 @@ In the case $g = X_a$ and $n_a = \{\}$ the graph is obviously unchanged. \end{proof} +\begin{lemma} + When $Y_a$ is measured with a result $s \in \{0, 1\}$ the state after the measurement is + + \begin{equation} + \ket{(-1)^s_Y}_a \otimes U_{Y,s} \ket{\bar{G}'} + \end{equation} + \begin{equation} + U_{Y,s} = \prod\limits_{l \in n_a} \sqrt{(-1)^s iZ_l} + \end{equation} + + Where $\ket{\bar{G}'}$ is associated with the graph $G' = (V \setminus \{a\}, E')$ + with $E'$ being changed according to $\forall i \in n_a$ $n_i' = (n_i \cup n_a) \setminus (n_i \cap n_a) \setminus \{a\}$. +\end{lemma} + +\begin{proof} + It is known that $(-1)^s Y_a$ has to stabilize the state after measurement, + further $\forall j \in n_a$ $S^{(j)} = K_G^{(a)} K_G^{(j)}$, are the stabilizers of the new + state around the measured qbit and the stabilizers $K_G^{(j)}, j \notin n_a$ are unchanged. + Let $j \in n_a$ and $n_a =: \{j\} \cup I$, $n_j =: \{a\} \cup J$. + + \begin{equation} + \begin{aligned} + S^{(i)} &= K_G^{(a)} K_G^{(j)} \\ + &= X_a \left(\prod\limits_{l \in n_a} Z_l\right)X_j\left(\prod\limits_{l\in n_j} Z_l\right) \\ + &= X_a Z_a Z_j X_j \left(\prod\limits_{l \in I} Z_l\right)\left(\prod\limits_{l \in J} Z_l\right)\\ + &= X_a Z_a Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J)} Z_l\right) \\ + &= iY_a Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J) \setminus \{a\}} Z_l\right) \\ + &= (-1)^si S^{(a)} Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J)\setminus\{a\}} Z_l\right) \\ + &= S^{(a)} \sqrt{(-1)^siZ_j} K_{G'}^{(j)} \sqrt{(-1)^siZ_j}^\dagger + \end{aligned} + \end{equation} + with $G'$ as above, the rest of the argument is analogous to lemma \ref{lemma:Z_measurement}. +\end{proof} +