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Daniel Knüttel 2019-12-14 10:29:34 +01:00
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159
thesis/chapters/graph_simulator.tex
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@ -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

2
thesis/chapters/introduction_qc.tex
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@ -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}

125
thesis/chapters/stabilizer.tex
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@ -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}