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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{equation}
\end{proof} \end{proof}
\begin{lemma} %\begin{lemma}
One cannot measure phases by projecting states. % One cannot measure phases by projecting states.
\end{lemma} %\end{lemma}
\begin{proof} %\begin{proof}
Let $\ket{\psi}$ be a state, $\ket{\varphi}\bra{\varphi}$ a projector. $\ket{\psi'} := \exp(i\phi)\ket{\psi}$ for some $\phi \in [0, 2\pi)$. % 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{equation}
\begin{aligned} % \begin{aligned}
\bra{\psi'}\ket{\varphi}\bra{\varphi}\ket{\psi'} &= \exp(-i\phi)\bra{\psi}\ket{\varphi}\bra{\varphi}\exp(i\phi)\ket{\psi} \\ % \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} \\ % & = \exp(-i\phi)\exp(i\phi)\bra{\psi}\ket{\varphi}\bra{\varphi}\ket{\psi} \\
& = \bra{\psi}\ket{\varphi}\bra{\varphi}\ket{\psi} % & = \bra{\psi}\ket{\varphi}\bra{\varphi}\ket{\psi}
\end{aligned} % \end{aligned}
\end{equation} % \end{equation}
\end{proof} %\end{proof}
%
\begin{definition} %\begin{definition}
A phase $\phi \in [0, 2\pi)$ is called qbit-global, if for some qbit states $\ket{\psi}, \ket{\varphi}$ $\ket{\psi} = \exp(i\phi)\ket{\varphi}$. % 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} %\end{definition}
%
\begin{lemma} %\begin{lemma}
When entangling qbits via projections one can disregard qbit-global phases. % When entangling qbits via projections one can disregard qbit-global phases.
%
Two qbits are entangled via projection, if for some single qbit gates $M,N$ % Two qbits are entangled via projection, if for some single qbit gates $M,N$
and two orthonormal states $\ket{a}, \ket{b}$ % and two orthonormal states $\ket{a}, \ket{b}$
%
\begin{equation} % \begin{equation}
C^{M,N}(i,j) = \ket{a}\bra{a}_j \otimes M_i + \ket{b}\bra{b}_j \otimes N_i % C^{M,N}(i,j) = \ket{a}\bra{a}_j \otimes M_i + \ket{b}\bra{b}_j \otimes N_i
\end{equation} % \end{equation}
%
\textbf{Remark.} % \textbf{Remark.}
In particular when entangling states using $CX$ and $CZ$ one can disregard qbit-global phases. % 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}. % This is immideatly clear when recalling \eqref{eq:CX_pr} and \eqref{eq:CZ_pr}.
\end{lemma} %\end{lemma}
%
\begin{proof} %\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, % 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. % $M, N, \ket{a}, \ket{b}, C^{M,N}(i,j)$ as defined above.
%
\begin{equation} %\begin{equation}
\begin{aligned} %\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'} \\ % 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)\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))(\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}) % & = \exp(i(\beta + \alpha))C^{M,N}(1, 0) (\ket{\psi}\otimes\ket{\varphi})
\end{aligned} %\end{aligned}
\end{equation} %\end{equation}
%
Where $\exp(i(\beta + \alpha))$ is a multi-qbit-global phase which can be (following the above argumentation) disregarded. % Where $\exp(i(\beta + \alpha))$ is a multi-qbit-global phase which can be (following the above argumentation) disregarded.
%
\end{proof} %\end{proof}
%
\begin{corrolary} %\begin{corrolary}
One can disregard global phases of elements of the $C_1$ group. % One can disregard global phases of elements of the $C_1$ group.
\end{corrolary} %\end{corrolary}
%
\begin{proof} %\begin{proof}
As it has been shown above a quantum computer cannot measure global phases. Also % 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 % 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 % be measured. It has been shown above that one can choose the $C_1$ operators such that they do not yield
a phase. % a phase.
\end{proof} %\end{proof}
\begin{definition} \begin{definition}
\begin{equation} \begin{equation}
@ -100,9 +100,7 @@ generated using $CZ$ and $C_1$ gates.
\end{definition} \end{definition}
\textbf{Remark.} When computing the elements of $C_L$ and their products one will realize that $C_L$ is not a group. \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 If one however disregards a global phase the product of two $C_L$ elements will be in $C_L$ again.
can be disregarded as discussed above $C_L$ will be used from now on instead of $C_1$.
\begin{theorem} \begin{theorem}
\begin{equation} \begin{equation}
| C_L | = 24 | 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} \begin{proof}
It is clear that $\forall a \in C_L$ a is a group isomorphism $P \circlearrowleft$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$. 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 Therefore $a$ will preserve the (anti-)commutator relations of $P$.
does not matter for anticommutator relations. Further note that $Y = iXZ$, so one has to consider the anti-commutator relations
This means that $X$ can be mapped to any $p \in P$ which are six elements disregarding 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} \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} \subsection{Introduction to the Graph Formalism}
The first step towards the simulation in the graph formalism has been 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 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}. 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. 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}. 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 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 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}. 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 The matrix representation of $CX$ and $CZ$ for two qbits is given by
\begin{equation} \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, 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. if the control-qbit is set.
The following notation\cite{dahlberg_ea2019} can be more handy when discussing more qbits: The following notation\cite{dahlberg_ea2019} can be more handy when discussing more qbits:
\begin{equation}\label{eq:CX_pr} \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} \begin{definition}
\label{def:vop_free_g_state} \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 by the $n$ operators
\begin{equation} \begin{equation}
@ -222,6 +222,94 @@ were derived from the vertex operator-free graph states.
as $X, Z$ anticommute and $Z\ket{1} = -1\ket{1}$. as $X, Z$ anticommute and $Z\ket{1} = -1\ket{1}$.
\end{proof} \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}: 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 $g \in \{Z_a, Y_a, X_a\}$ 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. 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. In the case $g = X_a$ and $n_a = \{\}$ the graph is obviously unchanged.
\begin{lemma} \begin{lemma}
\label{lemma:Z_measurement}
\begin{enumerate} \begin{enumerate}
\item{For a result $+Z_a$ the new state is \item{For a result $+Z_a$ the new state is
$\ket{+_Z}_a \otimes \ket{\bar{G}'}$ $\ket{+_Z}_a \otimes \ket{\bar{G}'}$
@ -281,3 +370,37 @@ In the case $g = X_a$ and $n_a = \{\}$ the graph is obviously unchanged.
\end{proof} \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}