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Some changes from Andreas

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Daniel Knüttel 2020-03-09 17:25:38 +01:00
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thesis/chapters/stabilizer.tex
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@ -312,13 +312,30 @@ graph from graph theory that are implied in this definition.
\end{definition} \end{definition}
It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute
follows from the fact that $\{i,j\} \in E \Leftrightarrow \{j,i\} \in E$ therefore for two is clear if $\{a,b\} \notin E$. If $\{a, b\} \in E$
operators $K_G^{(i)}$ and $K_G^{(j)}$ either $\{i, j\} \notin E$ so they commute trivially.
If $\{i,j\} \in E$ $X_i$, $Z_j$ and $X_j$, $Z_i$ anticommute which yields that the \begin{equation}
operators commute. \begin{aligned}
K_G^{(a)} K_G^{(b)} &= X_a \left(\prod\limits_{i \in n_a} Z_i\right)
X_b \left(\prod\limits_{j\in n_b} Z_j\right)\\
&= X_a \left(\prod\limits_{i \in \setminus \{b\}} Z_i\right) Z_b
X_b \left(\prod\limits_{j\in n_b\setminus \{b\}} Z_j\right) Z_a\\
&= X_a Z_b X_b Z_a
\left(\prod\limits_{j\in n_b\setminus \{b\}} Z_j\right)
\left(\prod\limits_{i \in \setminus \{b\}} Z_i\right)\\
&= -X_b Z_b X_a Z_a
\left(\prod\limits_{j\in n_b\setminus \{b\}} Z_j\right)
\left(\prod\limits_{i \in \setminus \{b\}} Z_i\right)\\
&= X_b Z_a X_a Z_b
\left(\prod\limits_{j\in n_b\setminus \{b\}} Z_j\right)
\left(\prod\limits_{i \in \setminus \{b\}} Z_i\right)\\
&= K_G^{(b)} K_G^{(a)}.\\
\end{aligned}
\end{equation}
This definition of a graph state might not seem to be straight forward This definition of a graph state might not seem to be straight forward
but recalling theorem \ref{thm:unique_s_state} it is clear that $\ket{\bar{G}}$ but recalling Theorem \ref{thm:unique_s_state} it is clear that $\ket{\bar{G}}$
is unique. The following lemma will provide a way to construct this state is unique. The following lemma will provide a way to construct this state
from the graph. from the graph.
@ -413,7 +430,8 @@ that will be used later\cite{andersbriegel2005}.
\begin{proof} \begin{proof}
$\ket{\bar{G}'}$ is stabilized by $\langle M_a K_G^{(i)} M_a^\dagger \rangle_i$, so it is sufficient $\ket{\bar{G}'}$ 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 $M_a$. to study how the $ K_G^{(i)}$ change under $M_a$.
At first note that $[K_G^{(a)}, M_a] = 0$ and $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$, At first note that $M_a^2 \alpha K_G^{(a)} \Rightarrow [K_G^{(a)}, M_a] = 0$.
Further $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$,
so the first two equations follow trivially. For $j \in n_a$ set so the first two equations follow trivially. For $j \in n_a$ set
\begin{equation} \begin{equation}
@ -435,19 +453,20 @@ that will be used later\cite{andersbriegel2005}.
\end{aligned} \end{aligned}
\end{equation} \end{equation}
One can now construct a new set of $K_{G'}^{(i)}$ s.t. $M_a \ket{\bar{G}}$ is the $+1$ eigenstate One can now construct a new set of $K_{G'}^{(i)}$ such that $M_a \ket{\bar{G}}$ is the $+1$ eigenstate
of the new $K_{G'}^{(i)}$. It is clear that $\forall j \notin n_a$ $K_{G'}^{(j)} = K_G^{(j)}$. of the new $K_{G'}^{(i)}$. Because $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$
To construct the $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a = \{j\} \cup I$ and $n_j = \{a\} \cup J$. it is clear that $\forall j \notin n_a$ $K_{G'}^{(j)} = K_G^{(j)}$.
To construct the $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a =: \{j\} \cup F$ and $n_j =: \{a\} \cup D$.
Then follows: Then follows:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
S^{(j)} &= Z_j X_a X_j Z_a \prod\limits_{l \in J} Z_l\\ S^{(j)} &= Z_j X_a X_j Z_a \prod\limits_{l \in D} Z_l\\
&= Z_j X_a X_j Z_a \left(\prod\limits_{l \in J} Z_l\right) &= Z_j X_a X_j Z_a \left(\prod\limits_{l \in D} Z_l\right)
\left(\prod\limits_{l \in I}Z_l\right) \left(\prod\limits_{l \in F}Z_l\right)
\left(\prod\limits_{l \in I}Z_l\right) \\ \left(\prod\limits_{l \in F}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) &= Z_j X_a X_j Z_a \left(\prod\limits_{l \in ((F\cup D) \setminus (F\cap D))} Z_L\right)
\left(\prod\limits_{l \in I}Z_l\right) \\ \left(\prod\limits_{l \in F}Z_l\right) \\
&= K_{G'}^{(a)} K_{G'}^{(j)} \\ &= K_{G'}^{(a)} K_{G'}^{(j)} \\
&= K_{G}^{(a)} K_{G'}^{(j)} &= K_{G}^{(a)} K_{G'}^{(j)}
\end{aligned} \end{aligned}
@ -477,7 +496,7 @@ The answer is straight forward: No. The most simple cases are the single qbit
stated $\ket{0},\ket{1}$ and $\ket{+_Y}, \ket{-_Y}$. But there is an extension stated $\ket{0},\ket{1}$ and $\ket{+_Y}, \ket{-_Y}$. But there is an extension
to the VOP-free graph states that allows the representation of an arbitrary to the VOP-free graph states that allows the representation of an arbitrary
stabilizer state. The proof that indeed any state can be represented is stabilizer state. The proof that indeed any state can be represented is
purely constructive. As seen in theorem \ref{thm:clifford_group_approx} any $c \in C_n$ purely constructive. As seen in Theorem \ref{thm:clifford_group_approx} any $c \in C_n$
can be constructed from $CZ$ and $C_L$. In the following discussion it will become can be constructed from $CZ$ and $C_L$. In the following discussion it will become
clear that both $C_L$ and $CZ$ can be applied to a general graph state. clear that both $C_L$ and $CZ$ can be applied to a general graph state.
@ -544,10 +563,10 @@ $C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in one specific
So far the graphical representation of stabilizer states is just another way to store So far the graphical representation of stabilizer states is just another way to store
basically a stabilizer tableaux that might require less memory than the tableaux used in basically a stabilizer tableaux that might require less memory than the tableaux used in
CHP. The true power of this formalism is seen when studying its dynamics. The simplest case CHP\cite{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 is a local Clifford operator $c_j$ acting on a qbit $j$: The stabilizers are changed to
$\langle c_j S^{(i)} c_j^\dagger\rangle_i$. Using the definition of the graphical representation $\langle c_j S^{(i)} c_j^\dagger\rangle_i$. 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 one sees that just the vertex operators are changed and the new vertex operators are given by
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -570,11 +589,12 @@ In this case the CZ can be pulled past the vertex operators and just the edges
are changed to $E' = E \Delta \left\{\{a,b\}\right\}$. are changed to $E' = E \Delta \left\{\{a,b\}\right\}$.
\textbf{Case 2}(\textit{Sub-Sub-Case 2.2.1})\textbf{:}\\ \textbf{Case 2}(\textit{Sub-Sub-Case 2.2.1})\textbf{:}\\
The two qbits are isolated: From the definition of the graph state it is clear that The two qbits are isolated: From the definition of the graph state one can derive that
any isolated clique of the graph can be treated independently. Therefore the two isolated qbits any isolated clique of the graph can be treated independently. Therefore the two isolated qbits
can be treated as an independent state and the set of two qbit stabilizer states is finite. An can be treated as an independent state and the set of two qbit stabilizer states is finite. An
upper bound to the number of two qbit stabilizer states is given by $2\cdot24^2$: With and without upper bound to the number of two qbit stabilizer states is given by $2\cdot24^2$:
an edge between the qbits and $24$ Clifford operators on each vertex. A factor of two for the options with and withoit an edge between the
qbits and $24$ Clifford operators on each vertex.
All those states and the resulting state after a $CZ$ application can be computed which leads to All those states and the resulting state after a $CZ$ application can be computed which leads to
another interesting result that will be useful later: If one vertex has the vertex operator $I$ the another interesting result that will be useful later: If one vertex has the vertex operator $I$ the
@ -583,8 +603,9 @@ the identity on the vertex can be preserved under the application of a $CZ$.
\textbf{Case 3}(\textit{Case 2, but one sub-case has been handled})\textbf{:}\\ \textbf{Case 3}(\textit{Case 2, but one sub-case has been handled})\textbf{:}\\
At least one vertex operator does not commute with $CZ$ and at least one vertex At least one vertex operator does not commute with $CZ$ and at least one vertex
has non-operand neighbours. In this case one can try to clear the vertex operators has non-operand (i.e. neighbours that are neither $a$ nor $b$) neighbours. In
which will succeed for at least one qbit. this case one can try to clear the vertex operators which will succeed for at
least one qbit.
The transformation given in The transformation given in
Lemma \ref{lemma:M_a} is used to "clear" the vertex operators. Recalling that Lemma \ref{lemma:M_a} is used to "clear" the vertex operators. Recalling that
@ -605,7 +626,7 @@ to the following theorem:
As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$ and $\sqrt{iZ}$. As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$ and $\sqrt{iZ}$.
This yields an algorithm to reduce the vertex operator of a non-isolated qbit $j$ to the identity. 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 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 $M_j^\dagger$ is called $L_j$ transformation, which transforms $(V, E, O)$ into a so-called
local Clifford equivalent graphical representation. The algorithm is given by the following steps: local Clifford equivalent graphical representation. The algorithm is given by the following steps:
\begin{enumerate} \begin{enumerate}
@ -622,19 +643,22 @@ local Clifford equivalent graphical representation. The algorithm is given by th
} }
\end{enumerate} \end{enumerate}
This algorithm has the important properties that if the algorithm succeeds This algorithm has the important properties that if the algorithm succeeds $o_a
$o_a = I$ and $o_b$ has picked up powers of $\sqrt{iZ}^\dagger$. = I$ and $o_b$ has picked up powers of $\sqrt{iZ}^\dagger$. If $b$ has
If $b$ has non-operand neighbours after clearing the vertex operators on $a$ the vertex operators on $b$ non-operand neighbours after clearing the vertex operators on $a$, then the
can be cleared using the same algorithm which gives $o_b = I$ and $o_a = (\sqrt{iZ}^\dagger)^l = S^l \in \mathcal{Z}$ vertex operators on $b$ can be cleared using the same algorithm which gives
for some $l \in \{1, ..., 5\}$. Therefore Case 1 can now be applied. $o_b = I$ and $o_a = (\sqrt{iZ}^\dagger)^l = S^l \in \mathcal{Z}$ for some $l
\in \{1, ..., 5\}$. Therefore Case 1 can now be applied.
If $o_a$ could not be cleared $o_b$ can be cleared using the same procedure and after 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 $o_a$. after 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 In any case at least one vertex operator has been cleared. If both vertex
cleared Case 1 will be applied. If there is just one cleared vertex operator it operators have been cleared Case 1 will be applied. If there is just one
is the vertex with non-operand neighbours. Using this one can still apply a $CZ$: Without loss of generality cleared vertex operator it is the vertex with non-operand neighbours. Using
assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form \cite{andersbriegel2005} 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}
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -653,23 +677,25 @@ As $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicat
\end{aligned} \end{aligned}
\end{equation} \end{equation}
This allows to re-use the method 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 it has been shown how both $CZ$ and $C_L$ act on a graph state $\ket{G}$ and the resulting state As it has been shown how both $CZ$ and $C_L$ act on a graph state $\ket{G}$
is a graph state as well this proofs constructively that and the resulting state is a graph state as well this proofs constructively
the graphical representation of a stabilizer state is indeed able to represent any stabilizer state. that the graphical representation of a stabilizer state is indeed able to
If one wants to do computations using this formalism it is however also necessary to perform measurements. 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} \subsubsection{Measurements on Graph States}
This is adapted from \cite{andersbriegel2005}; measurement results and updating the graph after This is adapted from \cite{andersbriegel2005}; measurement results and updating
a measurement is described in \cite{hein_eisert_briegel2008}. 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 Recalling \ref{ref:meas_stab} it is clear that one has to compute the
the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes commutator of the observable $g_a = Z_a$ with the stabilizers to get the
which is a quite expensive computation in theory. It is possible to simplify probability amplitudes which is a quite expensive computation in theory. It is
the problem by pulling the observable behind the vertex operators. For this consider possible to simplify the problem by pulling the observable behind the vertex
the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$ operators. For this consider the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}