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Daniel Knüttel 2020-02-25 11:19:09 +01:00
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thesis/chapters/stabilizer.tex
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@ -382,7 +382,7 @@ is done by using the symmetric set difference:
\end{definition} \end{definition}
Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$. Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$.
Another transformation on the VOP-free graph states is for a vertex $a \in V$: Another transformation on the VOP-free graph states for a vertex $a \in V$ is:
\begin{equation} \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}
@ -599,15 +599,15 @@ As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$ and
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 and it transforms $(V, E, O)$ into a so-called
local Clifford equivalent graphical representation. local Clifford equivalent graphical representation. The algorithm is given by the following steps:
\begin{enumerate} \begin{enumerate}
\item{Check whether $n_a \setminus \{b\} \neq \{\}$. If this condition does not hold \item{Check whether $n_a \setminus \{b\} \neq \{\}$. If this condition does not hold
the vertex operator on $a$ cannot be cleared } the vertex operator on $a$ cannot be cleared. }
\item{Express the vertex operator $o_a$ as a product of $\sqrt{-iX}$ and $\sqrt{iZ}$.} \item{Express the vertex operator $o_a$ as a product of $\sqrt{-iX}$ and $\sqrt{iZ}$.}
\item{Move from right to left through the product \item{Move from right to left through the product:
\begin{enumerate} \begin{enumerate}
\item{If the current matrix is $\sqrt{-iX}$ apply $L_a$} \item{If the current matrix is $\sqrt{-iX}$ apply $L_a$.}
\item{If the current matrix is $\sqrt{iZ}$ select a vertex \item{If the current matrix is $\sqrt{iZ}$ select a vertex
$j \in n_a \setminus \{b\}$ (which is possible as it has been checked before) $j \in n_a \setminus \{b\}$ (which is possible as it has been checked before)
and apply $L_j$.} and apply $L_j$.}
@ -625,9 +625,9 @@ 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 $o_a$. 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 operators have been
cleared Case 1 will be applied, if there is just one cleared vertex operator it is sure that it 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 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{equation}
\begin{aligned} \begin{aligned}
@ -637,7 +637,7 @@ assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket
\end{aligned} \end{aligned}
\end{equation} \end{equation}
as $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicates whether there is an edge between $a$ and $b$. As $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicates whether there is an edge between $a$ and $b$.
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -646,25 +646,26 @@ 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}
Using this one can re-use the method used 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 both $CZ$ and $C_L$ can be applied to a graph state $\ket{G}$ this proofs constructively that As it has been shown how both $CZ$ and $C_L$ act on a graph state $\ket{G}$ and the resulting state
is a graph state as well this proofs constructively that
the graphical representation of a stabilizer state is indeed able to represent any stabilizer state. the graphical representation of a stabilizer state is indeed able to represent any stabilizer state.
If one wants to do computations using this formalism it is however also necessary to perform measurements. 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}
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 commutator of
the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes. the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes
This is a quite expensive computation in theory however it is possible to simplify 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 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{equation}
\begin{aligned} \begin{aligned}
P_{a,s} \ket{\psi} &= P_{a,s} \left(\prod\limits_{o_i \in O} o_i \right) \ket{\bar{G}} \\ P_{a,s} \ket{\psi} &= P_{a,s} \left(\prod\limits_{o_i \in O} o_i \right) \ket{\bar{G}} \\
&= \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)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 \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) 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{aligned}
@ -682,10 +683,10 @@ as $o_a$ is a Clifford operator:
\end{aligned} \end{aligned}
\end{equation} \end{equation}
Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. It is therefore enough to study the measurements 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 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 wither commute or anticommute with the $K_G^{(i)}$ are quite easy to compute. Note that Pauli matrices either commute or anticommute
so it is easier to list the operators that anticommute: and it is easier to list the operators that anticommute:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -697,8 +698,8 @@ so it is easier to list the operators that anticommute:
This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a (-X_a)$ This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a (-X_a)$
is measured the result $s=0(1)$ is obtained with probability $1$ and $(V, E, O)$ is unchanged. is measured the result $s=0(1)$ is obtained with probability $1$ and $(V, E, O)$ is unchanged.
In any other case the results $s=1$ and $s=0$ both have probability $\frac{1}{2}$ and both In any other case the results $s=1$ and $s=0$ have probability $\frac{1}{2}$ and both
graph and vertex operators have to be updated. Further it is clear that measurement of $-\tilde{g}_a$ graph and vertex operators have to be updated. Further it is clear that measurements of $-\tilde{g}_a$
and $\tilde{g}_a$ are related by just inverting the result $s$. 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 calculations to obtain the transformation on graph and vertex operators are lengthy and follow
@ -712,10 +713,10 @@ the steps required to obtain the following results:
\end{aligned} \end{aligned}
\end{equation} \end{equation}
These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement} These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}.
and the second part makes sure that the qbit $a$ is diagonal in the correct state of the measured state. 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 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{equation}
\begin{aligned} \begin{aligned}
@ -725,7 +726,7 @@ $K_G^{(a)}$ is chosen. The graph is changed according to
\end{equation} \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{equation}
\begin{aligned} \begin{aligned}