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Daniel Knüttel 2020-02-25 11:19:09 +01:00
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47
thesis/chapters/stabilizer.tex
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@ -382,7 +382,7 @@ is done by using the symmetric set difference:
\end{definition}
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}
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.
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
local Clifford equivalent graphical representation.
local Clifford equivalent graphical representation. The algorithm is given by the following steps:
\begin{enumerate}
\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{Move from right to left through the product
\item{Move from right to left through the product:
\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
$j \in n_a \setminus \{b\}$ (which is possible as it has been checked before)
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 $o_a$.
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
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{aligned}
@ -637,7 +637,7 @@ assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket
\end{aligned}
\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{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{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.
If one wants to do computations using this formalism it is however also necessary to perform measurements.
\subsubsection{Measurements on Graph States}
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.
This is a quite expensive computation in theory however it is possible to simplify
the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes
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 projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$:
\begin{equation}
\begin{aligned}
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) 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)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) \tilde{P}_{a,s} \ket{\bar{G}} \\
\end{aligned}
@ -682,10 +683,10 @@ as $o_a$ is a Clifford operator:
\end{aligned}
\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
with the $K_G^{(i)}$ are quite easy to compute, note that Pauli matrices wither commute or anticommute
so it is easier to list the operators that anticommute:
with the $K_G^{(i)}$ are quite easy to compute. Note that Pauli matrices either commute or anticommute
and it is easier to list the operators that anticommute:
\begin{equation}
\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 $\tilde{g}_a = X_a (-X_a)$
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
graph and vertex operators have to be updated. Further it is clear that measurement of $-\tilde{g}_a$
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 measurements of $-\tilde{g}_a$
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
@ -712,10 +713,10 @@ the steps required to obtain the following results:
\end{aligned}
\end{equation}
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.
These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}.
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
$K_G^{(a)}$ is chosen. The graph is changed according to
$K_G^{(a)}$ is chosen. The graph is changed according to:
\begin{equation}
\begin{aligned}
@ -725,7 +726,7 @@ $K_G^{(a)}$ is chosen. The graph is changed according to
\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{aligned}