some work on the graph states

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Daniel Knüttel 2020-02-10 20:11:13 +01:00
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@ -281,6 +281,7 @@ This section will discuss the vertex operator(VOP)-free graph states. Why they a
vertex operator-free will be clear in the following section about graph states.
\begin{definition}
\label{def:graph}
The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$.
In the following $V = \{0, ..., n-1\}$ will be used.
$E$ is the set of edges $E = \left\{\{i, j\} | i,i \in V, i \neq j\right\}$.
@ -452,5 +453,55 @@ just constructive. As seen in theorem \ref{thm:clifford_group_approx} any $c \in
can be constructed from $CZ$ and $C_L$ and in the following discussion it will become
clear that both $C_L$ and $CZ$ can be applied to a general graph state.
\subsubsection{Graph States and Vertex operators}
\subsubsection{Graph States and Vertex Operators}
\begin{definition}
\label{def:g_state}
A tuple $(V, E, O)$ is called the graphical representation of a stabilizer state
if $(V, E)$ is a graph as in Definition \ref{def:graph} and $O = \{o_1, ..., o_n\}$ where $o_i \in C_L$.
The state $\ket{G}$ is defined by the eigenvalue relation
\begin{equation}
+1 \ket{G} = \left(\prod\limits_{j=1}^no_j\right) K_G^{(i)} \left(\prod\limits_{j=1}^no_j\right)^\dagger \ket{G}
\end{equation}
$o_i$ are called the vertex operators of $\ket{G}$.
\end{definition}
Recalling the dynamics of stabilizer states the following relation follows immediately:
\begin{equation}
\ket{G} = \left(\prod\limits_{j=1}^no_j\right) \ket{\bar{G}}
\end{equation}
The great advantage of this representation of a stabilizer state is its space requirement:
Instead of storing $n^2$ $P_1$ matrices only some vertices (which often are implicit),
the edges and some vertex operators ($n$ matrices) have to be stored. The following theorem
will improve this even further: instead of $n$ matrices it is enough to store $n$ integers
representing the vertex operators is enough:
\begin{theorem}
$C_L$ has $24$ degrees of freedom.
\end{theorem}
\begin{proof}
It is clear that $\forall a \in C_L$ a is a group isomorphism $P_1 \circlearrowleft$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$.
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 $Z$ has four possible images under the transformation.
This gives another $4$ degrees of freedom and a total of $24$.
\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
$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which will be used in one specific operation on graph states.
$$ S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)$$
$$ \sqrt{-iX} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right)$$
$$ \sqrt{-iZ} = \exp(-i\frac{\pi}{4})\left(\begin{array}{cc} 1 & 0 \\ 0 & -i \end{array}\right)$$