some more work

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Daniel Knüttel 2020-01-29 17:13:16 +01:00
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commit ddf9094feb

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@ -31,7 +31,7 @@ the elements of $P$ either commute or anticommute.
For $n$ qbits
\begin{equation}
P_n := \{\bigotimes\limits_{i=0}^{n-1} p_i | p_i \in P\}
P_n := \left\{\bigotimes\limits_{i=0}^{n-1} p_i | p_i \in P\right\}
\end{equation}
is called the multilocal Pauli group on $n$ qbits.
@ -204,4 +204,94 @@ product of $n$ Pauli matrices. This has led to the simulation using stabilizer t
\cite{gottesman_aaronson2008}.
Interestingly also measurements are dynamics covered by the stabilizers.
When an observable $g_i \in \{\pm X_i, \pm Y_i \pm Z_i\}$ is measured
When an observable $g_a \in \{\pm X_a, \pm Y_a \pm Z_a\}$ acting on qbit $a$ is measured
one has to consider the projector
\begin{equation}
P_{g_a,s} = \frac{I + (-1)^s g_a}{2}
\end{equation}
If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with probability $1$
and the stabilizers are left unchanged:
\begin{equation}
\begin{aligned}
\ket{\psi'} &= \frac{I + g_a}{2}\ket{\psi} \\
&= \frac{I + g_a}{2}S^{(i)} \ket{\psi} \\
&= S^{(i)} \frac{I + g_a}{2}\ket{\psi} \\
&= S^{(i)}\ket{\psi'} \\
\end{aligned}
\end{equation}
As the state that is stabilized by $S$ is unique $\ket{\psi'} = \ket{\psi}$.
If $g_a$ does not commute with all stabilizers the following lemma gives
the result of the measurement.
\begin{lemma}
\label{lemma:stab_measurement}
Let $J := \{ S^{(i)} | [g_a, S^{(i)}] \neq 0\} \neq \{\}$. When measuring
$\frac{I + (-1)^s g_a}{2} $
$1$ and $0$ are obtained with probability $\frac{1}{2}$ and after choosing
a $j \in J$ the new state $\ket{\psi'}$ is stabilized by
\begin{equation}
\langle \{(-1)^s g_a\} \cup \{S^{(i)} S^{(j)} | S^{(i)} \in J \setminus \{S^{(j)}\} \} \cup J^c\rangle
\end{equation}
\end{lemma}
\begin{proof}
As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli operators,
$S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then
\begin{equation}
\begin{aligned}
P(s=+1) &= \hbox{Tr}\left(\frac{I + g_a}{2}\ket{\psi}\bra{\psi}\right) \\
&= \hbox{Tr}\left(\frac{I + g_a}{2}S^{(j)} \ket{\psi}\bra{\psi}\right)\\
&= \hbox{Tr}\left(S^{(j)}\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
&= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}S^{(j)}\right)\\
&= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
&= P(s=-1)
\end{aligned}
\notag
\end{equation}
With $P(s=+1) + P(s=-1) = 1$ follows $P(s=+1) = \frac{1}{2} = P(s=-1)$.
Further for $S^{(i)},S^{(j)} \in J$
\begin{equation}
\begin{aligned}
\frac{I + (-1)^sg_a}{2}\ket{\psi} &= \frac{I + (-1)^sg_a}{2}S^{(j)}S^{(i)} \ket{\psi} \\
&= S^{(j)}\frac{I + (-1)^{s + 1}g_a}{2}S^{(i)} \ket{\psi} \\
&= S^{(j)}S^{(i)}\frac{I + (-1)^{s + 2}g_a}{2}\ket{\psi} \\
&= S^{(j)}S^{(i)}\frac{I + (-1)^sg_a}{2}\ket{\psi}
\end{aligned}
\notag
\end{equation}
the state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$, and by
$S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$.
\end{proof}
\subsection{The VOP-free Graph States}
This section will discuss the vertex operator(VOP)-free graph states. Why they are called
vertex operator-free will be clear in the following section about graph states.
\begin{definition}
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 $E = \left\{\{i, j\} | i,i \in V, i \neq j\right\}$.
\end{definition}
This definition of a graph is way less general than the definition of a mathematical graph.
Using this definition will however allow to avoid an extensive list of constraints on the
mathematical graph that are implied in this definition.
\begin{definition}
For a graph $G = (V = \{0, ..., n-1\}, E)$ the associated stabilizers are
\begin{equation}
K_G^{(i)} := X_i \prod\limits_{\{i,j\} \in E} Z_i
\end{equation}
for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is the state stabilized by
$\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$.
\end{definition}