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some more work on stabilizers

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Daniel Knüttel 2019-12-09 17:45:43 +01:00
parent 47cd875151
commit c26f1b19a0
2 changed files with 122 additions and 6 deletions
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8
computations/projector_after_commute_C_L.ipynb
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@ -2,7 +2,7 @@
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@ -53,7 +53,7 @@
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@ -108,7 +108,7 @@
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@ -123,7 +123,7 @@
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120
thesis/chapters/stabilizer.tex
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@ -80,6 +80,66 @@ This is an important insight that is used for simulations\cite{gottesman_aaronso
updating the $n$ stabilizers that are a tensor product of $n$ Pauli matrices scales with roughly
$\mathcal{O}(n^2)$ instead of $\mathcal{O}(2^n)$ for the state vector approach.
Every qbit can be measured in the $X, Y$ or $Z$ basis which is a projection using
\begin{equation}
\frac{I + (-1)^s g_a}{2}
\end{equation}
Where $g_a \in \{Z_a, Y_a, X_a\}$ and $s \in \{0, 1\}$. How the stabilizers change when
measuring a qbit is given by the following lemma:
\begin{lemma}
\label{lemma:stab_measurement}
Let $J := \{ S_i | [g_a, S_i] \neq 0\}$. Then
\begin{enumerate}
\item{If $J = \{\}$, one value is measured with probability $1$ and the stabilizers are unchanged.}
\item{If $J \neq \{\}$, $1$ and $0$ are measured with probability $\frac{1}{2}$ and the new state
$\ket{\psi'}$ is stabilized by
\begin{equation}
\langle \{(-1)^s g_a\} \cup \{K_G^{(i)} K_G^{(j)} | j \in J, i \in J \setminus \{j\} \} \cup \{K_G^{(i)} | i \in J^c\}\rangle
\end{equation}}
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}
\item{As $g_a$ commutes with all stabilizers, $\ket{\psi}$ is an eigenstate of $g_a$,
so the result of the measurement is deterministic and $\ket{\psi}$ is left unchanged.}
\item{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
$S^{(i)} \in J^c\setminus\{a\}$. $g_a$ trivially stabilizes $\ket{\psi'}$.
}
\end{enumerate}
\end{proof}
\subsection{The Vertex Operator-Free Graph States}
@ -98,7 +158,7 @@ were derived from the vertex operator-free graph states.
for all $i \in V$ where for some operator $O$ $O_i$ indicates that it acts on the $i$-th qbit.
A state $\ket{\overline{G}}$ is a $+1$ eigenstate of all $n$ $K^{(i)}_G$.
$\ket{\overline{G}}$ is the $+1$ eigenstate of all $n$ $K^{(i)}_G$.
\end{definition}
\begin{corrolary}
@ -163,5 +223,61 @@ were derived from the vertex operator-free graph states.
\end{proof}
These insights can be used to understand how measurement works on the vop-free graph state \cite{nielsen_chuang_2010}:
Consider a state $\ket{\psi}$ that is stabilized by $g_1, ... g_n$ and a hermitian $g$ that is to be measured.
Consider a state $\ket{\psi}$ that is stabilized by $g_1, ... g_n$ and a Pauli operator $g \in \{Z_a, Y_a, X_a\}$
that is to be measured.
Recalling lemma \ref{lemma:stab_measurement} the following relations follow immideately:
\begin{enumerate}
\item{For $g = X_a$ if $a$ is an isolated qbit $+1$ is measured with probability $1$
and the state $\ket{\bar{G}}$ is unchanged.}
\item{For $g = X_a$ and $a$ is non-isolated, choose $b \in n_a$ and the new stabilizers are
\begin{equation}
\langle \{(-1)^sX_a\}
\cup \{K_G^{(b)}K_G^{(i)} | i \in n_a \setminus \{b\}\}
\cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle
\end{equation}}
\item{For $g = Z_a$ the new stabilizers are
\begin{equation}
\langle \{(-1)^sZ_a\}
\cup \{K_G^{(i)} | i \in V \setminus n_a\} \rangle
\end{equation}}
\item{For $g = Y_a$ the new stabilizers are
\begin{equation}
\langle \{(-1)^sY_a\}
\cup \{K_G^{(a)}K_G^{(i)} | i \in n_a\}
\cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle
\end{equation}}
\end{enumerate}
The states after the measurement are in general no vop-free graph states anymore,
the following discussion will allow to construct new vop-free graph states and
Clifford transformations from the vop-free graph state to the resulting state.
In the case $g = X_a$ and $n_a = \{\}$ the graph is obviously unchanged.
\begin{lemma}
\begin{enumerate}
\item{For a result $+Z_a$ the new state is
$\ket{+_Z}_a \otimes \ket{\bar{G}'}$
with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for
$i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$.
}
\item{For a result $-Z_a$ the new state is $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$
with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for $i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$
and $U = \prod\limits_{i \in n_a}Z_i$.}
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}
\item{
It is trivial that $Z_a$ and $K_G^{(i)} i \neq a$ stabilize $\ket{+_Z}_a \otimes \ket{\bar{G}'}$.
$Z_aK_G^{(i)}$ for $i \in n_a$ do not act on $a$, so $\ket{\bar{G}'}$ is well-defined.
}
\item{
The state $\ket{-_Z}_a \otimes \ket{\bar{G}'}$ is the $-1$ eigenstate of $K_G^{(i)}$ for all
$i \in n_a$. This can be corrected by transforming $K_G^{(i)}$ to $-K_G^{(i)} = Z_i K_G^{(i)} Z_i^\dagger$
which stabilizes the state $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$.
}
\end{enumerate}
\end{proof}