some corrections in the notation

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Daniel Knüttel 2019-12-17 11:19:36 +01:00
parent ebbd86ea93
commit 40b979835b

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@ -23,31 +23,34 @@ Where $X = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right)$,
\end{definition} \end{definition}
\begin{definition} \begin{definition}
For a group $G$, $g_1, ..., g_n$ are called the generators of $G$ iff $\forall g \in G: g = \prod\limits_{i \in I} g_i$ for a For a group $G$, $g_1, ..., g_n$ are called the generators of $G$ iff $\forall g \in G: g = \prod\limits_{i \in J} g_i$ for a
subsed $I$ of $\{1, ..., n\}$. We write $G = \langle g_i \rangle_i$ if G is generated by the $g_i$. The generators $g_i$ are chosen subsed $J$ of $I = \{1, ..., n\}$. We write $G = \langle g_i \rangle_i$ if G is generated by the $g_i$. The generators $g_i$ are chosen
to be the smallest set of generators of $G$. to be the smallest set of generators of $G$.
\end{definition} \end{definition}
The notation $\langle g_i \rangle_i \equiv \langle g_i \rangle_{i \in I}$ is used used as a shorthand
notation for $\langle \{g_i\}_{i \in I} \rangle$.
\begin{definition} \begin{definition}
\label{def:stabilizer} \label{def:stabilizer}
For a $n$ qbit state $\ket{\psi}$ $\langle S_i \rangle_i$ is called the stabilizer of $\ket{\psi}$ if For a $n$ qbit state $\ket{\psi}$ $\langle S^{(i)} \rangle_{i}$ is called the stabilizer of $\ket{\psi}$ if
\begin{enumerate} \begin{enumerate}
\item{$\forall i = 1, ..., n$ $S_i \in P_n$} \item{$\forall i = 1, ..., n$ $S^{(i)} \in P_n$}
\item{$\forall i,j = 1, ..., n$ $[S_i, S_j] = 0$ $S_i$ and $S_j$ commute} \item{$\forall i,j = 1, ..., n$ $[S^{(i)}, S^{(j)}] = 0$ $S^{(i)}$ and $S^{(j)}$ commute}
\item{$\forall i = 1, ..., n$ $S_i\ket{\psi} = +1 \ket{\psi}$} \item{$\forall i = 1, ..., n$ $S^{(i)}\ket{\psi} = +1 \ket{\psi}$}
\end{enumerate} \end{enumerate}
\end{definition} \end{definition}
\begin{lemma} \begin{lemma}
For every $\langle S_i \rangle_i$ fulfilling the first two conditions in definition \ref{def:stabilizer} there exists For every $\langle S^{(i)} \rangle_i$ fulfilling the first two conditions in definition \ref{def:stabilizer} there exists
a (up to a global phase) unique state $\ket{\psi}$ fulfilling the third condition. This state is called stabilizer state. a (up to a global phase) unique state $\ket{\psi}$ fulfilling the third condition. This state is called stabilizer state.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
All multi-local Pauli operators are hermitian (observables in terms of quantum mechanics), as the $S_i$ All multi-local Pauli operators are hermitian (observables in terms of quantum mechanics), as the $S^{(i)}$
commute they have a common set of eigenstates. Because each $S_i$ has eigenvalues $+1, -1$, there commute they have a common set of eigenstates. Because each $S^{(i)}$ has eigenvalues $+1, -1$, there
exist $2^n$ eigenstates, one state $\ket{\psi}$ with eigenvalue $+1$ for all $S_i$. As the dimension of $n$ qbits is $2^n$ exist $2^n$ eigenstates, one state $\ket{\psi}$ with eigenvalue $+1$ for all $S^{(i)}$. As the dimension of $n$ qbits is $2^n$
the state $\ket{psi}$ is unique up to a global phase. the state $\ket{psi}$ is unique up to a global phase.
\end{proof} \end{proof}
@ -67,13 +70,13 @@ transformations that map stabilizers to other stabilizers: the Clifford group.
The properties of this group will be discussed later, for the time being is existence is enough. The properties of this group will be discussed later, for the time being is existence is enough.
\begin{lemma} \begin{lemma}
Let $\ket{\psi}$ be stabilized by $\langle S_i \rangle_i$, then $U\ket{\psi}$ is stabilized Let $\ket{\psi}$ be stabilized by $\langle S^{(i)} \rangle_i$, then $U\ket{\psi}$ is stabilized
by $\langle US_iU^\dagger \rangle_i$. by $\langle US^{(i)}U^\dagger \rangle_i$.
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
$$ U\ket{\psi} = US_i\ket{\psi} = US_iU^\dagger U\ket{\psi}$$ $$ U\ket{\psi} = US^{(i)}\ket{\psi} = US^{(i)}U^\dagger U\ket{\psi}$$
So $U\ket{\psi}$ is a $+1$ eigenstate of $US_iU^\dagger$. So $U\ket{\psi}$ is a $+1$ eigenstate of $US^{(i)}U^\dagger$.
\end{proof} \end{proof}
This is an important insight that is used for simulations\cite{gottesman_aaronson2008}, as This is an important insight that is used for simulations\cite{gottesman_aaronson2008}, as
@ -91,14 +94,14 @@ measuring a qbit is given by the following lemma:
\begin{lemma} \begin{lemma}
\label{lemma:stab_measurement} \label{lemma:stab_measurement}
Let $J := \{ S_i | [g_a, S_i] \neq 0\}$. Then Let $J := \{ S^{(i)} | [g_a, S^{(i)}] \neq 0\}$. Then
\begin{enumerate} \begin{enumerate}
\item{If $J = \{\}$, one value is measured with probability $1$ and the stabilizers are unchanged.} \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 after choosing \item{If $J \neq \{\}$, $1$ and $0$ are measured with probability $\frac{1}{2}$ and after choosing
a $j \in J$ the new state $\ket{\psi'}$ is stabilized by a $j \in J$ the new state $\ket{\psi'}$ is stabilized by
\begin{equation} \begin{equation}
\langle \{(-1)^s g_a\} \cup \{S_i S_j | i \in J \setminus \{j\} \} \cup \{S_i | i \in J^c\}\rangle \langle \{(-1)^s g_a\} \cup \{S^{(i)} S^{(j)} | S^{(i)} \in J \setminus \{S^{(j)}\} \} \cup J^c\rangle
\end{equation}} \end{equation}}
\end{enumerate} \end{enumerate}
\end{lemma} \end{lemma}
@ -108,7 +111,7 @@ measuring a qbit is given by the following lemma:
\item{As $g_a$ commutes with all stabilizers, $\ket{\psi}$ is an eigenstate of $g_a$, \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.} 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, \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 $S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}