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}
\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
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
to be the smallest set of generators of $G$.
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 $J$ 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 $G$.
\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}
\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}
\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{$\forall i = 1, ..., n$ $S_i\ket{\psi} = +1 \ket{\psi}$}
\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{$\forall i = 1, ..., n$ $S^{(i)}\ket{\psi} = +1 \ket{\psi}$}
\end{enumerate}
\end{definition}
\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.
\end{lemma}
\begin{proof}
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
exist $2^n$ eigenstates, one state $\ket{\psi}$ with eigenvalue $+1$ for all $S_i$. As the dimension of $n$ qbits is $2^n$
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
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.
\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.
\begin{lemma}
Let $\ket{\psi}$ be stabilized by $\langle S_i \rangle_i$, then $U\ket{\psi}$ is stabilized
by $\langle US_iU^\dagger \rangle_i$.
Let $\ket{\psi}$ be stabilized by $\langle S^{(i)} \rangle_i$, then $U\ket{\psi}$ is stabilized
by $\langle US^{(i)}U^\dagger \rangle_i$.
\end{lemma}
\begin{proof}
$$ U\ket{\psi} = US_i\ket{\psi} = US_iU^\dagger U\ket{\psi}$$
So $U\ket{\psi}$ is a $+1$ eigenstate of $US_iU^\dagger$.
$$ U\ket{\psi} = US^{(i)}\ket{\psi} = US^{(i)}U^\dagger U\ket{\psi}$$
So $U\ket{\psi}$ is a $+1$ eigenstate of $US^{(i)}U^\dagger$.
\end{proof}
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}
\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}
\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
a $j \in J$ the new state $\ket{\psi'}$ is stabilized by
\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{enumerate}
\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 $g_a$,
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
$S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then
\begin{equation}
\begin{aligned}