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