some work
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@ -21,19 +21,13 @@ common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ wit
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\label{ref:singleqbitgates}
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\begin{equation}
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X := \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right)
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\end{equation}
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\begin{equation}
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Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right)
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\end{equation}
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\begin{equation}
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H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right)
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\end{equation}
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\begin{equation}
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R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)
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\end{equation}
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\begin{equation}
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I := \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right)
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\begin{aligned}
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& X := \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right) \\
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& Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right) \\
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& H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right) \\
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& R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)\\
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& I := \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right) \\
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\end{aligned}
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\end{equation}
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Note that $X = HZH$ and $Z = R_{\pi}$, so the set of $H, R_\phi$ is sufficient.
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@ -47,6 +41,20 @@ transforming to the other Pauli eigenstates is done using $H$ and $SH$:
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S H Z H^\dagger S^\dagger = S X S^\dagger = Y
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\end{equation}
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The following states are the $\pm 1$ eigenstates of the $X$, $Y$, $Z$ operators
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and will be used in some calculations later.
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\begin{equation}
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\begin{aligned}
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&\ket{+_X} \equiv \ket{+} := \frac{1}{\sqrt{2}} (\ket{0} + \ket{1}) \\
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&\ket{-_X} \equiv \ket{-} := \frac{1}{\sqrt{2}} (\ket{0} - \ket{1}) \\
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&\ket{+_Y} := \frac{1}{\sqrt{2}} (\ket{0} + i\ket{1}) \\
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&\ket{-_Y} := \frac{1}{\sqrt{2}} (\ket{0} - i\ket{1}) \\
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&\ket{+_Z} := \ket{0} \\
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&\ket{-_Z} := \ket{1} \\
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\end{aligned}
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\end{equation}
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\subsubsection{Many Qbits}
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\begin{postulate}
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@ -130,8 +138,6 @@ matrices to the basis states)
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\subsubsection{Measurements}
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\textbf{FIXME} I don't like this at all.
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\begin{postulate}
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Let
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$$\ket{\psi} = \alpha\ket{\phi_1} \otimes \ket{1}_j + \beta\ket{\phi_0} \otimes \ket{0}_j$$
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@ -145,6 +151,9 @@ matrices to the basis states)
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\end{postulate}
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Measuring a qbit will also yield a classical result $0$ or $1$ with the respective probabilities.
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Measurements are always performed in the computational basis, i.e. for a qbit
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$i$ $Z_i$ is measured. With the eigenstate of the $+1$ eigenvalue being the $\ket{0}$
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and $\ket{1}$ for the $-1$ eigenvalue of $Z$.
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\begin{corrolary}
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In general the measurement of a qbit is not invertible, in particular it cannot be represented as a
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@ -162,27 +171,31 @@ As a measurement is not unitary it is not a gate as in the definition above.
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In the following discussion the term \textit{measurement gate} will be used from time
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to time as a measurement can be treated similarely while doing numerics.
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Measurements are always performed in the computational basis, i.e. for a qbit
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$i$ $Z_i$ is measured. Let the state to be measured be
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\begin{equation}
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\ket{\psi} = \alpha\ket{0}\otimes\ket{\psi_0} + \beta\ket{1}\otimes\ket{\psi_1}
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\end{equation}
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then a result $0$ is measured with probability $|\alpha|^2$ and $1$ with probability
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$|\beta|^2 = 1 - |\alpha|^2$. The wave function is then collapsed to
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\begin{equation}
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\begin{aligned}
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\ket{\psi'} = \left\{\begin{array}{c}\ket{0}\otimes\ket{\psi_0}, \mbox{ for a result 0 } \\
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\ket{1}\otimes\ket{\psi_1}, \mbox{ for a result 1 }
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\end{array}\right\}
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\end{aligned}
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\end{equation}
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\subsection{Quantum Circuits}
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Quantum circuits are a simple and well-readable way to express the application
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of several gates on a state.
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\textbf{FIXME: citation needed}
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\textbf{TODO}
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Quantum circuits are a simple and well-readable way to express the application
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of several gates on a state. Qbits are represented by horizontal line on which
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time goes from left to right. On these lines single qbit gates are
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represented by a box:
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\[
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\Qcircuit @C=1em @R=.7em {
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& \gate{H} & \gate{X} & \gate{Z} &\qw \\
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}
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\]
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The controlled gates (such as $CX$ and $CZ$) have a vertical line from the control qbit to
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the gate, for instance the circuit for $CZ_{2, 1}CX_{2,0}$ is
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\[
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\Qcircuit @C=1em @R=.7em {
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& \qw & \ctrl{2} & \qw & \qw &\qw \\
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& \qw & \qw & \qw & \ctrl{1} &\qw \\
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& \qw & \gate{X} & \qw & \gate{Z} &\qw \\
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}
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\]
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\textbf{TODO: more info about quantum circuits}
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@ -14,3 +14,107 @@ performance has since been improved to $n\log(n)$ time on average \cite{andersbr
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\subsection{Stabilizers and Stabilizer States}
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\subsubsection{Local Pauli Group and Multilocal Pauli Group}
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\begin{definition}
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\begin{equation}
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P := \{\pm 1, \pm i\} \cdot \{I, X, Y, Z\}
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\end{equation}
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Is called the Pauli group.
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\end{definition}
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The group property of $P$ can be verified easily. Note that
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the elements of $P$ either commute or anticommute.
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\begin{definition}
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For $n$ qbits
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\begin{equation}
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P_n := \{\bigotimes\limits_{i=0}^{n-1} p_i | p_i \in P\}
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\end{equation}
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is called the multilocal Pauli group on $n$ qbits.
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\end{definition}
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The group property of $P_n$ follows directly from its definition
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via the tensor product as do the (anti-)commutator relationships.
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Further are $p \in P_n$ hermitian and have the eigenvalues $\pm 1$ for
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$p \neq \pm I$, $+1$ for $p = I$ and $-1$ for $p = -I$.
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\subsubsection{Stabilizers}
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\begin{definition}
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\label{def:stabilizer}
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An abelian subgroup $S = \{S^{(0)}, ..., S^{(N)}\}$ of $P_n$ is called a set of stabilizers iff
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\begin{enumerate}
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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{$-I \notin S$}
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\end{enumerate}
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\end{definition}
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\begin{lemma}
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If $S$ is a set of stabilizers, the following statements are follow
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directly
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\begin{enumerate}
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\item{$\pm iI \notin S$}
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\item{$(S^{(i)})^2 = I$ for all $i$}
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\item{$S^{(i)}$ are hermitian for all $i$}
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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\begin{enumerate}
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\item{$(iI)^2 = (-iI)^2 = -I$. Which contradicts the definition of $S$.}
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\item{From the definition of $S$ ($G_n$ respectively) follows that any
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$S^{(i)} \in S$ has the form $\pm i^l (\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j)$ where
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$\tilde{p}_j \in \{X, Y, Z, I\}$ and $l \in \{0, 1\}$. As $(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j)$
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is hermitian $(S^{(i)})^2$ is either $+I$ or $-I$. As $-I \notin S$ $(S^{(i)})^2 = I$ follows directly.
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}
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\item{Following the argumentation above $(S^{(i)})^2 = -I \Leftrightarrow l=1$
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therefore $(S^{(i)})^2 = -I \Leftrightarrow (S^{(i)})^\dagger \neq (S^{(i)})$.}
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\end{enumerate}
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\end{proof}
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As considering all elements of a group can be unpractical for some calculations
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the generators of a group are introduced. It is usually enough to discuss the generator's
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properties to understand the properties of the group.
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\begin{definition}
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For a finite group $G$ and some $m \in \mathbb{N}$ one denotes the generators
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of G
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$$ \langle g_1, ..., g_m \rangle \equiv \langle g_i \rangle_{i=1,...,m}$$
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where $g_i \in G$, every element in $G$ can be written as a product of the $g_i$
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and $m$ is the smallest integer for which these statements hold.
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\end{definition}
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In the following discussions $\rangle S^{(i)} \rangle_{i=0, ..., n-1}$ will be used as
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the properties of a set of stabilizers that are used in the discussions can be studied using only its
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generators.
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\subsubsection{Stabilizer States}
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One important basic property of quantum mechanics is that hermitian operators have real eigenvalues
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and eigenspaces associated with these eigenvalues. Finding these eigenvalues and eigenvectors
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is what one calls solving a quantum mechanical system. One of the most fundamental insights of
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quantum mechanics is that operators that commute have a common set of eigenvectors, i.e. they
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can be diagonalized simultaneously. This motivates and justifies the following definition
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\begin{definition}
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For a set of stabilizers $S$ the vector space
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\begin{equation}
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V_S := \{\ket{\psi} | S^{(i)}\ket{\psi} = +1\ket{\psi} \forall S^{(i)} \in S\}
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\end{equation}
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is called the space of stabilizer states associated with $S$ and one says
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$\ket{\psi}$ is stabilized by $S$.
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\end{definition}
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It is clear that it is sufficient to show the stabilization property for the generators of
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$S$, as all the generators forming an element in $S$ can be absorbed into $\ket{\psi}$.
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The dimension of $V_S$ is not immediately
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