fixed some ingris mistakes
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@ -7,8 +7,8 @@
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\begin{definition}
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A qbit is a two level quantum mechanical system, i.e. it has the eigenbasis
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$ \{\ket{\uparrow} \equiv \ket{1}, \ket{\downarrow} \equiv \ket{0}\} $
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with $\braket{\uparrow}{\downarrow} = 0$. In the folling this basis will be called
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the $Z$ basis in analogy to the conventions used in spin systems. For some computations
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with $\braket{\uparrow}{\downarrow} = 0$. In the following this basis will be called
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the $Z$ basis in analogy to the conventions used in spin systems ($\sigma_Z$). For some computations
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it can be useful to have component vectors, $\ket{\uparrow} \equiv \left(\begin{array}{c} 0 \\ 1\end{array} \right)$ and
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$\ket{\downarrow} \equiv \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$
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are used in these cases.
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@ -16,7 +16,7 @@
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A gate acting on a qbit is a unitary operator $G \in SU(2)$. One can show that
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$\forall G \in SU(2)$ $G$ can be approximated arbitrarily good as a product of unitary generator matrices
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\cite[Chapter 4.3]{kaye_ea2007}\cite[Chapter 2]{marquezino_ea_2019},
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\cite[Chapter 4.3]{kaye_ea2007}\cite[Chapter 2]{marquezino_ea_2019};
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common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ with
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\label{ref:singleqbitgates}
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@ -31,18 +31,18 @@ common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ wit
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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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Further note that the basis vectors are chosen s.t. $Z\ket{0} = +\ket{0}$ and $Z\ket{1} = -\ket{1}$,
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Further note that the basis vectors are chosen s.t. $Z\ket{0} = +\ket{0}$ and $Z\ket{1} = -\ket{1}$;
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transforming to the other Pauli eigenstates is done using $H$ and $SH$:
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\begin{equation}
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S = R_{\frac{\pi}{2}} = \left(\begin{array}{cc} 1 & 0 \\ 0 & i\end{array}\right)
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S := R_{\frac{\pi}{2}} = \left(\begin{array}{cc} 1 & 0 \\ 0 & i\end{array}\right)
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\end{equation}
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\begin{equation}
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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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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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@ -55,16 +55,16 @@ and will be used in some calculations later.
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\end{aligned}
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\end{equation}
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\subsubsection{Many Qbits}
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\subsubsection{Many-Qbit Systems}
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\label{ref:many_qbits}
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\begin{postulate}
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A $N$ qbit quantum mechanical state is the tensor product\cite[Definition 14.3]{wuest1995} of the $N$ single qbit
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A $n$-qbit quantum mechanical state is the tensor product\cite[Definition 14.3]{wuest1995} of the $n$ one-qbit
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states.
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\end{postulate}
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Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$
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be the basis of the one-qbit systems. Then two-qbit basis states are
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be the basis of the one-qbit systems. Then two-qbit basis states are:
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\begin{equation}
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\ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)
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@ -79,15 +79,17 @@ be the basis of the one-qbit systems. Then two-qbit basis states are
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\ket{3} := \ket{0b11} := \ket{1}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right)
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\end{equation}
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The $N$ qbit basis states can then be constructed in a similar manner.
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A general $N$ qbit state can then be written as a superposition of the
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The $n$-qbit basis states can be constructed in a similar manner.
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A general $n$-qbit state can then be written as a superposition of the
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integer states:
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\begin{equation}
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\ket{\psi} = \sum\limits_{i = 0}^{2^N - 1} c_i \ket{i}
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\ket{\psi} = \sum\limits_{i = 0}^{2^n - 1} c_i \ket{i}
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\end{equation}
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With the normation condition:
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\begin{equation}
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\sum\limits_{i = 0}^{2^N - 1} |c_i|^2 = 1
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\sum\limits_{i = 0}^{2^n - 1} |c_i|^2 = 1
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\end{equation}
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The states $\ket{i}$ for $i = 0, ..., 2^{n}-1$ are called integer states. Note
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@ -95,16 +97,19 @@ that integer states are eigenstates of the $Z$ operators. The computational basi
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is $\{\ket{i_0} \otimes ... \otimes \ket{i_{n-1}} | i_0, ..., i_{n-1} = 0, 1\}$.
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\begin{definition}
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For a single qbit gate $U$ and a qbit $j = 0, 1, ..., n - 1$
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For a single-qbit gate $U$ and a qbit $j = 0, 1, ..., n - 1$
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\begin{equation}
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U_j := \left(\bigotimes\limits_{i = 0}^{j - 1} I\right)
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\otimes U
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\otimes \left(\bigotimes\limits_{i = j + 1}^{n - 1} I \right)
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\end{equation}
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is the $U$ gate acting on qbit $j$.
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is acting on qbit $j$.
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\end{definition}
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% XXX
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\newpage
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\begin{definition}\label{def:CU}
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For two qbits $i,j = 0, 1, ..., n - 1$, $i \neq j$ and a gate $U_i$ acting on $i$
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the controlled version of $U$ is defined by
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@ -141,8 +146,8 @@ In Definition \ref{def:CU} $i$ is called the act-qbit and $j$ the control-qbit.
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$CU$ applies the gate $U$ to the act-qbit if the control-qbit is in its $\ket{1}$ state.
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One can show that the gates in \ref{ref:singleqbitgates} together with an entanglement gate, such as $CX$ or $CZ$ are enough
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to generate an arbitrary $N$ qbit gate\cite[Chapter 4.3]{kaye_ea2007}\cite{barenco_ea_1995}.
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The matrix representation of $CX$ and $CZ$ for two qbits is given by
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to generate an arbitrary $n$-qbit gate\cite[Chapter 4.3]{kaye_ea2007}\cite{barenco_ea_1995}.
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The matrix representation of $CX$ and $CZ$ for two qbits is given by:
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\begin{equation}
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CX_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)
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@ -156,7 +161,7 @@ The matrix representation of $CX$ and $CZ$ for two qbits is given by
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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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be a $n$ qbit state
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be a $n$-qbit state
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where $\ket{1}_j, \ket{0}_j$ denote the $j$-th qbit state and $|\alpha|^2 + |\beta|^2 = 1$.
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Then the measurement of the $j$-th qbit will yield
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$$\ket{\phi_1} \otimes \ket{1}_j$$
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@ -167,8 +172,7 @@ The matrix representation of $CX$ and $CZ$ for two qbits is given by
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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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$i$ $Z_i$ is measured. The $+1$ eigenvalue of $Z_i$ is $\ket{0}_i$, $\ket{1}_i$ for $-1$.
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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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@ -182,21 +186,21 @@ and $\ket{1}$ for the $-1$ eigenvalue of $Z$.
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Any unitary matrix $U$ has the inverse $U^\dagger \equiv U^{-1}$.
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\end{proof}
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As a measurement is not unitary it is not a gate as in the definition above.
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Because a measurement is not unitary it is not a gate in the sense 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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\subsection{Quantum Circuits}
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As mentioned in \ref{ref:many_qbits} one can approximate an arbitrary $n$
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qbit gate $U$ as a product of some single qbit gates and either $CX$ or $CZ$.
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As mentioned in \ref{ref:many_qbits} one can approximate an arbitrary $n$-qbit
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gate $U$ as a product of some single-qbit gates and either $CX$ or $CZ$.
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Writing (possibly huge) products of matrices is quite unpractical and very much
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unreadable. To address this problem quantum circuits have been introduced.
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These represent the qbits as a horizontal line, a gate acting on a qbit is
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These represent the qbits as a horizontal line and a gate acting on a qbit is
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a box with a name on the respective line. Quantum circuits are read from
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left to right. This means that a gate $U_i = Z_i X_i H_i$ has the
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circuit representation
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circuit representation:
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\[
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\Qcircuit @C=1em @R=.7em {
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@ -204,8 +208,8 @@ circuit representation
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}
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\]
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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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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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\[
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\Qcircuit @C=1em @R=.7em {
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@ -230,7 +234,7 @@ The great hope behind quantum computing is that it will speed up some computatio
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exponentially using algorithms that utilize the laws of quantum mechanics. Current
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algorithms are based upon quantum search and quantum fourier transform \cite{nielsen_chuang_2010}.
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The latter is particular interesting for physical problems as it is used in the phase estimation
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algorithm that can be used to analyze the spectrum of the transfer matrix
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algorithm that can be used to analyze the spectrum of the transfer matrix:
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\begin{equation}
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T = \exp(itH)
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@ -246,6 +250,6 @@ The eigenvalue of $T$ can now be estimated by using the phase estimation circuit
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\]
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Where $T\ket{\varphi} = \exp(2\pi i\varphi) \ket{\varphi}$ and the measurement result $\tilde{\varphi} = \frac{x}{2^n}$ is an
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estimate for $\varphi$. If a success rate of $1-\epsilon$ and an accuracy of $| \varphi - \tilde{\varphi} | < 2^{-m}$
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is wanted $N = m + \log_2(2 + \frac{1}{2\epsilon})$ qbits are required\cite{nielsen_chuang_2010}\cite{lehner2019}.
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estimation for $\varphi$. If a success rate of $1-\epsilon$ and an accuracy of $| \varphi - \tilde{\varphi} | < 2^{-m}$
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is wanted, then $N = m + \log_2(2 + \frac{1}{2\epsilon})$ qbits are required\cite{nielsen_chuang_2010}\cite{lehner2019}.
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@ -37,8 +37,8 @@ the elements of $P$ either commute or anticommute.
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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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The group property of $P_n$ and the (anti-)commutator relationships follow directly from its definition
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via the tensor product.
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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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\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{$\forall i,j = 1, ..., N$ $[S^{(i)}, S^{(j)}] = 0$%: $S^{(i)}$ and $S^{(j)}$ commute
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}
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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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If $S$ is a set of stabilizers, the following statements 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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\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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Considering all the elements of a group can be impractical for some calculations,
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the generators of a group are introduced. Often it is enough to discuss the generator's
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properties in order 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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\end{definition}
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In the following discussions $\langle S^{(i)} \rangle_{i=0, ..., n-1}$ 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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the required properties of a set of stabilizers that can be studied on 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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and eigenspaces which are 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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quantum mechanics is that commuting operators 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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$\ket{\psi}$ 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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It is clear that to show the stabilization property of
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$S$ the proof for the generators is sufficient,
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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 clear. One can however show that
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for a set of stabilizers $\langle S^{(i)} \rangle_{i=1, ..., n-m}$ the dimension
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$dim V_S = 2^m$ \cite[Chapter 10.5]{nielsen_chuang_2010}. This yields the following important
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