422 lines
12 KiB
TeX
422 lines
12 KiB
TeX
\section{HXH}
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Let's return to the quantum circuit diagrams we discussed a few pages ago. \par
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Keep in mind that we're working with quantum gates and proper half-qubits---not classical bits, as we were before.
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\definition{Controlled Inputs}
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A \textit{control input} or \textit{inverted control input} may be attached to any gate. \par
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These are drawn as filled and empty circles in our circuit diagrams:
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\null\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{0}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{2}{$X$}
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\node[gray] at ($([shift={(0,-0.75)}] dot)$) {Non-inverted control input};
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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;
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\draw[wireijoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{2}{$X$}
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\node[gray] at ($([shift={(0,-0.75)}] dot)$) {Inverted control input};
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill\null
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\vspace{2mm}
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An $X$ gate with a (non-inverted) control input behaves like an $X$ gate if \textit{all} its control inputs are $\ket{1}$,
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and like $I$ otherwise. An $X$ gate with an inverted control inputs does the opposite, behaving like $I$ if its input is $\ket{1}$
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and like $X$ otherwise. The two circuits above illustrate this fact---take a look at their inputs and outputs.
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\vspace{2mm}
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Of course, we can give a gate multiple controls. \par
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An $X$ gate with multiplie controls behaves like an $X$ gate if...
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\begin{itemize}
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\item all non-inverted controls are $\ket{1}$, and
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\item all inverted controls are $\ket{0}$
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\end{itemize}
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...and like $I$ otherwise.
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\problem{}
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What are the final states of the qubits in the diagram below?
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\begin{center}
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\begin{tikzpicture}[scale = 1.0]
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\node[qubit] (a) at (0, 0) {$\ket{1}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\node[qubit] (c) at (0, -2) {$\ket{1}$};
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\node[qubit] (d) at (0, -3) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {?};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {?};
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\draw[wire] (c) -- ([shift={(3, 0)}] c.center) node[qubit] {?};
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\draw[wire] (d) -- ([shift={(3, 0)}] d.center) node[qubit] {?};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] d)!0.5!([shift={(2,0)}] d)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$)
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circle[radius=0.1] coordinate(dot)
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;
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\draw[wireijoin]
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($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
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circle[radius=0.1] coordinate(dot)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] d)!0.5!([shift={(2,0)}] d)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{b}{1}{b}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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Consider the diagram below, with one controlled $X$ gate: \par
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\note[Note]{The CNOT gate from \ref{cnot} is a controlled $X$ gate.}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{a}$};
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\node[qubit] (b) at (0, -1) {$\ket{b}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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Find a matrix $\text{X}_\text{c}$ that represents this gate, so that $\text{X}_\text{c}\ket{ab}$ works as expected.
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\begin{solution}
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\begin{equation*}
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\text{X}_\text{c} = \begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 \\
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0 & 0 & 1 & 0 \\
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0 & 1 & 0 & 0
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\end{bmatrix}
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\end{equation*}
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Note that this is also the solution to \ref{cnotflipped}.
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\end{solution}
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\vfill
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\problem{}
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Now, evaluate the following. Remember that
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$\ket{+} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} + \ket{1}\Bigr)$ and
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$\ket{-} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} - \ket{1}\Bigr)$
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\null\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{+}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill\null
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\vspace{5mm}
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\null\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{-}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{+}$};
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\node[qubit] (b) at (0, -1) {$\ket{-}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill\null
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\hint{
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Note that some of these states are entangled. The circuit diagrams are a bit misleading:
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we can't write an entangled state as two distinct qubits!
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\vspace{2mm}
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So, don't try to find $\ket{a}$ and $\ket{b}$. \par
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Instead find $\ket{ab} = \psi_0\ket{00} + \psi_1\ket{01} + \psi_2\ket{10} + \psi_3\ket{11}$, and factor it into $\ket{a} \otimes \ket{b}$ if you can.
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}
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\begin{solution}
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In all the below equations, let $\tau = \frac{1}{\sqrt{2}}$.
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\begin{itemize}[itemsep = 1mm]
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\item $
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\text{X}_\text{c}\ket{01}
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= \ket{11}
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$
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\item $
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\text{X}_\text{c}\ket{0+}
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= \tau\ket{00} + \tau\ket{11}
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$ \note[Note]{This state is entangled!}
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\item $
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\text{X}_\text{c}\ket{-1}
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= -\tau\ket{10} + \tau\ket{11}
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= (-\ket{-}) \otimes \ket{1}
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$
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\item $
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\text{X}_\text{c}\ket{+-}
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= \frac{1}{2}(\ket{00} - \ket{01} + \ket{10} - \ket{11})
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= \ket{+-}
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$
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\end{itemize}
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\end{solution}
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\vfill
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\pagebreak
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\generic{Remark:}
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Now, consider the following circuit:
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{b}$};
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\qubox{b}{1}{b}{2}{$H$}
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\qubox{b}{3}{b}{4}{$H$}
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\draw[wire]
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($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{2}{a}{3}{$X$}
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\end{tikzpicture}
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\end{center}
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We already know that $H$ is its own inverse: $HH = I$. \par
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Applying $H$ to a qubit twice does not change its state.
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\note{
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Recall that $H = \frac{1}{\sqrt{2}}\left[\begin{smallmatrix} 1 & 1 \\ 1 & -1 \end{smallmatrix}\right]$
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}
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\vspace{2mm}
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So, we might expect that the two circuits below are equivalent: \par
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After all, we $H$ the second bit, use it to control an $X$ gate, and then $H$ it back to its previous state.
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\null\hfill
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\begin{minipage}{0.48\textwidth}\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{b}$};
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\qubox{b}{1}{b}{2}{$H$}
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\qubox{b}{3}{b}{4}{$H$}
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\draw[wire]
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($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{2}{a}{3}{$X$}
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\end{tikzpicture}
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\end{center}\end{minipage}
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\hfill
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\begin{minipage}{0.48\textwidth}\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {$\ket{c}$};
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\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{d}$};
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\draw[wire]
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($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{2}{a}{3}{$X$}
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\end{tikzpicture}
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\end{center}\end{minipage}
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\hfill\null
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\vspace{2mm}
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This, however, isn't the case: \par
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If we compute the state $\ket{ab}$ in the left circuit, we get $[0.5, ~0.5, -0.5, ~0.5]$ (which is entangled), \par
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but the state $\ket{cd}$ on the right is $\ket{11} = [0,0,0,1]$. \par
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\note{This is easy to verify with a few matrix multiplications.}
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\vspace{4mm}
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How does this make sense? \par
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Remember that a two-bit quantum state is \textit{not} equivalent to a pair of one-qubit quantum states.
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We must treat a multi-qubit state as a single unit.
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Recall that a two-bit state $\ket{ab}$ comes with four probabilities:
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$\mathcal{P}(\texttt{00})$, $\mathcal{P}(\texttt{01})$, $\mathcal{P}(\texttt{10})$, and $\mathcal{P}(\texttt{11})$.
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If we change the probabilities of only $\ket{a}$, \textit{all four of these change!}
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\vfill
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Because of this fact, \say{controlled gates} may not work as you expect. They may seem
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to \say{read} their controlling qubit without affecting its state, but remember---a
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controlled gate still affects the \textit{entire} state. As we noted before, it is
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not possible to apply a transformation to one bit of a quantum state.
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\begin{center}
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\begin{tikzpicture}[scale=1]
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\node[qubit] (a) at (0, 0) {\texttt{1}};
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\node[qubit] (b) at (0, -1) {\texttt{0}};
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\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {\texttt{0}};
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\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {\texttt{1}};
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\ghostqubox{a}{1}{b}{2}{$T_1$}
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\ghostqubox{a}{2}{b}{3}{$T_2$}
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\ghostqubox{a}{3}{b}{4}{$T_3$}
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\end{tikzpicture}
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\end{center}
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\vfill
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\pagebreak
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