245 lines
5.7 KiB
TeX
245 lines
5.7 KiB
TeX
\section{Quantum Gates}
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In the previous section, we stated that a quantum gate is a linear map. \par
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Let's complete that definition.
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\definition{}
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A quantum gate is a \textit{orthonormal matrix}, which means any gate $G$
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satisfies $GG^\text{T} = I$. \par
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This implies the following: \par
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\begin{itemize}
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\item $G$ is square \par
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\note{
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If we think of $G$ as a map, this means that $G$ has as many inputs as it has outputs. \\
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This is to be expected: we stated earlier that quantum gates do not destroy or create qubits.
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}
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\item $G$ preserves lengths; i.e $|x| = |Gx|$. \par
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\note{This ensures that $G\ket{\psi}$ is always a valid state.}
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\end{itemize}
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(You will prove all these properties in any introductory linear algebra course. \\
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This isn't a lesson on linear algebra, so you may take them as given today.)
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\definition{}
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Let $\mathbb{U} \subset \mathbb{R}^2$ be the set of points in the unit circle. \par
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We can restate the above definition as follows: \par
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A quantum gate is an invertible map from $\mathbb{U}^n$ to $\mathbb{U}^n$.
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\definition{}<qgateislinear>
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Let $G$ be a quantum gate. \par
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Since quantum gates are, by definition, \textit{linear} maps,
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the following holds: \par
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\begin{equation*}
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G\bigl(a_0 \ket{0} + a_1\ket{1}\bigr) = a_0G\ket{0} + a_1G\ket{1}
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\end{equation*}
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\problem{}
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Consider the \textit{controlled not} (or \textit{cnot}) gate, defined by the following table: \par
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\begin{itemize}
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\item $\text{X}_\text{c}\ket{00} = \ket{00}$
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\item $\text{X}_\text{c}\ket{01} = \ket{11}$
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\item $\text{X}_\text{c}\ket{10} = \ket{10}$
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\item $\text{X}_\text{c}\ket{11} = \ket{01}$
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\end{itemize}
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In other words, the cnot gate inverts its first bit if its second bit is $\ket{1}$. \par
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Find the matrix that applies the cnot gate.
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\begin{solution}
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\begin{equation*}
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\text{CNOT} = \left[\begin{smallmatrix}
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1 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 0 & 1 \\
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0 & 0 & 1 & 0 \\
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\end{smallmatrix}\right]
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\end{equation*}
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\vspace{4mm}
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If $\ket{a}$ is $\ket{0}$, $\ket{a} \otimes \ket{b}$ is
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$
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\left[
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\begin{smallmatrix}
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\left[
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\begin{smallmatrix}
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b_1 \\ b_2
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\end{smallmatrix}
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\right]
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\\ 0 \\ 0
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\end{smallmatrix}
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\right]
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$, and the \say{not} portion of the matrix is ignored.
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\vspace{4mm}
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If $\ket{a}$ is $\ket{1}$, $\ket{a} \otimes \ket{b}$ is
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$
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\left[
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\begin{smallmatrix}
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0 \\ 0 \\
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\left[
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\begin{smallmatrix}
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b_1 \\ b_2
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\end{smallmatrix}
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\right]
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\end{smallmatrix}
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\right]
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$, and the \say{identity} portion of the matrix is ignored.
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The state of $\ket{a}$ is always preserved, since it's determined by the position of
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$\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ in the tensor product.
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If $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on top, $\ket{a}$ is $\ket{0}$,
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and if $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on the bottom, $\ket{a}$ is $\ket{1}$.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}<applycnot>
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Evaluate the following:
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\begin{equation*}
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\text{X}_\text{C}
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\Bigl(
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\frac{1}{2}\ket{00} +
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\frac{1}{2}\ket{01} -
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\frac{1}{2}\ket{10} -
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\frac{1}{2}\ket{11}
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\Bigr)
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\end{equation*}
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\vfill
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\problem{}
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If we measure the result of \ref{applycnot}, what are the probabilities of getting each state?
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\vfill
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\cgeneric{Remark}
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As we just saw, a quantum gate is fully defined by the place it maps our basis states $\ket{0}$ and $\ket{1}$ \par
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(or, $\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates). This directly follows from \ref{qgateislinear}.
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\pagebreak
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%\problem{}
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%Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
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%
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%\begin{solution}
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% \begin{equation*}
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% \text{CNOT}_{\text{mod}} = \begin{bmatrix}
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% 0 & 1 & 0 & 0 \\
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% 1 & 0 & 0 & 0 \\
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% 0 & 0 & 1 & 0 \\
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% 0 & 0 & 0 & 1
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% \end{bmatrix}
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% \end{equation*}
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%\end{solution}
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%\problem{}
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%Finally, modify the original CNOT gate so that the roles of its bits are reversed: \par
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%$\text{CNOT}_{\text{flip}} \ket{ab}$ should invert $\ket{a}$ iff $\ket{b}$ is $\ket{1}$.
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%
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%
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%\begin{solution}
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% \begin{equation*}
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% \text{CNOT}_{\text{flip}} = \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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%\end{solution}
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%
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%\vfill
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\definition{}
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The \textit{Hadamard Gate} is given by the following matrix: \par
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\begin{equation*}
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H = \frac{1}{\sqrt{2}}\begin{bmatrix}
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1 & 1 \\
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1 & -1
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\end{bmatrix}
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\end{equation*}
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\note{Note that we divide by $\sqrt{2}$, since $H$ must be orthonormal.}
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\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black!65!white}
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Matrix multiplication works as follows:
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\begin{equation*}
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AB =
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\begin{bmatrix}
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1 & 2 \\
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3 & 4 \\
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\end{bmatrix}
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\begin{bmatrix}
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a_0 & b_0 \\
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a_1 & b_1 \\
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\end{bmatrix}
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=
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\begin{bmatrix}
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1a_0 + 2a_1 & 1b_0 + 2b_1 \\
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3a_0 + 4a_1 & 3b_0 + 4b_1 \\
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\end{bmatrix}
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\end{equation*}
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Note that this is very similar to multiplying each column of $B$ by $A$. \par
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The product $AB$ is simply $Ac$ for every column $c$ in $B$:
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\begin{equation*}
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Ac_0 =
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\begin{bmatrix}
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1 & 2 \\
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3 & 4 \\
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\end{bmatrix}
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\begin{bmatrix}
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a_0 \\ a_1
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\end{bmatrix}
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=
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\begin{bmatrix}
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1a_0 + 2a_1 \\
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3a_0 + 4a_1
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\end{bmatrix}
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\end{equation*}
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This is exactly the first column of the matrix product.
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\end{ORMCbox}
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\problem{}
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What is $HH$? \par
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Using this result, find $H^{-1}$.
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\begin{solution}
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$HH = I$, so $H^{-1} = H$
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\end{solution}
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\vfill
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\problem{}
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What are $H\ket{0}$ and $H\ket{1}$? \par
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Are these states entangled?
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\begin{solution}
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$H\ket{0} = \frac{1}{\sqrt{2}}\bigl(\ket{0} + \ket{1}\bigr)$ and $H\ket{1} = \frac{1}{\sqrt{2}}\bigl(\ket{0} - \ket{1}\bigr)$ \par
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Both of these are entangled states.
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\end{solution}
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\vfill
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\pagebreak
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