Quantum edits
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@ -44,6 +44,55 @@
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\input{parts/02.00 half a qubit}
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\input{parts/02.01 two halves}
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\input{parts/03.00 logic gates}
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%\input{parts/03.01 quantum gates}
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\input{parts/03.01 quantum gates}
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\end{document}
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\end{document}
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\problem{}
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The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par
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Find its matrix.
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\begin{solution}
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\begin{equation*}
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\text{SWAP} = \begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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0 & 1 & 0 & 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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\vfill
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% \problem{}
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% The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par
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% In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par
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% Find the $T$ gate's matrix. \par
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% \note{
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% This gate is particularly interesting because it's a \textit{universal quantum gate}: \\
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% like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates.
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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{T} = \begin{bmatrix}
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% 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
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% 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
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% 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
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% 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
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% 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
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% 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
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% 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
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% 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
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% \end{bmatrix}
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% \end{equation*}
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% \end{solution}
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@ -108,5 +108,38 @@ Factor the following product state:
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\end{solution}
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\vfill
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\problem{}
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Show that the following is an entangled state.
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\begin{equation*}
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\frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}
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\end{equation*}
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\begin{solution}
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$
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\left[
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\begin{smallmatrix}
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a_0 \\ a_1
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\end{smallmatrix}
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\right]
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\otimes
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\left[
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\begin{smallmatrix}
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b_0 \\ b_1
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\end{smallmatrix}
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\right]
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=
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a_0b_0\ket{00} + a_0b_1\ket{01} + a_1b_0\ket{10} + a_1b_1\ket{11}
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$
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\vspace{2mm}
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So, we have that $a_1b_1 = a_0b_0 = \sqrt{2}^{-1}$ \par
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But $a_0b_1 = a_1b_0 = 0$, so one of $a_0$ and $b_1$ must be zero. \par
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We thus have a contradiction.
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\end{solution}
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\vfill
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\pagebreak
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@ -1,38 +1,78 @@
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\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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The quantum analog of a logic gate is a \textit{quantum gate,} \par
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which we'll define as a linear map from $\mathbb{B}^n$ to $\mathbb{B}^n$. \par
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\note[Note]{This definition is temporary, we'll extend it soon.}
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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 implies it has as many inputs as it has outputs. \\
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This makes sense---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 CNOT (controlled not) gate. \par
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When applied to a two-bit state $\ket{ab}$, CNOT inverts $b$ iff $a$ is $\ket{1}$. \par
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Find the matrix that represents the CNOT gate. \par
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\hint{what are the dimensions of this matrix?}
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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} = \begin{bmatrix}
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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{bmatrix}
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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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\begin{bmatrix}
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\begin{bmatrix}
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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{bmatrix} \\ 0 \\ 0
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\end{bmatrix}
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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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@ -40,12 +80,16 @@ Find the matrix that represents the CNOT gate. \par
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If $\ket{a}$ is $\ket{1}$, $\ket{a} \otimes \ket{b}$ is
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$
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\begin{bmatrix}
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\left[
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\begin{smallmatrix}
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0 \\ 0 \\
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\begin{bmatrix}
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\left[
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\begin{smallmatrix}
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b_1 \\ b_2
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\end{bmatrix}
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\end{bmatrix}
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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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@ -55,479 +99,141 @@ Find the matrix that represents the CNOT gate. \par
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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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\problem{}
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Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
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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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\vfill
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\pagebreak
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\iffalse
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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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\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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\vfill
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\fi
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\problem{}
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The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par
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Find its matrix.
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\begin{solution}
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\begin{equation*}
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\text{SWAP} = \begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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0 & 1 & 0 & 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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\vfill
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\generic{Remark:}
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Note that a quantum gate is fully defined by the place it maps
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our basis states $\ket{0}$ and $\ket{1}$ (or,$\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates).
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This directly follows from \ref{qgateislinear}.
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\problem{}
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The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par
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In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par
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Find the $T$ gate's matrix. \par
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\note{
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This gate is particularly interesting because it's a \textit{universal quantum gate}: \\
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like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates.
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}
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\begin{solution}
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\begin{equation*}
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\text{T} = \begin{bmatrix}
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1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
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0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
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\end{bmatrix}
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\end{equation*}
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\end{solution}
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\vfill
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\pagebreak
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The last thing we need is a way to draw complex sequences of gates. \par
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We already know how to do this with classical gates, \'a la logic circuit:
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\begin{center}
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\begin{tikzpicture}[circuit logic US, scale=1.5]
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\node[and gate] (and) at (0,-0.8) {\tiny\texttt{and}};
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\draw ([shift={(-0.5, 0)}] and.input 1) node[left] {\texttt{1}} -- (and.input 1);
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\draw ([shift={(-0.5, 0)}] and.input 2) node[left] {\texttt{0}} -- (and.input 2);
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\draw (and.output) -- ([shift={(0.5, 0)}] and.output) node[right] {\texttt{0}};
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\end{tikzpicture}
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\end{center}
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We draw quantum circuits in a very similar way. \par
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For example, here a simple three-bit circuit consisting of a CNOT gate on the first bit, \par
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controlled by the third. The first bit is inverted iff the third bit is $\ket{1}$:
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\begin{center}
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\begin{tikzpicture}[scale=1]
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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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\node[qubit] (c) at (0, -2) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{1}$};
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\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{0}$};
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\draw[wire] (c) -- ([shift={(4, 0)}] c.center) node[qubit] {$\ket{1}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{3}{CNOT}
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\end{tikzpicture}
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\end{center}
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\problem{}
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Draw the CNOT gate as a classical logic circuit. \par
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\hint{This can be done with one gate.}
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\begin{solution}
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\begin{center}
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\begin{tikzpicture}[circuit logic US, scale=2]
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\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
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\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
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\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
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\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$a$};
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\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$b$};
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\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
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\draw (dot) -- (dot -| 1,0) node[right] {$b_\text{out}$};
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\draw (xor.output) -- (xor.output -| 1,0) node[right] {$a_\text{out}$};
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\end{tikzpicture}
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\end{center}
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\end{solution}
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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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Gate controls may be marked with a filled circle or an empty circle. \par
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Empty circles denote \textit{inverse controls,} which (of course) have an inverse effect. \par
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For example, the two circuits below are identical:
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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=1]
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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={(4, 0)}] a.center) node[
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qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
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qubit] {$\ket{b}$};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
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;
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\draw[wireijoin]
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($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{3}{CNOT}
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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=1]
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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={(4, 0)}] a.center) node[
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qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
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qubit] {$\ket{b}$};
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\draw[wire]
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($([shift={(1.5,0)}] a)!0.5!([shift={(2.5,0)}] a)$) --
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($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{1}{a}{3}{CNOT}
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\qubox{b}{0.5}{b}{1.5}{X}
|
||||
\qubox{b}{2.5}{b}{3.5}{X}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill\null
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\problem{}
|
||||
What are $\ket{a}$ and $\ket{b}$ in the diagrams above?
|
||||
If we measure the result of \ref{applycnot}, what are the probabilities of getting each state?
|
||||
|
||||
\begin{solution}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\node[qubit] (a) at (0, 0) {$\ket{0}$};
|
||||
\node[qubit] (b) at (0, -1) {$\ket{0}$};
|
||||
\vfill
|
||||
|
||||
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[
|
||||
qubit] {$\ket{1}$};
|
||||
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
|
||||
qubit] {$\ket{0}$};
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
|
||||
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
|
||||
;
|
||||
|
||||
\draw[wireijoin]
|
||||
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
|
||||
\qubox{a}{1}{a}{3}{CNOT}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{solution}
|
||||
%\problem{}
|
||||
%Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
|
||||
%
|
||||
%\begin{solution}
|
||||
% \begin{equation*}
|
||||
% \text{CNOT}_{\text{mod}} = \begin{bmatrix}
|
||||
% 0 & 1 & 0 & 0 \\
|
||||
% 1 & 0 & 0 & 0 \\
|
||||
% 0 & 0 & 1 & 0 \\
|
||||
% 0 & 0 & 0 & 1
|
||||
% \end{bmatrix}
|
||||
% \end{equation*}
|
||||
%\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
%\problem{}
|
||||
%Finally, modify the original CNOT gate so that the roles of its bits are reversed: \par
|
||||
%$\text{CNOT}_{\text{flip}} \ket{ab}$ should invert $\ket{a}$ iff $\ket{b}$ is $\ket{1}$.
|
||||
%
|
||||
%
|
||||
%\begin{solution}
|
||||
% \begin{equation*}
|
||||
% \text{CNOT}_{\text{flip}} = \begin{bmatrix}
|
||||
% 1 & 0 & 0 & 0 \\
|
||||
% 0 & 0 & 0 & 1 \\
|
||||
% 0 & 0 & 1 & 0 \\
|
||||
% 0 & 1 & 0 & 0 \\
|
||||
% \end{bmatrix}
|
||||
% \end{equation*}
|
||||
%\end{solution}
|
||||
%
|
||||
%\vfill
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
As we noted before, quantum gates don't \textit{consume} bits, they \textit{transform} them. \par
|
||||
Thus, quantum circuits are drawn with a fixed set of bits, whose states change with time: \par
|
||||
\note[Note]{
|
||||
In this diagram, CNOT and SWAP are drawn as $\oplus$ and \rotatebox[origin=c]{90}{$\leftrightarrows$} to save space.
|
||||
}
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\node[qubit] (a) at (0, 0) {$\ket{a}$};
|
||||
\node[qubit] (b) at (0, -1) {$\ket{1}$};
|
||||
\node[qubit] (c) at (0, -2) {$\ket{c}$};
|
||||
\node[qubit] (d) at (0, -3) {$\ket{d}$};
|
||||
\node[qubit] (e) at (0, -4) {$\ket{1}$};
|
||||
|
||||
\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[qubit] {$\ket{0}$};
|
||||
\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[qubit] {$\ket{b}$};
|
||||
\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {$\ket{1}$};
|
||||
\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {$\ket{0}$};
|
||||
\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[qubit] {$\ket{e}$};
|
||||
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
|
||||
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\draw[wireijoin]
|
||||
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{a}{1}{a}{2}{T}
|
||||
|
||||
\qubox{a}{4}{a}{5}{X}
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
|
||||
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
|
||||
;
|
||||
\draw[wireijoin]
|
||||
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{e}{2}{e}{3}{$\oplus$}
|
||||
|
||||
\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}}
|
||||
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
|
||||
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{b}{5}{b}{6}{T}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
In a quantum circuit, the ONLY way two bits can interact is through a gate. \par
|
||||
We cannot add \say{branches} in quantum circuits like we do in classical circuits:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[circuit logic US, scale=1.2]
|
||||
\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
|
||||
\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
|
||||
\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
|
||||
\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$x$};
|
||||
\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$y$};
|
||||
\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
|
||||
\draw (dot) -- (dot -| 1,0) node[right] {$y_\text{out}$};
|
||||
\draw (xor.output) -- (xor.output -| 1,0) node[right] {$x_\text{out}$};
|
||||
|
||||
\draw[->, line width = 1, ogrape]
|
||||
([shift={(0.3,-0.5)}] dot) node[right] {This is a branch}
|
||||
-| ([shift={(0,-0.2)}] dot)
|
||||
;
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
\definition{}
|
||||
The \textit{Hadamard Gate} $H$, is given by the following matrix: \par
|
||||
\begin{equation*}
|
||||
H = \frac{1}{\sqrt{2}}\begin{bmatrix}
|
||||
1 & 1 \\
|
||||
1 & -1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
\note[Note]{Note that we divide by $\sqrt{2}$, since $H$ must be orthonormal}
|
||||
|
||||
|
||||
\problem{}
|
||||
Find the values of $\ket{a}$ through $\ket{e}$ in the above circuit.
|
||||
What is $HH$? \par
|
||||
Using this result, find $H^{-1}$.
|
||||
|
||||
\begin{solution}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\node[qubit] (a) at (0, 0) {$\ket{1}$};
|
||||
\node[qubit] (b) at (0, -1) {$\ket{1}$};
|
||||
\node[qubit] (c) at (0, -2) {$\ket{1}$};
|
||||
\node[qubit] (d) at (0, -3) {$\ket{0}$};
|
||||
\node[qubit] (e) at (0, -4) {$\ket{1}$};
|
||||
|
||||
\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[
|
||||
qubit] {$\ket{0}$};
|
||||
\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[
|
||||
qubit] {$\ket{1}$};
|
||||
\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[
|
||||
qubit] {$\ket{1}$};
|
||||
\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[
|
||||
qubit] {$\ket{0}$};
|
||||
\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[
|
||||
qubit] {$\ket{0}$};
|
||||
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
|
||||
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\draw[wireijoin]
|
||||
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{a}{1}{a}{2}{T}
|
||||
|
||||
\qubox{a}{4}{a}{5}{X}
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
|
||||
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
|
||||
;
|
||||
\draw[wireijoin]
|
||||
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{e}{2}{e}{3}{$\oplus$}
|
||||
|
||||
\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}}
|
||||
|
||||
|
||||
\draw[wire]
|
||||
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
|
||||
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{b}{5}{b}{6}{T}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
$HH = I$, so $H^{-1} = H$
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black}
|
||||
Matrix multiplication works as follows:
|
||||
|
||||
\begin{equation*}
|
||||
AB =
|
||||
\begin{bmatrix}
|
||||
1 & 2 \\
|
||||
3 & 4 \\
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
a_0 & b_0 \\
|
||||
a_1 & b_1 \\
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1a_0 + 2a_1 & 1b_0 + 2b_1 \\
|
||||
3a_0 + 4a_1 & 3b_0 + 4b_1 \\
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
|
||||
|
||||
Note that this is very similar to multiplying each column of $B$ by $A$. \par
|
||||
The product $AB$ is simply $Ac$ for every column $c$ in $B$:
|
||||
|
||||
\begin{equation*}
|
||||
Ac_0 =
|
||||
\begin{bmatrix}
|
||||
1 & 2 \\
|
||||
3 & 4 \\
|
||||
\end{bmatrix}
|
||||
\begin{bmatrix}
|
||||
a_0 \\ a_1
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
1a_0 + 2a_1 \\
|
||||
3a_0 + 4a_1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
|
||||
This is exactly the first column of the matrix product.
|
||||
\end{ORMCbox}
|
||||
|
||||
\pagebreak
|
||||
|
||||
Loading…
x
Reference in New Issue
Block a user