Edits
This commit is contained in:
parent
56b72a0531
commit
bd33aeb739
GPG Key ID:
C6D63995FE72FD80
@ -8,7 +8,7 @@ To keep things simple, we'll use regular (usually called \textit{classical}) bit
|
||||
|
||||
\definition{Binary Digits}
|
||||
$\mathbb{B}$ is the set of binary digits. In other words, $\mathbb{B} = \{\texttt{0}, \texttt{1}\}$. \par
|
||||
\note[Note]{We've seen $\mathbb{B}$ before: It's $(\mathbb{Z}_2, +)$, the addition group mod 2.}
|
||||
\note[Note]{We've seen $\mathbb{B}$ before---it's the set of integers mod 2.}
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
@ -28,7 +28,8 @@ In this handout, we'll often see the following sets:
|
||||
\begin{itemize}
|
||||
\item $\mathbb{R}^2$, a two-dimensional plane
|
||||
\item $\mathbb{R}^n$, an n-dimensional space
|
||||
\item $\mathbb{B}^2$, the set $\{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$
|
||||
\item $\mathbb{B}^2$, the set
|
||||
$\{(\texttt{0},\texttt{0}), (\texttt{0},\texttt{1}), (\texttt{1},\texttt{0}), (\texttt{1},\texttt{1})\}$
|
||||
\item $\mathbb{B}^n$, the set of all possible states of $n$ bits.
|
||||
\end{itemize}
|
||||
|
||||
@ -61,35 +62,35 @@ We'll therefore say that \texttt{0} and \texttt{1} \textit{orthogonal} (or equiv
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
We can draw $\vec{0}$ and $\vec{1}$ as perpendicular axis on a plane to represent this:
|
||||
We can draw $\vec{e}_0$ and $\vec{e}_1$ as perpendicular axis on a plane to represent this:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1.5]
|
||||
\fill[color = black] (0, 0) circle[radius=0.05];
|
||||
|
||||
\draw[->] (0, 0) -- (1.5, 0);
|
||||
\node[right] at (1.5, 0) {$\vec{0}$ axis};
|
||||
\node[right] at (1.5, 0) {$\vec{e}_0$ axis};
|
||||
\fill[color = oblue] (1, 0) circle[radius=0.05];
|
||||
\node[below] at (1, 0) {\texttt{0}};
|
||||
|
||||
\draw[->] (0, 0) -- (0, 1.5);
|
||||
\node[above] at (0, 1.5) {$\vec{1}$ axis};
|
||||
\node[above] at (0, 1.5) {$\vec{e}_1$ axis};
|
||||
\fill[color = oblue] (0, 1) circle[radius=0.05];
|
||||
\node[left] at (0, 1) {\texttt{1}};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
|
||||
The point marked $1$ is at $[0, 1]$. It is no parts $\vec{0}$, and all parts $\vec{1}$. \par
|
||||
The point marked $1$ is at $[0, 1]$. It is no parts $\vec{e}_0$, and all parts $\vec{e}_1$. \par
|
||||
Of course, we can say something similar about the point marked $0$: \par
|
||||
It is at $[1, 0] = (1 \times \vec{0}) + (0 \times \vec{1})$, and is thus all $\vec{0}$ and no $\vec{1}$. \par
|
||||
It is at $[1, 0] = (1 \times \vec{e}_0) + (0 \times \vec{e}_1)$, and is thus all $\vec{e}_0$ and no $\vec{e}_1$. \par
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
Naturally, the coordinates $[0, 1]$ and $[1, 0]$ denote how much of each axis a point \say{contains.} \par
|
||||
We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts $\vec{0}$ and $\vec{1}$: \par
|
||||
We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts $\vec{e}_0$ and $\vec{e}_1$: \par
|
||||
\note[Note]{
|
||||
We could also write $\texttt{x} = \vec{0} + \vec{1}$ explicitly. \\
|
||||
We could also write $\texttt{x} = \vec{e}_0 + \vec{e}_1$ explicitly. \\
|
||||
I've drawn \texttt{x} as a point on the left, and as a sum on the right.
|
||||
}
|
||||
|
||||
@ -100,10 +101,10 @@ We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts
|
||||
\fill[color = black] (0, 0) circle[radius=0.05];
|
||||
|
||||
\draw[->] (0, 0) -- (1.5, 0);
|
||||
\node[right] at (1.5, 0) {$\vec{0}$ axis};
|
||||
\node[right] at (1.5, 0) {$\vec{e}_0$ axis};
|
||||
|
||||
\draw[->] (0, 0) -- (0, 1.5);
|
||||
\node[above] at (0, 1.5) {$\vec{1}$ axis};
|
||||
\node[above] at (0, 1.5) {$\vec{e}_1$ axis};
|
||||
|
||||
\fill[color = oblue] (1, 0) circle[radius=0.05];
|
||||
\node[below] at (1, 0) {\texttt{0}};
|
||||
@ -141,7 +142,7 @@ We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts
|
||||
\vspace{4mm}
|
||||
|
||||
But \texttt{x} isn't a member of $\mathbb{B}$, it's not a valid state. \par
|
||||
Our bit is fully $\vec{0}$ or fully $\vec{1}$. There's nothing in between.
|
||||
Our bit is fully $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, there's nothing in between.
|
||||
|
||||
|
||||
\vspace{8mm}
|
||||
@ -158,11 +159,11 @@ Our bit is fully $\vec{0}$ or fully $\vec{1}$. There's nothing in between.
|
||||
|
||||
|
||||
\definition{Orthonormal Basis}
|
||||
The unit vectors $\vec{0}$ and $\vec{1}$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$. \par
|
||||
The unit vectors $\vec{e}_0$ and $\vec{e}_1$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$. \par
|
||||
\note{
|
||||
\say{ortho-} means \say{orthogonal}; normal means \say{normal,} which means length $= 1$. \\
|
||||
}{
|
||||
Note that $\vec{0}$ and $\vec{1}$ are orthonormal by \textit{definition}. \\
|
||||
Note that $\vec{e}_0$ and $\vec{e}_1$ are orthonormal by \textit{definition}. \\
|
||||
We don't have to prove anything, we simply defined them as such.
|
||||
} \par
|
||||
|
||||
@ -215,11 +216,11 @@ Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them
|
||||
|
||||
\null\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
\[ \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = (1 \times \vec{0}) + (0 \times \vec{1}) \]
|
||||
\[ \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = (1 \times \vec{e}_0) + (0 \times \vec{e}_1) \]
|
||||
\end{minipage}
|
||||
\hfill
|
||||
\begin{minipage}{0.48\textwidth}
|
||||
\[ \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = (0 \times \vec{0}) + (1 \times \vec{1}) \]
|
||||
\[ \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = (0 \times \vec{e}_0) + (1 \times \vec{e}_1) \]
|
||||
\end{minipage}
|
||||
\hfill\null
|
||||
|
||||
@ -250,10 +251,10 @@ Write \texttt{x} and \texttt{y} in the diagram below in terms of $\ket{0}$ and $
|
||||
\fill[color = black] (0, 0) circle[radius=0.05];
|
||||
|
||||
\draw[->] (0, 0) -- (1.5, 0);
|
||||
\node[right] at (1.5, 0) {$\vec{0}$ axis};
|
||||
\node[right] at (1.5, 0) {$\vec{e}_0$ axis};
|
||||
|
||||
\draw[->] (0, 0) -- (0, 1.5);
|
||||
\node[above] at (0, 1.5) {$\vec{1}$ axis};
|
||||
\node[above] at (0, 1.5) {$\vec{e}_1$ axis};
|
||||
|
||||
\fill[color = oblue] (1, 0) circle[radius=0.05];
|
||||
\node[below] at (1, 0) {$\ket{0}$};
|
||||
|
||||
@ -61,7 +61,7 @@ $\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \
|
||||
\vspace{1mm}
|
||||
|
||||
The same is true of any other state set: if $a$ takes values in $A$ and $b$ takes values in $B$, \par
|
||||
the compound state $(a,b)$ takes values in $A \times B$. This is trivial.
|
||||
the compound state $(a,b)$ takes values in $A \times B$.
|
||||
|
||||
|
||||
\vspace{4mm}
|
||||
@ -69,8 +69,6 @@ the compound state $(a,b)$ takes values in $A \times B$. This is trivial.
|
||||
We would like to do the same in vector notation. Given $\ket{a}$ and $\ket{b}$, \par
|
||||
how should we represent the state of $\ket{ab}$?
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
|
||||
@ -303,7 +301,10 @@ Write $\ket{5}$ as three-bit state vector. \par
|
||||
|
||||
\problem{}
|
||||
Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par
|
||||
What do you see?
|
||||
\hint{
|
||||
You do not need to compute every tensor product. \\
|
||||
Do a few, you should quickly see the pattern.
|
||||
}
|
||||
|
||||
|
||||
|
||||
|
||||
@ -165,13 +165,11 @@ Our measurement again returns either $\ket{0}$ or $\ket{1}$, with the following
|
||||
|
||||
\vspace{2mm}
|
||||
|
||||
In addition $\ket{\psi}$ \textit{collapses} to the state we measure: it instantly jumps to the state we measure, \par
|
||||
leaving no trace of its previous state. If we measure $\ket{\psi}$ and get $\ket{1}$, $\ket{\psi}$ becomes $\ket{1}$---and
|
||||
In addition, $\ket{\psi}$ \textit{collapses} when it is measured: it instantly changes its state to the result of the measurement,
|
||||
leaving no trace of its previous state. \par
|
||||
If we measure $\ket{\psi}$ and get $\ket{1}$, $\ket{\psi}$ becomes $\ket{1}$---and
|
||||
it will remain in that state until it is changed.
|
||||
\vspace{2mm}
|
||||
|
||||
Quantum bits \textit{cannot} be measured without their state collapsing. \par
|
||||
We cannot certainly know the state of a qubit unless that state is $\ket{0}$ or $\ket{1}$.
|
||||
|
||||
\pagebreak
|
||||
|
||||
|
||||
@ -107,7 +107,7 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
|
||||
|
||||
\generic{Remark:}
|
||||
The way a quantum circuit handles information is a bit different than the way a classical circuit does.
|
||||
We usually think of logic gates as \textit{functions}: they consume some set of bits, and return another:
|
||||
We usually think of logic gates as \textit{functions}: they consume one set of bits, and return another:
|
||||
|
||||
|
||||
\begin{center}
|
||||
@ -143,10 +143,10 @@ We'll also add a horizontal time axis, moving from left to right:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\node[qubit] (a) at (0, 0) {\texttt{0}};
|
||||
\node[qubit] (b) at (0, -1) {\texttt{0}};
|
||||
\node[qubit] (b) at (0, -1) {\texttt{1}};
|
||||
|
||||
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {\texttt{0}};
|
||||
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {\texttt{0}};
|
||||
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {\texttt{1}};
|
||||
|
||||
\draw[
|
||||
color = oblue,
|
||||
|
||||
@ -12,12 +12,12 @@ This implies the following: \par
|
||||
\begin{itemize}
|
||||
\item $G$ is square \par
|
||||
\note{
|
||||
If we think of $G$ as a map, this implies it has as many inputs as it has outputs. \\
|
||||
This makes sense---we stated earlier that quantum gates do not destroy or create qubits.
|
||||
If we think of $G$ as a map, this means that $G$ has as many inputs as it has outputs. \\
|
||||
This is to be expected: we stated earlier that quantum gates do not destroy or create qubits.
|
||||
}
|
||||
|
||||
\item $G$ preserves lengths; i.e $|x| = |Gx|$. \par
|
||||
\note{This ensures that $G\ket{\psi}$ is always a valid state!}
|
||||
\note{This ensures that $G\ket{\psi}$ is always a valid state.}
|
||||
\end{itemize}
|
||||
|
||||
(You will prove all these properties in any introductory linear algebra course. \\
|
||||
@ -103,12 +103,6 @@ Find the matrix that applies the cnot gate.
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
\generic{Remark:}
|
||||
As we just saw, a quantum gate is fully defined by the place it maps our basis states $\ket{0}$ and $\ket{1}$ \par
|
||||
(or, $\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates). This directly follows from \ref{qgateislinear}.
|
||||
|
||||
|
||||
\pagebreak
|
||||
|
||||
|
||||
@ -133,7 +127,11 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti
|
||||
|
||||
\vfill
|
||||
|
||||
\generic{Remark:}
|
||||
As we just saw, a quantum gate is fully defined by the place it maps our basis states $\ket{0}$ and $\ket{1}$ \par
|
||||
(or, $\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates). This directly follows from \ref{qgateislinear}.
|
||||
|
||||
\pagebreak
|
||||
|
||||
%\problem{}
|
||||
%Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
|
||||
@ -149,9 +147,6 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti
|
||||
% \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 $\ket{1}$.
|
||||
@ -238,11 +233,11 @@ Using this result, find $H^{-1}$.
|
||||
|
||||
\problem{}
|
||||
What are $H\ket{0}$ and $H\ket{1}$? \par
|
||||
Are these states superpositions?
|
||||
Are these states entangled?
|
||||
|
||||
\begin{solution}
|
||||
$H\ket{0} = \frac{1}{\sqrt{2}}\bigl(\ket{0} + \ket{1}\bigr)$ and $H\ket{1} = \frac{1}{\sqrt{2}}\bigl(\ket{0} - \ket{1}\bigr)$ \par
|
||||
Both of these are superpositions.
|
||||
Both of these are entangled states.
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
|
||||
Loading…
x
Reference in New Issue
Block a user