handouts/Advanced/Introduction to Quantum/parts/02 two bits.tex

\section{Two Bits}


\problem{}<compoundclassicalbits>
As we already know, the set of states a single bit can take is $\mathbb{B} = \{\texttt{0}, \texttt{1}\}$. \par
What is the set of compound states \textit{two} bits can take? How about $n$ bits? \par
\hint{Cartesian product.}


\vspace{5cm}



Of course, \ref{compoundclassicalbits} is fairly easy: \par
If $a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$,
the values $ab$ can take are
$\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$.

\vspace{2mm}

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 $A \times B$.

\vspace{2mm}

We would like to do the same in vector notation. Given bits $\ket{a}$ and $\ket{b}$,
how should we represent the state of $\ket{ab}$? We'll spend the rest of this section solving this problem.




\problem{}
When we have two bits, we have four orthogonal states:
$\overrightarrow{00}$, $\overrightarrow{01}$, $\overrightarrow{10}$, and $\overrightarrow{11}$. \par

\vspace{2mm}

Write $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$ as column vectors \par
with respect to the orthonormal basis $\{\overrightarrow{00}, \overrightarrow{01}, \overrightarrow{10}, \overrightarrow{11}\}$.

\vfill
\pagebreak












\definition{Tensor Products}
The \textit{tensor product} of two vectors is defined as follows:
\begin{equation*}
	\begin{bmatrix}
		x_1 \\ x_2
	\end{bmatrix}
	\otimes
	\begin{bmatrix}
		y_1 \\ y_2
	\end{bmatrix}
=
	\begin{bmatrix}
		x_1
		\begin{bmatrix}
			y_1 \\ y_2
		\end{bmatrix}

		\\[4mm]

		x_2
		\begin{bmatrix}
			y_1 \\ y_2
		\end{bmatrix}
	\end{bmatrix}
=
	\begin{bmatrix}
		x_1y_1 \\[1mm]
		x_1y_2 \\[1mm]
		x_2y_1 \\[1mm]
		x_2y_2 \\[0.5mm]
	\end{bmatrix}
\end{equation*}


That is, we take our first vector, multiply the second
vector by each of its components, and stack the result.
You could think of this as a generalization of scalar
mulitiplication, where scalar mulitiplication is a
tensor product with a vector in $\mathbb{R}^1$:
\begin{equation*}
	a
	\begin{bmatrix}
		x_1 \\ x_2
	\end{bmatrix}
=
	\begin{bmatrix}
		a_1
	\end{bmatrix}
	\otimes
	\begin{bmatrix}
		y_1 \\ y_2
	\end{bmatrix}
=
	\begin{bmatrix}
		a_1
		\begin{bmatrix}
			y_1 \\ y_2
		\end{bmatrix}
	\end{bmatrix}
=
	\begin{bmatrix}
		a_1y_1 \\[1mm]
		a_1y_2
	\end{bmatrix}
\end{equation*}


\vspace{2mm}

Also, note that the tensor product is very similar to the
Cartesian product: if we take $x$ and $y$ as sets, with
$x = \{x_1, x_2\}$ and $y = \{y_1, y_2\}$, the Cartesian product
contains the same elements as the tensor product---every possible
pairing of an element in $x$ with an element in $y$:
\begin{equation*}
	x \times y = \{~(x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2y_2)~\}
\end{equation*}


In fact, these two operations are (in a sense) essentially identical. \par
Let's quickly demonstrate this.













\problem{}
Say $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. \par
What is the dimension of $x \otimes y$?

\vfill

\problem{}<basistp>
What is the pairwise tensor product
$
\Bigl\{
	\left[
	\begin{smallmatrix}
		1 \\ 0 \\ 0
	\end{smallmatrix}
	\right],
	\left[
	\begin{smallmatrix}
		0 \\ 1 \\ 0
	\end{smallmatrix}
	\right],
	\left[
	\begin{smallmatrix}
		0 \\ 0 \\ 1
	\end{smallmatrix}
	\right]
\Bigr\}
\otimes
\Bigl\{
	\left[
	\begin{smallmatrix}
		1 \\ 0
	\end{smallmatrix}
	\right],
	\left[
	\begin{smallmatrix}
		0 \\ 1
	\end{smallmatrix}
	\right]
\Bigr\}
$?

\note{in other words, distribute the tensor product between every pair of vectors.}

\vfill










\problem{}
What is the \textit{span} of the vectors we found in \ref{basistp}? \par
In other words, what is the set of vectors that can be written as linear combinations of the vectors above?

\vfill

Look through the above problems and convince yourself of the following fact: \par
If $a$ is a basis of $A$ and $b$ is a basis of $B$, $a \otimes b$ is a basis of $A \times B$.
\pagebreak










\problem{}
The compound state of two vector-form bits is their tensor product. \par
Compute the following. Is the result what we'd expect?
\begin{itemize}
	\item $\ket{0} \otimes \ket{0}$
	\item $\ket{0} \otimes \ket{1}$
	\item $\ket{1} \otimes \ket{0}$
	\item $\ket{1} \otimes \ket{1}$
\end{itemize}
\hint{
	Remember that the coordinates of
	$\ket{0}$ are $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$,
	and the coordinates of
	$\ket{1}$ are $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
}


\vfill









\problem{}<fivequant>
Of course, writing $\ket{0} \otimes \ket{1}$ is a bit excessive. We'll shorten this notation to $\ket{01}$. \par

\vspace{2mm}

In fact, we could go further: if we wanted to write the set of bits $\ket{1} \otimes \ket{1} \otimes \ket{0} \otimes \ket{1}$, \par
we could write $\ket{1101}$---but a shorter alternative is $\ket{13}$, since $13$ is \texttt{1101} in binary.

\vspace{2mm}

Write $\ket{5}$ as three-bit state vector. \par

\begin{solution}
	$\ket{5} = \ket{101} = \ket{1} \otimes \ket{0} \otimes \ket{1} = [0,0,0,0,0,1,0,0]^T$ \par
	Notice how we're counting from the top, with $\ket{000} = [1,0,...,0]$ and $\ket{111} = [0, ..., 0, 1]$.
\end{solution}

\vfill






\problem{}
Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par
\hint{You do not need to compute every tensor product. Do a few and find the pattern.}






\vfill
\pagebreak