\section{Two Bits}
How do we represent multi-bit states using vectors? \par
Unfortunately, this is hard to visualize---but the idea is simple.




\problem{}
What is the set of possible states of two bits (i.e, $\mathbb{B}^2$)?


\vspace{2cm}






\generic{Remark:}
When we have two bits, we have four orthogonal states:
$\overrightarrow{00}$, $\overrightarrow{01}$, $\overrightarrow{10}$, and $\overrightarrow{11}$. \par
We need four dimensions to draw all of these vectors, so I can't provide a picture... \par
but the idea here is the same as before.







\problem{}
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









\generic{Remark:}
So, we represent each possible state as an axis in an $n$-dimensional space. \par
A set of $n$ bits gives us $2^n$ possible states, which forms a basis in $2^n$ dimensions.

\vspace{1mm}

Say we now have two seperate bits: $\ket{a}$ and $\ket{b}$. \par
How do we represent their compound state? \par

\vspace{4mm}

If we return to our usual notation, this is very easy:
$a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$, \par
so the possible compound states of $ab$ are
$\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$

\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 $A \times B$.


\vspace{4mm}

We would like to do the same in vector notation. Given $\ket{a}$ and $\ket{b}$, \par
how should we represent the state of $\ket{ab}$?

\pagebreak












\definition{Tensor Products}
The \textit{tensor product} between 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{}
The vectors we found in \ref{basistp} are a basis of what space? \par

\vfill
\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. \par
We'll shorten this notation to $\ket{01}$. \par
Thus, the two-bit kets we saw on the previous page are, by definition, tensor products.

\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, you should quickly see the pattern.
}






\vfill
\pagebreak