316 lines
5.7 KiB
TeX
316 lines
5.7 KiB
TeX
\section{Two Bits}
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How do we represent multi-bit states using vectors? \par
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Unfortunately, this is hard to visualize---but the idea is simple.
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\problem{}
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What is the set of possible states of two bits (i.e, $\mathbb{B}^2$)?
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\vspace{2cm}
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\cgeneric{Remark}
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When we have two bits, we have four orthogonal states:
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$\overrightarrow{00}$, $\overrightarrow{01}$, $\overrightarrow{10}$, and $\overrightarrow{11}$. \par
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We need four dimensions to draw all of these vectors, so I can't provide a picture... \par
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but the idea here is the same as before.
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\problem{}
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Write $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$ as column vectors \par
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with respect to the orthonormal basis $\{\overrightarrow{00}, \overrightarrow{01}, \overrightarrow{10}, \overrightarrow{11}\}$.
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\vfill
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\cgeneric{Remark}
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So, we represent each possible state as an axis in an $n$-dimensional space. \par
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A set of $n$ bits gives us $2^n$ possible states, which forms a basis in $2^n$ dimensions.
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\vspace{1mm}
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Say we now have two seperate bits: $\ket{a}$ and $\ket{b}$. \par
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How do we represent their compound state? \par
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\vspace{4mm}
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If we return to our usual notation, this is very easy:
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$a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$, \par
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so the possible compound states of $ab$ are
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$\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$
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\vspace{1mm}
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The same is true of any other state set: if $a$ takes values in $A$ and $b$ takes values in $B$, \par
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the compound state $(a,b)$ takes values in $A \times B$.
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\vspace{4mm}
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We would like to do the same in vector notation. Given $\ket{a}$ and $\ket{b}$, \par
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how should we represent the state of $\ket{ab}$?
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\pagebreak
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\definition{Tensor Products}
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The \textit{tensor product} between two vectors
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is defined as follows:
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\begin{equation*}
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\begin{bmatrix}
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x_1 \\ x_2
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\end{bmatrix}
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\otimes
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\begin{bmatrix}
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y_1 \\ y_2
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\end{bmatrix}
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=
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\begin{bmatrix}
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x_1
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\begin{bmatrix}
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y_1 \\ y_2
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\end{bmatrix}
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\\[4mm]
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x_2
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\begin{bmatrix}
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y_1 \\ y_2
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\end{bmatrix}
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\end{bmatrix}
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=
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\begin{bmatrix}
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x_1y_1 \\[1mm]
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x_1y_2 \\[1mm]
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x_2y_1 \\[1mm]
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x_2y_2 \\[0.5mm]
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\end{bmatrix}
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\end{equation*}
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That is, we take our first vector, multiply the second
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vector by each of its components, and stack the result.
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You could think of this as a generalization of scalar
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mulitiplication, where scalar mulitiplication is a
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tensor product with a vector in $\mathbb{R}^1$:
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\begin{equation*}
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a
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\begin{bmatrix}
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x_1 \\ x_2
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\end{bmatrix}
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=
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\begin{bmatrix}
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a_1
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\end{bmatrix}
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\otimes
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\begin{bmatrix}
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y_1 \\ y_2
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\end{bmatrix}
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=
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\begin{bmatrix}
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a_1
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\begin{bmatrix}
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y_1 \\ y_2
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\end{bmatrix}
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\end{bmatrix}
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=
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\begin{bmatrix}
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a_1y_1 \\[1mm]
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a_1y_2
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\end{bmatrix}
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\end{equation*}
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\vspace{2mm}
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Also, note that the tensor product is very similar to the
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Cartesian product: if we take $x$ and $y$ as sets, with
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$x = \{x_1, x_2\}$ and $y = \{y_1, y_2\}$, the Cartesian product
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contains the same elements as the tensor product---every possible
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pairing of an element in $x$ with an element in $y$:
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\begin{equation*}
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x \times y = \{~(x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2y_2)~\}
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\end{equation*}
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In fact, these two operations are (in a sense) essentially identical. \par
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Let's quickly demonstrate this.
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\problem{}
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Say $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. \par
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What is the dimension of $x \otimes y$?
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\vfill
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\problem{}<basistp>
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What is the pairwise tensor product
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$
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\Bigl\{
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\left[
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\begin{smallmatrix}
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1 \\ 0 \\ 0
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\end{smallmatrix}
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\right],
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\left[
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\begin{smallmatrix}
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0 \\ 1 \\ 0
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\end{smallmatrix}
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\right],
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\left[
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\begin{smallmatrix}
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0 \\ 0 \\ 1
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\end{smallmatrix}
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\right]
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\Bigr\}
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\otimes
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\Bigl\{
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\left[
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\begin{smallmatrix}
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1 \\ 0
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\end{smallmatrix}
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\right],
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\left[
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\begin{smallmatrix}
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0 \\ 1
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\end{smallmatrix}
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\right]
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\Bigr\}
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$?
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\note{in other words, distribute the tensor product between every pair of vectors.}
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\vfill
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\problem{}
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What is the \textit{span} of the vectors we found in \ref{basistp}? \par
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In other words, what is the set of vectors that can be written as weighted sums of the vectors above?
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\vfill
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\pagebreak
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\problem{}
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The compound state of two vector-form bits is their tensor product. \par
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Compute the following. Is the result what we'd expect?
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\begin{itemize}
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\item $\ket{0} \otimes \ket{0}$
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\item $\ket{0} \otimes \ket{1}$
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\item $\ket{1} \otimes \ket{0}$
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\item $\ket{1} \otimes \ket{1}$
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\end{itemize}
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\hint{
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Remember that the coordinates of
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$\ket{0}$ are $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$,
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and the coordinates of
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$\ket{1}$ are $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
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}
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\vfill
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\problem{}<fivequant>
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Of course, writing $\ket{0} \otimes \ket{1}$ is a bit excessive. \par
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We'll shorten this notation to $\ket{01}$. \par
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Thus, the two-bit kets we saw on the previous page are, by definition, tensor products.
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\vspace{2mm}
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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
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we could write $\ket{1101}$---but a shorter alternative is $\ket{13}$, since $13$ is \texttt{1101} in binary.
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\vspace{2mm}
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Write $\ket{5}$ as three-bit state vector. \par
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\begin{solution}
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$\ket{5} = \ket{101} = \ket{1} \otimes \ket{0} \otimes \ket{1} = [0,0,0,0,0,1,0,0]^T$ \par
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Notice how we're counting from the top, with $\ket{000} = [1,0,...,0]$ and $\ket{111} = [0, ..., 0, 1]$.
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\end{solution}
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\vfill
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\problem{}
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Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par
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\hint{
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You do not need to compute every tensor product. \\
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Do a few, you should quickly see the pattern.
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}
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\vfill
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\pagebreak |