Quantum edits
This commit is contained in:
@ -27,7 +27,7 @@ We'll represent this probabilistic bit's \textit{state} as a vector:
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$\left[\begin{smallmatrix}
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p_0 \\ p_1
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\end{smallmatrix}\right]$ \par
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We do \textbf{not} assume this coin is fair, and thus $p_0$ might not equal $p_1$.
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We do \textbf{not} assume this coin is fair---$p_0$ might not equal $p_1$.
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\note{
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This may seem a bit redundant: since $p_0 + p_1$, we can always calculate one probability given the other. \\
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@ -45,15 +45,16 @@ The simplest probabilistic bit states are of course $[0]$ and $[1]$, defined as
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\item $[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
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\item $[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$
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\end{itemize}
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That is, $[0]$ represents a bit that we known to be \texttt{0}, \par
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That is, $[0]$ represents a bit that we know to be \texttt{0}, \par
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and $[1]$ represents a bit we know to be \texttt{1}.
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\vfill
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\pagebreak
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\definition{}
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$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states: \par
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Every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$:
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$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states. This means that \par
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every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$:
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\begin{equation*}
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\begin{bmatrix} p_0 \\ p_1 \end{bmatrix}
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@ -66,7 +67,6 @@ Every other probabilistic bit can be written as a \textit{linear combination} of
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\vfill
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\pagebreak
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\problem{}
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Every possible state of a probabilistic bit is a two-dimensional vector. \par
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@ -118,6 +118,10 @@ Draw all possible states on the axis below.
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%
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% MARK: measure
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%
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\section{Measuring Probabilistic Bits}
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@ -125,7 +129,7 @@ Draw all possible states on the axis below.
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\definition{}
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As we noted before, a probabilistic bit represents a coin we've tossed but haven't looked at. \par
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We do not know whether the bit is \texttt{0} or \texttt{1}, but we do know the probability of both of these outcomes. \par
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We do not know whether the bit is \texttt{0} or \texttt{1}, but we do know the probability of each outcome. \par
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\vspace{2mm}
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@ -135,7 +139,7 @@ knowledge of its state is updated to either $[0]$ or $[1]$, since we now certain
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\vspace{2mm}
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Since measurement changes what we know about a probabilistic bit, it changes the probabilistic bit's state.
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When we measure a bit, it's state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the
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When we measure a bit, its state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the
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bit vanishes. We \textit{cannot} recover the state $[x_0, x_1]$ from a measured probabilistic bit.
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@ -165,7 +169,10 @@ Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nic
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\item If we measure $y$ first and observe \texttt{1}, \par
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what is the probability of getting each of \texttt{00}, \texttt{01}, \texttt{10}, and \texttt{11}?
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\end{itemize}
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\note[Note]{$[x]$ and $[y]$ are column vectors, but I've written them horizontally to save space.}
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\note[Note]{
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$[x]$ and $[y]$ are column vectors in the previous pages, and are still column vectors here. \par
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I've written them horizontally to save space.
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}
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\vfill
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@ -189,7 +196,9 @@ What is the probability that $x$ and $y$ produce different outcomes?
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%
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% MARK: tensor product
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%
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\section{Tensor Products}
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@ -315,7 +324,7 @@ $?
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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 linear combinations of the vectors above?
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In other words, what is the set of all vectors that can be written as linear combinations of the vectors above?
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\vfill
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@ -387,7 +396,7 @@ Compute the following. Is the result what we'd expect?
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\problem{}<fivequant>
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Of course, writing $[0] \otimes [1]$ is a bit excessive. We'll shorten this notation to $[01]$. \par
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Writing $[0] \otimes [1]$ is tedious. We'll shorten this notation to $[01]$. \par
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\vspace{2mm}
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@ -429,9 +438,9 @@ Write the three-bit states $[0]$ through $[7]$ as column vectors. \par
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%
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% MARK: ops
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%
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@ -443,25 +452,6 @@ Write the three-bit states $[0]$ through $[7]$ as column vectors. \par
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Now that we can write probabilistic bits as vectors, we can represent operations on these bits
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with linear transformations---in other words, as matrices.
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\definition{}
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Consider the NOT gate, which operates as follows: \par
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\begin{itemize}
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\item $\text{NOT}[0] = [1]$
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\item $\text{NOT}[1] = [0]$
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\end{itemize}
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What should NOT do to a probabilistic bit $[x_0, x_1]$? \par
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If we return to our coin analogy, we can think of the NOT operation as
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flipping a coin we have already tossed, without looking at its state.
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Thus,
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\begin{equation*}
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\text{NOT} \begin{bmatrix}
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x_0 \\ x_1
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\end{bmatrix} = \begin{bmatrix}
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x_1 \\ x_0
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\end{bmatrix}
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\end{equation*}
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\begin{hobox}{Review: Multiplying Vectors by Matrices}{black!10!white}{black!65!white}
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\begin{equation*}
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Av =
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@ -482,6 +472,10 @@ Thus,
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Note that each element of $Av$ is the dot product of a row in $A$ and a column in $v$.
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\end{hobox}
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\generic{Remark:}
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Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par
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We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably.
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\problem{}
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Compute the following product:
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\begin{equation*}
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@ -496,12 +490,23 @@ Compute the following product:
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\vfill
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\generic{Remark:}
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Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par
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We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably.
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\pagebreak
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\definition{}
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Consider the NOT gate, which operates as follows: \par
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\begin{itemize}
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\item $\text{NOT}[0] = [1]$
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\item $\text{NOT}[1] = [0]$
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\end{itemize}
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What should NOT do to a probabilistic bit $[x_0, x_1]$? \par
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If we return to our coin analogy, we can think of the NOT operation as
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flipping a coin we have already tossed, without looking at its state.
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Thus,
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\begin{equation*}
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\text{NOT} \begin{bmatrix}
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x_0 \\ x_1
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\end{bmatrix} = \begin{bmatrix}
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x_1 \\ x_0
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\end{bmatrix}
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\end{equation*}
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\problem{}
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Find the matrix that represents the NOT operation on one probabilistic bit.
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@ -516,6 +521,8 @@ Find the matrix that represents the NOT operation on one probabilistic bit.
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\vfill
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\pagebreak
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\problem{Extension by linearity}
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Say we have an arbitrary operation $M$. \par
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@ -4,7 +4,8 @@ Quantum bits (or \textit{qubits}) are very similar to probabilistic bits, but ha
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probabilities are replaced with \textit{amplitudes}.
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\vspace{2mm}
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Of course, a qubit can take the values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par
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As before a qubit can take the extremal values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par
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Like probabilistic bits, a quantum bit is written as a linear combination of $\ket{0}$ and $\ket{1}$:
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\begin{equation*}
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\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}
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@ -19,8 +20,8 @@ $\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one
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\vspace{2mm}
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This is very similar to the \say{box} $[~]$ notation we used for probabilistic bits. \par
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As before, we will write $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
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and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
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When we work with qubits, we will write $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ as $\ket{0}$
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and $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ as $\ket{1}$.
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\vspace{8mm}
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@ -28,7 +29,7 @@ and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
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Recall that probabilistic bits are subject to the restriction that $p_0 + p_1 = 1$. \par
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Quantum bits have a similar condition: $\psi_0^2 + \psi_1^2 = 1$. \par
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Note that this implies that $\psi_0$ and $\psi_1$ are both in $[-1, 1]$. \par
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Quantum amplitudes may be negative, but probabilistic bit probabilities cannot.
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Quantum amplitudes may be negative!
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\vspace{2mm}
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@ -89,8 +90,8 @@ Write $\ket{\psi}$ as a linear combination of $\ket{0}$ and $\ket{1}$.
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\definition{Measurement I}
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Just like a probabilistic bit, we must observed $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par
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If we were to measure $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$, we'd observe either $\ket{0}$ or $\ket{1}$, \par
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Just like a probabilistic bit, we must observe $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par
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If we measure $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$, we will see either $\ket{0}$ or $\ket{1}$,
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with the following probabilities:
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\begin{itemize}[itemsep = 2mm, topsep = 2mm]
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\item $\mathcal{P}(\ket{1}) = \psi_1^2$
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@ -117,7 +118,7 @@ leaving no trace of the previous superposition. \par
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\problem{}
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\begin{itemize}
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\item What is the probability we observe $\ket{0}$ when we measure $\ket{\psi}$? \par
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\item What is the probability that we observe $\ket{0}$ when we measure $\ket{\psi}$? \par
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\item What can we observe if we measure $\ket{\psi}$ a second time? \par
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\item What are these probabilities for $\ket{\varphi}$?
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\end{itemize}
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@ -235,6 +236,10 @@ Consider the following qubit states:
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\pagebreak
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%
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% MARK: ops
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%
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\section{Operations on One Qubit}
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We may apply transformations to qubits just as we apply transformations to probabilistic bits.
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@ -277,7 +282,7 @@ Find the matrix $X$.
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\vfill
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\problem{}
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What is $X\ket{+}$ and $X\ket{-}$? \par
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What are $X\ket{+}$ and $X\ket{-}$? \par
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\hint{Remember that all matrices are linear maps. What does this mean?}
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\begin{solution}
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@ -331,11 +336,21 @@ As we noted earlier, any rotation about the center is a valid quantum gate. \par
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Let's derive all transformations of this form.
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\begin{itemize}[itemsep = 1mm]
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\item Let $U_\phi$ be the matrix that represents a counterclockwise rotation of $\phi$ degrees. \par
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What is $U\ket{0}$ and $U\ket{1}$?
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What are $U\ket{0}$ and $U\ket{1}$?
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\item Find the matrix $U_\phi$ for an arbitrary $\phi$.
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\end{itemize}
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\begin{solution}
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\begin{equation*}
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\begin{bmatrix}
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cos(\theta) & -sin(\theta) \\
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sin(\theta) & cos(\theta)
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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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@ -343,5 +358,10 @@ Let's derive all transformations of this form.
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Say we have a qubit that is either $\ket{+}$ or $\ket{-}$. We do not know which of the two states it is in. \par
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Using one operation and one measurement, how can we find out, for certain, which qubit we received? \par
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\begin{solution}
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Use a 45 degree ccw rotation.
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\end{solution}
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\vfill
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\pagebreak
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@ -60,7 +60,7 @@ $\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3
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Again, consider the two-qubit state
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$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$ \par
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If we measure the first qubit of $\ket{\psi}$ and get $\ket{0}$, what is the resulting state of $\ket{\psi}$? \par
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What would the state be if we'd measured $\ket{1}$ instead?
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What would that state be if we'd measured $\ket{1}$ instead?
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\vfill
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@ -85,7 +85,7 @@ Say we measure the first two qubits and get $\ket{00}$. What is the resulting st
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\definition{Entanglement}
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Some product states can be factored into a tensor product of individual qubit states. For example,
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Some compound states can be factored into a tensor product of individual qubit states. For example,
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\begin{equation*}
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\frac{1}{2} \bigl(\ket{00} + \ket{01} + \ket{10} + \ket{11}\bigr)
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= \frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr) \otimes
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@ -1,40 +1,42 @@
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\section{Logic Gates}
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\definition{Matrices}
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Throughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix,
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and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear
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map, we can write it as follows:
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\begin{equation*}
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f\left(
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\ket{x}
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\right)
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=
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\begin{bmatrix}
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m_1 & m_2 \\
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m_3 & m_4
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\end{bmatrix}
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\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
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=
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\left[
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\begin{matrix}
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m_1x_1 + m_2x_2 \\
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m_3x_1 + m_4x_2
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\end{matrix}
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\right]
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\end{equation*}
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%\definition{Matrices}
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%cThroughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix,
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%and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear
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%map, we can write it as follows:
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%\begin{equation*}
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% f\left(
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% \ket{x}
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% \right)
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% =
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% \begin{bmatrix}
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% m_1 & m_2 \\
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% m_3 & m_4
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% \end{bmatrix}
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% \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
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% =
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% \left[
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% \begin{matrix}
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% m_1x_1 + m_2x_2 \\
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% m_3x_1 + m_4x_2
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% \end{matrix}
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% \right]
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%\end{equation*}
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%
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%A classical logic gate is a linear map from $\{0,1\}^m$ to $\{0,1\}^n$
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\definition{}
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Before we discussing multi-qubit quantum gates, we need to review to classical logic. \par
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Of course, a classical logic gate is a linear map from $\{0,1\}^m$ to $\{0,1\}^n$
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Before we discussing multi-qubit quantum gates, we need to review classical logic. \par
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In this section, let's return to probabilistic bits.
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\problem{}<notgatex>
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The \texttt{not} gate is a map defined by the following table: \par
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\begin{itemize}
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\item $X\ket{0} = \ket{1}$
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\item $X\ket{1} = \ket{0}$
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\item $X[0] = [1]$
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\item $X[1] = [0]$
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\end{itemize}
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Write the \texttt{not} gate as a matrix that operates on single-bit vector states. \par
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@ -93,11 +95,11 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
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\vspace{2mm}
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For example, if we look at the first column of $A$ (which is $[1, 0]$), we see: \par
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$A\ket{00} = A[1,0,0,0] = [1,0] = \ket{0}$
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$A[00] = A[1,0,0,0] = [1,0] = [0]$
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\vspace{2mm}
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Also with the last column (which is $[0,1]$): \par
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$A\ket{00} = A[0,0,0,1] = [0,1] = \ket{1}$
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$A[00] = A[0,0,0,1] = [0,1] = [1]$
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\end{instructornote}
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\end{solution}
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@ -244,10 +246,10 @@ Say we want to invert the first bit of a two-bit state. That is, we want a trans
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In other words, we want a matrix $T$ satisfying the following equalities:
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\begin{itemize}
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\item $T\ket{00} = \ket{10}$
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\item $T\ket{01} = \ket{11}$
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\item $T\ket{10} = \ket{00}$
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\item $T\ket{11} = \ket{01}$
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\item $T[00] = [10]$
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\item $T[01] = [11]$
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\item $T[10] = [00]$
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\item $T[11] = [01]$
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\end{itemize}
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@ -256,8 +258,8 @@ In other words, we want a matrix $T$ satisfying the following equalities:
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Find the matrix that corresponds to the above transformation. \par
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\hint{
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Remember that
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$\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and
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$\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\
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$[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and
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$[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\
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Also, we found earlier that $X = \left[\begin{smallmatrix} 0 && 1 \\ 1 && 0 \end{smallmatrix}\right]$,
|
||||
and of course $I = \left[\begin{smallmatrix} 1 && 0 \\ 0 && 1 \end{smallmatrix}\right]$.
|
||||
}
|
||||
@ -279,25 +281,26 @@ Find the matrix that corresponds to the above transformation. \par
|
||||
We could draw the above transformation as a combination $X$ and $I$ (identity) gate:
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
\node[qubit] (a) at (0, 0) {$\ket{0}$};
|
||||
\node[qubit] (b) at (0, -1) {$\ket{0}$};
|
||||
\node[qubit] (a) at (0, 0) {$[0]$};
|
||||
\node[qubit] (b) at (0, -1) {$[0]$};
|
||||
|
||||
\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
|
||||
\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
|
||||
\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$[1]$};
|
||||
\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$[0]$};
|
||||
|
||||
\qubox{a}{1}{a}{2}{$X$}
|
||||
\qubox{b}{1}{b}{2}{$I$}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
We can even omit the $I$ gate, since we now know that transformations affect the whole state: \par
|
||||
We can even omit the $I$ gate. We know that transformations affect the whole state,
|
||||
and can assume the empty space uncer $X$ implies $I$. \par
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
\node[qubit] (a) at (0, 0) {$\ket{0}$};
|
||||
\node[qubit] (b) at (0, -1) {$\ket{0}$};
|
||||
\node[qubit] (a) at (0, 0) {$[0]$};
|
||||
\node[qubit] (b) at (0, -1) {$[0]$};
|
||||
|
||||
\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
|
||||
\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
|
||||
\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$[1]$};
|
||||
\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$[0]$};
|
||||
|
||||
\qubox{a}{1}{a}{2}{$X$}
|
||||
\end{tikzpicture}
|
||||
|
@ -1,7 +1,7 @@
|
||||
\section{HXH}
|
||||
|
||||
Let's return to the quantum circuit diagrams we discussed a few pages ago. \par
|
||||
Keep in mind that we're working with quantum gates and proper half-qubits---not classical bits, as we were before.
|
||||
Keep in mind that we're working with quantum gates and proper qubits---not classical bits, as we were before.
|
||||
|
||||
\definition{Controlled Inputs}
|
||||
A \textit{control input} or \textit{inverted control input} may be attached to any gate. \par
|
||||
@ -289,7 +289,7 @@ $\ket{-} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} - \ket{1}\Bigr)$
|
||||
\pagebreak
|
||||
|
||||
\generic{Remark:}
|
||||
Now, consider the following circuit:
|
||||
Consider the following circuit:
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=0.8]
|
||||
@ -386,21 +386,21 @@ but the state $\ket{cd}$ on the right is $\ket{11} = [0,0,0,1]$. \par
|
||||
\vspace{4mm}
|
||||
|
||||
How does this make sense? \par
|
||||
Remember that a two-bit quantum state is \textit{not} equivalent to a pair of one-qubit quantum states.
|
||||
We must treat a multi-qubit state as a single unit.
|
||||
Remember that a two-bit quantum state is \textit{not} equivalent to a pair of disjoint one-qubit quantum states.
|
||||
We cannot treat a multi-qubit state as a combination of $n$ independent bits.
|
||||
|
||||
Recall that a two-bit state $\ket{ab}$ comes with four probabilities:
|
||||
$\mathcal{P}(\texttt{00})$, $\mathcal{P}(\texttt{01})$, $\mathcal{P}(\texttt{10})$, and $\mathcal{P}(\texttt{11})$.
|
||||
\vspace{2mm}
|
||||
|
||||
A two-bit state $\ket{ab}$ comes with four probabilities:
|
||||
$\mathcal{P}(\texttt{00})$, $\mathcal{P}(\texttt{01})$, $\mathcal{P}(\texttt{10})$, and $\mathcal{P}(\texttt{11})$. \par
|
||||
If we change the probabilities of only $\ket{a}$, \textit{all four of these change!}
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
Because of this fact, \say{controlled gates} may not work as you expect. They may seem
|
||||
to \say{read} their controlling qubit without affecting its state, but remember---a
|
||||
controlled gate still affects the \textit{entire} state. As we noted before, it is
|
||||
not possible to apply a transformation to one bit of a quantum state.
|
||||
|
||||
Because of this fact, \say{controlled gates} behave in a somewhat counterintuitive way.
|
||||
They do not simply \say{read} their controlling qubit without affecting its state---
|
||||
they mutate the entire state they are applied to, and may change all its bits!
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
|
@ -12,7 +12,7 @@ Consider the following entangled two-qubit states, called the \textit{bell state
|
||||
The probabilistic bits we get when measuring any of the above may be called \textit{anticorrelated bits}. \par
|
||||
If we measure the first bit of any of these states and observe $1$, what is the resulting compound state? \par
|
||||
What if we observe $0$ instead? \par
|
||||
Do you see why we can call these bits anticorrelated?
|
||||
Do you see why we call these bits \say{anticorrelated}?
|
||||
|
||||
\vfill
|
||||
|
||||
|
Reference in New Issue
Block a user