From f88160b5f53cf86ec11b6ecfdf7539f4e656e77d Mon Sep 17 00:00:00 2001 From: Mark Date: Thu, 2 Oct 2025 09:02:01 -0700 Subject: [PATCH] Quantum edits --- .../src/parts/01 bits.tex | 85 +++++++++-------- .../src/parts/02 qubit.tex | 38 ++++++-- .../src/parts/03 two qubits.tex | 4 +- .../src/parts/04 logic gates.tex | 91 ++++++++++--------- .../src/parts/06 hxh.tex | 22 ++--- .../src/parts/07 superdense.tex | 2 +- 6 files changed, 136 insertions(+), 106 deletions(-) diff --git a/src/Advanced/Introduction to Quantum/src/parts/01 bits.tex b/src/Advanced/Introduction to Quantum/src/parts/01 bits.tex index c9857b9..320c76c 100644 --- a/src/Advanced/Introduction to Quantum/src/parts/01 bits.tex +++ b/src/Advanced/Introduction to Quantum/src/parts/01 bits.tex @@ -27,7 +27,7 @@ We'll represent this probabilistic bit's \textit{state} as a vector: $\left[\begin{smallmatrix} p_0 \\ p_1 \end{smallmatrix}\right]$ \par -We do \textbf{not} assume this coin is fair, and thus $p_0$ might not equal $p_1$. +We do \textbf{not} assume this coin is fair---$p_0$ might not equal $p_1$. \note{ This may seem a bit redundant: since $p_0 + p_1$, we can always calculate one probability given the other. \\ @@ -45,15 +45,16 @@ The simplest probabilistic bit states are of course $[0]$ and $[1]$, defined as \item $[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ \item $[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \end{itemize} -That is, $[0]$ represents a bit that we known to be \texttt{0}, \par +That is, $[0]$ represents a bit that we know to be \texttt{0}, \par and $[1]$ represents a bit we know to be \texttt{1}. \vfill +\pagebreak \definition{} -$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states: \par -Every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$: +$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states. This means that \par +every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$: \begin{equation*} \begin{bmatrix} p_0 \\ p_1 \end{bmatrix} @@ -66,7 +67,6 @@ Every other probabilistic bit can be written as a \textit{linear combination} of \vfill -\pagebreak \problem{} Every possible state of a probabilistic bit is a two-dimensional vector. \par @@ -118,6 +118,10 @@ Draw all possible states on the axis below. +% +% MARK: measure +% + \section{Measuring Probabilistic Bits} @@ -125,7 +129,7 @@ Draw all possible states on the axis below. \definition{} As we noted before, a probabilistic bit represents a coin we've tossed but haven't looked at. \par -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 +We do not know whether the bit is \texttt{0} or \texttt{1}, but we do know the probability of each outcome. \par \vspace{2mm} @@ -135,7 +139,7 @@ knowledge of its state is updated to either $[0]$ or $[1]$, since we now certain \vspace{2mm} Since measurement changes what we know about a probabilistic bit, it changes the probabilistic bit's state. -When we measure a bit, it's state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the +When we measure a bit, its state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the bit vanishes. We \textit{cannot} recover the state $[x_0, x_1]$ from a measured probabilistic bit. @@ -165,7 +169,10 @@ Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nic \item If we measure $y$ first and observe \texttt{1}, \par what is the probability of getting each of \texttt{00}, \texttt{01}, \texttt{10}, and \texttt{11}? \end{itemize} -\note[Note]{$[x]$ and $[y]$ are column vectors, but I've written them horizontally to save space.} +\note[Note]{ + $[x]$ and $[y]$ are column vectors in the previous pages, and are still column vectors here. \par + I've written them horizontally to save space. +} \vfill @@ -189,7 +196,9 @@ What is the probability that $x$ and $y$ produce different outcomes? - +% +% MARK: tensor product +% \section{Tensor Products} @@ -315,7 +324,7 @@ $? \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? +In other words, what is the set of all vectors that can be written as linear combinations of the vectors above? \vfill @@ -387,7 +396,7 @@ Compute the following. Is the result what we'd expect? \problem{} -Of course, writing $[0] \otimes [1]$ is a bit excessive. We'll shorten this notation to $[01]$. \par +Writing $[0] \otimes [1]$ is tedious. We'll shorten this notation to $[01]$. \par \vspace{2mm} @@ -429,9 +438,9 @@ Write the three-bit states $[0]$ through $[7]$ as column vectors. \par - - - +% +% MARK: ops +% @@ -443,25 +452,6 @@ Write the three-bit states $[0]$ through $[7]$ as column vectors. \par Now that we can write probabilistic bits as vectors, we can represent operations on these bits with linear transformations---in other words, as matrices. -\definition{} -Consider the NOT gate, which operates as follows: \par -\begin{itemize} - \item $\text{NOT}[0] = [1]$ - \item $\text{NOT}[1] = [0]$ -\end{itemize} -What should NOT do to a probabilistic bit $[x_0, x_1]$? \par -If we return to our coin analogy, we can think of the NOT operation as -flipping a coin we have already tossed, without looking at its state. -Thus, -\begin{equation*} - \text{NOT} \begin{bmatrix} - x_0 \\ x_1 - \end{bmatrix} = \begin{bmatrix} - x_1 \\ x_0 - \end{bmatrix} -\end{equation*} - - \begin{hobox}{Review: Multiplying Vectors by Matrices}{black!10!white}{black!65!white} \begin{equation*} Av = @@ -482,6 +472,10 @@ Thus, Note that each element of $Av$ is the dot product of a row in $A$ and a column in $v$. \end{hobox} +\generic{Remark:} +Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par +We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably. + \problem{} Compute the following product: \begin{equation*} @@ -496,12 +490,23 @@ Compute the following product: \vfill -\generic{Remark:} -Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par -We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably. - -\pagebreak - +\definition{} +Consider the NOT gate, which operates as follows: \par +\begin{itemize} + \item $\text{NOT}[0] = [1]$ + \item $\text{NOT}[1] = [0]$ +\end{itemize} +What should NOT do to a probabilistic bit $[x_0, x_1]$? \par +If we return to our coin analogy, we can think of the NOT operation as +flipping a coin we have already tossed, without looking at its state. +Thus, +\begin{equation*} + \text{NOT} \begin{bmatrix} + x_0 \\ x_1 + \end{bmatrix} = \begin{bmatrix} + x_1 \\ x_0 + \end{bmatrix} +\end{equation*} \problem{} Find the matrix that represents the NOT operation on one probabilistic bit. @@ -516,6 +521,8 @@ Find the matrix that represents the NOT operation on one probabilistic bit. \vfill +\pagebreak + \problem{Extension by linearity} Say we have an arbitrary operation $M$. \par diff --git a/src/Advanced/Introduction to Quantum/src/parts/02 qubit.tex b/src/Advanced/Introduction to Quantum/src/parts/02 qubit.tex index f302e19..52b642e 100644 --- a/src/Advanced/Introduction to Quantum/src/parts/02 qubit.tex +++ b/src/Advanced/Introduction to Quantum/src/parts/02 qubit.tex @@ -4,7 +4,8 @@ Quantum bits (or \textit{qubits}) are very similar to probabilistic bits, but ha probabilities are replaced with \textit{amplitudes}. \vspace{2mm} -Of course, a qubit can take the values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par + +As before a qubit can take the extremal values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par Like probabilistic bits, a quantum bit is written as a linear combination of $\ket{0}$ and $\ket{1}$: \begin{equation*} \ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1} @@ -19,8 +20,8 @@ $\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one \vspace{2mm} This is very similar to the \say{box} $[~]$ notation we used for probabilistic bits. \par -As before, we will write $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ -and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$. +When we work with qubits, we will write $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ as $\ket{0}$ +and $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ as $\ket{1}$. \vspace{8mm} @@ -28,7 +29,7 @@ and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$. Recall that probabilistic bits are subject to the restriction that $p_0 + p_1 = 1$. \par Quantum bits have a similar condition: $\psi_0^2 + \psi_1^2 = 1$. \par Note that this implies that $\psi_0$ and $\psi_1$ are both in $[-1, 1]$. \par -Quantum amplitudes may be negative, but probabilistic bit probabilities cannot. +Quantum amplitudes may be negative! \vspace{2mm} @@ -89,8 +90,8 @@ Write $\ket{\psi}$ as a linear combination of $\ket{0}$ and $\ket{1}$. \definition{Measurement I} -Just like a probabilistic bit, we must observed $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par -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 +Just like a probabilistic bit, we must observe $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par +If we measure $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$, we will see either $\ket{0}$ or $\ket{1}$, with the following probabilities: \begin{itemize}[itemsep = 2mm, topsep = 2mm] \item $\mathcal{P}(\ket{1}) = \psi_1^2$ @@ -117,7 +118,7 @@ leaving no trace of the previous superposition. \par \problem{} \begin{itemize} - \item What is the probability we observe $\ket{0}$ when we measure $\ket{\psi}$? \par + \item What is the probability that we observe $\ket{0}$ when we measure $\ket{\psi}$? \par \item What can we observe if we measure $\ket{\psi}$ a second time? \par \item What are these probabilities for $\ket{\varphi}$? \end{itemize} @@ -235,6 +236,10 @@ Consider the following qubit states: \pagebreak +% +% MARK: ops +% + \section{Operations on One Qubit} We may apply transformations to qubits just as we apply transformations to probabilistic bits. @@ -277,7 +282,7 @@ Find the matrix $X$. \vfill \problem{} -What is $X\ket{+}$ and $X\ket{-}$? \par +What are $X\ket{+}$ and $X\ket{-}$? \par \hint{Remember that all matrices are linear maps. What does this mean?} \begin{solution} @@ -331,11 +336,21 @@ As we noted earlier, any rotation about the center is a valid quantum gate. \par Let's derive all transformations of this form. \begin{itemize}[itemsep = 1mm] \item Let $U_\phi$ be the matrix that represents a counterclockwise rotation of $\phi$ degrees. \par - What is $U\ket{0}$ and $U\ket{1}$? + What are $U\ket{0}$ and $U\ket{1}$? \item Find the matrix $U_\phi$ for an arbitrary $\phi$. \end{itemize} + +\begin{solution} + \begin{equation*} + \begin{bmatrix} + cos(\theta) & -sin(\theta) \\ + sin(\theta) & cos(\theta) + \end{bmatrix} + \end{equation*} +\end{solution} + \vfill @@ -343,5 +358,10 @@ Let's derive all transformations of this form. Say we have a qubit that is either $\ket{+}$ or $\ket{-}$. We do not know which of the two states it is in. \par Using one operation and one measurement, how can we find out, for certain, which qubit we received? \par + +\begin{solution} + Use a 45 degree ccw rotation. +\end{solution} + \vfill \pagebreak \ No newline at end of file diff --git a/src/Advanced/Introduction to Quantum/src/parts/03 two qubits.tex b/src/Advanced/Introduction to Quantum/src/parts/03 two qubits.tex index d62b60f..bd53235 100644 --- a/src/Advanced/Introduction to Quantum/src/parts/03 two qubits.tex +++ b/src/Advanced/Introduction to Quantum/src/parts/03 two qubits.tex @@ -60,7 +60,7 @@ $\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3 Again, consider the two-qubit state $\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$ \par If we measure the first qubit of $\ket{\psi}$ and get $\ket{0}$, what is the resulting state of $\ket{\psi}$? \par -What would the state be if we'd measured $\ket{1}$ instead? +What would that state be if we'd measured $\ket{1}$ instead? \vfill @@ -85,7 +85,7 @@ Say we measure the first two qubits and get $\ket{00}$. What is the resulting st \definition{Entanglement} -Some product states can be factored into a tensor product of individual qubit states. For example, +Some compound states can be factored into a tensor product of individual qubit states. For example, \begin{equation*} \frac{1}{2} \bigl(\ket{00} + \ket{01} + \ket{10} + \ket{11}\bigr) = \frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr) \otimes diff --git a/src/Advanced/Introduction to Quantum/src/parts/04 logic gates.tex b/src/Advanced/Introduction to Quantum/src/parts/04 logic gates.tex index 53f927a..7b1ed65 100644 --- a/src/Advanced/Introduction to Quantum/src/parts/04 logic gates.tex +++ b/src/Advanced/Introduction to Quantum/src/parts/04 logic gates.tex @@ -1,40 +1,42 @@ \section{Logic Gates} -\definition{Matrices} -Throughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix, -and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear -map, we can write it as follows: -\begin{equation*} - f\left( - \ket{x} - \right) - = - \begin{bmatrix} - m_1 & m_2 \\ - m_3 & m_4 - \end{bmatrix} - \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} - = - \left[ - \begin{matrix} - m_1x_1 + m_2x_2 \\ - m_3x_1 + m_4x_2 - \end{matrix} - \right] -\end{equation*} - +%\definition{Matrices} +%cThroughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix, +%and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear +%map, we can write it as follows: +%\begin{equation*} +% f\left( +% \ket{x} +% \right) +% = +% \begin{bmatrix} +% m_1 & m_2 \\ +% m_3 & m_4 +% \end{bmatrix} +% \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} +% = +% \left[ +% \begin{matrix} +% m_1x_1 + m_2x_2 \\ +% m_3x_1 + m_4x_2 +% \end{matrix} +% \right] +%\end{equation*} +% +%A classical logic gate is a linear map from $\{0,1\}^m$ to $\{0,1\}^n$ \definition{} -Before we discussing multi-qubit quantum gates, we need to review to classical logic. \par -Of course, a classical logic gate is a linear map from $\{0,1\}^m$ to $\{0,1\}^n$ +Before we discussing multi-qubit quantum gates, we need to review classical logic. \par +In this section, let's return to probabilistic bits. + \problem{} The \texttt{not} gate is a map defined by the following table: \par \begin{itemize} - \item $X\ket{0} = \ket{1}$ - \item $X\ket{1} = \ket{0}$ + \item $X[0] = [1]$ + \item $X[1] = [0]$ \end{itemize} Write the \texttt{not} gate as a matrix that operates on single-bit vector states. \par @@ -93,11 +95,11 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par \vspace{2mm} For example, if we look at the first column of $A$ (which is $[1, 0]$), we see: \par - $A\ket{00} = A[1,0,0,0] = [1,0] = \ket{0}$ + $A[00] = A[1,0,0,0] = [1,0] = [0]$ \vspace{2mm} Also with the last column (which is $[0,1]$): \par - $A\ket{00} = A[0,0,0,1] = [0,1] = \ket{1}$ + $A[00] = A[0,0,0,1] = [0,1] = [1]$ \end{instructornote} \end{solution} @@ -244,10 +246,10 @@ Say we want to invert the first bit of a two-bit state. That is, we want a trans In other words, we want a matrix $T$ satisfying the following equalities: \begin{itemize} - \item $T\ket{00} = \ket{10}$ - \item $T\ket{01} = \ket{11}$ - \item $T\ket{10} = \ket{00}$ - \item $T\ket{11} = \ket{01}$ + \item $T[00] = [10]$ + \item $T[01] = [11]$ + \item $T[10] = [00]$ + \item $T[11] = [01]$ \end{itemize} @@ -256,8 +258,8 @@ In other words, we want a matrix $T$ satisfying the following equalities: Find the matrix that corresponds to the above transformation. \par \hint{ Remember that - $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and - $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\ + $[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and + $[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\ 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} diff --git a/src/Advanced/Introduction to Quantum/src/parts/06 hxh.tex b/src/Advanced/Introduction to Quantum/src/parts/06 hxh.tex index ee3b0f9..d3e6014 100644 --- a/src/Advanced/Introduction to Quantum/src/parts/06 hxh.tex +++ b/src/Advanced/Introduction to Quantum/src/parts/06 hxh.tex @@ -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] diff --git a/src/Advanced/Introduction to Quantum/src/parts/07 superdense.tex b/src/Advanced/Introduction to Quantum/src/parts/07 superdense.tex index cf60ded..02daf7d 100644 --- a/src/Advanced/Introduction to Quantum/src/parts/07 superdense.tex +++ b/src/Advanced/Introduction to Quantum/src/parts/07 superdense.tex @@ -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