Quantum Chapter Changes (#208)

* fixes

* fix unintended changes

* bruh?

* dont expand tabs

* more spacing

* GHZ text

* Text on Shor's

* another typo

* a

* Some other shor comments

* more?

* typo by me

* changes

* Trailing whitespace

* Rewrite comment on Nobel prize

* Cleanup shor.tex

---------

Co-authored-by: Evan Chen <[email protected]>

tex/quantum/circuits.tex

@@ -658,7 +658,7 @@ \section{Deutsch-Jozsa algorithm}
 		\]
 		since $H\ket0 = \frac{1}{\sqrt2}(\ket0+\ket1)$
 		while $H\ket1 = \frac{1}{\sqrt2}(\ket0-\ket1)$,
-		so minus signs arise exactly if $x_i = 0$ and $y_i = 0$ simultaneously,
+		so minus signs arise exactly if $x_i = 1$ and $y_i = 1$ simultaneously,
 		hence the term $(-1)^{x_1 y_1 + \dots + x_n y_n}$.
 		Swapping the order of summation, we get
 		\[
@@ -682,7 +682,7 @@ \section{Deutsch-Jozsa algorithm}
 			\]
 			To see this, note that the result is clear for $y_1 = \dots = y_n = 0$;
 			otherwise, if WLOG $y_1 = 1$, then the terms for $x_1 = 0$ exactly cancel
-			the terms for $x_1 = 0$, pair by pair.
+			the terms for $x_1 = 1$, pair by pair.
 			Thus in this state, the measurements all result in $\ket0 \dots \ket0$.
 
 			\ii On the other hand if $f$ is balanced, we derive that

tex/quantum/shor.tex

@@ -2,6 +2,10 @@ \chapter{Shor's algorithm}
 OK, now for Shor's Algorithm:
 how to factor $M = pq$ in $O\left( (\log M)^2 \right)$ time.
 
+This is arguably the reason agencies such as the US's National
+Security Agency have been diverting millions of dollars toward
+quantum computing.
+
 \section{The classical (inverse) Fourier transform}
 The ``crux move'' in Shor's algorithm is the so-called
 quantum Fourier transform.
@@ -32,7 +36,14 @@ \section{The classical (inverse) Fourier transform}
 	\]
 \end{definition}
 The reason this operation is important is because it lets
-us detect if the $x_i$ are periodic:
+us detect if the $x_i$ are periodic.
+More generally, given a sequence of $1$'s appearing with period $r$,
+the amplitudes will peak at inputs which are divisible by $\frac{N}{\gcd(N,r)}$.
+Mathematically, we have that
+\[
+	x_k = \sum_{j=0}^{N-1} y_j \omega_N^{-jk}.
+\]
+
 \begin{example}
 	[Example of discrete inverse Fourier transform]
 	Let $N = 6$, $\omega = \omega_6 = \exp(\frac{2\pi i}{6})$
@@ -52,8 +63,6 @@ \section{The classical (inverse) Fourier transform}
 	thus this reveals the periodicity of the original
 	sequence by $\frac N3 = 2$.
 \end{example}
-More generally, given a sequence of $1$'s appearing with period $r$,
-the amplitudes will peak at inputs which are divisible by $\frac{N}{\gcd(N,r)}$.
 \begin{remark}
 	The fact that this operation is called the ``inverse''
 	Fourier transform is mostly a historical accident
@@ -62,8 +71,9 @@ \section{The classical (inverse) Fourier transform}
 	(not-inverted) Fourier transform.
 \end{remark}
 If we apply the definition as written, computing the transform takes $O(N^2)$ time.
-It turns out that by an algorithm called the \vocab{fast Fourier transform}
-(whose details we won't discuss), one can reduce this to $O(N \log N)$ time.
+It turns out that by a classical algorithm called the \vocab{fast Fourier transform}
+(whose details we won't discuss, but it effectively ``reuses'' calculations),
+one can reduce this to $O(N \log N)$ time.
 However, for Shor's algorithm this is also insufficient;
 we need something like $O\left( (\log N)^2 \right)$ instead.
 This is where the quantum Fourier transform comes in.
@@ -82,6 +92,7 @@ \section{The quantum Fourier transform}
 	where $x = x_n x_{n-1} \dots x_1$ in binary.
 	For example, if $n = 3$
 	then $\ket{6}$ really means $\ket1 \otimes \ket1 \otimes \ket 0$.
+	Likewise, we refer to $0.x_1x_2 \dots x_n$ as binary.
 \end{abuse}
 Observe that the $n$-qubit space now has an
 orthonormal basis $\ket0$, $\ket1$, \dots, $\ket{N-1}$
@@ -128,8 +139,8 @@ \section{The quantum Fourier transform}
 	In short, expand everything.
 \end{proof}
 
-So by using mixed states, we can deal with the quantum Fourier transform
-using this ``multiplication by tensor product'' trick that isn't possible classically.
+So by using mixed states, the quantum Fourier transform
+can use this ``multiplication by tensor product'' trick that isn't possible classically.
 
 Now, without further ado, here's the circuit.
 Define the rotation matrices
@@ -143,8 +154,7 @@ \section{The quantum Fourier transform}
 	}
 \]
 \begin{exercise}
-	Show that in this circuit, the image of $\ket{x_3x_2x_1}$
-	(for binary $x_i$) is
+	Show that in this circuit, the image of $\ket{x_3x_2x_1}$ is
 	\[
 		\Big(\ket0+\exp(2\pi i \cdot 0.x_1) \ket1\Big)
 		\otimes \Big(\ket0+\exp(2\pi i \cdot 0.x_2x_1) \ket1\Big)
@@ -155,15 +165,15 @@ \section{The quantum Fourier transform}
 
 For general $n$, we can write this as inductively as
 \[
- \Qcircuit @C=1em @R=.7em {
-  \lstick{\ket{x_n}}     & \multigate{5}{\text{QFT}_{n-1}} & \gate{R_n} & \qw                    & \qw & \cdots & & \qw                    & \qw & \cdots & & \qw              & \qw      & \rstick{\ket{y_1}} \qw \\
-  \lstick{\ket{x_{n-1}}}     & \ghost{\text{QFT}_{n-1}}        & \qw                    & \gate{R_{n-1}} & \qw & \cdots & & \qw                    & \qw & \cdots & & \qw              & \qw      & \rstick{\ket{y_2}} \qw \\
-  \lstick{\vdots\ \ }    & \pureghost{\text{QFT}_{n-1}}    &                        &                        &     &        & &                        &     &        & &                  &          & \rstick{\ \ \vdots} \\
-  \lstick{\ket{x_i}}     & \ghost{\text{QFT}_{n-1}}        & \qw                    & \qw                    & \qw & \cdots & & \gate{R_i} & \qw & \cdots & & \qw              & \qw      & \rstick{\ket{y_{n-i+1}}} \qw \\
-  \lstick{\vdots\ \ }    & \pureghost{\text{QFT}_{n-1}}    &                        &                        &     &        & &                        &     &        & &                  &          & \rstick{\ \ \vdots} \\
-  \lstick{\ket{x_2}} & \ghost{\text{QFT}_{n-1}}        & \qw                    & \qw                    & \qw & \cdots & & \qw                    & \qw & \cdots & & \gate{R_2} & \qw      & \rstick{\ket{y_{n-1}}} \qw \\
-  \lstick{\ket{x_1}}     & \qw                             & \ctrl{-6}               & \ctrl{-5}               & \qw & \cdots & & \ctrl{-3}               & \qw & \cdots & & \ctrl{-1}         & \gate{H} & \rstick{\ket{y_n}} \qw
- }
+	\Qcircuit @C=1em @R=.7em {
+	\lstick{\ket{x_n}}     & \multigate{5}{\text{QFT}_{n-1}} & \gate{R_n} & \qw                    & \qw & \cdots & & \qw                    & \qw & \cdots & & \qw              & \qw      & \rstick{\ket{y_1}} \qw \\
+	\lstick{\ket{x_{n-1}}}     & \ghost{\text{QFT}_{n-1}}        & \qw                    & \gate{R_{n-1}} & \qw & \cdots & & \qw                    & \qw & \cdots & & \qw              & \qw      & \rstick{\ket{y_2}} \qw \\
+	\lstick{\vdots\ \ }    & \pureghost{\text{QFT}_{n-1}}    &                        &                        &     &        & &                        &     &        & &                  &          & \rstick{\ \ \vdots} \\
+	\lstick{\ket{x_i}}     & \ghost{\text{QFT}_{n-1}}        & \qw                    & \qw                    & \qw & \cdots & & \gate{R_i} & \qw & \cdots & & \qw              & \qw      & \rstick{\ket{y_{n-i+1}}} \qw \\
+	\lstick{\vdots\ \ }    & \pureghost{\text{QFT}_{n-1}}    &                        &                        &     &        & &                        &     &        & &                  &          & \rstick{\ \ \vdots} \\
+	\lstick{\ket{x_2}} & \ghost{\text{QFT}_{n-1}}        & \qw                    & \qw                    & \qw & \cdots & & \qw                    & \qw & \cdots & & \gate{R_2} & \qw      & \rstick{\ket{y_{n-1}}} \qw \\
+	\lstick{\ket{x_1}}     & \qw                             & \ctrl{-6}               & \ctrl{-5}               & \qw & \cdots & & \ctrl{-3}               & \qw & \cdots & & \ctrl{-1}         & \gate{H} & \rstick{\ket{y_n}} \qw
+}
 \]
 \begin{ques}
 	Convince yourself that when $n=3$ the two circuits displayed are equivalent.
@@ -209,7 +219,7 @@ \section{Shor's algorithm}
 	and we randomly select $x = 2$, and want to find its order $r$.
 	Let $n = 13$ and $N = 2^{13}$, and start by initializing the state
 	\[ \ket\psi = \frac{1}{\sqrt N} \sum_{k=0}^{N-1} \ket k. \]
-	Now, build a circuit $U_x$ (depending on $x=2$!)
+	Now, build a circuit $U_x$ (depending on $x$)
 	which takes $\ket k \ket 0$ to $\ket k \ket{x^k \bmod M}$.
 	Applying this to $\ket\psi \otimes \ket0$ gives
 	\[ U(\ket\psi\ket0) =
@@ -287,12 +297,15 @@ \section{Shor's algorithm}
 		\frac{1152}{2033}, \;
 		\dots
 	\]
-	So $\frac{17}{30}$ is a very good approximation,
+	So $\frac{17}{30}$ is a good approximation,
 	hence we deduce $s = 17$ and $r = 30$ as candidates.
 	And indeed, one can check that $r = 30$ is the desired order.
 \end{example}
 
-This won't work all the time (for example, we could get unlucky and
+This won't work all the time\footnote{%
+	Not to mention the general issue of noise, but that's for
+	engineers to worry about.
+} (for example, we could get unlucky and
 measure $j=0$, i.e.\ $s=0$, which would tell us no information at all).
 
 But one can show that we succeed any time that \[ \gcd(s,r) = 1. \]

tex/quantum/vectors.tex

@@ -244,12 +244,15 @@ \section{Observations}
 	\qquad
 	\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}.
 \]
+
 These matrices are important because:
 \begin{ques}
 	Show that these three matrices, plus the identity matrix,
 	form a basis for the set of Hermitian $2 \times 2$ matrices.
 \end{ques}
-So the Pauli matrices are a natural choice of basis.
+So the Pauli matrices are a natural choice of basis.\footnote{Well,
+	natural due to physics reasons.
+}
 
 Their normalized eigenvectors are
 \[ \zup = \ket0 = \pair10 \qquad \zdown = \ket1 = \pair01 \]
@@ -258,7 +261,7 @@ \section{Observations}
 \[ \yup = \frac{1}{\sqrt2}\pair1i
 	\qquad \ydown = \frac{1}{\sqrt2}\pair1{-i} \]
 which we call ``$z$-up'', ``$z$-down'',
-``$x$-up'', ``$x$-down'', ``$y$-up'', ``$y$-down''.
+``$x$-up'', ``$x$-down'', ``$y$-up'', ``$y$-down'' respectively.
 (The eigenvalues are $+1$ for ``up'' and $-1$ for ``down''.)
 So, given a state $\ket\psi \in \CC^{\oplus 2}$
 we can make a measurement with respect to any of these three bases
@@ -476,12 +479,15 @@ \section{Entanglement}
 	\]
 	As for the paradox: what happens if you multiply all these measurements together?
 	\begin{hint}
-		$-1$, $1$, $1$, $1$.
+		$1$, $1$, $1$, $-1$ respectively.
 		When we multiply them all together,
 		we get that $\id^A \otimes \id^B \otimes \id^C$
 		has measurement $-1$, which is the paradox.
 		What this means is that the values of the measurements
-		are created when we make the observation,
-		and not prepared in advance.
+		can't be prepared in advance independently.
+		In other words, this contradicts certain local hidden-variable theories.
+
+		This was one of several results for which Zeilinger won a (shared)
+		Nobel Prize in 2022.
 	\end{hint}
 \end{problem}