Caught up pqm, today's lectures.

Principles of Quantum Mechanics/pqm.tex

@@ -0,0 +1,19 @@
+%&..\preamble
+
+\def\npart {II}
+\def\nterm {Michaelmas}
+\def\nyear {2023}
+\def\nlecturer {Dr E. Pajer}
+\def\ncourse {Principles of Quantum Mechanics}
+
+\input{../preamble-dynamic.tex}
+
+\begin{document}
+% \maketitle
+
+\includepdf[pages=-]{chap1.pdf}
+\includepdf[pages=-]{chap2.pdf}
+\includepdf[pages=-]{chap3.pdf}
+\includepdf[pages=-]{chap4.pdf}
+
+\end{document}

ProbAndMeasure/02_measurable_functions.tex

@@ -78,96 +78,130 @@ \subsection{Monotone class theorem}
 \end{proof}
 
 \subsection{Image measures}
-\begin{definition}
-	Let $f \colon (E,\mathcal E) \to (G,\mathcal G)$ be a measurable function, and $\mu$ is a measure on $(E, \mathcal E)$.
-	Then the \emph{image measure} $\nu = \mu \circ f^{-1}$ is obtained from assigning $\nu(A) = \mu(f^{-1}(A))$ for all $A \in \mathcal G$.
+
+\begin{definition}[Image Measure]
+	Let $f \colon (E,\mathcal E) \to (G,\mathcal G)$ be a measurable function and $\mu$ a measure on $(E, \mathcal E)$.
+	Then the \vocab{image measure} $\nu = \mu \circ f^{-1}$ is obtained from assigning $\nu(A) = \mu(f^{-1}(A))$ for all $A \in \mathcal G$.
 \end{definition}
+
+\begin{remark}
+	This is well defined as $f\inv(A) \in \mathcal{E}$ as $f$ measurable. $\nu$ is countably additive because the preimage satisfies set operations and $\mu$ countably additive (See Sheet 1).
+\end{remark}
+
+Starting from the Lebesgue measure, we can get all probability measures (in fact we can get all Radon measures) in this way.
+
 % TODO: Define right-continuous
+\begin{definition}[Right-Continuous]
+	A function $f$ is \vocab{right-continuous} if $x_n \downarrow x \implies f(x_n) \to f(x)$.
+\end{definition}
+
 \begin{lemma}
-	Let $g \colon \mathbb R \to \mathbb R$ be an increasing, right-continuous function, and set $g(\pm\infty) = \lim_{z \to \pm \infty} g(z)$.
-	On $I = (g(-\infty), g(+\infty))$ we define the \emph{generalised inverse}
-	\[ f(x) = \inf \qty{y \in \mathbb R \mid x \leq g(y)} \]
-	for $x \in I$.
-	Then $f$ is increasing, left-continuous, and $f(x) \leq y$ if and only if $x \leq g(y)$ for all $x \in I, y \in \mathbb R$.
+	Let $g \colon \mathbb R \to \mathbb R$ be a non-constant, increasing, right-continuous function, and set $g(\pm\infty) = \lim_{z \to \pm \infty} g(z)$.
+	On $I = (g(-\infty), g(+\infty))$ we define the \vocab{generalised inverse} $f : I \to \mathbb{R}$ by
+	\[ f(x) = \inf \qty{y \in \mathbb R : g(y) \geq x}. \]
+	Then $f$ is increasing, left-continuous, and $f(x) \leq y$ iff $x \leq g(y)$ for all $x \in I, y \in \mathbb R$.
 \end{lemma}
+
 \begin{remark}
 	$f$ and $g$ form a Galois connection.
 \end{remark}
+
 \begin{proof}
-	Let $J_x = \qty{y \in \mathbb R \mid x \leq g(y)}$.
+	Fix $x \in I$. \\
+	Let $J_x = \qty{y \in \mathbb R : g(y) \geq x}$.
 	Since $x > g(-\infty)$, $J_x$ is nonempty and bounded below.
-	Hence $f(x)$ is a well-defined real number.
+	Hence $f(x)$ is a well-defined real number. \\
 	If $y \in J_x$, then $y' \geq y$ implies $y' \in J_x$ since $g$ is increasing.
-	Further, if $y_n$ converges from the right to $y$, and all $y_n \in J_x$, we can take limits in $x \leq g(y_n)$ to find $x \leq \lim_n g(y_n) = g(y)$ since $g$ is right-continuous.
-	Hence $y \in J_x$.
+	Since $g$ is right-continuous, if $y_n \downarrow y$, and all $y_n \in J_x$, then $g(y) = \lim_n g(y_n) \geq x$ so $y \in J_x$. \\
 	So $J_x = [f(x), \infty)$.
 	Hence $f(x) \leq y \iff x \leq g(y)$ as required.
 
-	If $x \leq x'$, we have $J_x \supseteq J_{x'}$ by definition, so $f(x) \leq f(x')$.
-	Similarly, if $x_n$ converges from the left to $x$, we have $J_x = \bigcap_n J_{x_n}$, so $f(x_n) \to f(x)$ as $x_n \to x$.
+	If $x \leq x'$, we have $J_x \supseteq J_{x'}$ (as $y \in J_x \Longleftarrow y \in J_x'$), i.e. $[f(x), \infty) \supseteq [f(x'), \infty)$ so $f(x) \leq f(x')$. \\
+	Similarly, if $x_n \uparrow x$, we have $J_x = \bigcap_n J_{x_n}$\footnote{As $y \in \bigcap_n J_{x_n} \iff g(y) \geq x_n \ \forall \; n \iff g(y) \geq x \iff y \in J_x$.} so $[f(x), \infty) = \bigcap_n [f(x_n), \infty)$ so $f(x_n) \to f(x)$ as $x_n \to x$.
 \end{proof}
+
 \begin{theorem}
-	Let $g \colon \mathbb R \to \mathbb R$ be an increasing, right-continuous function, and set $g(\pm\infty) = \lim_{z \to \pm \infty} g(z)$.
-	Then there exists a unique Radon measure $\mu_g$ on $\mathbb R$ such that $\mu_g((a,b]) = g(b) - g(a)$ for all $a < b$.
-	Further, all Radon measures can be obtained in this way.
+	Let $g \colon \mathbb R \to \mathbb R$ as in the previous lemma.
+	Then $\exists$ a unique Radon measure $\mu_g$ on $\mathbb R$ such that $\mu_g((a,b]) = g(b) - g(a)$ for all $a < b$.
+	Further, all Radon measures on $\mathbb{R}$ can be obtained in this way.
 \end{theorem}
+
 \begin{proof}
-	We will show that the generalised inverse $f$ as defined above is measurable.
-	For all $z \in \mathbb R$, we find $f^{-1}((-\infty,z]) = \qty{x \colon f(x) \leq z} = \qty{x \colon x \leq g(z)} = [-g(\infty),g(z)]$ which is measurable.
-	Since $\mathcal B$ is generated by these such sets, $f$ is $\mathcal B(I)$-$\mathcal B$ measurable as required.
-	Therefore, the image measure $\mu_g = \mu \circ f^{-1}$, where $\mu$ is the Lebesgue measure on $I$, exists.
+	Define $I, f$ as in the previous lemma and $\lambda$ the Lebesgue measure on $I$.
+
+	$f$ is Borel measurable since $f^{-1}((-\infty,z]) = \qty{x \in I \colon f(x) \leq z} = \qty{x \in I \colon x \leq g(z)} = (-g(\infty),g(z)] \in \mathcal{B}$. As $\qty{(-\infty,z] : z \in \mathbb{R}}$ generate $\mathcal{B}$, $f$ measurable.
+
+	Therefore, the image measure $\mu_g = \lambda \circ f^{-1}$ exists on $\mathcal{B}$.
 	Then for any $-\infty < a < b < \infty$, we have
 	\begin{align*}
-		\mu_g((a,b]) &= \mu(f^{-1}((a,b])) \\
-		&= \mu(\qty{x \colon a < f(x) \leq f(b)}) \\
-		&= \mu(\qty{x \colon g(a) < x \leq g(b)}) \\
+		\mu_g((a,b]) &= \lambda \left( f^{-1}\left( (a,b] \right) \right) \\
+		&= \lambda \left( \qty{x \colon a < f(x) \leq f(b)} \right) \\
+		&= \lambda \left( \qty{x \colon g(a) < x \leq g(b)} \right) \\
 		&= g(b) - g(a)
 	\end{align*}
-	This uniquely determines $\mu_g$ by the same argument as shown previously for the Lebesgue measure $\mu$ on $\mathbb R$.
-	Since $g$ maps into $\mathbb R$, $g(b) - g(a) \in \mathbb R$ so any compact set has finite measure as it is a subset of a closed bounded interval.
+	By the \nameref{thm:uni} for $\sigma$-finite measures, $\mu_g$ is uniquely defined.
+	% Since $g$ maps into $\mathbb R$, $g(b) - g(a) \in \mathbb R$ so any compact set has finite measure as it is a subset of a closed bounded interval.
 
 	Conversely, let $\nu$ be a Radon measure on $\mathbb R$.
-	Define
+	Define $g : \mathbb{R} \to \mathbb{R}$ as
 	\[ g(y) = \begin{cases}
 		\nu((0,y]) & \text{if } y \geq 0 \\
 		-\nu((y,0]) & \text{if } y < 0
 	\end{cases} \]
+	$\nu$ Radon tells us that $g$ is finite.
+	Easy to check $g$ is right-continuous\footnote{For $y_n \downarrow y$ where $y \geq 0$, $(0, y_n] \downarrow (0, y]$ and then $\nu((0, y_n]) \downarrow \nu((0, y])$ by countably additivity. Similarly for $y < 0$.}.
 	This is an increasing function in $y$, since $\nu$ is a measure.
-	Since we are using right-closed intervals, $g$ is right-continuous.
 	Finally, $\nu((a,b]) = g(b) - g(a)$ which can be seen by case analysis and additivity of the measure $\nu$.
 	By uniqueness as before, this characterises $\nu$ in its entirety.
 \end{proof}
+
 \begin{remark}
-	Such image measures $\mu_g$ are called \emph{Lebesgue--Stieltjes measures}, where $g$ is the \emph{Stieltjes distribution}.
+	Such image measures $\mu_g$ are called \vocab{Lebesgue--Stieltjes measures} associated with $g$, where $g$ is the \vocab{Stieltjes distribution}.
 \end{remark}
+
 \begin{example}
-	The \emph{Dirac measure at $x$}, written $\delta_x$, is defined by
+	Fix $x \in \mathbb{R}$ and take $g = 1_{[x,\infty)}$.
+	Then $\mu_g = \delta_x$ the \emph{dirac measure at $x$} defined for all $A \in \mathcal{B}$ by
 	\[ \delta_x(A) = \begin{cases}
 		1 & \text{if } x \in A \\
 		0 & \text{otherwise}
 	\end{cases} \]
-	This has Stieltjes distribution $g(x) = 1_{[x,\infty)}$.
 \end{example}
 
 \subsection{Random variables}
-\begin{definition}
+
+\begin{definition}[Random Variable]
 	Let $(\Omega, \mathcal F, \mathbb P)$ be a probability space, and $(E, \mathcal E)$ be a measurable space.
-	An \emph{$E$-valued random variable} $X$ is an $\mathcal F$-$\mathcal E$ measurable map $X \colon \Omega \to E$.
-	When $E = \mathbb R$ or $\mathbb R^d$ with the Borel $\sigma$-algebra, we simply call $X$ a random variable or random vector.
+	If $X : \Omega \to E$ a measurable function then $X$ is a \vocab{random variable} in $E$.
+\end{definition}
+When $E = \mathbb R$ or $\mathbb R^d$ with the Borel $\sigma$-algebra, we simply call $X$ a random variable or random vector.
+
+\begin{example}
+	$X$ models a ``random'' outcome of an experiment, e.g. when tossing a coin $\Omega = \{H, T\}, X = \text{\# heads} : \Omega \to \{0, 1\}$.
+\end{example}
 
-	The \emph{law} or \emph{distribution} $\mu_X$ of a random variable $X$ is given by the image measure $\mu_X = \mathbb P \circ X^{-1}$.
-	When $E$ is the real line, this measure has a distribution function
-	\[ F_X(z) = \mu_X((-\infty, z]) = \mathbb P(X^{-1}(-\infty,z]) = \prob{\qty{\omega \in \Omega \mid X(\omega) \leq z}} = \prob{X \leq z} \]
-	This uniquely determines $\mu_X$ by the $\pi$-system argument given above.
+\begin{definition}[Distribution]
+	The \vocab{law} or \vocab{distribution} $\mu_X$ of a random variable $X$ is given by the image measure $\mu_X = \mathbb P \circ X^{-1}$.
+	It is a measure on $(E, \mathcal{E})$.
+
+	When $(E, \mathcal{E}) = (\mathbb{R}, \mathcal{B})$, $\mu_X$ is uniquely determined by its values on any $\pi$-system, we shall take $\qty{(-\infty, x] : x \in \mathbb{R}}$ and
+	\begin{align*}
+		F_X(z) = \mu_X((-\infty, z]) = \mathbb P(X^{-1}(-\infty,z]) = \prob{\qty{\omega \in \Omega : X(\omega) \leq z}} = \prob{X \leq z}
+	\end{align*}
+	The function $F_x$ is called the \vocab{distribution function} of $X$, because it uniquely determines the distribution of $X$.
 \end{definition}
+
 Using the properties of measures, we can show that any distribution function satisfies:
+
 \begin{enumerate}
 	\item $F_X$ is increasing;
-	\item $F_X$ is right-continuous;
-	\item $\lim_{z \to -\infty} F_X(z) = \mu_X(\varnothing) = 0$;
-	\item $\lim_{z \to \infty} F_X(z) = \mu_X(\mathbb R) = \prob{\Omega} = 1$.
+	\item $F_X$ is right-continuous\footnote{$x_n \downarrow x \implies (-\infty, x_n] \downarrow (-\infty, x]$ hence by countable additivity of $\mathbb{P} \circ X\inv$.};
+	\item $F_X(-\infty) = \lim_{z \to -\infty} F_X(z) = \mu_X(\varnothing) = 0$;
+	\item $F_X(\infty) = \lim_{z \to \infty} F_X(z) = \mu_X(\mathbb R) = \prob{\Omega} = 1$.
 \end{enumerate}
-Given any function $F_X$ satisfying each property, we can obtain a random variable $X$ on $(\Omega, \mathcal F, \mathbb P) = ((0,1), \mathcal B((0,1)), \mu)$ by $X(\omega) = \inf\qty{x \mid \omega \leq f(x)}$, and then $F_X$ is the distribution function of $X$.
+
+Given any function $F_X : \mathbb{R} \to [0, 1]$ satisfying each property, we can obtain a random variable $X$ on $(\Omega, \mathcal F, \mathbb P) = ((0,1), \mathcal B((0,1)), \mu)$ by $X(\omega) = \inf\qty{x \mid \omega \leq f(x)}$, and then $F_X$ is the distribution function of $X$.
+
 \begin{definition}
 	Consider a countable collection $(X_i \colon (\Omega, \mathcal F, \mathbb P) \to (E, \mathcal E))$ for $i \in I$.
 	This collection of random variables is called \emph{independent} if the $\sigma$-algebras $\sigma\qty(\qty{X_i^{-1}(A) \colon A \in \mathcal E})$ are independent.