\documentstyle[11pt,reduce]{article}

\title{SETS: A Basic Set Theory Package}

\author{Francis J. Wright \\
School of Mathematical Sciences \\
Queen Mary and Westfield College \\
University of London \\
Mile End Road, London E1 4NS, UK. \\
Email: {\tt F.J.Wright@QMW.ac.uk}}

\begin{document}
\maketitle

\begin{abstract}
  The SETS package for \REDUCE 3.5 and later versions provides
  algebraic-mode support for set operations on lists regarded as sets
  (or representing explicit sets) and on implicit sets represented by
  identifiers.  It provides the set-valued infix operators (with
  synonyms) {\tt union}, {\tt intersection} ({\tt intersect}) and {\tt
  setdiff} (\verb|\|, {\tt minus}) and the Boolean-valued infix
  operators (predicates) {\tt member}, {\tt subset\_eq}, {\tt subset},
  {\tt set\_eq}.  The union and intersection operators are n-ary and
  the rest are binary.  A list can be explicitly converted to the
  canonical set representation by applying the operator {\tt mkset}.
  (The package also provides an operator not specifically related to
  set theory called {\tt evalb} that allows the value of any
  Boolean-valued expression to be displayed in algebraic mode.)
\end{abstract}

\section{Introduction}

REDUCE has no specific representation for a set, neither in algebraic
mode nor internally, and any object that is mathematically a set is
represented in REDUCE as a list.  The difference between a set and a
list is that in a set the ordering of elements is not significant and
duplicate elements are not allowed (or are ignored).  Hence a list
provides a perfectly natural and satisfactory representation for a set
(but not vice versa).  Some languages, such as Maple, provide
different internal representations for sets and lists, which may allow
sets to be processed more efficiently, but this is not {\em
necessary}.

This package supports set theoretic operations on lists and represents
the results as normal algebraic-mode lists, so that all other REDUCE
facilities that apply to lists can still be applied to lists that have
been constructed by explicit set operations.  The algebraic-mode set
operations provided by this package have all been available in
symbolic mode for a long time, and indeed are used internally by the
rest of REDUCE, so in that sense set theory facilities in REDUCE are
far from new.  What this package does is make them available in
algebraic mode, generalize their operation by extending the arity of
union and intersection, and allow their arguments to be implicit sets
represented by unbound identifiers.  It performs some simplifications
on such symbolic set-valued expressions, but this is currently rather
{\it ad hoc\/} and is probably incomplete.

For examples of the operation of the SETS package see (or run) the
test file {\tt sets.tst}.  This package is experimental and
developments are under consideration; if you have suggestions for
improvements (or corrections) then please send them to me (FJW),
preferably by email.  The package is intended to be run under
\REDUCE 3.5 and later versions; it may well run correctly under earlier
versions although I cannot provide support for such use.

\section{Infix operator precedence}

The set operators are currently inserted into the standard REDUCE
precedence list (see page 28, \S2.7, of the REDUCE 3.6 manual) as
follows:
\begin{verbatim}
or and not member memq = set_eq neq eq >= > <= < subset_eq
subset freeof + - setdiff union intersection * / ^ .
\end{verbatim}

\section{Explicit set representation and {\tt mkset}}

Explicit sets are represented by lists, and this package does not
require any restrictions at all on the forms of lists that are
regarded as sets.  Nevertheless, duplicate elements in a set
correspond by definition to the same element and it is conventional
and convenient to represent them by a single element, i.e.\ to remove
any duplicate elements.  I will call this a normal representation.
Since the order of elements in a set is irrelevant it is also
conventional and may be convenient to sort them into some standard
order, and an appropriate ordering of a normal representation gives a
canonical representation.  This means that two identical sets have
identical representations, and therefore the standard REDUCE equality
predicate ({\tt =}) correctly determines set equality; without a
canonical representation this is not the case.

Pre-processing of explicit set-valued arguments of the set-valued
operators to remove duplicates is always done because of the obvious
efficiency advantage if there were any duplicates, and hence explicit
sets appearing in the values of such operators will never contain any
duplicate elements.  Such sets are also currently sorted, mainly
because the result looks better.  The ordering used satisfies the {\tt
ordp} predicate used for most sorting within REDUCE, except that
explicit integers are sorted into increasing numerical order rather
than the decreasing order that satisfies {\tt ordp}.

Hence explicit sets appearing in the result of any set operator are
currently returned in a canonical form.  Any explicit set can also be
put into this form by applying the operator {\tt mkset} to the list
representing it.  For example
\begin{verbatim}
mkset {1,2,y,x*y,x+y};

{x + y,x*y,y,1,2}
\end{verbatim}

The empty set is represented by the empty list \verb|{}|.

\section{Union and intersection}

The operator {\tt intersection} (the name used internally) has the
shorter synonym {\tt intersect}.  These operators will probably most
commonly be used as binary infix operators applied to explicit sets,
e.g.
\begin{verbatim}
{1,2,3} union {2,3,4};

{1,2,3,4}

{1,2,3} intersect {2,3,4};

{2,3}
\end{verbatim}
They can also be used as n-ary operators with any number of arguments,
in which case it saves typing to use them as prefix operators (which
is possible with all REDUCE infix operators), e.g.
\begin{verbatim}
{1,2,3} union {2,3,4} union {3,4,5};

{1,2,3,4,5}

intersect({1,2,3}, {2,3,4}, {3,4,5});

{3}
\end{verbatim}
For completeness, they can currently also be used as unary operators,
in which case they just return their arguments (in canonical form),
and so act as slightly less efficient versions of {\tt mkset} (but
this may change), e.g.
\begin{verbatim}
union {1,5,3,5,1};

{1,3,5}
\end{verbatim}

\section{Symbolic set expressions}

If one or more of the arguments evaluates to an unbound identifier
then it is regarded as representing a symbolic implicit set, and the
union or intersection will evaluate to an expression that still
contains the union or intersection operator.  These two operators are
symmetric, and so if they remain symbolic their arguments will be
sorted as for any symmetric operator.  Such symbolic set expressions
are simplified, but the simplification may not be complete in
non-trivial cases.  For example:
\begin{verbatim}
a union b union {} union b union {7,3};

{3,7} union a union b

a intersect {};

{}
\end{verbatim}

In implementations of REDUCE that provide fancy display using
mathematical notation, such as PSL-REDUCE~3.6 for MS-Windows, the
empty set, union, intersection and set difference are all displayed
using their conventional mathematical symbols, namely $\emptyset$,
$\cup$, $\cap$, $\setminus$.

A symbolic set expression is a valid argument for any other set
operator, e.g.
\begin{verbatim}
a union (b intersect c);

b intersection c union a
\end{verbatim}

Intersection distributes over union, which is not applied by default
but is implemented as a rule list assigned to the variable {\tt
set\_distribution\_rule}, e.g.
\begin{verbatim}
a intersect (b union c);

(b union c) intersection a

a intersect (b union c) where set_distribution_rule;

a intersection b union a intersection c
\end{verbatim}

\section{Set difference}

The set difference operator is represented by the symbol \verb|\| and
is always output using this symbol, although it can also be input using
either of the two names {\tt setdiff} (the name used internally) or
{\tt minus} (as used in Maple).  It is a binary operator, its operands
may be any combination of explicit or implicit sets, and it may be
used in an argument of any other set operator.  Here are some
examples:
\begin{verbatim}
{1,2,3} \ {2,4};

{1,3}

{1,2,3} \ {};

{1,2,3}

a \ {1,2};

a\{1,2}

a \ a;

{}

a \ {};

a

{} \ a;

{}
\end{verbatim}

\section{Predicates on sets}

These are all binary infix operators.  Currently, like all REDUCE
predicates, they can only be used within conditional statements ({\tt
if}, {\tt while}, {\tt repeat}) or within the argument of the {\tt
evalb} operator provided by this package, and they cannot remain
symbolic -- a predicate that cannot be evaluated to a Boolean value
causes a normal REDUCE error.

The {\tt evalb} operator provides a convenient shorthand for an {\tt
if} statement designed purely to display the value of any Boolean
expression (not only predicates defined in this package).  It has some
similarity with the {\tt evalb} function in Maple, except that the
values returned by {\tt evalb} in REDUCE (the identifiers {\tt true}
and {\tt false}) have no significance to REDUCE itself.  Hence, in
REDUCE, use of {\tt evalb} is {\em never\/} necessary.
\begin{verbatim}
if a = a then true else false;

true

evalb(a = a);

true

if a = b then true else false;

false

evalb(a = b);

false

evalb 1;

true

evalb 0;

false
\end{verbatim}
I will use the {\tt evalb} operator in preference to an explicit {\tt
if} statement for purposes of illustration.

\subsection{Set membership}

Set membership is tested by the predicate {\tt member}.  Its left
operand is regarded as a potential set element and its right operand
{\em must\/} evaluate to an explicit set.  There is currently no sense
in which the right operand could be an implicit set; this would
require a mechanism for declaring implicit set membership (akin to
implicit variable dependence) which is currently not implemented.  Set
membership testing works like this:
\begin{verbatim}
evalb(1 member {1,2,3});

true

evalb(2 member {1,2} intersect {2,3});

true

evalb(a member b);

***** b invalid as list
\end{verbatim}

\subsection{Set inclusion}

Set inclusion is tested by the predicate {\tt subset\_eq} where {\tt a
subset\_eq b} is true if the set $a$ is either a subset of or equal to
the set $b$; strict inclusion is tested by the predicate {\tt subset}
where {\tt a subset b} is true if the set $a$ is {\em strictly\/} a
subset of the set $b$ and is false is $a$ is equal to $b$.  These
predicates provide some support for symbolic set expressions, but this
is not yet correct as indicated below.  Here are some examples:
\begin{verbatim}
evalb({1,2} subset_eq {1,2,3});

true

evalb({1,2} subset_eq {1,2});

true

evalb({1,2} subset {1,2});

false

evalb(a subset a union b);

true

evalb(a\b subset a);

true

evalb(a intersect b subset a union b);  %%% BUG

false
\end{verbatim}

An undecidable predicate causes a normal REDUCE error, e.g.
\begin{verbatim}
evalb(a subset_eq {b});

***** Cannot evaluate a subset_eq {b} as Boolean-valued set
 expression

evalb(a subset_eq b);  %%% BUG

false
\end{verbatim}

\subsection{Set equality}

As explained above, equality of two sets in canonical form can be
reliably tested by the standard REDUCE equality predicate ({\tt =}).
This package also provides the predicate {\tt set\_eq} to test
equality of two sets not represented canonically.  The two predicates
behave identically for operands that are symbolic set expressions
because these are always evaluated to canonical form (although
currently this is probably strictly true only in simple cases).  Here
are some examples:
\begin{verbatim}
evalb({1,2,3} = {1,2,3});

true

evalb({2,1,3} = {1,3,2});

false

evalb(mkset{2,1,3} = mkset{1,3,2});

true

evalb({2,1,3} set_eq {1,3,2});

true

evalb(a union a = a\{});

true
\end{verbatim}

\section{Installation}

The source file {\tt sets.red} can be read into REDUCE when required
using {\tt IN}.  If the ``professional'' version is being used this
should be done with {\tt ON COMP} set, but it is much better to
compile the code as a {\tt FASL} file using {\tt FASLOUT} and then
load it with {\tt LOAD\_PACKAGE} (or {\tt LOAD}).  See the REDUCE
manual and implementation-specific guide for further details.

This package has to redefine the REDUCE internal procedure {\tt
mk!*sq} and a warning about this can be expected and ignored.  I
believe (and hope!) that this redefinition is safe and will not have
any unexpected consequences for the rest of REDUCE.

\section{Possible future developments}

\begin{itemize}
\item Unary union/intersection to implement repeated
  union/intersection on a set of sets.
\item More symbolic set algebra, canonical forms for set expressions,
  more complete simplification.
\item Better support for Boolean variables via a version (evalb10?)
  of {\tt evalb} that returns 1/0 instead of {\tt true}/{\tt false},
  or predicates that return 1/0 directly.
\end{itemize}

\end{document}


\documentclass[12pt]{amsart}

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% TITLE AUTHOR DATE
\title{Controlled natural language for type theory}

\date{November 17, 2019}
\author{Thomas Hales}

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% DOCUMENT

\begin{document}
\maketitle

\section{Introduction}

This abstract describes the current designs and development of a
controlled natural language for mathematics that compiles to the Lean
proof assistant.  We call this language Colada (short for
\emph{Co}ntrolled \emph{la}nguage \emph{da}ta).

The design of language grows out of previous controlled natural
languages for mathematics (specifically, Forthel-Naproche-SAD), as
described in Peter Koepke's AITP 2019 talk.  We use
Forthel-Naproche-SAD (or simply Forthel) as a generic name for any of
the dialect inspired by Forthel, and our language Colada is one of
those dialects. This document will refer to the Colada language as
\emph{our dialect}.

Documents in our dialect are written in a specially prepared \LaTeX\ file.
The output of the file will eventually be a type-checked Lean file,
although parts of this process are still under development.  (Checking
mathematical proofs is not currently part of our intended project;
type-checked Lean we exclude proof-checking.)

\subsection{Controlled Natural Languages (CNL)}\label{sub:CNL}

By controlled natural language for mathematics (CNL), we mean an
artificial language for the communication of mathematics that is (1)
designed in a deliberate and explicit way with precise
computer-readable syntax and semantics, (2) based on a single natural
language (which for us will be  English), and (3) broadly
understood at least in an intuitive way by mathematically literate
speakers of the natural language.

CNLs can achieve a much higher degree of English fluency than other
proof-checking languages such as the \emph{Mizar} and DeBruijn's
\emph{vernacular}.  Some other proof languages are purely stylistic,
such as \emph{structured derivations}.

At AITP 2019 Peter Koepke displayed a short proof from Rudin's {\it
  Principles of mathematical analysis} that he modified with Steffen
Frerix so that it can be read and checked by their system. Their
modified proof is written in fluent English, is typeset by \LaTeX, and
yet is fully checkable. (The target of their language is first-order
logic.)

It is our belief that controlled natural languages are undervalued
technologies in AITP.  Following a divide-and-conquer strategy, our
basic aim is to develop a technology that lies roughly at the midpoint
between current practice of research mathematicians and the current
practice within the proof assistant community.

\subsection{Lean}

Lean theorem prover is a proof assistant built on the
foundations of calculus of inductive constructions.

The eventual target of our CNL is Lean.

Why is Lean the target?  We wanted something more powerful than
first-order logic as used in other Forthel dialects.
Many mathematicians are finding it a good system
for research-level mathematics.  There is also the M.K. argument
that if we succeed in bridging the gap between English and Lean,
then automated translation tools will eventually give us translations
from Lean to other proof assistants.

\subsection{Research to Date}

This abstract describes the current stage of a project that is
intended to continue over a period of years.

Our specific research contributions to date are as follows.

\subsubsection{A design and specification of a controlled natural
  language}

Like other Forthel dialects, our grammar is not a context-free.
However, it is similar to a context-free grammar by being described as
a collection of terminal and nonterminal symbols and production rules.
However, the grammar grows as a document is parsed by the addition of
new production rules, which are described in the document.

Our dialect can be viewed as a fusion of three different syntaxes:
Forthel-Naproche-SAD syntax, \LaTeX\ syntax, and Lean theorem-prover
syntax.

The lexical structure of our dialect is specified in sedlex, a lexical
generator tool for OCaml.

Our dialect has been specified in menhir, an OCaml-based
parser-generator tool for LR(1) grammars.  (Although our dialect is
not an LR(1) grammar, which prevents menhir from automatically
generating a parser, the software checks that we have a well-formed
grammar.)

Our grammar is both complex and recursive to an extraordinary degree.
The grammar has about 350 nonterminals and about 700 production
rules.  The grammar has about 150 context-dependent key words.  (This
is before any of the user-defined grammar extensions.)  However, we
feel that a some complexity is justified (and even required) to
capture widespread mathematical idioms and formulas, the syntax of type
theory (for us, the calculus of inductive constructions), and their
interactions.

We keep most features of Forthel, such its handling of synonyms, noun
phrases, verbs, and adjectives; and its grammar extension mechanisms.
We add additional features such as operator precedence parsing (with
user-specified precendence levels and associativities); scoping;
syntax for \LaTeX\ macros; and dependent type theory including
inductive types and mutual inductive types, structures, and lambda
terms.

A parser for our grammar has been implemented in OCaml, building on
the parser combinator library that John Harrison wrote to parse HOL
Light.  However, this parser is still at an early stage. For example,
it is still not capable of transforming a syntax tree of a parsed
document into expressions that can be processed Lean. (Building this
capability will be a major project.)

We try to make the grammar unambiguous by always taking the longest
match possible in a greedy way.  By greedy, we mean that we take
the longest match of a given production rule when it occurs
even if doing so produces a match for the enclosing production rule
of suboptimal length. Whenever this is not the desired behavior,
the author of the document must insert parentheses.

And if, for example, a symbol has been assigned several meanings, it
is always the most recently declared that wins out.

\subsubsection{\TeX to raw controlled natural language syntax}

We have written a software program takes a specially prepared
\LaTeX\ file as input and strips away the non-semantic content and
outputs raw (such as headers, spaces and other layout, graphics,
remarks, and dollar signs).

To create a specially prepared \LaTeX\ file, the author imports a
package \verb!\usepackage{cnl}!. Then any text that appears within the
\emph{cnl} \LaTeX\ environment is handled as controlled natural
language.  Arbitrary \LaTeX\ and graphics may appear outside the CNL
environment.  The key to beautifully typeset \TeX\ documents is an
dual expansion system for macros.  The \TeX\ engine expands macros in
the usual way, but the CNL engine expands some macros according to an
independent semantic specification.

\section{examples}

One project we have started is the translation of the number theory
Planet Math files to our CNL dialect.

We have been using a 30-page CNL file (for debugging) that
gives many elementary definitions of groups, rings, fields, real numbers, etc.
It goes as far as statements of Sylow's theorems in group theory.

\section{motivation and summary}

This is part of the Formal Abstracts project.

A CNL fixes a problem we encountered early on in the Formal Abstracts
project.   If we leave out proofs, we need more eyes, but Lean
is not written for human eyes (or only as a secondary objective).
Lean is optimized for writing
computer-checked proofs.
One can often not tell what the idea of a proof is by reading a Lean file.
Therefore we need a language that all mathematicians can read.

Our language is already finding uses at this early stage of
development.  For example, our tool can reveal to the author of a
\LaTeX\ document which terms and symbols are being used without
definition.  Ultimately each symbol, word, and phrase can be traced
its source either as a keyword of our dialect or as author-supplied.

I thank the Thang Long University Formal Abstracts team for feedback,
in their use of the software.  I thank Jesse Han, who wrote a
preliminary version of the parser using Haskell's megaparsec parser
combinator library.  I thank Peter Koepke for inspiring this project
and for many conversations.  I thank many others at the project github page.

bibliography

Lean proof assistant.

CiC.

My blog post on CNL and github.

Koepke AITP and github
\footnote{\href{http://aitp-conference.org/2019/aitp19-proceedings.pdf}{AITP
    2019 proceedings}, page 84}

Paskevich thesis and paper

Glushkov and history.

\end{document}


%
% revised at Jan 21th, 2009
% does not change too much on the content, only get
% mild modification on the multi-system discussion. However,
% this part is still not very satisfied. I do not think it
% has been clearly discussed.
%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% outline for this chapter:
%
%  1  the angular operator; expression, some concrete characters and relation with H
%  2  the eigen value and eigen function of angular operator
%  3  angular operator in multi-particle system
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{Angular Momentum}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Definition of angular momentum operator}
%
%  1  the expression of  l, l^{2} and l_{x},l_{y},l_{z},l_{+},l_{z}
%  2  hermite character
%  3  commute relationship
%  4  expression under sphere coordinates

In classical mechanics, the angular momentum is defined as
$\overrightarrow{l} = \overrightarrow{r}\times\overrightarrow{p}$,
in quantum mechanics; the angular momentum operator is defined as
$\hei{l} = \hei{r}\times\hei{p}$. So we have:
\begin{equation}\label{}
\begin{aligned}
         \hat{l}_{x} &= y\hat{p}_{z} -  z\hat{p}_{y} \nonumber \\
         \hat{l}_{y} &= z\hat{p}_{x} -  x\hat{p}_{z} \nonumber \\
         \hat{l}_{z} &= x\hat{p}_{y} -  y\hat{p}_{x}
 \end{aligned}
\end{equation}

First we can prove that \heit{l} is hermite. Actually if the
$\hat{l}_{x}, \hat{l}_{y}$ and $\hat{l}_{z}$ are hermite, the
\heit{l} is hermite.

For $\hat{l}_{\alpha} = x_{\beta}\hat{p}_{\gamma} -
x_{\gamma}\hat{p}_{\beta}$; we have:
\begin{align}\label{}
\hat{l}_{\alpha}^{+} &= \{x_{\beta}\hat{p}_{\gamma}\}^{+} -
\{x_{\gamma}\hat{p}_{\beta}\}^{+} \nonumber \\
&=\hat{p}_{\gamma}x_{\beta} - \hat{p}_{\beta}x_{\gamma} \nonumber \\
&=x_{\beta}\hat{p}_{\gamma} - x_{\gamma}\hat{p}_{\beta} \nonumber \\
&=\hat{l}_{\alpha}
\end{align}
Here we note that $[x_{\alpha}, p_{\beta}] =
-i\hbar\delta_{\alpha\beta}$. Therefore, the \heit{l} is hermite.

Then we concentrate on the commutative relationship between
$\hat{l}_{\alpha}$ and $x_{\beta}$:
\begin{align}\label{}
[\hat{l}_{x}, x] &= 0  &  [\hat{l}_{x}, y] &= i\hbar z  &
[\hat{l}_{x}, z] &= -i\hbar y  \nonumber \\
[\hat{l}_{y}, x] &= -i\hbar z  &  [\hat{l}_{y}, y] &= 0  &
[\hat{l}_{y}, z] &=  i\hbar x  \nonumber \\
[\hat{l}_{z}, x] &= i\hbar y  &  [\hat{l}_{z}, y] &= -i\hbar x  &
[\hat{l}_{z}, z] &= 0  \nonumber \\
\end{align}
In short, we use the formula below to totally express the above
equations:
\begin{equation}\label{}
[\hat{l}_{\alpha}, x_{\beta}] =
i\hbar\varepsilon_{\alpha\beta\gamma} x_{\gamma}
\end{equation}
$\varepsilon_{\alpha\beta\gamma}$ is a three rank antisymmetric unit
tensor, within it if two index number are changed from the order of
$\alpha, \beta, \gamma$ it equals to -1; if changed twice it equals
to $+1$. Thus this is similar to the sign changing in the
determinant.

For example, for the tensor we have:
\begin{equation}\label{}
\varepsilon_{\alpha\beta\gamma} = 1, \quad
\varepsilon_{\alpha\gamma\beta} =\varepsilon_{\beta\alpha\gamma}
=\varepsilon_{\gamma\beta\alpha}= -1, \quad
\varepsilon_{\gamma\alpha\beta}= 1
\end{equation}

For the commutation between $\hat{l}_{\alpha}$ and
$\hat{p}_{\beta}$; we also have similar expressions:
\begin{align}\label{}
[\hat{l}_{\alpha}, \hat{p}_{\beta}] &=
i\hbar\varepsilon_{\alpha\beta\gamma} \hat{p}_{\gamma} \nonumber \\
[\hat{l}_{\alpha}, \hat{l}_{\beta}] &=
i\hbar\varepsilon_{\alpha\beta\gamma} \hat{l}_{\gamma}
\end{align}

For the expression above, we can see that $\hat{l}_{x}, \hat{l}_{y},
\hat{l}_{z}$ can not commuted with each other, thus they can not
share the same eigen states. This is an important extinguishment
between angular momentum and momentum, where the $\hat{p}_{x},
\hat{p}_{y}, \hat{p}_{z}$ could commute with each other.

On the other hand, we can see that for some specific systems (which
includes all the quantum system in quantum chemistry, so that's
enough) the \heit{l} commutes with the $\hat{H}$. Since that we can
write the $\hat{H}$ as the sum of $\hat{T}$ and $\hat{V}(r)$; they
can be taken into account respectively.
\begin{align}\label{}
[\hat{V}(r), \hei{l}] &= [\hat{V}(r), \hei{r}\times\hei{p}] \nonumber \\
&=[\hat{V}(r), \hei{r}]\times\hei{p} + \hei{r}\times[\hat{V}(r), \hei{p}] \nonumber \\
&=\hei{r}\times[\hat{V}(r), \hei{p}]
\end{align}
If $\hat{V}(r)$ is some polynomial function of $\hat{R}$ (such as in
molecule Hamiltonian, the $\hat{V}(r) = \hat{R}^{-1}$), the
$[\hat{V}(r), \hei{p}]$ is some function of $\hat{R}$ and $\hei{r}$
according to the (\ref{OPERATORMOREeq:8}); thus this term is $0$.

On the other hand, for the kinetic operator of $\hat{T}$:
\begin{align}\label{}
[\hat{T}, \hei{l}] &= [\hat{T}, \hei{r}\times\hei{p}] \nonumber \\
&=[\hat{T}, \hei{r}]\times\hei{p} + \hei{r}\times[\hat{T}, \hei{p}] \nonumber \\
&=[\hat{T}, \hei{r}]\times\hei{p}
\end{align}
From the (\ref{OPERATORMOREeq:9}) we have known that $[\hei{r},
\hei{p}\cdot\hei{p}] =-2i\hbar\hei{p}$, thus this term is $0$.

Consequently, the angular momentum is able to commute with the
$\hat{H}$.

On the other hand, it's easy to get this only from physical guess.
Since the angular momentum describes the rotation states of the
quantum particle, while if the rotation does not affect the overall
energy state, it can be expected that the Hamiltonian operator
commutes with the angular momentum operator.

 Similar to the kinetic operator, we can define the operator
of the square of the modulus of the angular momentum vector, that
is:
\begin{equation}\label{}
\hat{l}^{2} = \hei{l}\cdot\hei{l} = \hat{l}^{2}_{x} +
\hat{l}^{2}_{y} + \hat{l}^{2}_{z}
\end{equation}

It's easy to see that $\hat{l}^{2}$ is a hermite operator. For the
$\hat{l}^{2}$, it can prove that:
\begin{equation}\label{}
[\hat{l}^{2}, \hat{l}_{\alpha}] = 0 \quad \text{$\alpha = x, y, z$}
\end{equation}
Since for any arbitrary $\alpha$ and $\beta$, we have:
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  [\hat{l}_{\alpha}, \hat{l}^{2}_{\beta}] &=&  \hat{l}_{\beta}[\hat{l}_{\alpha}, \hat{l}_{\beta}] + [\hat{l}_{\alpha},
  \hat{l}_{\beta}]\hat{l}_{\beta} \nonumber \\
   &=& i\hbar\varepsilon_{\alpha\beta\gamma}\hat{l}_{\beta} \hat{l}_{\gamma}  + i\hbar
   \varepsilon_{\alpha\beta\gamma}\hat{l}_{\gamma}\hat{l}_{\beta} \nonumber \\
\end{eqnarray}
If we have $\alpha = x$, as the $\beta = y, z$, it has:
\begin{align}\label{}
  [\hat{l}_{x}, \hat{l}^{2}_{y}] &= i\hbar\hat{l}_{y} \hat{l}_{z}  + i\hbar
   \hat{l}_{z}\hat{l}_{y} \nonumber \\
  [\hat{l}_{x}, \hat{l}^{2}_{z}] &= -i\hbar\hat{l}_{z} \hat{l}_{y}  - i\hbar
   \hat{l}_{y}\hat{l}_{z}
\end{align}

Therefore, we can know that:
\begin{equation}\label{}
[\hat{l}_{x}, \hat{l}^{2}] = [\hat{l}_{x}, \hat{l}^{2}_{y}] +
[\hat{l}_{x}, \hat{l}^{2}_{z}] + [\hat{l}_{x}, \hat{l}^{2}_{x}] = 0
\end{equation}

Similarly, we know that $[\hat{l}^{2}, \hat{l}_{\alpha}] = 0$.

on the other hand, it can see that if $[\hei{l}, \hat{H}] = 0$, the
$[\hat{l}^{2}, \hat{H}] = 0$.
\begin{align}\label{}
[\hat{H}, \hat{l}^{2}] &= [\hat{H}, \hei{l}^{2}] \nonumber \\
&=[\hat{H}, \hei{l}]\cdot\hei{l} + \hei{l}\cdot[\hat{H}, \hei{l}]
\end{align}

Thus, within the quantum chemistry; we can arrive two important
conclusions:
\begin{align}\label{}
[\hat{H}, \hei{l}] &= 0 \nonumber \\
[\hat{H}, \hat{l}^{2}] &= 0
\end{align}

Next, let's introduce another two operators based on the $\hei{l}$:
\begin{align}\label{}
\hat{l}_{+} &= \hat{l}_{x} + i\hat{l}_{y}  \nonumber \\
\hat{l}_{-} &= \hat{l}_{x} - i\hat{l}_{y}
\end{align}
For both of the new operators, it's easy to prove the commutative
relationship for them:
\begin{equation}\label{}
[\hat{l}_{+}, \hat{l}_{-}] = 2\hbar\hat{l}_{z}, \quad [\hat{l}_{z},
\hat{l}_{+}] = \hbar\hat{l}_{+}, \quad [\hat{l}_{z}, \hat{l}_{-}] =
-\hbar\hat{l}_{-}
\end{equation}

Furthermore, we can show that:
\begin{equation}\label{ANGULAReq:3}
\hat{l}^{2} = \hat{l}_{+}\hat{l}_{-} + \hat{l}^{2}_{z} - \hat{l}_{z}
=\hat{l}_{-}\hat{l}_{+} + \hat{l}^{2}_{z} + \hat{l}_{z}
\end{equation}

In the following content, we can see that why we introduce both of
the two new operators.

Lastly, since that we always use the spheral coordinates to
represent the angular momentum operator, its form is listed below:
\begin{align}\label{ANGULAReq:1}
\hat{l}_{x} &=i\hbar\left\{\sin\varphi \frac{\partial}{\partial
\theta} + \cot\theta\cos\varphi \frac{\partial}{\partial\varphi} \right\}       \nonumber \\
\hat{l}_{y} &=i\hbar\left\{-\cos\varphi \frac{\partial}{\partial
\theta} + \cot\theta\sin\varphi \frac{\partial}{\partial\varphi}
\right\} \nonumber \\
\hat{l}_{z} &= -i\hbar
\frac{\partial}{\partial\varphi}
\end{align}

\begin{equation}\label{ANGULAReq:2}
\hat{l}^{2}
=-\hbar^{2}\left\{\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}
\left(\sin\theta\frac{\partial}{\partial\theta}\right)
+\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\varphi^{2}}
\right\}
\end{equation}

To get the result has to change from the Cartesian coordinate to the
spheral coordinates. Thus we have to use some basic differential
relationships. For the $\frac{\partial}{\partial x}$, we have:
\begin{equation}\label{}
\frac{\partial}{\partial x} = \frac{\partial}{\partial
r}\frac{\partial r}{\partial x} +
\frac{\partial}{\partial\theta}\frac{\partial\theta}{\partial x} +
\frac{\partial}{\partial\varphi}\frac{\partial\varphi}{\partial x}
\end{equation}
The $\frac{\partial}{\partial y}$ and $\frac{\partial}{\partial z}$
are similar to the expression above. By inserting the above
transformation into the $\hat{l}_{\alpha}$, we can immediately get
the (\ref{ANGULAReq:1}).

However, if we adopt the same procedure to deduce the
$\frac{\partial^{2}}{\partial x^{2}}$, it will be very complicated.
However, we can use the $\hat{l}_{+}$ and $\hat{l}_{+}$ to avoid
such trouble.

Directly from the (\ref{ANGULAReq:1}), we can get the expression for
the $\hat{l}_{+}$ and $\hat{l}_{+}$:
\begin{equation}\label{}
\hat{l}_{\pm} = e^{\pm i\varphi}\left\{\pm \frac{\partial}{\partial
\theta} + i\cot\theta \frac{\partial}{\partial\varphi} \right\}
\end{equation}
Thus from the (\ref{ANGULAReq:3}) we can easily get the expression
for the $\hat{l}^{2}$ as shown in (\ref{ANGULAReq:2}).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Eigen value of angular momentum operator}
%
% eigen value for l^{2} and l_{z}
% since the CSCCO has pickup the l^{2} and l_{z}, thus it's
%  impossible to discuss the the eigen value for l_{x} and l_{y}.
%  but we can talk about their matrix elements
%
Here in this section we begin to concentrate on the eigen value of
the angular momentum operator. There has some classical methods
tackle down this problem, so we will follow these methods.

First, let's concentrate on the eigen value of $\hat{l}_{z}$. Since
that $\hat{l}_{z}$ has a simple expression which is shown in
(\ref{ANGULAReq:1}), also the $\hat{l}_{z}$ can commute with
$\hat{H}$; it's picked up into the CSCCO for $\hat{H}$. Thus its
eigen function is:
\begin{align}\label{}
-i\hbar\frac{\partial \Psi}{\partial\varphi} &= l_{z}\Psi
\Rightarrow
\nonumber \\
\frac{\partial \ln\Psi}{\partial\varphi} &= \frac{il_{z}}{\hbar}
\Rightarrow
\nonumber \\
\Psi &= C exp(il_{z}\varphi/\hbar)
\end{align}

Now we have to determine the $l_{z}$ and $C$. $C$ can be fixed by
normalization condition; yet for the $l_{z}$, to ensure that
$\hat{l}_{z}$ is a hermite operator (see the book by JinYan
Zeng(Volume I)\cite{ZengJinYan} to get more details), the eigen
state of $\Psi$ has to satisfy: $\Psi(\varphi + 2\pi) =
\Psi(\varphi)$. Therefore the $l_{z}$ has to be:
\begin{equation}\label{}
\frac{l_{z}}{\hbar} = m \quad (m=0, \pm 1, \pm 2, \cdots)
\end{equation}
Thus the eigen state for the $\hat{l}_{z}$ is:
\begin{align}\label{ANGULAReq:6}
\Psi &= C exp(im\varphi) \nonumber \\
&= \frac{1}{\sqrt{2 \pi}} exp(im\varphi)
\end{align}

Here there's something so interesting that need to elucidate.
Although judging from the normal experience that there should has no
difference between $\hat{l}_{x}$, $\hat{l}_{y}$ and $\hat{l}_{z}$;
therefore the $\hat{l}_{y}$ and $\hat{l}_{x}$ should be able to get
their corresponding eigen states. However, there's an important
point we have previously made; that the $\hat{l}_{x}$, $\hat{l}_{y}$
and $\hat{l}_{z}$ can not commute with each other. Thus, if the
$\hat{l}_{z}$ is picked out into the complete set of dynamic
operators for $\hat{H}$, we can not use this Hilbert space to
compose the eigen states for $\hat{l}_{y}$ and $\hat{l}_{z}$.

Secondly, we are going to turn to the $\hat{l}^{2}$. Since that
$\hat{l}^{2}$ also commutes with $\hat{H}$, so it's selected into
the complete set of dynamic operators for $\hat{H}$, just like
$\hat{l}_{z}$.

However, because the expression of the $\hat{l}^{2}$ (see the
\ref{ANGULAReq:2}) is very complicated, it's hard to get the eigen
values directly from solving the eigen function like the procedure
done to the $\hat{l}_{z}$. Nevertheless, from the expression of
(\ref{ANGULAReq:3}), we can see that $\hat{l}^{2}$ can de directly
related to the $\hat{l}_{z}$ and $\hat{l}_{z}^{2}$, while their
eigen values are easily gained. Therefore, can we find out a way to
achieve the eigen value but avoid to solve the eigen function of
$\hat{l}^{2}$?

Here, to get through the answer we must use the operator of
$\hat{l}_{+}$ and $\hat{l}_{-}$, and that's the reason why we
introduce both of the two operators.

Now let's go to see more details for $\hat{l}_{+}$ and
$\hat{l}_{-}$. First we suggest some wave functions of $\psi_{m}$
whose eigen values for $\hat{l}_{z}$ are labeled as $m$:
\begin{align}\label{}
\hat{l}_{z}\hat{l}_{+} &= \hat{l}_{+}\hat{l}_{z} + \hbar\hat{l}_{+}
\Rightarrow \nonumber \\
\hat{l}_{z}\hat{l}_{+}\psi_{m} &= (\hat{l}_{+}\hat{l}_{z} +
\hbar\hat{l}_{+})\psi_{m} \Rightarrow \nonumber \\
&=m\hbar\hat{l}_{+}\psi_{m} + \hbar\hat{l}_{+}\psi_{m} \nonumber \\
&=(m+1)\hbar\hat{l}_{+}\psi_{m}
\end{align}
Thus we can see that $\hat{l}_{+}\psi_{m}$ is also some eigen
function for the $\hat{l}_{z}$, while whose eigen value is
$(m+1)\hbar$. That means, after the operation on the $\hat{l}_{+}$,
the $\hat{l}_{+}\psi_{m}$ increase its $m$ value by $1$.

the $\hat{l}_{-}$ has likewise the same procedure to reach the
result:
\begin{equation}\label{}
\hat{l}_{z}\hat{l}_{-}\psi_{m} = (m-1)\hbar\hat{l}_{-}\psi_{m}
\end{equation}
Thus the $\hat{l}_{+}\psi_{m}$ decrease its $m$ value by $1$.

For this reason, the $\hat{l}_{+}$ is called ``up operator'', and
the $\hat{l}_{-}$ is called ``down operator''.

Now let's go to see the eigen values for the $\hat{l}^{2}$. Since
that $\hat{l}_{z}$ possesses no speciality among $\hat{l}_{x}$,
$\hat{l}_{y}$ and $\hat{l}_{z}$; thus for a common system, different
$l_{z}$ value for $\hat{l}_{z}$ may correspond to the same eigen
value for $\hat{l}^{2}$. This is similar to the case of the three
dimensional wave function in the cubic box (see the previous
chapter), where whose wave function is:
\begin{align}\label{}
\Psi (x,y,z) &=
\frac{\sqrt{8}}{l^{3}}\sin\left(\frac{n_{1}\pi}{l}x\right)
\sin\left(\frac{n_{2}\pi}{l}y\right)
\sin\left(\frac{n_{3}\pi}{l}z\right) \nonumber \\
E (n_{1},n_{2},n_{3}) &=
\frac{\pi^{2}\hbar^{2}}{2ml^{2}}(n_{1}^{2}+n_{2}^{2}+n_{3}^{2})
\end{align}
In this case there's also no speciality between x, y and z, so for
the $n^{2} = n_{1}^{2}+n_{2}^{2}+n_{3}^{2}$; different $n_{i}$ may
correspond to the same $n^{2}$.

Now let's consider a system, whose wave function are the eigen
functions both for the $\hat{l}_{z}$ and $\hat{l}^{2}$.  Now we can
see that if the eigen value for the $\hat{l}^{2}$ is fixed up, there
must have maximum number of $\overline{m}$: $l_{z} =
\overline{m}\hbar$. Because that $\hat{l}^{2}-\hat{l}^{2}_{z} =
\hat{l}^{2}_{y} + \hat{l}^{2}_{x}$, the eigen values for the
$\hat{l}^{2}$ should always larger than (or at least equal to)
$\overline{m}^{2}$. Similarly there must have a minimum number of
$\underline{m}$, $l_{z} = \underline{m}\hbar$. Here, we do not know
whether they equal to each other, but actually their absolute value
should equal to each other; since that both of the two directions
for the $\hat{l}_{z}$ are identical.

Suggest that as the eigen value of $\hat{l}^{2}$ is fixed up to
$l^{2}$, and the eigen state of $\psi$ gives $\overline{m}$
($\hat{l}_{z}\psi = \overline{m}\hbar\psi$); here we have:
\begin{align}\label{}
\hat{l}_{+}\psi &= 0 \rightarrow \nonumber \\
\hat{l}_{-}\hat{l}_{+}\psi
&=(\hat{l}^{2}-\hat{l}_{z}^{2}-\hat{l}_{z})\psi= 0 \rightarrow \nonumber \\
l^{2} &= \overline{m}(\overline{m}+1)
\end{align}

Similarly for the $\psi^{'}$ which gives $\underline{m}$
($\hat{l}_{z}\psi^{'} = \underline{m}\hbar\psi^{'}$); we have:
\begin{align}\label{}
\hat{l}_{-}\psi^{'} &= 0 \rightarrow \nonumber \\
\hat{l}_{+}\hat{l}_{-}\psi^{'}
&=(\hat{l}^{2}-\hat{l}_{z}^{2}+\hat{l}_{z})\psi^{'}= 0 \rightarrow \nonumber \\
l^{2} &= \underline{m}(\underline{m}-1)
\end{align}

Thus we have $\overline{m}(\overline{m}+1) =
\underline{m}(\underline{m}-1)$. This yields that $\overline{m} =-
\underline{m}$, another solution is abandoned.

Anyway, for a specific eigen value of $l$, $l^{2} =
\overline{m}(\overline{m}+1)$; the eigen function corresponding to
the $\hat{l}_{z}$ are $2\overline{m}+1$ folds of degenerate; their
$m$ can adopt $m=-\overline{m}, -\overline{m}+1, \cdots, 0, 1,
\cdots, \overline{m}$.

Next let's consider the matrix elements for the $\hat{l}_{x}$ and
$\hat{l}_{y}$ for the same system, where we choose the
$(\hat{H},\hat{l}_{z})$ as the CSCCO. Therefore, the wave functions
$\psi_{1}, \psi_{2}, \cdots, \psi_{n}$ are the eigen vectors for
them $(\hat{H},\hat{l}_{z})$.

First for the different $\psi$ and $\psi^{'}$, which hold different
energy level, since $\hat{l}_{x}$ commuted with the Hamiltonian
operator, it's easy to see that $\bra{\psi^{'}}\hat{l}_{x}\ket{\psi}
= 0$ ($\hat{l}_{y}$ is same).
\begin{align}\label{}
\bra{\psi^{'}}\hat{H}\hat{l}_{x}\ket{\psi} &=
\bra{\psi^{'}}\hat{l}_{x}\hat{H}\ket{\psi} \nonumber \\
&=E\bra{\psi^{'}}\hat{l}_{x}\ket{\psi}
\end{align}
While on the other hand, we have:
\begin{align}\label{}
\bra{\psi^{'}}\hat{H}\hat{l}_{x}\ket{\psi} &=
\bra{\psi}\hat{H}\hat{l}_{x}\ket{\psi^{'}} \nonumber \\
&=\bra{\psi}\hat{l}_{x}\hat{H}\ket{\psi^{'}} \nonumber \\
&=E^{'}\bra{\psi}\hat{l}_{x}\ket{\psi^{'}} \nonumber \\
&=E^{'}\bra{\psi^{'}}\hat{l}_{x}\ket{\psi}
\end{align}
Thus we can have that :
\begin{equation}\label{}
(E-E^{'})\bra{\psi^{'}}\hat{l}_{x}\ket{\psi} = 0 \Rightarrow
\bra{\psi^{'}}\hat{l}_{x}\ket{\psi} = 0
\end{equation}

Hence, we will consider the wave functions which occupy the same
energy level and have different z directional angular momentum
($l_{z}(m)$ is different).

Now we will achieve this through $\hat{l}_{+}$ and $\hat{l}_{-}$.
Since $\hat{l}_{+}\ket{\psi_{m}} = c\ket{\psi_{m+1}}$, and
$\hat{l}_{-}\ket{\psi_{m}} = c\ket{\psi_{m-1}}$; only these
$\bra{\psi_{m}}\hat{l}_{+}\ket{\psi_{m-1}}$ and
$\bra{\psi_{m}}\hat{l}_{-}\ket{\psi_{m+1}}$ do not equal to $0$.

So far we have:
\begin{equation}\label{}
\hat{l}^{2} = \hat{l}_{+}\hat{l}_{-} + \hat{l}^{2}_{z} - \hat{l}_{z}
\end{equation}
we can multiply the operator with $\ket{\psi_{m}}$, whose eigen
value for $\hat{l}^{2}$ is $l(l+1)$, and $l_{z}$ is $m$; we
integrate it:
\begin{align}\label{}
\bra{\psi_{m}}\hat{l}^{2}\ket{\psi_{m}} &= \sum_{i}^{n}
\bra{\psi_{m}}\hat{l}_{+}\ket{\psi_{i}}
\bra{\psi_{i}}\hat{l}_{-}\ket{\psi_{m}} +
\bra{\psi_{m}}\hat{l}^{2}_{z}\ket{\psi_{m}} -
\bra{\psi_{m}}\hat{l}_{z}\ket{\psi_{m}} \nonumber \\
l(l+1) &= \bra{\psi_{m}}\hat{l}_{+}\ket{\psi_{m-1}}
\bra{\psi_{m-1}}\hat{l}_{-}\ket{\psi_{m}} + m^{2} -m
\end{align}
Since we have $\hat{l}_{+} = \hat{l}_{-}^{+}$, we have:
\begin{equation}\label{}
\bra{\psi_{m}}\hat{l}_{+}\ket{\psi_{m-1}} =
\bra{\psi_{m-1}}\hat{l}_{-}\ket{\psi_{m}}^{*}
\end{equation}
Hence we have:
\begin{equation}\label{}
|\bra{\psi_{m}}\hat{l}_{+}\ket{\psi_{m-1}}|^{2} = l(l+1)-m^{2}+m
\end{equation}
$\bra{\psi_{m}}\hat{l}_{+}\ket{\psi_{m-1}}= \sqrt{(l+m)(l-m+1)}$.

Since that $\hat{l}_{+} = \hat{l}_{x} + i\hat{l}_{y}$, and
$\hat{l}_{-} = \hat{l}_{x} - i\hat{l}_{y}$; now we can get:
\begin{align}\label{}
\bra{\psi_{m}}\hat{l}_{x}\ket{\psi_{m-1}} &=
\frac{1}{2}\sqrt{(l+m)(l-m+1)} \nonumber \\
\bra{\psi_{m}}\hat{l}_{y}\ket{\psi_{m-1}} &=
-\frac{1}{2}i\sqrt{(l+m)(l-m+1)}
\end{align}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Eigen functions for the $\hat{l}_{z}$ and $\hat{l}^{2}$}
%
% Eigen functions for the $\hat{l}_{z}$ and $\hat{l}^{2}$
%
%
%

In the last section, the eigen functions for the $\hat{l}_{z}$ has
been easily got:
\begin{align}\label{}
-i\hbar\frac{\partial \Psi}{\partial\varphi} &= l_{z}\Psi
\Rightarrow \nonumber
\\
\Psi &= C exp(il_{z}\varphi/\hbar)
\end{align}
Now we begin to derive the eigen function for the $\hat{l}^{2}$.

For the $\hat{l}^{2}$(whose expression under sphere coordinate can
see \ref{ANGULAReq:2}), we can safely presume that it's eigen
function is $Y(\theta, \varphi)$($\hat{l}^{2}$ is only function of
$\theta$ and  $\varphi$); furthermore, the $Y(\theta, \varphi)$ can
be split into:
\begin{equation}\label{ANGULAReq:5}
Y(\theta, \varphi) = \Theta (\theta)\Phi_{m}(\varphi)
\end{equation}
We can have this because judging from (\ref{ANGULAReq:2}) the
variables of $\theta$ and  $\varphi$ can be separated from each
other. Here the $\Phi_{m}(\varphi)$ adopts the form in
(\ref{ANGULAReq:6}).

Moreover, suggest we have such the eigen functions for the
$\hat{l}^{2}$:
\begin{equation}\label{ANGULAReq:4}
\hat{l}^{2}Y(\theta, \varphi) = \lambda \hbar^{2}Y(\theta, \varphi)
\end{equation}
Now by bring the sphere coordinate expression of the $\hat{l}^{2}$
and (\ref{ANGULAReq:5}) into (\ref{ANGULAReq:4}), we can have:
\begin{align}\label{}
-\Phi_{m}\times\frac{\hbar^{2}}{\sin\theta}\frac{d}{d
\theta}\left(\sin\theta\frac{d\Theta}{d \theta}\right) +
\frac{\Theta}{\sin^{2}\theta}(\hat{l}^{2}_{z}\Phi_{m}) &=
\lambda \hbar^{2}\Theta \Phi_{m} \nonumber \\
-\Phi_{m}\times\frac{\hbar^{2}}{\sin\theta}\frac{d}{d
\theta}\left(\sin\theta\frac{d\Theta}{d \theta}\right) +
\frac{\Theta}{\sin^{2}\theta}m^{2}\hbar^{2}\Phi_{m} &=
\lambda \hbar^{2}\Theta \Phi_{m} \nonumber \\
\frac{1}{\sin\theta}\frac{d}{d
\theta}\left(\sin\theta\frac{d\Theta}{d \theta}\right) +
\left(\lambda - \frac{m^{2}}{\sin^{2}\theta}\right)\Theta &= 0
\end{align}

However, this function is not easy to gain its solution. So we make
$\xi = \cos \theta$ (here we have $0\leq\theta\leq\pi$). Thus:
\begin{align}\label{}
\frac{d\Theta}{d \xi} = -\frac{1}{\sin \theta}\frac{d\Theta}{d
\theta}
\end{align}
Then we have:
\begin{align}\label{}
\frac{1}{\sin\theta}\frac{d}{d
\theta}\left(-\sin^{2}\theta\frac{d\Theta}{d \xi}\right) +
\left(\lambda - \frac{m^{2}}{(1-\xi^{2})}\right)\Theta &= 0
\nonumber \\
\frac{d}{d \xi}\left((1-\xi^{2})\frac{d\Theta}{d \xi}\right) +
\left(\lambda -
\frac{m^{2}}{(1-\xi^{2})}\right)\Theta &= 0 \nonumber \\
(1-\xi^{2})\frac{d^{2}\Theta}{d \xi^{2}} - 2\xi\frac{d\Theta}{d \xi}
+ \left(\lambda - \frac{m^{2}}{(1-\xi^{2})}\right)\Theta &= 0
\end{align}

This is the associated Legendre function. Under the condition of
$0\leq\theta\leq\pi$, the function has only one meaningful physical
solution, and this solution requires that $\lambda = l(l+1)$ and
$|m| \leq l$; here the $\lambda$ is the eigen value for the
$\hat{l}^{2}$, which is in agreement with the answer we found out in
the last section.

In this case, the solution is a polynomial:
\begin{equation}\label{}
P^{m}_{l}(\cos\theta) \quad |m| \leq l
\end{equation}
By normalized condition, we can finally get the expression for the
$\Theta$:
\begin{equation}\label{}
\Theta(\theta) =
(-1)^{m}\sqrt{\frac{2l+1}{2}\frac{(l-m)!}{(l+m)!}}P^{m}_{l}(\cos\theta)
\end{equation}

Therefore, the eigen function for the $\hat{l}^{2}$ is:
\begin{equation}\label{}
Y^{m}_{l}(\theta, \varphi) =
(-1)^{m}\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P^{m}_{l}(\cos\theta)e^{im\pi}
\end{equation}

Here we note that different $Y^{m}_{l}$ are orthogonal with each
other:
\begin{equation}\label{}
\int_{0}^{2\pi}\int_{0}^{\pi}Y^{m^{'}}_{l^{'}}Y^{m}_{l}\sin\theta
d\theta d\varphi = \delta_{l^{'}l}\delta_{m^{'}m}
\end{equation}
This is important in understanding the orthogonal characters of
concrete wave functions.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{ Coupling between two angular momentums}
%
% 1  to choose the L^{2} and L_z into the CSCCO
%    because they can commute with the \hat{H}
%
%
Now let's consider a system, which is composed by two subsystems. If
there's no interactions between the two subsystems, the total
angular momentum $\hat{L}$ for the whole system will simply be the
addition between the two subsystems; each subsystem is a fully
conserved system:
\begin{equation}\label{ANGULAReq:7}
\hat{\mathbf{L}} = \hat{\mathbf{L}}_{1} + \hat{\mathbf{L}}_{2}
\end{equation}

On the other hand, if there exists some interactions between the two
subsystems, the (\ref{ANGULAReq:7}) can not be applied anymore;
that's because the angular momentum for each subsystem is not
conserved anymore.

However, the weight of $\hat{\mathbf{L}}$ on each specific
direction, namely the $\hat{\mathbf{L}}_{x}$, $\hat{\mathbf{L}}_{y}$
and $\hat{\mathbf{L}}_{z}$ can be still made certain:
\begin{align}\label{}
\hat{\mathbf{L}}_{x} &= \hat{\mathbf{L}}_{1x} +
\hat{\mathbf{L}}_{2x} \nonumber \\
\hat{\mathbf{L}}_{y} &= \hat{\mathbf{L}}_{1y} +
\hat{\mathbf{L}}_{2y} \nonumber \\
\hat{\mathbf{L}}_{z} &= \hat{\mathbf{L}}_{1z} +
\hat{\mathbf{L}}_{2z}
\end{align}
This relation is obvious because the angular momentum operator is
the vector operator, so it obeys the rules for the vectors. For this
reason, we can select the $\hat{\mathbf{L}}_{z}$ into the CSCCO.

On the other hand, even though the total angular momentum is not the
conserved physical quantity anymore, the $\hat{L}^{2}$ can still
commute with the Hamiltonian, thus we can select the $\hat{L}^{2}$
into the CSCCO, too:
\begin{equation}\label{}
\hat{L}^{2} = \hat{\mathbf{L}}_{x}^{2} + \hat{\mathbf{L}}_{y}^{2} +
\hat{\mathbf{L}}_{z}^{2}
\end{equation}

For the $\hat{\mathbf{L}}_{z}$, it's eigen value and eigen states
are straightforward. Here we remember that different subsystem has
its own $l_{z}$, and they do not affect the others:
\begin{equation}\label{}
[\hat{\mathbf{L}}_{1z}, \hat{\mathbf{L}}_{2z}] = 0 \Rightarrow L_{z}
= L_{z1} + L_{z2}
\end{equation}
Nevertheless, for the $\hat{L}^{2}$ there does not have such clear
relations. Therefore, we should solve this question in another way.

If we select the $\hat{L}_{1}^{2}$, $\hat{L}_{2}^{2}$,
$\hat{\mathbf{L}}_{1z}$ and $\hat{\mathbf{L}}_{2z}$ into the CSCCO,
(this can be done for that all of the four operators are commuting
with the Hamiltonian, and also they are commuting with each other),
any quantum state related to the angular momentum will be determined
by $L^{2}_{1},L^{2}_{2},L_{1z},L_{2z}$. Therefore for the fixed
value of $L_{1}, L_{2}$, they will be $(2L_{1}+1)(2L_{2}+1)$
different states.

On the other hand, we can express the quantum states by choosing
another representation, which is derived from the operators of
$\hat{L}^{2}_{1}$, $\hat{L}^{2}_{2}$, $\hat{L}^{2}$ and
$\hat{\mathbf{L}}_{z}$. Here the quantum states can be labeled by
$L^{2}_{1},L^{2}_{2},L^{2},L_{z}$. Similarly, for the fixed value of
$L_{1}, L_{2}$, they will be $(2L_{1}+1)(2L_{2}+1)$ different
states.

We can determine these states by the way below. By addition between
different $L_{1z}$ and $L_{2z}$, we can get the $L_{z}$:
\begin{center}\label{ANGULARTABLE:1}
\begin{tabular}{c c c}
  \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
  $L_{1z}$ & $L_{2z}$ & $L_{z}$ \\
  \hline
  $L_{1}$     & $L_{2 }$     & $L_{1}+L_{2}$ \\
  $L_{1}-1$   & $L_{2 }$     & $L_{1}+L_{2}-1$ \\
  $L_{1}$     & $L_{2 }-1$   & $L_{1}+L_{2}-1$ \\
  $L_{1}-2$   & $L_{2 }$     & $L_{1}+L_{2}-2$ \\
  $L_{1}-1$   & $L_{2 }-1$   & $L_{1}+L_{2}-2$ \\
  $L_{1}$     & $L_{2 }-2$   & $L_{1}+L_{2}-2$ \\
  $\cdots$    & $\cdots$     & $\cdots$ \\
  \hline
\end{tabular}
\end{center}

Here we can see that for same $L_{z}$, it's always corresponding to
many states. These states must have different $L$ value (since that
for a fixed $L$, its angular momentum on z axis must be different.
Hence there's no two states can share the same $L_{z}$ if they have
shared the same $L$), thus we can count the number of states as the
change of $L_{z}$. For example, as $L_{z} = L_{1}+L_{2}-2$, there
must have three states which possess different $L$, and because the
$L \geq L_{z}$; thus the three states must be $L = L_{1}+L_{2}$, $L
= L_{1}+L_{2}-1$ and $L = L_{1}+L_{2}-2$. Therefore, if the number
of states for the $L_{z}$ continues to grow, there implies it has
new states of different $L$.

Such continuing will be stopped at $L_{z} = |L_{1} - L_{2}|$. As the
$L_{z}$ decreases, the number of states do not grow. This implies
that the eigen value of $\hat{L}$ for the whole system will be:
\begin{equation}\label{}
L = L_{1}+L_{2}, L_{1}+L_{2}-1, L_{1}+L_{2}-2, \cdots, |L_{1} -
L_{2}|
\end{equation}
Here the $L_{1}$ and $L_{2}$ has to give to be certain.

This phenomenon can be well understood by the "vector model" of
angular momentum. Since that the angular momentum is some vector, as
the angular momentum in the subsystems take same direction, they
will give the maximum angular momentum addition for the whole; and
if they take opposite direction, they will give the minimum angular
momentum for the whole. This coincides with the conclusion above.

%%% Local Variables:
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%%% End:


\chapter{Functional Renormalization and Quantum Gravity}\label{chap:EHT}
After the formal introduction of the physical and mathematical concepts in the last chapters, we are now able to motivate and formulate the key idea of the Asymptotic Safety approach to quantum gravity, which aims at finding a quantum field theoretical description of gravity within the language of the Functional Renormalization Group. This chapter briefly discusses the requirements for the existence of such a theory. We proceed by solving the flow equation for quantum gravity within the Einstein-Hilbert truncation in a transverse-traceless spin-two graviton approximation and investigate its fixed point structure.
\vspace{-0.6cm}
\section{Asymptotic Safety}
Asymptotic Safety generalizes the concept of Asymptotic Freedom known as a property of certain gauge theories. The latter one is based on the idea, that the interaction of particles becomes asymptotically weak in the high-energy limit. In the language of the FRG this means, that the UV behavior of such theories is governed by a non-interacting, Gaussian fixed point,  allowing perturbative calculations. Maybe the most popular example for an asymptotically free theory is Quantum  Chromodynamics (QCD), the theory of the strong interaction. In 1978,  Steven Weinberg proposed the Asymptotic Safety scenario for gravity, based on the existence of an interacting, Non-Gaussian fixed point, rendering the theory non-perturbatively renormalizable in the UV \cite{Weinberg1980}. The main advantage compared to other approaches is, that there is no need to introduce new symmetries, extra dimensions or other additional complications. The key requirements for an asymptotically safe quantum field theory of gravity can be summarized as follows:
\begin{enumerate}
	\item The existence of an UV-attractive NGFP of the Renormalization Group flow of $\Gammak$ has to be guaranteed.
	\item For the predictivity of the theory, it is crucial to be able to fix the trajectory $\Gammak$ by a finite amount of measurements, i.\,e. the corresponding UV hypersurface $\Sigma_{\mathrm{UV}}$ of the NGFP should be of finite dimension:
	\begin{equation*}
		\operatorname{dim}\Sigma_{\mathrm{UV}} < \infty.
	\end{equation*}
\end{enumerate}
We want to probe these requirements in the following by computing the running of the Newton constant $G_k$ and the cosmological constant $\Lambda_k$ in a pure-gravity setting. \\
For a more detailed discussion of current Asymptotic Safety research, we refer the interested reader to \cite{Eichhorn2018}. Recently, very detailed textbooks covering both, the physical and the mathematical concepts of Asymptotic Safety, have been released. For a complete treatment of the subject it is worth to have a look at \cite{Percacci2017} or \cite{ReuterSaueressig_2019}.
\vspace{-0.6cm}
\section{Einstein-Hilbert Truncation}
Solving the flow equation analytically is nothing but impossible. Therefore it is unavoidable to truncate the initially infinite dimensional theory space to a finite subspace, to be able to find approximated solutions. It is important, that all terms, that are invariant under the imposed symmetry, i.\,e. invariant under diffeomorphism transformations, need to be taken into account. The easiest truncation fulfilling this requirement is the \textit{Einstein-Hilbert truncation}. In 1996, Martin Reuter was the first to successfully investigate quantum gravity within the Einstein-Hilbert truncation \cite{Reuter1996}. He was able to prove the existence of a NGFP for a pure-gravity setting. This truncation takes only the scalar curvature $\Ricci$ and the cosmological constant $\Lambda$ into account and therefore the truncated subspace is $2$-dimensional\footnote{Recently, more sophisticated truncations including higher-order curvature terms ($\Ricci^2$, $f(\Ricci)$, $R^{\mu\nu}R_{\mu\nu}$ \dots) have been investigated, see e.\,g. \cite{AlkoferSaueressig2018}.}.\\ The full Einstein-Hilbert  truncation reads:
\begin{align}
	\Gammak = 2\kappa^2Z_{h,k} \int_x \sqrt{g} \ [-\mathcal{R} + 2\Lambda_k] + \mathcal{S}_{\text{gf}} + \mathcal{S}_{\text{gh}},
\label{eqn:EHtruncation}
\end{align}
where we abbreviated $\kappa^2 = \left(32\pi G\right)^{-1}$. Since we don't need to consider the gauge fixing action $\mathcal{S}_{\text{gf}}$ and the Non-Abelian ghost action $\mathcal{S}_{\text{gh}}$ for the calculation performed in this chapter, we won't specify their explicit forms at this point. We will come back to this discussion at the beginning of chapter \ref{chap:Matter}.\\
 The index $k$ indicates the \textit{running} i.\,e. the scale dependence of the cosmological constant and the wave function renormalization $Z_{h,k}$. It is quite convenient to define the running Newton coupling as
 \begin{equation}
 	G_k = G \cdot \left(Z_{h,k}\right)^{-1}.
 \end{equation}
Up to this point, we did not care about the dimensionality of the two couplings. As already mentioned, one usually works with dimensionless couplings when solving the flow equation. \\
We therefore define the dimensionless renormalized cosmological constant
\begin{align}
	\lambda_k = \Lambdak\cdot k^{-2}
\end{align}
and Newton constant
\begin{align}
	g_k = G_k\cdot k^{d-2} \ \overset{(d=4)}{=} \ G_k\cdot k^{2}.
\end{align}
After these redefinitions, we can already compute the l.\,h.\,s. of the flow equation, i.\,e. the scale derivative w.\,r.\,t. the RG time $t$. The derivative only acts on $\lambda_k$ and $g_k$.
\begin{align}
	\partial_{t}\Gamma_{k} = 2\kappa^2 Z_{k,h}\int_x \sqrt{g} \left\{\eta_h\mathcal{R}+2\left(k^2(\partial_t\lambda_k) + \Lambda_k(2 - \eta_h)\right)\right\}.
	\label{eqn:LHS}
\end{align}
Here we introduced the \textit{anomalous dimension} $\eta_h$, defined as
\begin{align}
	\eta_h=-\partial_t \ln Z_{h,k} = -\frac{\partial_tZ_{h,k}}{Z_{h,k}}.
\end{align}
We find terms of order $\sim\sqrt{g}$ and order $\sim\sqrt{g}\Ricci$. The r.\,h.\,s. of the flow equation is assumed to admit an expansion in terms of invariants, which will also be of order $\sim\sqrt{g}$ and $\sim\sqrt{g}\mathcal{R}$ since we are working in the Einstein-Hilbert truncation\footnote{There will also appear terms of higher order in curvature, but we will drop them.}. This will allow us to determine the $\beta$-function for $\lambda_k$ and the explicit form of $\eta_h$ by a comparison of the terms of these orders on both sides of the flow equation.
The $\beta$-function for $g_k$ follows directly from the anomalous dimension:
\begin{align}
	\beta_g = \partial_t g_k = \left(2 + \eta_h\right)g_k.
	\label{eqn:beta_gk}
\end{align}
At this point, we want to use the idea of the background field approximation. We assume a linear split of the metric into a background field $\bar{g}$ and a fluctuation field $h$, as described in  equation (\ref{eqn:metric_split}). The approximation at this point is, that in the following, we set all fluctuations to zero, i.\,e. we evaluate all derivatives etc. at $g=\bar{g}$. We will critically review this approximation in chapter \ref{chap:BGindependence}. In general, there is no conceptual need to fix the background metric, it nevertheless is really useful to choose specific classes of backgrounds for certain computations, since this can simplify calculations a lot. We exploit the freedom of choosing a spherical background. On the four-sphere $\mathbb{S}^4$, which is a maximally symmetric space\footnote{Maximally symmetric spaces are characterized by the fact, that they have the same number of symmetries as ordinary Euclidean space.}, the Riemann tensor and the Ricci tensor can be written as multiples of the curvature scalar:
\begin{equation}
\begin{aligned}
	\bar{R}_{\mu\nu} &= \frac{1}{4} \ \bar{g}_{\mu\nu} \bar{\mathcal{R}}\\[10pt]
	\bar{R}_{\mu\nu\rho\sigma} &= \frac{1}{12} \ (\bar{g}_{\mu\rho}\bar{g}_{\nu\sigma} - \bar{g}_{\mu\sigma}\bar{g}_{\nu\rho}) \bar{\mathcal{R}}.
\end{aligned}
\end{equation}
The bar in this notation refers to the background $\bar{g}$. The inversion of the two-point function on general curved backgrounds is non-trivial. We need to find a suitable tensor basis for the decomposition of the metric fluctuation $h_{\mu\nu}$. We choose a \textit{York decomposition} and find:
\begin{equation}
	h_{\mu v}=h_{\mu v}^{\mathrm{TT}}+\bar{\nabla}_{\mu} \xi_{v}+\bar{\nabla}_{v} \xi_{\mu}+\left(\bar{\nabla}_{\mu} \bar{\nabla}_{v}-\frac{1}{d} \bar{g}_{\mu v} \bar{\Delta}\right) \sigma+\frac{1}{d} \bar{g}_{\mu v} h.
	\label{eqn:York}
\end{equation}
Here, $ h_{\mu\nu}^{\text{TT}}$ is a transverse-traceless, spin-two degree of freedom, $\xi_{\mu}$ is transverse and carries a spin-one d.\,o.\,f. and $\sigma$ and $h$ have spin zero. For a more detailed motivation of the York decomposition, we refer to appendix \ref{chap:AppA}.\\
As a further approximation, we only take the contribution from the spin-two graviton mode $h_{\mu\nu}^{\text{TT}}$ into account\footnote{This choice is motivated by the fact, that the spin-two mode carries the most degrees of freedom. From the $10$ initial d.\,o.\,f. in this decomposition, $4$ can be removed via gauge fixing, the remaining $6$ are divided into $5$ from the $h_{\mu\nu}^{\text{TT}}$ mode and $1$ for the trace mode $h$.}. The other modes are included later on in chapter \ref{chap:Matter}, when we couple matter to the system. \\
Our starting point for the computation of the r.\,h.\,s. of the flow equation is the transverse-traceless graviton two-point function:
\begin{align}
\Gamma_{k, h^{\text{TT}}h^{\text{TT}}}^{(2)} = \frac{Z_k}{32\pi}\left(\bar{\Delta} - 2\Lambda_k+\frac{2}{3}\bar{\mathcal{R}}\right).
\end{align}
Here, we used the definition of the Laplace operator, given by $\Delta = -\nabla^2$. We use a regulator of the form
\begin{align}
R_k  = \eval{\Gamma_{h^{\text{TT}}h^{\text{TT}}}^{(2)}}_{\Lambda_k=\bar{\mathcal{R}}=0} \cdot r_k\left(\frac{\bar{\Delta}}{k^2}\right),
\end{align}
with a Litim-type shape function (\ref{eqn:Litim}). This allows us to determine the full propagator
\begin{equation}
	G_{k, h^{\text{TT}}h^{\text{TT}}} := \left(\Gamma_{k, h^{\text{TT}}h^{\text{TT}}}^{(2)} + R_k\right)^{-1} = \frac{32\pi}{Z_{h,k}}\left(\bar{\Delta}\left(1+r_k\right) -2\Lambda_k + \frac{2}{3}\bar{\mathcal{R}}\right)^{-1}
\end{equation}
and the scale derivative of the regulator
\begin{equation}
	\partial_t R_k =  \frac{Z_{h,k}}{32\pi}\bar{\Delta}\left(\partial_tr_k - \eta_hr_k\right).
\end{equation}
The computation of the r.\,h.\,s. of the flow equation is in general very hard because it involves the computation of a functional trace of a function depending on the Laplacian on a curved background. We can use \textit{heat-kernel techniques} to solve such equations. Heat-kernel computations are based on a curvature expansion in powers of the curvature scalar $\mathcal{R}$. For more details, have a look at the second part of appendix \ref{chap:AppA}.\\
The previously obtained results lead us to
\begin{equation}
\frac{1}{2}\operatorname{Tr}\left[G_{k, h^{\text{TT}}h^{\text{TT}}}\ \partial_t R_k\right] = \frac{1}{2}\operatorname{Tr}\left[\frac{\bar{\Delta}\left(\partial_t r_k - \eta_h r_k\right)}{\bar{\Delta}(1+r_k)-2\Lambda_k+\frac{2}{3}\bar{\mathcal{R}}}\right].
\end{equation}
We proceed by expanding this expression around vanishing curvature:
\begin{align}
\resizebox{.9 \textwidth}{!}{$
\frac{1}{2}\operatorname{Tr}\left[G_{k, h^{\text{TT}}h^{\text{TT}}}\ \partial_t R_k\right] = \frac{1}{2}\operatorname{Tr}\left[\frac{\bar{\Delta}\left(\partial_t r_k - \eta_h r_k\right)}{\bar{\Delta}(1+r_k)-2\Lambda_k}\right] - \frac{1}{3}\bar{\mathcal{R}}\operatorname{Tr}\left[\frac{\bar{\Delta}\left(\partial_t r_k - \eta_h r_k\right)}{\left(\bar{\Delta}(1+r_k)-2\Lambda_k\right)^2}\right] + \mathcal{O}(\mathcal{R}^2).$
}
\end{align}
We evaluate these two terms separately using the heat-kernel formulas presented in appendix \ref{chap:AppA}. The result for the first term reads
\begin{align}
\frac{1}{2}\operatorname{Tr}\left[\frac{\bar{\Delta}\left(\partial_t r_k - \eta_h r_k\right)}{\bar{\Delta}(1+r_k)-2\Lambda_k}\right] &= \frac{1}{2}\frac{1}{(4\pi)^2}\int_x \sqrt{\bar{g}} \left[5\Phi_2^1(-2\Lambda_k) - \frac{5}{6}\bar{\mathcal{R}}\Phi^1_1(-2\Lambda_k)\right] \\[10pt]
&= \frac{1}{2}\frac{1}{(4\pi)^2}\int_x \sqrt{\bar{g}}\left[\frac{1}{1-2\lambda_k}\left(5\left(1-\frac{\eta_h}{6}\right) - \frac{5}{3}\bar{\Ricci}\left(1-\frac{\eta_h}{4}\right)\right)\right],\nonumber
\label{eqn:hk41_test}
\end{align}
with the definition of threshold functions
\begin{equation}
\begin{aligned}
	\Phi_n^p(\omega) &= \frac{1}{\Gamma(n)}\int_0^{\infty}\dd z \ z^{n-1} \frac{z(-2zr_k(z)-\eta_{\Psi}r_k(z))}{(z(1+r_k(z))+\omega)^p}\\[10pt]
	&= \frac{1}{\Gamma(n)}\frac{1}{\left(1+\omega\right)^p}\left(\frac{2}{n} - \frac{\eta_{\Psi}}{n(n+1)}\right).
\end{aligned}
\label{eqn:threshold}
\end{equation}
In the last step, we evaluated the threshold functions for the Litim-type shape function. We used $\eta_\Psi$, since we want to keep this formula as general as possible and we will use it multiple times for different fields throughout this thesis. Analogously, the second term in our expansion reads
\begin{align}
	-\frac{1}{3}\bar{\mathcal{R}}\operatorname{Tr}\left[\frac{\bar{\Delta}\left(\partial_t r_k - \eta_h r_k\right)}{\left(\bar{\Delta}(1+r_k)-2\Lambda_k\right)^2}\right] &= -\frac{1}{2}\frac{1}{(4\pi)^2}\int_x \sqrt{\bar{g}} \ \frac{10}{3}\bar{\Ricci}\ \Phi_2^2(-2\Lambda_k)\\
	&=  -\frac{1}{2}\frac{1}{(4\pi)^2}\int_x \sqrt{\bar{g}}\left[\frac{1}{(1-2\lambda_k)^2}\left( \frac{10}{3}\bar{\Ricci}\left(1-\frac{\eta_h}{6}\right)\right)\right].\nonumber
	\label{eqn:hk42}
\end{align}
With these results, we are able to determine the $\beta$-function for the cosmological constant, simply by comparing the order $\sim\sqrt{\bar{g}}$ terms occurring on the l.\,h.\,s. of the flow equation and the results from the heat-kernel expansion of the functional trace:
\begin{align}
	\beta_{\lambda} = \partial_t\lambda_k = -4\lambda_k + \frac{\lambda_k}{g_k} \partial_t g_k + \frac{5}{4\pi}g_k\frac{1-\frac{\eta_h}{6}}{1-2\lambda_k}.
\end{align}
Additionally, we find an expression for the anomalous dimension $\eta_h$ by comparing the terms of order $\sim\sqrt{\bar{g}}\bar{\Ricci}$:
\begin{align}
\eta_h = -\frac{5g_k}{3\pi} \left(\frac{1-\frac{\eta_h}{4}}{1-2\lambda_k} + 2\frac{1-\frac{\eta_h}{6}}{(1-2\lambda_k)^2}\right).
\end{align}
To find the fixed points for this truncated solution, we use \verb|Mathematica| to solve the equation
\begin{equation}
	\vec{\beta} = \begin{pmatrix}\beta_g\\ \beta_{\lambda}\end{pmatrix} \overset{!}{=}  \begin{pmatrix}0\\ 0\end{pmatrix}.
\end{equation}
We arrive at the following values for the Newton coupling and the cosmological constant at the NGFP:
\begin{align}
	(g_k^*, \lambda_k^*) = (0.86, 0.18).
\end{align}
The critical exponents of the fixed point are given by the complex conjugated pair
\begin{align}
	\theta_{1,2} = 2.9 \pm 2.6i.
\end{align}
\begin{figure}[t]
\centering
	\includegraphics[width=0.8\textwidth]{figs/Plots/EH_NoMatter}
	\caption[Flow diagram for the Einstein-Hilbert truncation in $h^{\mathrm{TT}}$ approximation]{Flow diagram  for the Einstein-Hilbert truncation in $h^{\mathrm{TT}}$ approximation as computed in this work. The flow points towards the infrared. The postulated UV-attractive Non-Gaussian fixed point (blue dot) is clearly visible. }
	\label{fig:flow_diag}
\hrulefill
\end{figure}
The corresponding flow diagram is depicted in figure  (\ref{fig:flow_diag}). The \enquote{swirl} around the NGFP is explained by the fact, that $\mathfrak{Im}\ \theta_{1,2} \neq 0$. The visualization and the plotting of the diagram was done in \verb|Python|.\\
Based on the results obtained in the scope of this calculation, we can confirm the existence of a UV-attractive Non-Gaussian fixed point for this truncated pure-gravity system, providing a good starting point for further analysis. As a first extension of this truncation, we want to go beyond the $h^{\mathrm{TT}}$ approximation by taking the contributions for the other modes into account. Then, we will investigate the impact of minimally coupled matter fields on the system. All these calculations will be explained in full detail in the next chapter.


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\pagestyle{main}
\renewcommand{\chaptermark}[1]{\markboth{\chaptername\ \thechapter\ (#1)}{}}
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\begin{document}

\chapter{Geometric Optics}
\section{Mirrors}
\begin{itemize}
    \item \marginnote{8/12:}Take what we know about reflection and refraction and apply it to mirrors and lenses.
    \item Bathroom mirror: A flat piece of glass with a shiny background behind it.
    \item Spherical mirror: Some part of a reflective sphere.
    \item Relating the \textbf{radius of curvature} and the distance to the \textbf{focal point}, or the \textbf{focal length}, of a concave spherical mirror.
    \begin{figure}[h!]
        \centering
        \begin{tikzpicture}[
            scale=1.3,
            pics/point/.style args={#1:#2}{code={
                \coordinate (#2);
                \node [circle,fill=pix,inner sep=1.5pt,label={#1:$#2$}] {};
            }}
        ]
            \footnotesize
            \def\r{30}
            \draw [name path=axis] (0,0) -- (4,0) node[right]{axis};

            \draw [blx,thick] (2.5,1) arc[start angle=45,end angle=-45,radius={2^0.5}];

            \draw [very thin,dashed] (1.5,0) pic{point=below:C} -- ++(\r:2^0.5) pic{point=right:A};
            \pic at ({1.5+2^0.5},0) {point=below right:V};

            \draw [
                rex,thick,-latex,name path=ray,postaction={decorate},decoration={
                    markings,mark=at position 0.1 with \arrow{latex}
                }
            ]
                (A) ++(-2.7,0) coordinate (1)
                -- (A) node[circle,fill=pix,inner sep=1.5pt]{}
                -- ++({180+2*\r}:2)
            ;

            \path [name intersections={of=axis and ray,by=F}] pic at (F) {point=below:F};
            \pic [draw,angle eccentricity=1.3,pic text={$\theta$}] {angle=1--A--C};
            \pic [draw,angle eccentricity=1.3,pic text={$\theta$}] {angle=C--A--F};
            \pic [draw,angle eccentricity=1.3,pic text={$\theta$}] {angle=F--C--A};
        \end{tikzpicture}
        \caption{Spherical mirror analysis.}
        \label{fig:sphericalMirror}
    \end{figure}
    \begin{itemize}
        \item For a distant source $s$, we can approximate the rays as parallel.
        \item Consider one specific ray.
        \item Drawing a normal to the surface of the spherical mirror, this normal will be a radius passing through the \textbf{center of curvature}.
        \item It follows by the law of reflection and the alternate interior angle theorem that all $\theta$ are equal.
        \item This makes $\triangle CFA$ isosceles.
        \item Now assume \textbf{paraxial rays}.
        \item Then $\triangle CFA$ converges to a line segment (a radius) with midpoint $F$.
        \item Therefore,
        \begin{equation*}
            f = \frac{R}{2}
        \end{equation*}
        for a concave spherical mirror.
    \end{itemize}
    \item \textbf{Radius of curvature}: The radius of the sphere into which the spherical mirror would fit. \emph{Denoted by} $\bm{R}$.
    \item \textbf{Focal point}: The point where parallel rays converge, after reflecting off of a curved mirror. \emph{Denoted by} $\bm{F}$.
    \item \textbf{Focal length}: The distance from the focal point to the mirror. \emph{Denoted by} $\bm{f}$.
    \item \textbf{Center of curvature}: The center of the sphere into which the spherical mirror would fit. \emph{Denoted by} $\bm{C}$.
    \item \textbf{Paraxial ray}: A ray that is close to the mirror axis.
    \item Alternatively, if you look at a convex mirror, it appears (via ray tracing) that all rays of light originated from the focal point.
    \begin{itemize}
        \item The image in this type of mirror will be a virtual image.
        \item In this case, we say that $f=-R/2$.
    \end{itemize}
    \item A spherical mirror does not focus all rays to a \emph{single} point --- the rays only go to \emph{approximately} the same point.
    \begin{itemize}
        \item The farther the rays get from being paraxial, the more they diverge from the focal point.
    \end{itemize}
    \item To have a true focal point, we need a parabolic mirror.
    \item Unfortunately, parabolic mirrors are hard to make.
    \item \textbf{Ray tracing}: Take a few \textbf{principal rays} and see where the image forms.
    \begin{figure}[h!]
        \centering
        \begin{tikzpicture}[
            scale=2.5,
            pics/point/.style args={#1:#2}{code={
                \coordinate (#2);
                \node [circle,fill=pix,inner sep=1.5pt,label={#1:${\color{black}#2}$}] {};
            }}
        ]
            \footnotesize
            \draw [name path=axis] (0,0) -- (3.5,0) node[right]{axis};

            \draw [blx,thick,name path=mirror] (2,1) arc[start angle=45,end angle=-45,radius={2^0.5}];

            \draw [grx,ultra thick,-stealth] (0.5,0) node[above left,black]{object} edge [black,thin,->,out=90,in=180,shorten >=2pt] (0.5,0.4) -- ++(0,{2^0.5/2}) coordinate (P1);

            \path [name path=Afinder] (P1) -- ++(4,0);
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                rex,thick,-latex,postaction={decorate},
                decoration={markings,mark=at position 0.2 with \arrow{latex}},
                name intersections={of=mirror and Afinder,by=A},name path=parallel
            ] (P1) -- (A) -- ++(-120:2);
            \draw [
                rex,thick,-latex,postaction={decorate},
                decoration={markings,mark=at position 0.2 with \arrow{latex}},
                name intersections={of=mirror and axis,by=V},name path=central
            ] (P1) -- (V) pic{point={[xshift=-2pt,yshift=2pt]below right:V}} -- (P1 |- 0,{-2^0.5/2});
            \path [name intersections={of=axis and parallel,by=F},name path=Fmirror] (P1) -- ($(P1)!1.5!(F)$);
            \draw [
                rex,thick,-latex,line join=bevel,postaction={decorate},
                decoration={markings,mark=at position 0.2 with \arrow{latex}},
                name intersections={of=mirror and Fmirror}
            ] (P1) -- (intersection-1) -- ++(-2,0);
            \path [name intersections={of=parallel and central,by={,,I}},name path=Imirror] (P1) -- ($(P1)!2!(I)$);
            \draw [
                rex,thick,postaction={decorate},
                decoration={
                    markings,
                    mark=at position 0.33 with \arrow{latex},
                    mark=at position 0.85 with \arrowreversed{latex}
                },
                name intersections={of=axis and Imirror,by=C},name intersections={of=mirror and Imirror}
            ] (P1) -- (intersection-1) pic at (C) {point={[xshift=2pt,yshift=2pt]below left:C}};

            \draw [grx,ultra thick,stealth-] (I) node[below left,yshift=-1mm,black,fill=white,inner sep=1.5pt]{image} edge[black,thin,->,out=90,in=180,shorten >=2pt] (I |- 0,-0.1) -- (I |- 0,0);
            \pic at (F) {point={[fill=white,circle,inner sep=0pt,xshift=1pt,yshift=0pt]below right:F}};
        \end{tikzpicture}
        \caption{Ray tracing a concave mirror.}
        \label{fig:rayTracing}
    \end{figure}
    \item \textbf{Principal ray}: A ray with well-understood geometry.
    \item \textbf{Focal ray}: A ray that goes through the focal point.
    \begin{itemize}
        \item Will be reflected out as a parallel ray.
    \end{itemize}
    \item \textbf{Central ray}: A ray that hits the center of the mirror.
    \begin{itemize}
        \item Will be reflected out with the same incident angle relative to the axis.
    \end{itemize}
    \item \textbf{Radial ray}: A ray that passes through the center of curvature.
    \begin{itemize}
        \item It follows a \emph{radius} of the sphere of curvature.
        \item Will be reflected such that it heads right back to where it started.
    \end{itemize}
    \item Calculating the location of the image in a spherical mirror.
    \begin{figure}[h!]
        \centering
        \begin{tikzpicture}[
            scale=2.5,
            pics/point/.style args={#1:#2}{code={
                \coordinate (#2);
                \node [circle,fill=pix,inner sep=1.5pt,label={#1:${\color{black}#2}$}] {};
            }}
        ]
            \footnotesize
            \draw [name path=axis] (0,0) -- (3.5,0) node[right]{axis};

            \draw [blx,thick,name path=mirror] (2,1) arc[start angle=45,end angle=-45,radius={2^0.5}];

            \draw [grx,ultra thick,-stealth] (0.5,0) coordinate (Q1) node[below left,black]{$Q$} node[above left,black]{object} edge [black,thin,->,out=90,in=180,shorten >=2pt] (0.5,0.4) -- ++(0,{2^0.5/2}) coordinate (P1) node[above left,black]{$P$};

            \path [name path=Afinder] (P1) -- ++(4,0);
            \path [name intersections={of=mirror and Afinder,by=A},name path=parallel] (P1) -- (A) -- ++(-120:2);
            \draw [
                rex,thick,-latex,postaction={decorate},
                decoration={markings,mark=at position 0.2 with \arrow{latex}},
                name intersections={of=mirror and axis,by=V},name path=central
            ] (P1) -- (V) pic{point={[xshift=-2pt,yshift=2pt]below right:V}} -- (P1 |- 0,{-2^0.5/2});
            \path [name intersections={of=parallel and central,by={,,I}},name path=Imirror] (P1) -- ($(P1)!2!(I)$);
            \draw [
                rex,thick,postaction={decorate},
                decoration={
                    markings,
                    mark=at position 0.33 with \arrow{latex},
                    mark=at position 0.85 with \arrowreversed{latex}
                },
                name intersections={of=axis and Imirror,by=C},name intersections={of=mirror and Imirror}
            ] (P1) -- (intersection-1) pic at (C) {point={[xshift=2pt,yshift=2pt]below left:C}};

            \draw [grx,ultra thick,stealth-] (I) coordinate (P2) node[below=1pt,xshift=-1pt,black]{$P'$} node[below right,xshift=4mm,black]{image} edge[black,thin,->,out=135,in=0,shorten >=2pt] (I |- 0,-0.1) -- (I |- 0,0) coordinate (Q2) node[above,black]{$Q'$};

            \path [name intersections={of=axis and parallel,by=F}] (F) pic {point={[fill=white,circle,inner sep=0pt,xshift=1pt,yshift=0pt]below right:F}};
            \pic [draw,angle radius=2mm] {right angle=C--Q1--P1};
            \pic [draw,angle radius=2mm] {right angle=C--Q2--P2};
            \pic [draw,angle radius=4mm,angle eccentricity=1,pic text={=},pic text options={rotate=160,yshift=-0.5pt}] {angle=P1--C--Q1};
            \pic [draw,angle radius=4mm,angle eccentricity=1,pic text={=},pic text options={rotate=-20,yshift=-0.5pt}] {angle=P2--C--Q2};
            \pic [draw,angle radius=6mm,angle eccentricity=1,pic text={\large--},pic text options={rotate=170}] {angle=P1--V--Q1};
            \pic [draw,angle radius=6mm,angle eccentricity=1,pic text={\large--},pic text options={rotate=190}] {angle=Q2--V--P2};

            \draw [very thin,|-|] (P1) ++(0,0.4) -- node[above]{$s$} ++(1.92,0);
            \draw [very thin,|-|] (P2) ++(0,-0.8) -- node[below]{$s'$} ++(0.76,0);
        \end{tikzpicture}
        \caption{Deriving the mirror equation.}
        \label{fig:mirrorEquation}
    \end{figure}
    \begin{itemize}
        \item The image in a spherical mirror will be an inverted, \textbf{real} image.
        \item Two principal rays define where the image is.
        \item Considering the image of the top point in our object, choose to analyze a central ray and a radial ray.
        \item Then $\triangle PVQ$ and $\triangle P'VQ'$ are similar, and $\triangle PCQ$ and $\triangle P'CQ'$ are similar.
        \item It follows that
        \begin{align*}
            \frac{\overline{PQ}}{{\color{white}(}\overline{P'Q'}{\color{white})}} &= \frac{\overline{QV}}{{\color{white}(}\overline{Q'V}{\color{white})}} = \frac{s}{s'}&
            \frac{\overline{PQ}}{{\color{white}(}\overline{P'Q'}{\color{white})}} &= \frac{\overline{QC}}{{\color{white}(}\overline{Q'C}{\color{white})}} = \frac{s-R}{R-s'}
        \end{align*}
        \item Thus, we have the following, which can be solved for the mirror equation.
        \begin{equation*}
            \frac{s}{s'} = \frac{s-R}{R-s'}
        \end{equation*}
    \end{itemize}
    \item \textbf{Real image}: An image that could be substituted for an image on a screen in real space.
    \item \textbf{Object distance}: The distance from an object to a spherical mirror. \emph{Denoted by} $\bm{s}$.
    \item \textbf{Image distance}: The distance from an object's image to the mirror. \emph{Denoted by} $\bm{s'}$.
    \item \textbf{Mirror equation}: The formula
    \begin{equation*}
        \frac{1}{s}+\frac{1}{s'} = \frac{1}{f}
    \end{equation*}
    \item \textbf{Lateral magnification}: The ratio of the distance from the object to the mirror and the distance from the image to the mirror. \emph{Denoted by} $\bm{m}$.
    \begin{itemize}
        \item Mathematically,
        \begin{equation*}
            m = \frac{-s'}{s}
        \end{equation*}
        \item If $m<0$, the image is inverted.
        \item If $m>0$, the image is upright.
        \item Sign convention: $s,s'$ are positive in front of the mirror and negative behind the mirror.
    \end{itemize}
    \item Check the above equations on the example of a plane mirror.
    \begin{itemize}
        \item For a plane mirror, $R=\infty$, so $\frac{1}{f}=0$.
        \item It follows by the mirror equation that $s'=-s$.
        \item Therefore, $m=1$, as desired.
    \end{itemize}
    \item \marginnote{8/13:}Ray tracing a convex mirror.
    \begin{figure}[H]
        \centering
        \begin{tikzpicture}[
            scale=2.5,
            pics/point/.style args={#1:#2}{code={
                \coordinate (#2);
                \node [circle,fill=pix,inner sep=1.5pt,label={#1:${\color{black}#2}$}] {};
            }}
        ]
            \footnotesize
            \draw [name path=axis] (0,0) -- (3.5,0) node[right]{axis};

            \draw [blx,thick,name path=mirror] (2,1) arc[start angle=135,end angle=225,radius={2^0.5}];

            \draw [grx,ultra thick,-stealth] (0.5,0) node[above left,black]{object} edge [black,thin,->,out=90,in=180,shorten >=2pt] (0.5,0.4) -- ++(0,{2^0.5/2}) coordinate (P1);

            \path [name path=Afinder] (P1) -- ++(4,0);
            \draw [
                rex,thick,-latex,postaction={decorate},
                decoration={markings,mark=at position 0.2 with \arrow{latex}},
                name intersections={of=mirror and Afinder,by=A},name path=parallel
            ] (P1) -- (A) -- ++(120:1);
            \draw [
                rex,thick,-latex,postaction={decorate},
                decoration={markings,mark=at position 0.2 with \arrow{latex}},
                name intersections={of=mirror and axis,by=V},name path=central
            ] (P1) -- (V) -- (P1 |- 0,{-2^0.5/2});
            \path [name path=Ffinder] (A) -- ++(-60:1);
            \draw [rex,thick,dashed,name intersections={of=Ffinder and axis,by=F}] (A) -- (F) pic{point=below:F};
            \path [name path=P2finder] (V) -- ++($(V)-(P1)+(0,2^0.5)$);
            \draw [rex,thick,dashed,name intersections={of=P2finder and Ffinder,by=P2}] (V) -- (P2);

            \draw [grx,ultra thick,-stealth] (P2 |- 0,0) -- (P2) node[above right,black]{image} edge [out=-90,in=0,->,black,thin,shorten >=2pt] (P2 |- 0,0.1);
        \end{tikzpicture}
        \caption{Ray tracing a convex mirror.}
        \label{fig:rayTracingConvex}
    \end{figure}
    \begin{itemize}
        \item Upright, smaller, virtual image.
    \end{itemize}
\end{itemize}

\section{Lenses}
\begin{itemize}
    \item \marginnote{8/12:}\textbf{Double convex} (lens): A lens with two convex exterior surfaces. \emph{Also known as} \textbf{convex} (lens), \textbf{converging} (lens), \textbf{positive} (lens).
    \item \textbf{Double concave} (lens): A lens with two concave exterior surfaces. \emph{Also known as} \textbf{concave} (lens), \textbf{diverging} (lens), \textbf{negative} (lens).
    \item \textbf{Plano convex} (lens): A lens with one concave exterior surface and one flat exterior surface.
    \begin{figure}[h!]
        \centering
        \begin{tikzpicture}[scale=1.3]
            \footnotesize
            \draw [name path={axis}] (0,0) -- (6,0) node[right]{axis};

            \draw [very thin,dashed]
                (1,0) coordinate (C) node[circle,fill=pix,inner sep=1.5pt,label={below:$C$}]{}
                ++(30:-0.5) -- ++(30:0.5) -- node[below]{$R$} ++(30:{2^0.5}) coordinate (V)
                -- ++(30:1) coordinate (B)
                (V) -- ++(1.5,0) coordinate (D)
            ;

            \draw [rex,thick,postaction={decorate},decoration={markings,mark=at position 0.1 with \arrow{latex}},-latex,name path={ray}] (V) ++(-2.2,0) coordinate (A) -- (V) -- ++(3.5,-1.3);

            \draw [blx,thick] (2,1) arc[start angle=45,end angle=-45,radius={2^0.5}] -- node[below left,black,text height=1.5ex,text depth=0.25ex]{$1$} node[below right,black,text height=1.5ex,text depth=0.25ex]{$n$} node[below right,xshift=5mm,black,text height=1.5ex,text depth=0.25ex]{$1$} cycle;

            \path [name intersections={of=axis and ray,by=F}] (F) node [circle,fill=pix,inner sep=1.5pt,label={below:$F$}] {};
            \draw [very thin,|-|] (-0.15,0) -- node[left]{$y$} (-0.15,0 |- V);
            \draw [very thin,|-|] (V |- 0,1.3) -- node[above]{$f$} (F |- 0,1.3);
            \pic [draw,angle eccentricity=1.3,angle radius=6mm,pic text={$\theta_1$}] {angle=A--V--C};
            \pic [draw,angle eccentricity=1.2,angle radius=1cm,pic text={$\theta_1$}] {angle=D--V--B};
            \pic [draw,angle eccentricity=1.3,angle radius=1cm,pic text={$\theta_\text{def}$}] {angle=F--V--D};
            \pic [draw,angle eccentricity=1.41,angle radius=5mm,pic text={$\theta_2$},pic text options={fill=white,inner sep=1.5pt}] {angle=F--V--B};
        \end{tikzpicture}
        \caption{Plano convex lens analysis.}
        \label{fig:planoConvexLens}
    \end{figure}
    \begin{itemize}
        \item Clearly, $\theta_\text{def}=\theta_2-\theta_1$ and $\sin\theta_1=y/R$.
        \item If this is a \textbf{thin lens}, then we can invoke the small angle approximation and say that $\theta_1\approx y/R$.
        \item By Snell's Law, $\sin\theta_2=n\sin\theta_1$.
        \item Invoking the SAA again, we have that $\theta_2=n\theta_1$.
        \item It follows that
        \begin{align*}
            \theta_\text{def} &= \theta_2-\theta_1\\
            &= n\theta_1-\theta_1\\
            &= (n-1)\theta_1\\
            &= (n-1)\cdot\frac{y}{r}
        \end{align*}
        \item Therefore, since $\sin\theta_\text{def}=\frac{y}{f}$ (i.e., with the SAA $\theta_\text{def}=y/f$), we have that
        \begin{align*}
            f &= \frac{y}{\theta_\text{def}}\\
            &= \frac{y}{(n-1)\cdot y/R}\\
            &= \frac{R}{n-1}
        \end{align*}
        \item More commonly, we express this with
        \begin{equation*}
            \frac{1}{f} = (n-1)\cdot\frac{1}{R}
        \end{equation*}
    \end{itemize}
    \item \textbf{Thin lens}: A lens where $\theta$'s are small.
    \item If you substitute a convex lens for the plano convex lens, everything gets doubled.
    \begin{itemize}
        \item Importantly, we now have
        \begin{equation*}
            \frac{1}{f} = (n-1)\cdot\frac{2}{R}
        \end{equation*}
    \end{itemize}
    \item We can also generalize a convex lens to a lens with two convex sides of varying radii of curvature on its two sides.
    \item \textbf{Lens maker's equation}: The following formula.
    \begin{equation*}
        \frac{1}{f} = (n-1)\cdot\left( \frac{1}{R_1}+\frac{1}{R_2} \right)
    \end{equation*}
    \begin{itemize}
        \item Notice that if, for instance, the first side is planar, then $R_1=\infty$ and the $1/R_1$ term disappears.
        \item Sign convention: $R_1,R_2$ are positive for convex lenses and negative for concave lenses.
    \end{itemize}
    \item Remember that knowing the focal length of a lens allows us to ray trace.
    \begin{figure}[h!]
        \centering
        \begin{tikzpicture}[
            pics/point/.style args={#1:#2}{code={
                \coordinate (#2);
                \node [circle,fill=pix,inner sep=1.5pt,label={#1:${\color{black}#2}$}] {};
            }}
        ]
            \footnotesize
            \fill [blz] (3,1) arc[start angle=30,end angle=-30,radius=2] arc[start angle=210,end angle=150,radius=2];
            \draw [name path=axis] (0,0) -- (6,0) node[right]{axis};
            \draw [blx,thick] (3,1) arc[start angle=30,end angle=-30,radius=2] arc[start angle=210,end angle=150,radius=2];

            \draw [grx,ultra thick,-stealth] (1,0) node[above left,black]{object} edge [black,thin,->,out=90,in=180,shorten >=2pt] (1,0.55) -- ++(0,0.9) coordinate (P1);
            \draw [grx,ultra thick,-stealth] (5,0) node[below right,black]{image} edge [black,thin,->,out=-90,in=0,shorten >=2pt] (5,-0.55) -- ++(0,-0.9) coordinate (P2);

            \draw [rex,thick,-latex,postaction={decorate},decoration={markings,mark=at position 0.2 with \arrow{latex}},name path=parallel] (P1) -- (3.05,0.9) -- ($(3.05,0.9)!1.5!(P2)$);
            \draw [rex,thick,-latex,postaction={decorate},decoration={markings,mark=at position 0.2 with \arrow{latex}}] (P1) -- ($(P1)!1.3!(P2)$);
            \draw [rex,thick,-latex,postaction={decorate},decoration={markings,mark=at position 0.2 with \arrow{latex}},name path=focal] (P1) -- (2.95,-0.9) -- ($(2.95,-0.9)!1.6!(P2)$);

            \path [name intersections={of=parallel and axis}] (intersection-1) pic{point=above right:F};
            \path [name intersections={of=focal and axis}] (intersection-1) pic{point=below left:F};
        \end{tikzpicture}
        \caption{Convex lens rays.}
        \label{fig:lensRaysConvex}
    \end{figure}
    \begin{itemize}
        \item Central rays in thin lenses have negligible lateral displacement.
        \item An image through a convex lens will be an inverted real image.
    \end{itemize}
    \item \marginnote{8/13:}Ray tracing a concave lens.
    \begin{itemize}
        \item Upright, smaller, virtual image.
    \end{itemize}
    \item Note that the lens maker's equation used by \textcite{bib:YoungFreedman} is $1/f=(n-1)(1/R_1-1/R_2)$, along with a complicated sign convention.
    \begin{itemize}
        \item The textbook uses this formula because it follows from the study of thick lenses, with which many older textbooks start.
    \end{itemize}
    \item A concave mirror and a plano convex lens with a mirrored background image the same way.
    \begin{itemize}
        \item A convex lens mirrors the same way as both, but with the image on the other side of the lens as opposed to the same side of the apparatus.
    \end{itemize}
    \item Therefore, the lens equation is also analogously
    \begin{equation*}
        \frac{1}{s}+\frac{1}{s'} = \frac{1}{f}
    \end{equation*}
    \begin{itemize}
        \item However, this equation comes with the sign convention that $s$ is positive in front of the lens and negative behind the lens, and $s'$ is positive behind the lens and negative in front of the lens.
    \end{itemize}
    \item Similarly,
    \begin{equation*}
        m = \frac{-s'}{s}
    \end{equation*}
    for lenses, with the new sign convention.
    \item Note that the focal length is positive for convex lenses and negative for concave lenses.
    \item Just like mirrors can image images from other mirrors, lenses can image images from other lenses.
    \begin{itemize}
        \item You do it the same way, too --- one at a time.
    \end{itemize}
    \item Example: Suppose that you have two identical thin lenses, $\SI{15}{\centi\meter}$ apart, with focal length $\SI{10}{\centi\meter}$ each. Place an object $\SI{15}{\centi\meter}$ to the left of the left lens. How does it image?
    \begin{figure}[h!]
        \centering
        \begin{tikzpicture}[
            pics/point/.style args={#1:#2}{code={
                \coordinate (#2);
                \node [circle,fill=pix,inner sep=1.5pt,label={#1:${\color{black}#2}$}] {};
            }},
            every node/.style={black}
        ]
            \footnotesize
            \fill [blz] (2,1) arc[start angle=30,end angle=-30,radius=2] arc[start angle=210,end angle=150,radius=2];
            \fill [blz] (3.5,1) arc[start angle=30,end angle=-30,radius=2] arc[start angle=210,end angle=150,radius=2];
            \draw [name path=axis] (0,0) -- (6,0) node[right]{axis};
            \draw [blx,thick] (2,1)   node[above]{$1$} arc[start angle=30,end angle=-30,radius=2] node[below]{$f=\SI{10}{\centi\meter}$} arc[start angle=210,end angle=150,radius=2];
            \draw [blx,thick] (3.5,1) node[above]{$2$} arc[start angle=30,end angle=-30,radius=2] node[below]{$f=\SI{10}{\centi\meter}$} arc[start angle=210,end angle=150,radius=2];

            \pic at (1,0) {point=below:F_1};
            \pic at (3,0) {point=below:F_1};
            \pic at (2.5,0) {point=below:F_2};
            \pic at (4.5,0) {point=below:F_2};

            \draw [grx,ultra thick,-stealth] (0.5,0) -- ++(0,0.8) node[above right=-2pt,black]{object} edge [black,thin,->,out=-90,in=0,shorten >=2pt] (0.5,0.4);
            \draw [grx,ultra thick,-stealth] (4.1,0) -- ++(0,-0.64) node[below right=-2pt,black]{image} edge [black,thin,->,out=90,in=0,shorten >=2pt] (4.1,-0.3);

            \draw (0,-1.7) -- node[below=6mm]{$\si{\centi\meter}$} ++(6,0);
            \foreach \x [evaluate=\x as \num using int(\x*10)] in {0,0.5,...,6} {
                \draw (\x,-1.6) -- ++(0,-0.2) node[below]{$\num$};
            }
        \end{tikzpicture}
        \caption{Imaging two lenses.}
        \label{fig:twoLensImage}
    \end{figure}
    \begin{itemize}
        \item Image location:
        \begin{itemize}
            \item For lens 1, we have $f=\SI{10}{\centi\meter}$ and $s=\SI{15}{\centi\meter}$, so $s'=\SI{30}{\centi\meter}$ by the lens equation.
            \item For lens 2, we still have $f=\SI{10}{\centi\meter}$ but now we have have $s=\SI{-15}{\centi\meter}$, so $s'=\SI{6}{\centi\meter}$ by the lens equation.
        \end{itemize}
        \item Magnification:
        \begin{align*}
            m_\text{total} &= m_1m_2\\
            &= \left( -\frac{\SI{30}{\centi\meter}}{\SI{15}{\centi\meter}} \right)\left( -\frac{\SI{6}{\centi\meter}}{\SI{-15}{\centi\meter}} \right)\\
            &= -\frac{4}{5}
        \end{align*}
        \item Thus, the total image is inverted, $4/5$ times the size of the original, and $\SI{6}{\centi\meter}$ to the right of the rightmost lens.
    \end{itemize}
    \item Mirrors and lenses have limitations.
    \begin{itemize}
        \item Thus, it does make sense to use multiple mirrors/lenses in some circumstances.
        \item Compensating for the paraxial approximation: When you need a sharper focus in good cameras.
        \item Compensating for the dependence of $n$ on frequency: Have any single lens do less work.
    \end{itemize}
\end{itemize}

\section{Chapter 34: Geometric Optics}
\emph{From \textcite{bib:YoungFreedman}.}
\begin{itemize}
    \item \marginnote{8/16:}\textbf{Point object}: A theoretical object consisting only of a zero-dimensional point.
    \item \textbf{Extended object}: An object that is not a point object.
    \item \textbf{Diffuse} (surface): A surface which scatters incoming rays in uncorrelated directions.
    \item \textbf{Virtual image}: An image for which the outgoing rays do not pass through the image point.
    \item \textbf{Real image}: An image for which the outgoing rays do pass through the image point.
    \item "When the center of curvature C is on the same side as the outgoing light, the radius of curvature is positive; otherwise, it is negative" \parencite[1112]{bib:YoungFreedman}.
    \item \textbf{Lateral magnification}: The ratio of image height to object height.
    \item \textbf{Erect} (image): An image that is oriented in the same direction as the object.
    \begin{itemize}
        \item Such as when an image arrow points in the same direction as the object arrow.
    \end{itemize}
    \item \textbf{Inverted} (image): An image that is oriented in the opposite direction relative to the object.
    \begin{itemize}
        \item Such as when an image arrow points in the opposite direction as the object arrow.
    \end{itemize}
    \item \textbf{Spherical aberration}: The smearing out of the image "point" from a zero-dimensional point due to the approximations made in deriving the mirror equation.
    \item If you remove part of a spherical mirror, a full image is still formed, just a dimmer one. This is because light rays from any object point reflect off of \emph{all} parts of the mirror to converge to the image point. However, without all of the mirror, fewer light rays will be reflected, but no paths will be entirely eliminated.
    \item Measure $s$ and $s'$ to where the mirror intersects the optic axis.
    \item \textbf{Virtual focal point}: The focal point of a convex mirror, lying behind it.
    \item "When the object distance $s$ is positive, a convex mirror \emph{always} forms an erect, virtual, reduced, reversed image" \parencite[1120]{bib:YoungFreedman}.
\end{itemize}

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