Weiter

4 files changed, 383 insertions, 109 deletions

diff --git a/fewstereveson.tex b/fewstereveson.tex
new file mode 100644
index 0000000..fdda174
--- /dev/null
+++ b/fewstereveson.tex
@@ -0,0 +1,11 @@
+\chapter{A Lower Bound on the Renormalized Energy Density by Fewster and Eveson}
+\label{chapter:fewster-eveson}
+
+\section{Test}
+
+Goal: make \cite{Fewster1998} rigorous based on the previous chapter
+
+\chapterbib
+\cleardoublepage
+
+% vim: syntax=mytex
diff --git a/main.tex b/main.tex
index d3badec..d09ac92 100644
--- a/main.tex
+++ b/main.tex
@@ -1,14 +1,17 @@
 \input{preamble}
-%\includeonly{standard}
+\includeonly{stresstensor,fewstereveson}
 \begin{document}
 \frontmatter
 \include{titlepage}
 \include{contents}
 \include{intro}
 \mainmatter
+\include{analytic}
 \include{distributions}
 \include{second}
 \include{standard}
+\include{stresstensor}
+\include{fewstereveson}
 \include{samplesection}
 \appendix
 \include{sampleappendix}
diff --git a/preamble.tex b/preamble.tex
index 37068a4..4a88ef7 100644
--- a/preamble.tex
+++ b/preamble.tex
@@ -72,6 +72,7 @@
 \SetCiteCommand{\textcite}
 \renewcommand*{\mkcitation}[1]{\trailingtext{---\mkbibnamegiven{#1}}}
 \renewcommand*{\mkccitation}[1]{\trailingtext{---#1}}
+\renewcommand*{\mkblockquote}[4]{\small#1#2#4#3}
 
 % ---------- selnolig
 \nolig{tz}{t|z}
@@ -231,12 +232,21 @@
 % TODO Why does this not work?
 \renewcommand{\Re}{\operatorname{Re}}
 \renewcommand{\Im}{\operatorname{Im}}
+\newcommand{\Imag}{\operatorname{Im}}
 
 % emphasis for defined terms
 \newcommand*{\defn}[1]{\textbf{\textit{#1}}}
 
 \newcommand*{\ts}[1]{\textnormal{#1}} % textual subscript
 
+% monomials
+\newcommand{\Mon}{\operatorname{Mon}}
+
+% Infinitesimal Weyl algebra
+\newcommand{\WeylAlg}{\mathcal{W}}
+\newcommand{\weylannihilator}{A}
+\newcommand{\weylcreator}{\weylannihilator^\dagger}
+
 % Fourier transformation
 % ----------------------
 \newcommand*{\ft}[1]{\hat{#1}}
diff --git a/stresstensor.tex b/stresstensor.tex
index 4cdc321..39c45cd 100644
--- a/stresstensor.tex
+++ b/stresstensor.tex
@@ -1,6 +1,10 @@
 \chapter{Construction of the Stress Tensor of a Free Scalar Quantum Field}
 \label{chapter:stress-tensor}
 
+\begin{center}
+  \emph{Note: Work in Progress}
+\end{center}
+
 \begin{equation*}
   H = \tfrac{1}{2} \parens*{(\partial_t \phi)^2 + \abs{\nabla_{\!\!\symbfit{x}} \phi}^2 + m^2 \phi^2} 
 \end{equation*}
@@ -10,7 +14,7 @@ we will have gained the ability to rigorously define
 arbitrary renormalized products of the free field and its derivatives
 as a densely defined quadratic-form valued tempered distribution,
 which on the dense subspace of the smooth vectors of the Hamiltonian
-is realized by essentially self-adjoint operators.
+is realized by essentially selfadjoint operators.
 
 \section{Choosing Conventions and Fixing Notation}
 \label{section:conventions}
@@ -67,7 +71,9 @@ as a service to the reader.
       \parens[\big]{a(g) a(g) \psi} {}_{n-2} (k_1, \ldots, k_{n-2}) = \\
       \sqrt{n} \sqrt{n-1} \int_M \int_M \! \bar{g}(p_1) \bar{g}(p_2) \, \psi_n (p_1,p_2,k_1, \ldots, k_{n-s}) \, d\Omega_m(p_1) d\Omega_m(p_2) \\
     \end{multline*}
+    For later use, we also give the action of an $s$-fold product of annihilation operators:
     \begin{multline}
+      \label{equation:multiple-annihilation-operators}
       \parens[\big]{a(g_1) \cdots a(g_s) \psi} {}_{n-s} (k_1, \ldots, k_{n-s}) = 
       \sqrt{n (n-1) \cdots (n-s+1)} \cdot {} \\
       \cdot \int_M \!\! d\Omega_m(p_1) \cdots \!\! \int_M \!\! d\Omega_m(p_s) \ \bar{g_1}(p_1) \cdots \bar{g_s}(p_s) \ \psi_n (k_1, \ldots, k_{n-2},p_1,\ldots,p_s)
@@ -114,7 +120,7 @@ as a service to the reader.
       \innerp[\big]{\psi'}{a(p)^\dagger \psi} =
       \innerp[\big]{a(p) \psi'}{\psi}
     \end{equation*}
-    Define ...
+    TODO(Explain why creation op at a point is actually not an op but a QF)
     \begin{equation*}
       a(p_1)^\dagger \cdots a(p_s)^\dagger a(p_{s+1}) \cdots a(p_r)
     \end{equation*}
@@ -122,7 +128,6 @@ as a service to the reader.
       \innerp[\big]{\psi'}{a(p_1)^\dagger \cdots a(p_s)^\dagger a(p_{s+1}) \cdots a(p_r) \psi}
       = \innerp[\big]{a(p_1) \cdots a(p_s) \psi'}{a(p_{s+1}) \cdots a(p_r) \psi}
     \end{equation*}
-    abc
 \end{itemize}
 
 \begin{equation*}
@@ -175,25 +180,181 @@ into annihilation operators on the left.
   \end{multlined}
 \end{align*}
 
-\begin{proposition}{}{}
-  asdf
-\end{proposition}
+\section{Normal Ordering}
+% The Renormalization Map?
+
+%\blockcquote{Wick1950}{%
+ %\textelp{} we then proceed to rearrange such a product so as to carry all
+ %creation operators to the left of all destruction operators \textelp{}. The
+ %main problem to be solved in carrying out this idea is one of algebraic
+ %technique \textelp{} 
+%}
+
+The process of renormalizing a product of field operators
+has the purpose of discarding infinite constants
+that occur when calculating the vacuum expectation value.
+(TODO: present physicists way of introducing normal ordering)
+
+Now let us extract the algebraic essence of the situation.
+The objects of our calculations are the field operators $\Phi(f)$,
+but it does not matter that these are realized as linear maps on Fock space.
+Forming the product $\Phi(f)\Phi(g)$ might as well be done purely symbolically,
+since none of what we want to do depends on this product
+having the meaning of operator composition;
+and similar for the other two arithmetic operations,
+addition and multiplication with a complex scalar.
+Thus we should calculate with abstract objects $\Phi(f)$ labeled by Hilbert space vectors $f \in \hilb{H}$.
+Considering that here $\Phi$ carries no meaning, we can use the label $f$ itself to represent the object.
+
+This leads us to consider formal expressions
+\begin{equation*}
+  \alpha^{(0)} e + \sum_{i} \alpha^{(1)}_i z^{(1)}_i + \sum_{j,k} \alpha^{(2)}_{j,k} z^{(2)}_j z^{(2)}_k + \cdots
+\end{equation*}
+where the $z^{(1)}_i,z^{(2)}_j,z^{(2)}_k,\ldots$ are in $\hilb{H}$,
+the $\alpha^{(0)},\alpha^{(1)}_i,\alpha^{(2)}_{j,k},\ldots$ are complex numbers,
+of which only finitely many are nonzero,
+and $e$ is a special object representing an empty product of $z$'s.
+To make this mathematically precise:
+we are speaking of the noncommutative associative algebra over $\CC$
+freely generated by the elements of $\hilb{H}$.
+The unit of the algebra is $e$.
+
+This in not quite what we want
+TODO(explain need for commutation relations)
+By abstract algebra, this is viable
+by forming the quotient of the free algebra
+with respect to the two-sided ideal
+generated by all elements $zz' - z'z = i \Imag \innerp{z}{z'} \, e$,
+where $z,z' \in \hilb{H}$.
+
+\begin{definition}{Infinitesimal Weyl Algebra}{}
+  Let $\hilb{H}$ be a complex Hilbert space.
+  The \emph{infinitesimal Weyl algebra} $\WeylAlg(\hilb{H})$ over $\hilb{H}$
+  is the noncommutative associative algebra over $\CC$
+  generated by the elements of $\hilb{H}$, with the relations
+  \begin{equation*}
+    zz' - z'z = i \Imag \innerp{z}{z'} \, e \qquad z,z' \in \hilb{H},
+  \end{equation*}
+  where $e$ is the unit of the algebra.
+\end{definition}
+
+TODO(introduce $\Phi$ as representation of $\WeylAlg$)
+
+\begin{definition}{Annihilator and Creator}{}
+  Suppose $\WeylAlg$ is the infinitesimal Weyl algebra
+  over some complex Hilbert space $\hilb{H}$.
+  For all $z \in \hilb{H}$,
+  we define, as elements of $\WeylAlg$, the \emph{annihilator}
+  \begin{equation*}
+    \weylannihilator(z) = \frac{1}{\sqrt{2}} \parens{z+iz},
+  \end{equation*}
+  and the \emph{creator}
+  \begin{equation*}
+    \weylcreator(z) = \frac{1}{\sqrt{2}} \parens{z-iz}.
+  \end{equation*}
+\end{definition}
+
+\begin{equation*}
+  z = \frac{1}{\sqrt{2}} \parens[\big]{\weylannihilator(z) + \weylcreator(z)}
+\end{equation*}
+
+A \emph{monomial} in the Weyl algebra $\WeylAlg$ over a complex Hilbert space $\hilb{H}$ is an element of the form
+$z_1 \cdots z_r \in \WeylAlg$, where $r \ge 0$ and $z_1,\ldots,z_r$ are in $\hilb{H}$.
+We allow $r=0$, meaning that the unit $e$ is a monomial.
+The set of all monomials in $\WeylAlg$ is denoted $\Mon(\WeylAlg)$.
+
+\begin{definition}{Normal Ordering}{}
+  Let $\hilb{H}$ be a complex Hilbert space and $\WeylAlg$ its associated infinitesimal Weyl algebra.
+  The mapping $\normord{\,\,}$, defined by
+  \begin{gather}
+    \Mon(\WeylAlg) \longrightarrow \WeylAlg \nonumber\\
+    \label{equation:normal-ordering}
+    \normord{z_1 \!\cdots z_r} =
+    \frac{1}{\sqrt{2^r}}
+    \sum_{I \subset \braces{1,\ldots,r}} \,
+    \prod_{i \in I\vphantom{\lbrace\rbrace}} \weylcreator(z_i)
+    \prod_{\mathclap{j \in \braces{1,\ldots,r} \setminus I}} \weylannihilator(z_j),
+  \end{gather}
+  is called the \emph{normal} (or \emph{Wick}) \emph{ordering} on $\hilb{H}$.
+  A monomial $z_1 \cdots z_r \in \Mon(\WeylAlg)$ is said to be in \emph{normal} (or \emph{Wick}) \emph{order},
+  if $\normord{z_1 \cdots z_r} = z_1 \cdots z_r$.
+\end{definition}
+
+The products in~\eqref{equation:normal-ordering} are well defined
+because creators commute with creators and annihilators commute with annihilators.
+Since an empty product equals, per convention, the neutral element of multiplication, which here is the unit $e$,
+the formula makes sense even for $r=0$ and asserts that $\normord{e} = e$.
+The cases $r=1$ and $r=2$ read
+\begin{align*}
+  \normord{z} &=
+  \frac{1}{\sqrt{2}} \parens[\big]{\weylannihilator(z) + \weylcreator(z)} = z \\
+  \normord{z_1 z_2} &= \frac{1}{2}
+  \parens[\big]{\weylannihilator(z_1) \weylannihilator(z_2) + \weylannihilator(z_1) \weylcreator(z_2)
+  + \weylcreator(z_1) \weylannihilator(z_2) + \weylcreator(z_1) \weylcreator(z_2) } 
+\end{align*}
+This suggests that the normally ordered product $\normord{z_1 \!\cdots z_r}$
+is symmetric in $z_1,\ldots,z_n$. This is in fact true, and becomes evident
+if one brings~\eqref{equation:normal-ordering} into the equivalent form
+\begin{gather}
+  \label{equation:normal-ordering-symmetric}
+  \normord{z_1 \!\cdots z_r} =
+  \frac{1}{\sqrt{2^r}}
+  \sum_{\sigma \in S_r}
+  \sum_{s=0\vphantom{S}}^{r}
+  \frac{1}{s!(r-s)!}
+  \prod_{i=1\vphantom{S}}^{s} \weylcreator(z_{\sigma(i)})
+  \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} \weylannihilator(z_{\sigma(j)}) 
+\end{gather}
+by basic combinatorial arguments (TODO: further explanation?).
+In~\cite{Klein1973}, the factor $\frac{1}{s!(r-s)!}$ is erroneously missing.
+
+
+\begin{equation*}
+  E(\normord{z_1 \!\cdots z_r}) = 0 \qquad \forall z_1,\ldots,z_r \in \hilb{H}, r \ge 1
+\end{equation*}
+\begin{equation*}
+  E\parens[\Big]{\prod_{i=1\vphantom{S}}^{s} \weylcreator(z_i)
+  \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} \weylannihilator(z_j)} = 0
+  \qquad \forall z_1,\ldots,z_r \in \hilb{H}, r \ge 1, 1 \le s \le r
+\end{equation*}
+
 
-\subsubsection{Linear Differential Operators and their Formal Adjoint}
+The normal ordered product is supposed to represent the identical quantity as before ordering,
+except that we have adjusted our point of reference, such that measurements yield finite results.
+It is therefore \emph{physically reasonable} that the commutation relations
+of the normal ordered product with the field are analogous.
+As it turns out, this additional property makes the construction of normal ordering
+\emph{mathematically unique}.
+
+\begin{theorem}{Uniqueness of the Normal Order}{}
+  Normal ordering is the unique mapping $N : \Mon(\WeylAlg) \to \WeylAlg$ such that
+  \begin{gather*}
+    E\parens[\big]{N(z_1 \!\cdots z_r)} = 0 \\
+    \bracks{N(z_1 \!\cdots z_r), z'} =
+    \sum_{s=1}^{r} \bracks{z_s,z'} N(z_1 \!\cdots \widehat{z_i} \cdots z_r)
+  \end{gather*}
+  for all $z_1,\ldots,z_r,z' \in \hilb{H}$ and all $r \ge 1$.
+\end{theorem}
+
+%\begin{theorem}{}{}
+  %The normal ordering is the renormalization with respect to the normal vacuum.
+%\end{theorem}
+
+\section{Linear Differential Operators and their Formal Adjoint}
 
 Before we turn to the problem of defining renormalized products of a quantum field and its derivatives
 we must clarify what is meant mathematically by the derivative of a field.
 For this, we recall that in Wightmans approach to quantum field theory,
 a quantum field $\varphi$ on a spacetime manifold $M$ is modeled by an operator valued tempered distribution,
-that is a mapping that assigns to each (Schwatz class) test function $f$ on $M$ an unbounded operator $\varphi(f)$
+that is a mapping that assigns to each (Schwartz class) test function $f$ on $M$ an unbounded operator $\varphi(f)$
 in the Fock space xxx over some Hilbert space $\hilb{H}$, such that for each fixed pair of states $\psi,\psi'$
 the mapping
 \begin{equation*}
   \schwartz{M} \to \CC, \quad
   f \mapsto \innerp{\psi'}{\varphi(f) \psi}
 \end{equation*}
-is a (scalar-valued) tempered distibution on $M$.
-It is well known that tempered distibutions have partial derivatives of any order.
+is a (scalar-valued) tempered distribution on $M$.
+It is well known that tempered distributions have partial derivatives of any order.
 Suppose we work with $M = \RR^d$ for simplicity,
 and let $\partial_i$ denote the partial derivative with respect to the $i$-th coordinate.
 Then a general \emph{linear differential operator with constant coefficients} on $M$ looks like
@@ -242,17 +403,37 @@ The operator corresponding to $D$ in Fourier space is the multiplication operato
 
 
 Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients.
-\begin{equation*}
-  \normord{D_1 \varphi(f) \cdots D_r \varphi(f)}
-\end{equation*}
+\begin{gather}
+  \label{equation:renormalized-product}
+  \normord{D_1 \varphi(f) \cdots D_r \varphi(f)} =
+  \frac{1}{\sqrt{2^r}}
+  \sum_{\sigma \in S_r}
+  \sum_{s=0\vphantom{S}}^{r}
+  \frac{1}{s!(r-s)!}
+  \prod_{i=1\vphantom{S}}^{s} a^\dagger(D^\dagger_{\sigma(i)}f)
+  \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} a(D^\dagger_{\sigma(j)}f) 
+\end{gather}
 
 \section{Renormalized Products of the Free Field and its Derivatives}
 
-
+For any given test function $f \in \schwartz{M}$ the renormalized product $\normord{D_1 \varphi(f) \cdots D_r \varphi(f)}$
+is well defined as a Fock space operator, but the product is \emph{not} an operator-valued distribution, unless $r=1$.
+This is because it has a multi-linear dependence on the test function.
+Conceptually, one wishes to treat any physical quantity derived from the quantum field
+on the same footing as the field itself.
+One construction to obtain an operator-valued distribution,
+is described in~\cite{Segal1969}, \cite{Segal1970} and \cite{Klein1973}.
+The idea is to take the limit $f \to \delta_x$, where $\delta_x$ is the delta distribution supported at a point $x \in M$,
+resulting in the renormalized product at the point $x$, now just a quadratic form,
+which is then smeared with \emph{one} instance of $f$.
+As we shall see,
+this approach incurs significant technical difficulties.
 
 \begin{lemma}{Integral Representation of the Renormalized Product}{renormalized-product-integral-representation}  
-  Let $\varphi$ be a free quantum field.
-  Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients. Then we have for all states $\psi,\psi'$
+  Let $\varphi$ be the free scalar quantum field with mass parameter $m > 0$.
+  Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients.
+  Then, for arbitrary Schwartz functions $f \in \schwartz{M}$ and Fock states $\psi,\psi' \in \BosonFock{L^2(X_m^+,d\Omega_m)}$,
+  we have
   \begin{equation*}
     \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} =
     \int dp_1 \!\cdots dp_r
@@ -271,41 +452,44 @@ Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex)
     \ \overline{\psi'_m(k_1,\ldots,k_{m-s},p_1,\ldots,p_s)}
     \ \psi_n(k_1,\ldots,k_{n-(r-s)},p_{s+1},\ldots,p_r)
   \end{multline*}
-  where $\chi(p) = \overline{\ft{f}(p)} / \ft{f}(p)$ and
   \begin{equation*}
-    P_s(p_1,\ldots,p_r) =
-    \frac{1}{\sqrt{2^r}} \sum_{\sigma \in S_r}
-    \ft{D}_{\sigma(1)}(p_1) \cdots \ft{D}_{\sigma(s)}(p_s)
-    \overline{\ft{D}_{\sigma(s+1)}(p_{s+1}) \cdots \ft{D}_{\sigma(r)}(p_r)}.
+    \text{where} \quad \chi(p) = \begin{cases*}
+      \overline{\ft{f}(p)} / \ft{f}(p) & if $\ft{f}(p) \ne 0$, \\
+      1 & otherwise,
+    \end{cases*}
   \end{equation*}
+  \begin{multline*}
+    \text{and} \quad P_s(p_1,\ldots,p_r) =
+    \frac{1}{\sqrt{2^r}}
+    \frac{1}{s!(r-s)!}
+    \sum_{\sigma \in S_r}
+    \ft{D}_{\sigma(1)}(p_1) \cdots \ft{D}_{\sigma(s)}(p_s) \hspace{1.5cm} \\[-1.5ex]
+    \cdot \overline{\ft{D}_{\sigma(s+1)}(p_{s+1}) \cdots \ft{D}_{\sigma(r)}(p_r)}.
+  \end{multline*}  
 \end{lemma}
 
-In the special case that $D_1 = \cdots = D_n = D$ we have
-\begin{equation*}
-  P_s(p_1,\ldots,p_r) =
-  \sqrt{2^r}
-  \ft{D}(p_1) \cdots \ft{D}(p_s)
-  \overline{\ft{D}(p_{s+1}) \cdots \ft{D}(p_r)}.
-\end{equation*}
-For squares, that is $r=2$
-\begin{equation*}
-  P_s(p_1,p_2) = \begin{cases}
-    2 \, \ft{D}(p_1)\ft{D}(p_2) & s=0 \\
-    2 \, \ft{D}(p_1)\overline{\ft{D}(p_2)} & s=1 \\
-    2 \, \overline{\ft{D}(p_1)\ft{D}(p_2)} & s=2
-  \end{cases}
-\end{equation*}
+TODO(Note about the remaining dependence of $K$ on $f$.)
 
 \begin{myproof}[lemma:renormalized-product-integral-representation]
-\begin{multline*}
-  \innerp{\psi'}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} =
-  \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}
-  \frac{1}{\sqrt{2^r}} \sum_{s=0}^{r} \sum_{\sigma \in S_r} \\
-  \big\langle
-  a(ED_{\sigma(1)}^{\dagger}f) \cdots a(ED_{\sigma(s)}^{\dagger}f) \psi_m,
-  a(ED_{\sigma(s+1)}^{\dagger}f) \cdots a(ED_{\sigma(r)}^{\dagger}f) \psi_n
-  \big\rangle
-\end{multline*}
+  From equation~\eqref{equation:renormalized-product},
+  applying the definition of the Fock space inner product,
+  and moving all creation operators to the left hand side,
+  we get
+  \begin{multline*}
+    \innerp{\psi'}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} =
+    \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}
+    \frac{1}{\sqrt{2^r}} \sum_{s=0}^{r} \sum_{\sigma \in S_r} \frac{1}{s!(r-s)!} \\
+    \cdot \big\langle
+    a(ED_{\sigma(1)}^{\dagger}f) \cdots a(ED_{\sigma(s)}^{\dagger}f) \psi_m,
+    a(ED_{\sigma(s+1)}^{\dagger}f) \cdots a(ED_{\sigma(r)}^{\dagger}f) \psi_n
+    \big\rangle
+  \end{multline*}
+  Notice that the inner product in the second line
+  can only be nonzero if the particle numbers match up
+  after the application of the annihilation operators in each argument,
+  that is if $m-s=n-(r-s)$.
+  With~\eqref{equation:multiple-annihilation-operators}
+  this expression may be further expanded into
   \begin{gather*}
     \sqrt{m(m-1) \cdots (m-s+1)}
     \sqrt{n(n-1) \cdots (n-(r-s)+1)}
@@ -317,14 +501,42 @@ For squares, that is $r=2$
     \ \overline{ED_{\sigma(s+1)}^{\dagger}f(p_{s+1}) \cdots ED_{\sigma(r)}^{\dagger}f(p_r)}
     \ \psi_n(k_1,\ldots,k_{n-(r-s)},p_{s+1},\ldots,p_r)
   \end{gather*}
+  Now recall that $E$ stands for Fourier transformation (followed by restriction to the mass shell)
+  and that in Fourier space the linear differential operator $D^\dagger$ corresponds to a
+  multiplication with the function $\hat{D}$, so that
+  \begin{equation*}
+    ED_{\sigma(i)}^{\dagger}f(p_i) = \hat{D}_{\sigma(i)}(p_i) \cdot \ft{f} (p_i) 
+    \qquad \forall i
+  \end{equation*}
+  By Fubini’s Theorem, the we may interchange the integrals with respect to the variables $p_i$
+  with the $k$-integrals.
+  This allows us to move all factors involving $\ft{f}$ in front of the $k$-integrals.
+  Finally, we introduce the $\chi$s through the substitution $\overline{\ft{f}} = \chi \ft{f}$,
+  and combine all terms depending on $\sigma$ into $P_s$.
 \end{myproof}
 
-The following assertion is key
+In the special case that $D_1 = \cdots = D_n = D$ we have
+\begin{equation*}
+  P_s(p_1,\ldots,p_r) =
+  \frac{1}{\sqrt{2^r}} \parens*{r \atop s\vphantom{y}}
+  \ft{D}(p_1) \cdots \ft{D}(p_s)
+  \overline{\ft{D}(p_{s+1}) \cdots \ft{D}(p_r)}.
+\end{equation*}
+For squares, that is $r=2$
+\begin{equation*}
+  P_s(p_1,p_2) = \begin{cases}
+    \tfrac{1}{2} \, \ft{D}(p_1)\ft{D}(p_2) & s=0 \\
+    \phantom{\tfrac{1}{2}} \, \ft{D}(p_1)\overline{\ft{D}(p_2)} & s=1 \\
+    \tfrac{1}{2} \, \overline{\ft{D}(p_1)\ft{D}(p_2)} & s=2
+  \end{cases}
+\end{equation*}
+
+The following assertion is key to realizing the idea of taking the limit $f \to \delta_x$.
 
 \begin{lemma}{}{integral-kernel-h-bound}
   In the setting of \cref{lemma:renormalized-product-integral-representation},
   there exist a constant $C$, and a positive integer $l$,
-  such that for arbitary states $\psi,\psi' \in xxx$,
+  such that for arbitrary states $\psi,\psi' \in \BosonFock{\hilb{H}}$,
   and test functions $f \in \schwartz{M}$,
   the function $K_{\psi'\!,\psi}$ is integrable (that is, $L^1$)
   and satisfies the $H$-bound
@@ -334,7 +546,14 @@ The following assertion is key
   \end{equation*}
 \end{lemma}
 
-\begin{myproof}
+The Hamilton operator $H$ acts on $n$-particle states $\psi_n$
+by multiplication with $\omega(p_1)$
+In the following proof it will we convenient to use the abbreviation
+\begin{equation*}
+  \omega(p_1,\ldots,p_s) = \omega(p_1) + \cdots + \omega(p_s)
+\end{equation*}
+
+\begin{myproof}[lemma:integral-kernel-h-bound]
   We have to find an estimate for
   \begin{equation*}
     \norm{K_{\psi'\!,\psi}}_1 =
@@ -349,16 +568,16 @@ The following assertion is key
     \quad \text{and} \quad 
     n(n-1) \cdots (n-(r-s)+1) \le n^r,
    \end{equation*}
-   and finally reorder the integration with Fubini’s theorem
+   and finally reorder the integration with Fubini’s Theorem
    to obtain
   \begin{equation}
-    \label{first-estimate}
+    \label{equation:first-estimate}
     \begin{multlined}[c]
       \norm{K_{\psi'\!,\psi}}_1 \le
       \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \sum_{s=0}^{r}
       \delta_{m-s}^{n-(r-s)} \sqrt{m^r n^r} \\
-      \hspace{2.5cm} \cdot \abs*{\int \!dk \int \!dp'\! \int \!dp \,
-      P_s(p',p) \, \psi'_m(k,p') \, \psi_n(k,p)},
+      \hspace{2.5cm} \cdot \int \!dk \int \!dp'\! \int \!dp \,
+      \abs*{P_s(p',p) \, \psi'_m(k,p') \, \psi_n(k,p)},
     \end{multlined}
   \end{equation}
   where we have used the abbreviations
@@ -369,6 +588,8 @@ The following assertion is key
     dk &= dk_1 \cdots dk_{m-s}
     \quad \text{and so on.}
   \end{align*}
+   For following discussion we will assume $n-(r-s)=m-s$.
+
   Observe that $P_s(p_1,\ldots,p_r)$ is a (complex) polynomial
   in the $4r$ variables $p_i^\mu$, $i=1,\ldots,r$, $\mu=0,\ldots,3$.
   Its degree is given by
@@ -377,54 +598,77 @@ The following assertion is key
   \end{equation*}
   that is the sum of the highest orders of differentiation
   occurring in each of the operators $D_1, \ldots, D_r$.
-  There is no reason to expect arbitary states $\psi,\psi'$ to temper fast enough
+  There is no reason to expect arbitrary states $\psi,\psi'$ to temper fast enough
   to counteract this polynomial growth.
-  Thus, the integral in \cref{first-estimate} will not converge, in general.
-  However, if $\psi$ lies in the domain of $H^l$ for some positive integer $l$,
-  then we can be sure that $(1+H)^l \psi$ is square integrable, and we have
-  \begin{equation*}
-    \psi_n(k,p) = \parens[\big]{1+\omega(k,p)} {}^{-l} (1+H)^l \psi_n(k,p)
-  \end{equation*}
-  \begin{equation*}
-    (1+H)^l psi
-  \end{equation*}
-  \begin{equation*}
-    \abs*{\int \!dk \int \!dp'\! \int \!dp \, F(k,p',p) \, G'(k,p') \, G(k,p)}
-  \end{equation*}
-  where
-  \begin{align}
-    F(k,p',p) &= \parens[\big]{1+\omega(k,p')} {}^{-l} \parens[\big]{1+\omega(k,p)} {}^{-l} P_s(p',p) \\
-    G'(k,p') &= \sqrt{m^r} (1+H)^l \psi_m(k,p') \\
-    G(k,p) &= \sqrt{n^r} (1+H)^l \psi_n(k,p)
-  \end{align}
+  Thus, the integral in~\eqref{equation:first-estimate} will not converge, in general.
+  However, if $\psi$ lies in the domain of $H^a$ for some positive integer $a$,
+  then we can be sure that $(1+H)^a \psi$ is square integrable, and we have
+  \begin{align*}
+    \psi_n(k,p) &= \parens[\big]{1+\omega(k,p)} {}^{-a} (1+H)^a \psi_n(k,p) \\
+    \psi'_m(k,p') &= \parens[\big]{1+\omega(k,p')} {}^{-a} (1+H)^a \psi'_m(k,p')
+  \end{align*}
+  We use this to rewrite the integral part of~\eqref{equation:first-estimate} as follows:
+  \begin{equation}
+    \label{equation:rewritten-integral}
+    \int \!dk \int \!dp'\! \int \!dp \, \abs*{F(k,p',p) \, G'(k,p') \, G(k,p)},
+  \end{equation}
+  where we have introduced the functions
+  \begin{align*}
+    F(k,p',p) &= \parens[\big]{1+\omega(k,p')} {}^{-a} \parens[\big]{1+\omega(k,p)} {}^{-a} P_s(p',p) \\
+    G'(k,p') &= \sqrt{m^r} (1+H)^a \psi'_m(k,p') \\
+    G(k,p) &= \sqrt{n^r} (1+H)^a \psi_n(k,p).
+  \end{align*}
+  Next, we derive an estimate of~\eqref{equation:rewritten-integral}.
+  By Cauchy-Schwarz, we have
   \begin{equation*}
-    \abs*{\int dp G(k,p) F(k,p',p)}^2
+    \abs*{\int dp \, \abs{F(k,p',p)G(k,p)}}^2
     \le \int dp \abs{F(k,p',p)}^2
     \cdot \int dp \abs{G(k,p)}^2
   \end{equation*}
-  \begin{equation*}
-    \int dp' \abs*{\int dp G(k,p) F(k,p',p)}^2
+  and this implies
+  \begin{equation}
+    \label{equation:estimate1}
+    \int dp' \abs*{\int \!dp \, \abs{F(k,p',p)G(k,p)}}^2
     \le \int dp \abs{G(k,p)}^2
-    \sup_{k} \norm{F(k,\cdot,\cdot)}_2^2
-  \end{equation*}
+    \cdot \int dp'\! \int dp \abs{F(k,p',p)}^2.
+  \end{equation}
+  Notice that the second factor is the $L^2$ norm of $F$ with its first argument held fixed:
+  \begin{equation}
+    \label{equation:norm}
+    \norm{F(k,\cdot,\cdot)}_2^2
+    = \int dp'\! \int dp \abs{F(k,p',p)}^2
+  \end{equation}
+  By another application of Cauchy-Schwarz, and using~\eqref{equation:estimate1} and~\eqref{equation:norm}, we obtain
   \begin{align*}
-    &\quad \abs*{\int \!dk \int \!dp'\! \int \!dp \, F(k,p',p) \, G'(k,p') \, G(k,p)} \\
-    &\le \int \!dk \int \!dp' \abs{G'(k,p')} \abs*{\int \!dp \, F(k,p',p) \, G(k,p)} \\
-    &\le \norm{G'}_2 \parens*{\int dk \int dp' \abs*{\int dp G(k,p) F(k,p',p)}^2} \\
+    \text{\eqref{equation:rewritten-integral}} \,
+    &= \int \!dk \int \!dp' \abs{G'(k,p')} \int \!dp \, \abs*{F(k,p',p) \, G(k,p)} \\
+    &\le \norm{G'}_2 \parens[\bigg]{\int dk \int dp' \abs*{\int \!dp \, \abs*{F(k,p',p)G(k,p)}}^2}^{\frac{1}{2}} \\
+    &\le \norm{G'}_2 \parens[\bigg]{\int dk \, \norm{F(k,\cdot,\cdot)}_2^2 \int dp \abs{G(k,p)}^2}^{\frac{1}{2}} \\
     &\le \norm{G'}_2 \norm{G}_2 \sup_{k} \norm{F(k,\cdot,\cdot)}_2
   \end{align*}
   We claim that there exists a positive constant $C_1$ independent of $m$ and $n$ such that
   \begin{equation*}
-    \norm{G}_2 \le C_1 \norm{(1+H)^{l+r/2} \psi_n}_2
+    \norm{G}_2 \le C_1 \norm{(1+H)^{a+r/2} \psi_n}_2
   \end{equation*}
-  and similary for $G'$.
-  This follows from $N \psi_n = n\psi_n$, where $N$ is the number operator, and the fact that $\omega(q)$ has a positive lower bound $M$
-
-  $H \psi_n(k,p) = (1+\omega(k,p) \psi_n(k,p)$ 
-
-  $1+\omega(k,p) \ge n \epsilon$ 
+  and similarly for $G'$.
+  Since $\omega$ has a positive lower bound on the mass shell,
+  there exists a constant $\epsilon > 0$ independent of $n$ such that
+  \begin{equation*}
+    1+\omega(k,p) = 1 + \underbrace{\omega(k_1) + \cdots + \omega(k_{m-s}) + \omega(p_{s+1}) + \cdots + \omega(p_r)}_{\text{$(m-s)+(r-s)=n$ terms}} \ge n \epsilon
+  \end{equation*}
+  Combine this with the fact that
+  $N \psi_n = n\psi_n$, where $N$ is the number operator,
+  to see
+  \begin{equation*}
+    \norm{G}_2 = \norm{N^{r/2} (1+H)^a \psi_n}_2 \le \epsilon^{-r/2} \norm{(1+H)^{a+r/2} \psi_n}_2.
+  \end{equation*}
+  Now set $C_1 = \epsilon^{-r/2}$.
+  The proof for $G'$ is analogous.
+  In particular, we have shown that both $\norm{G}_{2}$ and $\norm{G'}_{2}$ are finite,
+  provided that $\psi_n$ and $\psi'_m$ lie in the domain of $H^l$, where $l=a+r/2$.
 
-  $\norm{(1+H)\psi_n}_2 \ge n \epsilon \norm{\psi_n} = \epsilon \norm{N \psi_n}$ 
+  %$H \psi_n(k,p) = (1+\omega(k,p)) \psi_n(k,p)$ 
+  %$\norm{(1+H)\psi_n}_2 \ge n \epsilon \norm{\psi_n} = \epsilon \norm{N \psi_n}$ 
 
   In order to determine conditions for the finiteness of the remaining factor involving $F$,
   it is desireable to have an estimate of the growth of $P_s$ in terms of $\omega(p_1),\ldots,\omega(p_r)$.
@@ -435,27 +679,30 @@ The following assertion is key
     q^{0} &= \omega(q) \\
     \abs{q^{\mu}} &\le \omega(q) \quad \mu = 1,2,3.
   \end{align*}
-  Moreover, $\omega(q)$ has a positive lower bound, namely $m$, so that
+  Moreover, $\omega(q)$ has a positive lower bound on $X_m^+$, so that
   for all exponents $a,b \in \NN$ with $a < b$ there
   exists a constant $c_{a,b}$ such that $\omega(q)^a \le c_{a,b}\, \omega(q)^b$.
   This allows us to make the estimate
-  \begin{equation*}
-    \abs{P_s(p_1,\ldots,p_r)} \le C_s \prod_{i=1}^r \omega(p_i)^{d_i} \quad \text{where}\ d_i = \deg \ft{D}_i.
-  \end{equation*}
-
-  \begin{equation*}
+  \begin{equation}
+    \label{equation:polynomial-estimate}
+    \abs{P_s(p_1,\ldots,p_r)} \le C_s \prod_{i=1}^r \omega(p_i)^{d_i} \quad \text{where}\ d_i = \deg \ft{D}_i,
+  \end{equation}
+  and $C_s$ is a constant independent of $m$ and $n$.
+  By the Arithmetic Mean-Geometric Mean Inequality, we have
+  \begin{gather}
     \sqrt[s]{\omega(p_1) \cdots \omega(p_s)} 
     \le \frac{\omega(p_1) + \cdots + \omega(p_s)}{s} 
-    \le \omega(p') \le 1 + \omega(k,p') 
-  \end{equation*}
-  \begin{equation*}
-    \parens[\big]{1+\omega(k,p')} {}^{-l}
-    \le \parens[\big]{\omega(p_1) \cdots \omega(p_s)} {}^{-l/s}
-  \end{equation*}
+    \le \omega(p') \le 1 + \omega(k,p'), \nonumber\\
+    \shortintertext{hence}
+    \label{equation:one-plus-omega-estimate}
+    \parens[\big]{1+\omega(k,p')} {}^{-a}
+    \le \parens[\big]{\omega(p_1) \cdots \omega(p_s)} {}^{-a/s}.
+  \end{gather}
+  The estimates~\eqref{equation:polynomial-estimate} and~\eqref{equation:one-plus-omega-estimate} entail
   \begin{equation*}
-    \abs{F(k,p',p)} \le
-    \prod_{i=1}^{s} \omega(p_i)^{d_i-l/s} 
-    \prod_{j=s}^{r-s} \omega(p_j)^{d_j-l/(r-s)} 
+    \abs{F(k,p',p)} C_s \le
+    \prod_{i=1}^{s} \omega(p_i)^{d_i-a/s} 
+    \prod_{j=s}^{r-s} \omega(p_j)^{d_j-a/(r-s)} 
   \end{equation*}
 \end{myproof}
 
@@ -494,6 +741,7 @@ The following assertion is key
   which has finite integral as it is $L^1$
   by \cref{lemma:integral-kernel-h-bound}.
   Moreover, the integrand converges pointwise to $K_{\psi'\!,\psi}(p_1,\ldots,p_r)$, since $\ft{f} \to 1$ when $f \to \delta_x$.
+  TODO(With of choice of FT constants, $\ft{f} \to 1/(2\pi)^2$. Change here or change def?)
   The Dominated Convergence Theorem implies
 \end{proof}
 
@@ -533,10 +781,10 @@ The following assertion is key
   and $P_s(p_1,\ldots,p_r)$ is defined as before.
 \end{lemma}
 
-\[
-    f(T), f\left( T \right),
-    \int_{a}^{b} f\left( x \right) d x, \frac{1}{T},
-\]
+%\[
+    %f(T), f\left( T \right),
+    %\int_{a}^{b} f\left( x \right) d x, \frac{1}{T},
+%\]
 
 In the theory of a real scalar field $\phi$ of mass $m$,
 the Lagrangian density of the Klein-Gordon action is given by
@@ -590,9 +838,11 @@ where
   A \QFequal B
 \end{equation*}
 
-\section{Essential Self-Adjointness of Renormalized Products}
+\section{Essential Selfadjointness of Renormalized Products}
+
+TODO
 
-\nocite{*}
+%\nocite{*}
 
 \chapterbib
 \cleardoublepage