From b9e2609169709f8aad257fa5e3a92cb780dfad3f Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Sun, 19 May 2024 01:45:37 +0200
Subject: weiter

---
 stresstensor.tex | 302 ++++++++++++++++++++++++++++++++++++++++++++-----------
 1 file changed, 246 insertions(+), 56 deletions(-)

(limited to 'stresstensor.tex')

diff --git a/stresstensor.tex b/stresstensor.tex
index a4bb6fb..4c128c2 100644
--- a/stresstensor.tex
+++ b/stresstensor.tex
@@ -35,9 +35,13 @@ as a service to the reader.
   \item Given a complex-valued function $f$ on $M$, we define its \emph{Fourier transform} $\ft{f}\,$ by
     \begin{equation}
       \label{fourier-transform}
-      \hat{f}(p) = \frac{1}{(2 \pi)^2} \int_{M} e^{i p \cdot x} f(x) \, dx
+      \ft{f}(p) \defequal \int_{M} e^{i p \cdot x} f(x) \, dx
     \end{equation}
-    whenever the integral converges. The \emph{inverse Fourier transform} is TODO
+    whenever the integral converges. The \emph{inverse Fourier transform} is defined by
+    \begin{equation*}
+      \label{inverse-fourier-transform}
+      \ift{f}(p) \defequal \frac{1}{(2 \pi)^2} \int_{M} e^{-i p \cdot x} f(x) \, dx.
+    \end{equation*}
   \item To a mathematician $\overline{\phantom{z}}$ usually means complex conjugation and ${}^*$ indicates the Hilbert adjoint of an operator,
     while a physicist may read ${}^*$ as complex conjugation and
     denotes the Hilbert adjoint with ${}^{\dagger}$.
@@ -85,9 +89,15 @@ as a service to the reader.
     \begin{equation*}
       E : \schwartz{M} \to \hilb{H}, \quad f \mapsto Ef = \left.\ft{f}\,\right\vert {X_m^+}
     \end{equation*}
+    We define a $\RR$-linear mapping $\phi$ by
+    \begin{equation*}
+      \realschwartz{M} \ni f \mapsto \varphi(f) = \Phi_{\mathrm{S}}(Ef) = \frac{1}{\sqrt{2}} \parens*{a(Ef) + a(Ef)^\dagger}
+    \end{equation*}
+    This extedns to complex valued test functions $f \in \schwartz{M}$
     \begin{equation*}
-      \realschwartz{M} \ni f \mapsto \Phi(f) = \Phi_{\mathrm{S}}(Ef) = \frac{1}{\sqrt{2}} \parens*{a(Ef) + a(Ef)^\dagger}
+      \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(E\bar{f}) + a(Ef)^\dagger}
     \end{equation*}
+    called the \emph{massive free scalar quantum field} 
   \item
     annihilation and creation operators, $f \in \schwartz{M}$, $\psi \in \BosonFock{\hilb{H}}$
     \begin{align*}
@@ -176,7 +186,7 @@ leads to
   \parens[\big]{a(p)^\dagger \psi} {}_n (k_1, \ldots, k_n)
   = \frac{1}{\sqrt{n}} \sum_{i=1}^n \delta(p - k_i) \, \psi_{n-1} (k_1, \ldots, \widehat{k_i}, \ldots, k_n),
 \end{equation}
-where the symmetization is necessary to obtain an expression that
+where the symmetrization is necessary to obtain an expression that
 at least has a chance of being a $n$ Boson state.
 However, it clearly is not a $L^2$ function.
 Given any state $\psi'$, we can
@@ -207,7 +217,7 @@ For completeness, we give a precise definition of quadratic form.
     q : D(q) \times D(q) \to \CC,
   \end{equation*}
   where $D(q)$ is a linear subspace of $\hilb{H}$, called the \emph{form domain}\index{form domain}\index{quadratic form!domain of a},
-  such that $q$ is conjugate linear in its first agrument
+  such that $q$ is conjugate linear in its first argument
   and linear in its second argument (i.e.\ sesquilinear).
   We say that $q$ is \emph{densely defined}
   if $D(q)$ is dense in $\hilb{H}$.
@@ -221,13 +231,13 @@ but one may obtain an operator with trivial domain.
 
 \begin{definition}{Operator Associated to a Quadratic Form}{}
   Suppose $q$ is a densely defined quadratic form on a complex Hilbert space $\hilb{H}$.
-  The linear \emph{operator associated to}\index{quadratic form!operator associated to a} $q$, denoted $q_{\mathrm{op}}$,
+  The linear \emph{operator associated to}\index{quadratic form!operator associated to a} $q$, denoted $\QFop{q}$,
   is defined on the domain
   \begin{equation*}
-    D(q_{\mathrm{op}}) = \braces{\psi \in D(q) \mid \text{the map $q(\cdot,\psi) : D(q) \to \CC$ is bounded}},
+    D(\QFop{q}) = \braces{\psi \in D(q) \mid \text{the map $q(\cdot,\psi) \vcentcolon D(q) \to \CC$ is bounded}},
   \end{equation*}
-  and maps $\psi \in D(q_{\mathrm{op}})$ to the vector $q_{\mathrm{op}}\psi$ in $\hilb{H}$ satisfying
-  $q(\psi',\psi) = \innerp{\psi'}{q_{\mathrm{op}}\psi}$,
+  and maps $\psi \in D(\QFop{q})$ to the vector $\QFop{q}\psi$ in $\hilb{H}$ satisfying
+  $q(\psi'\!,\psi) = \innerp{\psi'\!}{\QFop{q}\psi}$,
   which exists and is unique by Riesz’s Representation Theorem.
 \end{definition}
 
@@ -299,7 +309,7 @@ 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 $\CC$
+we are speaking of the non-commutative associative algebra over $\CC$
 freely generated by the elements of $\hilb{H}$.
 The unit of the algebra is $e$.
 
@@ -314,7 +324,7 @@ where $z,z' \in \hilb{H}$.
 \begin{definition}{Infinitesimal Weyl Algebra}{}
   Let $\hilb{H}$ be a complex Hilbert space.
   The \emph{infinitesimal Weyl algebra}\index{infinitesimal Weyl algebra} $\WeylAlg(\hilb{H})$ over $\hilb{H}$
-  is the noncommutative associative algebra over $\CC$
+  is the non-commutative 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},
@@ -371,10 +381,10 @@ 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 \\
+  \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) } 
+  \parens[\big]{ \weylannihilator(z_1) \weylannihilator(z_2) + \weylcreator(z_1) \weylannihilator(z_2)
+  + \weylcreator(z_2) \weylannihilator(z_1) + \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
@@ -470,20 +480,19 @@ and may be obtained via integration by parts.
 
 Naturally, we now define the \emph{distributional derivative} of the field by
 \begin{equation*}
-  D \varphi(f) = \varphi(D^{\dagger} f) \qquad \forall f \in \schwartz{\RR^d}
+  D \varphi(f) = \varphi(D^{\dagger} f) \qquad \forall f \in \schwartz{\RR^4}
 \end{equation*}
-As one expects, $D\varphi$ is an operator-valued tempered distribution on $M=\RR^d$.
-TODO
+As one expects, $D\varphi$ is an operator-valued tempered distribution on $M=\RR^4$.
+In terms of creation and annihilation operators we have 
 \begin{equation}
   \label{derivative-free-field}
-  D \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(ED^{\dagger}f)^{\dagger} + a(ED^{\dagger}f)}
+  D \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(ED^{\dagger}f)^{\dagger} + a(E\overline{D^{\dagger}f})}.
 \end{equation}
-
-
-The operator corresponding to $D$ in Fourier space is the multiplication operator
+In Fourier space the operator $D^\dagger$ corresponds to muliplication with the polynomial
 \begin{equation*}
-  -i \sum_{\alpha} a_{\alpha} p_0^{\alpha_0} (-p_1)^{\alpha_1} (-p_2)^{\alpha_2} (-p_3)^{\alpha_3}
+  \ft{D}(p) \defequal \sum_{\alpha} i^{\abs{\alpha}} a_{\alpha} (+p^0)^{\alpha_0} (-p^1)^{\alpha_1} (-p^2)^{\alpha_2} (-p^3)^{\alpha_3}
 \end{equation*}
+If $D=\partial^{\mu}$, then $\ft{D}(p) = i @ p_{\!\mu}$, were the potential sign is concealed by lowering the index.
 
 
 Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients.
@@ -494,8 +503,8 @@ Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex)
   \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) 
+  \prod_{i=1\vphantom{S}}^{s} a^\dagger(ED^\dagger_{\sigma(i)}f)
+  \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} a(E\overline{D^\dagger_{\sigma(j)}f}) 
 \end{gather}
 
 \section{Renormalized Products of the Free Field and~its~Derivatives}
@@ -516,7 +525,7 @@ this approach incurs significant technical difficulties.
 \begin{lemma}{Integral Representation of the Renormalized Product}{renormalized-product-integral-representation}  
   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 $\psi,\psi' \in \BosonFock{L^2(X_m^+,d\Omega_m)}$,
+  Then, for arbitrary Schwartz functions $f \in \realschwartz{M}$ 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} =
@@ -552,7 +561,32 @@ this approach incurs significant technical difficulties.
   \end{multline*}  
 \end{lemma}
 
-TODO(Note about the remaining dependence of $K$ on $f$.)
+Note that $K$ has a remaining dependence on $f$ via $\chi$
+even thogh the notation does not indicate this.
+This is made explicit in the alternative integral representation
+  \begin{equation}
+    \label{equation:alternative-integral-representation}
+    \begin{multlined}
+      \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} =\\
+      \hspace{1cm} \int dp_1 \!\cdots dp_r
+      \sum_{s=0}^{r} 
+      \, \ft{f}(p_1) \cdots\! \ft{f}(p_s)
+      \, \overline{\ft{f}(p_{s+1}) \cdots\! \ft{f}(p_r)}
+      \, \tilde{K}^s_{\psi'\!,\psi}(p_1,\ldots,p_r)
+    \end{multlined}
+  \end{equation}
+  where
+  \begin{multline*}
+    \tilde{K}^s_{\psi'\!,\psi}(p_1,\ldots,p_r) =
+    P_s(p_1,\ldots,p_r)
+    \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}
+    \delta_{m-s}^{n-(r-s)} \\
+    \cdot \sqrt{m(m-1) \cdots (m-s+1)} \sqrt{n(n-1) \cdots (n-(r-s)+1)} \\
+    \cdot \int dk_1 \cdots dk_{m-s} 
+    \ \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*}
+  This will be more convenient for xxx
 
 \begin{myproof}[lemma:renormalized-product-integral-representation]
   From equation~\eqref{equation:renormalized-product},
@@ -617,24 +651,30 @@ In particular, for squares ($r=2$) we have
 
 The following assertion is key to realizing the idea of taking the limit $f \to \delta_x$.
 
-\begin{lemma}{}{integral-kernel-h-bound}
+\begin{lemma}{H-bounds for the Integral Kernel}{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 arbitrary states $\psi,\psi' \in \BosonFock{\hilb{H}}$,
-  and test functions $f \in \schwartz{M}$,
+  such that for arbitrary test functions $f \in \schwartz{M}$
+  and states $\psi,\psi' \in \Domain{H^l}$,
   the function $K_{\psi'\!,\psi}$ is integrable (that is, $L^1$)
   and satisfies the $H$-bound
   \begin{equation*}
     \norm{K_{\psi'\!,\psi}}_1 \le
     C \norm{(1+H)^l \psi'} \norm{(1+H)^l \psi}.
   \end{equation*}
+  More specifically, it is sufficient to choose $l > rd + r/2$,
+  where $d$ is the highest order of differentiation occuring in $D_1, \ldots, D_r$.
 \end{lemma}
 
-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
+The Hamilton operator $H$ acts on $n$-particle states $\psi_n$ as follows:
+\begin{gather*}
+  H \psi_n(p_1,\ldots,p_n) = \parens[\big]{\omega(p_1) + \cdots + \omega(p_n)} \psi_n(p_1,\ldots,p_n) \\
+  \shortintertext{where}
+  \omega(p) = \omega(\symbfit{p}) = \sqrt{m^2 + \abs{\symbfit{p}}^2} = \sqrt{m^2 + (p^1)^2 + (p^2)^2 + (p^3)^2}.
+\end{gather*}
+In the following proof it will be convenient to use the abbreviation
 \begin{equation*}
-  \omega(p_1,\ldots,p_s) = \omega(p_1) + \cdots + \omega(p_s)
+  \omega(p_1,\ldots,p_s) \defequal \omega(p_1) + \cdots + \omega(p_n).
 \end{equation*}
 
 \begin{myproof}[lemma:integral-kernel-h-bound]
@@ -778,28 +818,71 @@ In the following proof it will we convenient to use the abbreviation
     \le \frac{\omega(p_1) + \cdots + \omega(p_s)}{s} 
     \le \omega(p') \le 1 + \omega(k,p'), \nonumber\\
     \shortintertext{hence}
-    \label{equation:one-plus-omega-estimate}
+    \label{equation:one-plus-omega-estimate1}
     \parens[\big]{1+\omega(k,p')} {}^{-a}
-    \le \parens[\big]{\omega(p_1) \cdots \omega(p_s)} {}^{-a/s}.
+    \le \parens[\big]{\omega(p_1) \cdots \omega(p_s)} {}^{-a/s}, \\
+    \shortintertext{and similarly}
+    \label{equation:one-plus-omega-estimate2}
+    \parens[\big]{1+\omega(k,p)} {}^{-a}
+    \le \parens[\big]{\omega(p_1) \cdots \omega(p_{r-s})} {}^{-a/(r-s)}.
   \end{gather}
-  The estimates~\eqref{equation:polynomial-estimate} and~\eqref{equation:one-plus-omega-estimate} entail
+  The estimates~\eqref{equation:polynomial-estimate},~\eqref{equation:one-plus-omega-estimate1} and~\eqref{equation:one-plus-omega-estimate2} entail
   \begin{equation*}
     \abs{F(k,p',p)} \le C_s
     \prod_{i=1}^{s} \omega(p_i)^{d_i-a/s} 
-    \prod_{j=s+1}^{r} \omega(p_j)^{d_j-a/(r-s)} 
+    \prod_{j=s+1}^{r} \omega(p_j)^{d_j-a/(r-s)}.
   \end{equation*}
+  Since the right hand side does not depend on $k$, and the $p$-variables are separated,
+  the problem reduces to proving that
+  \begin{equation}
+    \label{equation:integral-finite}
+    \int \omega(q)^{-2b} \,d\Omega(q) < \infty
+  \end{equation}
+  for $b$ large enough.
+  Recall that $d \Omega(q) = \omega(q)^{-1} d^3 \symbfit{q}$.
+  By transformation to spherical coordinates, we find that~\eqref{equation:integral-finite}
+  is equivalent to
+  \begin{equation*}
+    \int \frac{r^2}{(m^2 + r^2)^{b+1/2}} \,dr < \infty
+  \end{equation*}
+  It it well known that this holds for $b > 1$.
+  $a > r d$ 
+
+  \begin{equation}
+    \label{equation:intermediate-result}
+    \norm{K_{\psi'\!,\psi}}_1 \le
+    \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}
+    \sum_{s=0}^{r} \delta_{m-s}^{n-(r-s)} C_s
+    \underbracket{\norm{(1+H)^l \psi'_m}_2}_{a'_m}
+    \underbracket{\norm{(1+H)^l \psi_n}_2}_{a_n}
+  \end{equation}
+  We introduce auxiliary variables $a'_m, a_n$ as shown above, and for convenience set $a_n = 0$ whenever $n < 0$.
+  Using this, we rewrite the right hand side of~\eqref{equation:intermediate-result}
+  and apply the Cauchy-Schwarz Inequality for sequences as follows:
+  \begin{equation*}
+    \sum_{s=0}^{r} C_s \sum_{m = 0}^{\infty} a'_m \cdot a_{m+r-2s} \le
+    \sum_{s=0}^{r} C_s \sqrt{\sum_{m=0}^{\infty} a'^2_m \sum_{n=0}^{\infty} a^2_n}
+  \end{equation*}
+  To complete the proof, observe that
+  \begin{equation*}
+    \norm{(1+H)^l \psi'}_2 =
+    \sqrt{\sum_{m=0}^{\infty} \norm{(1+H)^l \psi'_m}_2^2} =
+    \sqrt{\sum_{m=0}^{\infty} a'^2_m},
+  \end{equation*}
+  and similar for $\psi$, by definition of the inner prouct
+  and because $((1+H)^l \psi')_m = (1+H)^l \psi'_m$ for all $m$.
 \end{myproof}
 
 \begin{lemma}{Renormalized Product at a Point}{}
   In the setting of \cref{lemma:renormalized-product-integral-representation},
-  assume that $\psi,\psi'$ are in $D^l(H)$.
+  assume that $\psi,\psi'$ are in $\Domain{H^l}$.
   Let $x$ be any point in $M$ and let $\delta_x \in \tempdistrib{M}$ be the Dirac distribution supported in $x$.
   Then the limit
   \begin{equation*}
     \lim_{f \to \delta_x}
     \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi}
   \end{equation*}
-  exists and depends continously on $x$.
+  exists and depends continuously on $x$.
 \end{lemma}
 
 \begin{proof}
@@ -813,8 +896,22 @@ In the following proof it will we convenient to use the abbreviation
   The integrand is dominated by the function $\abs{K_{\psi'\!,\psi}(p_1,\ldots,p_r)}$,
   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?)
+
+  Since the Fourier transformation of tempered distribution
+  is a continuous mapping $\tempdistribnoarg \to \tempdistribnoarg$,
+  we have $\ft{f} \to \FT{\delta_x}$ whenever $f \to \delta_x$ in the topology of $\tempdistribnoarg$.
+  Recall that $\ft{\delta} = 1$, and thus $\FT{\delta_x}(p) = e^{ix \cdot p}$ for all $p \in M$.
+  This shows that the integrand converges pointwise to
+  \begin{equation*}
+    \sum_{s=0}^r
+    e^{ix \cdot (p_1 + \cdots + p_s)}
+    e^{-ix \cdot (p_{s+1} + \cdots + p_r)}
+    \tilde{K}_{\psi'\!,\psi}(p_1,\ldots,p_r)
+  \end{equation*}
+
   The Dominated Convergence Theorem implies
 \end{proof}
 
@@ -847,20 +944,21 @@ In the following proof it will we convenient to use the abbreviation
 \begin{lemma}{TODO}{}  
   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 $\psi,\psi' \in D^l(H)$
+  Suppose that $l$ is a positive integer large enough to satisfy the
+  Then we have for all states $\psi,\psi' \in \Domain{H^l}$
   \begin{multline*}
     \innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(f) \,\psi} = \\
     = \int dp_1 \!\cdots dp_r
     \sum_{s=0}^{r}
     \, \ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r)
-    \, L_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r)
+    \, \tilde{K}_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r)
   \end{multline*}
   where
   \begin{multline*}
-    L_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) =
+    \tilde{K}_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) =
+    P_s(p_1,\ldots,p_r)
     \sum_{m=0}^{\infty} \sum_{n=0}^{\infty}
-    \delta_{m-s}^{n-(r-s)}
-    \ P_s(p_1,\ldots,p_r) \\
+    \delta_{m-s}^{n-(r-s)} \\
     \cdot \sqrt{m(m-1) \cdots (m-s+1)} \sqrt{n(n-1) \cdots (n-(r-s)+1)} \\
     \cdot \int dk_1 \cdots dk_{m-s} 
     \ \overline{\psi'_m(k_1,\ldots,k_{m-s},p_1,\ldots,p_s)}
@@ -869,10 +967,12 @@ In the following proof it will we convenient to use the abbreviation
   and $P_s(p_1,\ldots,p_r)$ is defined as before.
 \end{lemma}
 
-%\[
-    %f(T), f\left( T \right),
-    %\int_{a}^{b} f\left( x \right) d x, \frac{1}{T},
-%\]
+\begin{proof}
+  a
+\end{proof}
+
+
+\section{Definition of the Stress Tensor}
 
 In the theory of a real scalar field $\phi$ of mass $m$,
 the Lagrangian density of the Klein-Gordon action is given by
@@ -890,7 +990,7 @@ Raising the index $\nu$ and inserting \cref{lagrangian-density} yields
 \end{equation*}
 The \emph{energy density}:
 \begin{equation*}
-  \rho = T^{00} = \frac{1}{2} \parens*{\sum_{\mu=0}^{3} (\partial^{\mu}\phi)^2  + m^2 \phi^2} 
+  \energydensity = T^{00} = \frac{1}{2} \parens*{\sum_{\mu=0}^{3} (\partial^{\mu}\phi)^2  + m^2 \phi^2} 
 \end{equation*}
 The discussion in the previous section enables us to define
 the \emph{renormalized stress-energy tensor} of a free scalar field $\varphi$ by
@@ -900,11 +1000,11 @@ the \emph{renormalized stress-energy tensor} of a free scalar field $\varphi$ by
 and this is a quadratic form.
 In particular, the energy density is
 \begin{equation*}
-  \rho  = \frac{1}{2} \sum_{\mu=0}^{3} \normord{(\partial^{\mu}\phi)^2}  + \frac{1}{2} m^2 \normord{\phi^2}
+  \energydensity  = \frac{1}{2} \sum_{\mu=0}^{3} \normord{(\partial^{\mu}\varphi)^2}  + \frac{1}{2} m^2 \normord{\varphi^2}
 \end{equation*}
 
 \begin{multline*}
-  \innerp{\psi'\!}{\rho(f) \,\psi} = \\
+  \innerp{\psi'\!}{\energydensity(f) \,\psi} = \\
   = \int dp_1 dp_2
   \parens{p_1^{\mu} p_2^{\mu} + m^2} 
   \sum_{s=0}^{r} (-1)^{s+1}
@@ -925,13 +1025,13 @@ where
 \begin{theorem}{TODO}{}  
   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 $\psi,\psi' \in D^l(H)$
+  Then we have for all test functions $f \in \schwartz{M}$
   \begin{multline*}
     \normord{D_1 \varphi \cdots D_r \varphi}(f) \QFequal \int dp_1 \!\cdots dp_r \sum_{s=0}^{r}
     P_s(p_1,\ldots,p_r) \, \ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r) \\
     \cdot a^\dagger(p_1) \cdots a^\dagger(p_s) a(p_{s+1}) \cdots a(p_r) 
   \end{multline*}
-  as quadratic forms on $D^l(H)$, where
+  as quadratic forms on $\Domain{H^l}$, where
   \begin{multline*}
     \quad P_s(p_1,\ldots,p_r) =
     \frac{1}{\sqrt{2^r}}
@@ -944,17 +1044,107 @@ where
 
 \begin{definition}{}{}
   \begin{multline*}
-    \rho(f) \QFequal \frac{1}{4} \int dp dp' (p \cdot p' + m^2)
+    \energydensity(f) \QFequal \frac{1}{4} \int dp dp' (p \cdot p' + m^2)
     \Big\lbrack \ft{f}(p+p') a(p) a(p') + {}\\
     + 2\ft{f}(p-p') a^\dagger(p) a(p') + \ft{f}(-p-p') a^\dagger(p) a^\dagger(p') \Big\rbrack
   \end{multline*}
 \end{definition}
 
+\begin{equation*}
+  \bar{p} := \eta p = (p^0,-\symbfit{p}) 
+\end{equation*}
+
+\begin{proposition}{}{}
+  \begin{multline*}
+    \innerp{\psi'}{\energydensity(f) \psi} =
+    \frac{1}{4} \int dp dp'
+    (\bar{p} \cdot p' + m^2)
+    \bracks[\big]{2 \ft{f}(p - p') L^1_{\psi'\!,\psi}(p,p')} \\
+    + (-\bar{p} \cdot p' + m^2)
+    \bracks[\big]{\ft{f}(- p - p') L^0_{\psi'\!,\psi}(p,p') + \ft{f}(p + p') L^2_{\psi'\!,\psi}(p,p')} 
+  \end{multline*}
+\end{proposition}
+
+\begin{proposition}{}{}
+  \begin{multline*}
+    \energydensity(f) \QFequal \frac{1}{4} \int dp dp'
+    (m^2 + \bar{p} \cdot p')
+    \bracks[\Big]{2\ft{f}(p-p') a^\dagger(p) a(p')} \\
+    + (m^2 - \bar{p} \cdot p')
+    \bracks[\Big]{\ft{f}(p+p') a(p) a(p') + \ft{f}(-p-p') a^\dagger(p) a^\dagger(p')} 
+  \end{multline*}
+\end{proposition}
+
+\begin{proposition}{}{}
+  The Fock vaccum $\fockvaccum$ lies in the domain of $\energydensity(f)\QFop{}$
+  for all test functions $f \in \schwartz{M}$
+  and $\energydensity(f)\QFop{}\fockvaccum$ is the vector $\psi$ defined by
+  \begin{equation*}
+    \psi_2(p,p') = \frac{\sqrt{2}}{4} (m^2 - \bar{p} \cdot p') \ft{f}(-p-p')
+  \end{equation*}
+  and $\psi_n \equiv 0$ for $n \ne 2$.
+\end{proposition}
+
+\begin{equation*}
+  \energydensity(f) \Omega = ?
+\end{equation*}
+
 \section{Essential Selfadjointness of Renormalized Products}
 
-a
+\begin{lemma}{H-Bounds for the Renormalized Product}{}
+  \begin{equation*}
+    \abs{\innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(f)\psi}} \le
+    C \norm{(I+H)^l \psi} \norm{(I+H)^l \psi'}
+  \end{equation*}
+\end{lemma}
 
-%\nocite{*}
+\begin{proof}
+  This proof is nearly identical to that of \cref{lemma:integral-kernel-h-bound}
+  and we will only cover the differences.
+  \begin{equation*}
+    \begin{multlined}[c]
+    \abs{\innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(f)\psi}} \le
+      \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \sum_{s=0}^{r}
+      \delta_{m-s}^{n-(r-s)} \\
+      \hspace{2.5cm} \cdot
+    \int \!dk \int \!dp'\! \int \!dp \, \abs*{F(k,p',p) \, G'(k,p') \, G(k,p)},
+    \end{multlined}
+  \end{equation*}
+  where
+  \begin{multline}
+    F(k,p',p) = \parens[\big]{1+\omega(k,p')} {}^{-a} \parens[\big]{1+\omega(k,p)} {}^{-a} P_s(p',p) \\
+    \cdot \ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r)
+  \end{multline}
+  and $G$ and $G'$ are defined as before.
+  All we have to do, is verifying that
+  \begin{equation*}
+    \sup_k \norm{F(k,\cdot,\cdot)}_2 < \infty
+  \end{equation*}
+  for a sufficiently large integer $a$.
+  Then we obtain the desired $H$-bound with $l=a+r/2$.
+
+  Recall that the Schwartz class is preserved by Fourier transform, translation and multiplication with polynomials.
+  Moreover, it is well known that Schwartz functions are square-integrable with repect to the Lorentz invariant measure on the mass shell.
+  Hence,
+  \begin{equation*}
+    \int dp_1 \abs{\ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r)}^2
+  \end{equation*}
+  is bounded by a constant independent of $p_2,\ldots,p_r$.
+\end{proof}
+
+\section{Covariance}
+
+\begin{equation*}
+  f_g(x) \defequal f(g^{-1} x) \qquad
+  x \in M, \quad g \in \ProperOrthochronousPoincareGroup.
+\end{equation*}
+
+\begin{theorem}{Covariance}{covariance-renormalized-product}
+  \begin{equation*}
+    U(g) \,\normord{D_1 \varphi \cdots D_r \varphi}(f)\, U(g)^{-1}
+    = \normord{D_1 \varphi \cdots D_r \varphi}(f_g)
+  \end{equation*}
+\end{theorem}
 
 \chapterbib
 \cleardoublepage
-- 
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