From c66c3bc73d5d627ec7051e9ada6316c98ae072e0 Mon Sep 17 00:00:00 2001
From: Justin Gassner <justin.gassner@mailbox.org>
Date: Wed, 29 May 2024 14:04:59 +0200
Subject: weiter

---
 stresstensor.tex | 36 +++++++++++++++++++-----------------
 1 file changed, 19 insertions(+), 17 deletions(-)

(limited to 'stresstensor.tex')

diff --git a/stresstensor.tex b/stresstensor.tex
index 4c128c2..79c0930 100644
--- a/stresstensor.tex
+++ b/stresstensor.tex
@@ -32,7 +32,9 @@ as a service to the reader.
       x \cdot y = g_{\mu \nu} x^{\mu} y^{\nu} = x^0y^0 - x^1 y^1 - x^2 y^2 - x^3 y^3
     \end{equation*}
     points $x = (x^0,x^1,x^2,x^3) \in M$ are sometimes written $x = (x^0,\symbfit{x})$ with separated time and space coordinates
-  \item Given a complex-valued function $f$ on $M$, we define its \emph{Fourier transform} $\ft{f}\,$ by
+  \item Given a complex-valued function $f$ on $M$,
+    we define its \emph{Fourier transform}\index{Fourier transform} $\ft{f}\,$ by
+    \nomenclature[f]{$\ft{f}$}{Fourier transform of $f$}
     \begin{equation}
       \label{fourier-transform}
       \ft{f}(p) \defequal \int_{M} e^{i p \cdot x} f(x) \, dx
@@ -93,7 +95,7 @@ as a service to the reader.
     \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 $f \in \schwartz{M}$
+    This extends to complex valued test functions $f \in \schwartz{M}$
     \begin{equation*}
       \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(E\bar{f}) + a(Ef)^\dagger}
     \end{equation*}
@@ -243,7 +245,7 @@ but one may obtain an operator with trivial domain.
 
 We will use the symbol $\QFequal$ between quadratic forms or operators
 to indicate their equality as quadratic forms.
-TODO(statement about domains?)
+\todo{statement about domains?}
 
 
 A natural question is how the smeared operators relate to the pointwise ones.
@@ -287,7 +289,7 @@ for all $\psi,\psi' \in D$.
 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)
+\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 $\Phi(f)$,
@@ -314,7 +316,7 @@ 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)
+\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
@@ -332,7 +334,7 @@ where $z,z' \in \hilb{H}$.
   where $e$ is the unit of the algebra.
 \end{definition}
 
-TODO(introduce $\Phi$ as representation of $\WeylAlg$)
+\todo{introduce $\Phi$ as representation of $\WeylAlg$}
 
 \begin{definition}{Annihilator and Creator}{}
   Suppose $\WeylAlg$ is the infinitesimal Weyl algebra
@@ -383,8 +385,8 @@ The cases $r=1$ and $r=2$ read
   \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) + \weylcreator(z_1) \weylannihilator(z_2)
-  + \weylcreator(z_2) \weylannihilator(z_1) + \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
@@ -399,7 +401,7 @@ if one brings~\eqref{equation:normal-ordering} into the equivalent form
   \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?).
+by basic combinatorial arguments \todo{further explanation?}.
 In~\cite{Klein1973}, the factor $\frac{1}{s!(r-s)!}$ is erroneously missing.
 
 
@@ -488,7 +490,7 @@ In terms of creation and annihilation operators we have
   \label{derivative-free-field}
   D \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(ED^{\dagger}f)^{\dagger} + a(E\overline{D^{\dagger}f})}.
 \end{equation}
-In Fourier space the operator $D^\dagger$ corresponds to muliplication with the polynomial
+In Fourier space the operator $D^\dagger$ corresponds to multiplication with the polynomial
 \begin{equation*}
   \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*}
@@ -562,7 +564,7 @@ this approach incurs significant technical difficulties.
 \end{lemma}
 
 Note that $K$ has a remaining dependence on $f$ via $\chi$
-even thogh the notation does not indicate this.
+even though the notation does not indicate this.
 This is made explicit in the alternative integral representation
   \begin{equation}
     \label{equation:alternative-integral-representation}
@@ -663,7 +665,7 @@ The following assertion is key to realizing the idea of taking the limit $f \to
     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 $D_1, \ldots, D_r$.
+  where $d$ is the highest order of differentiation occurring in $D_1, \ldots, D_r$.
 \end{lemma}
 
 The Hamilton operator $H$ acts on $n$-particle states $\psi_n$ as follows:
@@ -869,7 +871,7 @@ In the following proof it will be convenient to use the abbreviation
     \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 similar for $\psi$, by definition of the inner product
   and because $((1+H)^l \psi')_m = (1+H)^l \psi'_m$ for all $m$.
 \end{myproof}
 
@@ -898,7 +900,7 @@ In the following proof it will be convenient to use the abbreviation
   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?)
+  \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 $\tempdistribnoarg \to \tempdistribnoarg$,
@@ -1076,9 +1078,9 @@ where
 \end{proposition}
 
 \begin{proposition}{}{}
-  The Fock vaccum $\fockvaccum$ lies in the domain of $\energydensity(f)\QFop{}$
+  The Fock vacuum $\FockVacuum$ 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
+  and $\energydensity(f)\QFop{}\FockVacuum$ 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*}
@@ -1124,7 +1126,7 @@ where
   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.
+  Moreover, it is well known that Schwartz functions are square-integrable with respect 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
-- 
cgit v1.2.3-70-g09d2

