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

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+\chapter{A quantum energy inequality involving local modular data}
+
+
+\cite{Much2022}
+
+\begin{equation*}
+  \innerp{\psi}{\energydensity(f)\psi} \ge
+  - \epsilon - \norm{\smash[b]{\Delta}_{\smash[t]{\sharp}}^{-1/2} \ft{g}_{\lambda}(K_{\raisebox{5pt}{\footnotesize$\sharp$}}) \energydensity(f) \fockvaccum}
+\end{equation*}
+
+
+\section{Misc}
+
+\todo{Put this somwhere else.}
+
+A \emph{Lorentz transform} is a linear automorphism of Minkowski spacetime
+which preserves the Lorentz bilinear form.
+Lorentz transforms are usually represented by (real) $4 \times 4$ matrices,
+with respect to the standard basis.
+the \emph{Lorentz group} $\FullLorentzGroup$.
+\begin{equation*}
+  \FullPoincareGroup = \RR^4 \ltimes \FullLorentzGroup
+\end{equation*}
+
+The relativistic transformation law for one-particle states is given by
+\begin{equation*}
+  \parens[\big]{U(a,\Lambda) \psi}(p) = e^{ia \cdot p} \psi(\Lambda^{-1} p),
+  \quad \psi \in \hilb{H}, (a,\Lambda) \in \ProperOrthochronousPoincareGroup.
+\end{equation*}
+The mapping $(a,\Lambda) \mapsto U(a,\Lambda)$ is a (irreducible) unitary representation
+of the proper orthochronous Poincaré group $\ProperOrthochronousPoincareGroup$
+on the one-particle Hilbert space.
+By applying any of the second quantization functors we obtain a representation on the multi-particle state space.
+\begin{equation*}
+  \parens[\big]{U(a,\Lambda) \psi}{}_n(p_1,\ldots,p_n) = e^{ia \cdot (p_1 + \cdots + p_n)} \psi_n(\Lambda^{-1} p_1, \ldots, \Lambda^{-1} p_n),
+\end{equation*}
+
+Poincaré covariance
+\begin{equation}
+  \label{equation:poincare-covariance-local-algebras}
+  U(g) \localalg{\spacetimeregion{O}} U(g)^* = \localalg{g\spacetimeregion{O}}
+  \qquad g \in \ProperOrthochronousPoincareGroup
+\end{equation}
+
+\begin{definition}{Von Neumann Algebra of Local Observables}{}
+  \begin{equation*}
+    \localalg{\spacetimeregion{O}} = \braces{b(\varphi(f)) \mid b, f \in \realschwartz{M}, \supp f \subset \spacetimeregion{O}}''
+  \end{equation*}
+\end{definition}
+
+\section{Basic Concepts of Modular Theory}
+\index{modular!theory}
+
+If $\hilb{H}$ is a Hilbert space
+we shall denote the $C^*$-algebra of all bounded linear operators on $\hilb{H}$ by $B(\hilb{H})$.
+
+\begin{definition}{Cyclic and Separating Vectors}{}
+  Suppose $\hilb{H}$ is a Hilbert space and $\mathcal{A}$ is a $C^*$-subalgebra of $B(\hilb{H})$.
+  A vector $\Omega \in \hilb{H}$ is called
+  \begin{itemize}
+    \item \emph{cyclic}\index{cyclic vector} for $\mathcal{A}$ if the vector set $\mathcal{A} \Omega$ is dense in $\hilb{H}$.
+    \item \emph{separating}\index{separating vector} for $\mathcal{A}$ if the map $A \mapsto A \Omega$ from $\mathcal{A}$ into $\hilb{H}$ is injective.
+  \end{itemize}
+\end{definition}
+Occasionally, a vector that is both cyclic and separating is called \emph{standard}\index{standard vector}.
+
+Recall that the commutant of a set $\mathcal{S} \subset B(\hilb{H})$ of operators
+is defined as the set of all operators $T \in B(\hilb{H})$ which commute with all operators $S$ in $\mathcal{S}$.
+We shall denote the commutant of $\mathcal{S}$ by $\mathcal{S}'$.\nomenclature{$\mathcal{A}'$}{commutant of $\mathcal{A}$}
+
+\begin{proposition}{}{cyclic-separating}
+  \begin{enumerate}[label=(\roman*),nosep,leftmargin=*,widest=ii]
+    \item A vector is cyclic for $\mathcal{A}$ if and only if it is separating for $\mathcal{A}'$.
+    \item If $\vNa{M}$ is a von Neumann algebra, then a vector is cyclic and separating for $\vNa{M}$
+  if and only if it is cyclic and separating for $\vNa{M}'$.
+  \end{enumerate}
+\end{proposition}
+
+\begin{proof}
+  \todo{xxx}
+  The second assertion directly follows from the first and the fact that $\vNa{M}'' = \vNa{M}$.
+\end{proof}
+
+If $\Omega$ is separating for $\mathcal{A}$,
+then every element of $\mathcal{A}\Omega$ is of the form $A\Omega$
+with a unique $A \in \mathcal{A}$.
+This allows us to define an (anti-linear) operator $S_0$ in $\hilb{H}$ with domain $\mathcal{A}\Omega$ by
+\begin{equation}
+  \label{equation:definition-s0}
+  \quad S_0 A\Omega \defequal S_0 A^*\Omega \qquad A \in \mathcal{A}.
+\end{equation}
+The operator $S_0$ is densely defined if and only if $\Omega$ is cyclic for $\mathcal{A}$.
+Since the $*$-operation on $\mathcal{A}$ is involutive,
+the range of $S_0$ coincides with its domain.
+
+\begin{lemma}{}{}
+  If $\Omega$ is a cyclic and separating vector for a von Neumann algebra $\mathcal{A}$,
+  then the operator $S_0$ defined by~\eqref{equation:definition-s0} is closable.
+\end{lemma}
+\begin{proof}
+  By \cref{proposition:cyclic-separating},
+  $\Omega$ is also cyclic and separating for the commutant $\vNa{A}'$.
+  Hence we may, analogously to $S_0$,
+  define another anti-linear operator $F_0$ in $\hilb{H}$ with dense domain $\mathcal{A}' \Omega$ by
+  \begin{equation*}
+    \quad F_0 B\Omega \defequal F_0 B^*\Omega \qquad B \in \mathcal{A'}.
+  \end{equation*}
+  By definition of $S_0$ and $F_0$ we have for every $A \in \mathcal{A}$ and $B \in \mathcal{A}'$
+  \begin{equation*}
+    \innerp{S_0 A \Omega}{B \Omega} = 
+    \innerp{\Omega}{AB \Omega} = 
+    \innerp{\Omega}{BA \Omega} = 
+    \innerp{F_0 B\Omega}{A \Omega}.
+  \end{equation*}
+  This adjoint identity establishes that $S_0 \subset F_0^*$.
+  (The \enquote{twisted} appearance of the identity is correct,
+  since it involves anti-linear operators on both sides.)
+  The Hilbert adjoint $F_0^*$ of $F_0$ is closed.
+  Hence, we have shown that $S_0$ has a closed extension, and
+  this implies that $S_0$ is closable.
+\end{proof}
+
+\begin{definition}{Tomita operator}{}
+  Suppose $\Omega$ is a cyclic and separating vector for a von Neumann algebra $\mathcal{A}$.
+  The closure $S = \operatorclosure{S_0}$
+  of the operator $S_0$ defined on $\mathcal{A}\Omega$ by
+  $S_0 A\Omega = S_0 A^*\Omega$
+  for $A \in \mathcal{A}$
+  is called the
+  \emph{Tomita operator}\index{Tomita operator}\index{operator!Tomita}\nomenclature{$S$}{Tomita operator}
+  for the pair $(\mathcal{A},\Omega)$.
+\end{definition}
+
+It is a well-known fact that closed operators can be decomposed
+in a similar fashion to the polar coordinate representation $z = e^{i\arg z} \abs{z}$
+of a complex number.
+We state the theorem in its somewhat uncommon variant for anti-linear operators,
+as this is our only use case.
+
+\begin{theorem}{Polar Decomposition for Anti-Linear Closed Operators}{polar-decomposition}
+  \index{polar decomposition}
+  Let $T$ be an arbitrary closed anti-linear operator in a Hilbert space $\hilb{H}$.
+  Then there exist
+  a positive selfadjoint linear operator $\abs{T}$ and
+  a partial anti-linear isometry $U$
+  such that
+  \begin{equation*}
+    T = U \abs{T} \qquad \bracks[\big]{\text{in particular, $\Domain{T} = \Domain{\abs{T}}$}}.
+  \end{equation*}
+  The operators $U$ and $\abs{T}$ are uniquely determined given the additional conditions
+  \begin{equation*}
+    \ker\abs{T} = \ker T \qquad
+    (\ker U)^\perp = (\ker T)^\perp \qquad
+    \ran U = \overline{\ran T}.
+  \end{equation*}
+\end{theorem}
+
+Proofs of this statement are contained in~\cite{ReedSimon1} and~\cite{Schmüdgen2012}.
+When we speak of \emph{the} polar composition we tacitly assume that the additional conditions
+ensuring uniqueness are satisfied.
+
+Now we are able to introduce the fundamental objects of modular theory.
+
+\begin{definition}{Modular Conjugation, Modular Operator}{}
+  Suppose $\vNa{M}$ is a von Neumann algebra acting on a Hilbert space $\hilb{H}$,
+  and suppose $\Omega \in \hilb{H}$ is a cyclic and separating vector for $\vNa{M}$.
+  Let $S$ be the Tomita operator for $(\vNa{M},\Omega)$ and let
+  \begin{equation*}
+    S = J \Delta^{1/2}
+  \end{equation*}
+  be its polar decomposition.
+  The anti-unitary operator $J$ is called
+  \emph{modular conjugation}\index{modular!conjugation}\nomenclature{$J$}{modular conjugation}.
+  The positive selfadjoint operator $\Delta$ is called
+  \emph{modular operator}\index{modular!operator}\index{operator!modular}\nomenclature{$\Delta$}{modular operator}.
+  The pair $(J,\Delta)$ is said to be the \emph{modular data}\index{modular!data}\index{modular!objects} associated to
+  the pair $(\vNa{M},\Omega)$.
+\end{definition}
+
+\todo{clarify why $J$ is anti-unitary}
+
+\begin{definition}{Modular Group}{}
+  Adopt the notation of the foregoing definition.
+  The mapping $\RR \ni t \mapsto \Delta^{it}$ is called the \emph{modular group}\index{modular!group} associated to 
+  $(\vNa{M},\Omega)$.
+\end{definition}
+
+The modular group is a strongly continuous one-parameter unitary group on $\hilb{H}$.
+
+\newpage
+
+\begin{proposition}{}{modular-data-unitary}
+  Suppose $\vNa{M}$ is a von Neumann algebra acting on a Hilbert space $\hilb{H}$.
+  Let $U$ be a unitary operator on $\hilb{H}$.
+  Then $U\vNa{M}U^*$ is a von Neumann algebra on $\hilb{H}$.
+  Suppose further that $\Omega \in \hilb{H}$ is a cyclic and separating vector for $\vNa{M}$.
+  Then $U \Omega$ is cyclic and separating for $U\vNa{M}U^*$.
+  Let $(J,\Delta)$ be the modular data associated to $(\vNa{M},\Omega)$.
+  Then $(UJU^*,U{\Delta}U^*)$ is the modular data associated to $(U\vNa{M}U^*,U\Omega)$.
+\end{proposition}
+
+\begin{proof}
+  To prove the first assertion,
+  consider any $A \in (U\vNa{M}U^*)''$.
+  By the double commutant theorem,
+  it suffices to show that $A \in U\vNa{M}U^*$.
+  As $\vNa{M}$ is a von Neumann algebra,
+  this is equivalent to $U^*\! AU \in \vNa{M}''$,
+  again by the double commutant theorem.
+  Let $B \in \vNa{M}'$.
+  It is easy to check that $UBU^* \in (U\vNa{M}U^*)'$.
+  By assumption, $A$ lies in the commutant of $(U\vNa{M}U^*)'$.
+  Thus we find that $[U^*\! AU,B] = U^* [A,UBU^*] U = 0$, as desired.
+
+  The set of vectors $U\vNa{M}U^* U\Omega = U\vNa{M}\Omega$ is dense in $\hilb{H}$,
+  since it is the image of $\vNa{M} \Omega$ under the homeomorphism $U$.
+  Thus, the vector $U\Omega$ is cyclic for $U\vNa{M}U^*$.
+  Let us show that it is also separating.
+  Suppose $A$ is in $\vNa{M}$ and $UAU^*U\Omega = UA\Omega = 0$
+  Since unitaries are injective, $A\Omega = 0$.
+  Now $A=0$ follows from the assumption that $\Omega$ is separating for $\vNa{M}$.
+  We have shown that the mapping $UAU^*U\Omega = UA\Omega$ from $U\vNa{M}U^* \to \hilb{H}$ is injective.
+
+  Let $S = \overline{S_0}$ be the Tomita operator associated to $(\vNa{M},\Omega)$,
+  and let $S' = \overline{S'_0}$ be the Tomita operator associated to $(U\vNa{M}U^*,U\Omega)$.
+  Then we have
+  \begin{equation*}
+    (S'_0 U) A \Omega =
+    S'_0 (U A U^*)  U \Omega =
+    (U A^* U^*)  U \Omega =
+    U A^* \Omega =
+    U S_0 A \Omega
+  \end{equation*}
+  for all $A \in \vNa{M}$. Consequently, $S'_0 = U S_0 U^*$ as operators with domain $U\vNa{M}\Omega$.
+  Taking the closure, we obtain $S' = U S U^*$.
+  We can write this as $S' = UJU^* U\Delta^{1/2} U^*$,
+  where $S = J \Delta^{1/2}$ is the polar decomposition of the Tomita operator.
+  It is straightforward to check that
+  $UJU^*$ is anti-unitary and $U\Delta^{1/2} U^*$ is positive selfadjoint,
+  and satisfy the additional condition of \cref{theorem:polar-decomposition}.
+  The uniqueness of the polar decomposition implies that
+  $UJU^*$ is the modular conjugation and $U\Delta U^*$ is the modular conjugation
+  associated to the pair $(U\vNa{M}U^*,U\Omega)$.
+\end{proof}
+
+\newpage
+
+Finally, let us outline how modular theory enters into algebraic quantum field theory.
+
+\begin{theorem}{Reeh-Schlieder Theorem}{reeh-schlieder}
+  \todo{spell it out}
+\end{theorem}
+
+By Reeh-Schlieder (\cref{theorem:reeh-schlieder}), the vacuum $\Omega$ is cyclic and separating for $\localalg{\spacetimeregion{O}}$.
+Thus, modular theory
+
+
+\section{The Geometric Action of the Modular Operator Associated With a Wedge Domain}
+
+\begin{definition}{Right and Left Wedge, General Wedges}{}
+  The \emph{right wedge}\index{wedge!right}\nomenclature[WR]{$\rightwedge$}{right wedge}
+  and \emph{left wedge}\index{wedge!left}\nomenclature[WL]{$\leftwedge$}{left wedge}
+  in Minkowski space $M$ are the open subsets
+  \begin{equation*}
+    \rightwedge \defequal \braces[\big]{x \in M \vcentcolon x^1 > \abs{x^0}} 
+    \quad \text{and} \quad
+    \leftwedge \defequal \braces[\big]{x \in M \vcentcolon x^1 < -\abs{x^0}}.
+  \end{equation*}
+We say that a spacetime region $W \subset M$ is a \emph{wedge}\index{wedge}
+  if there exists an element $g$ of the Poincaré group
+  such that $W = g \rightwedge$.
+\end{definition}
+
+Instead of the right wedge,
+we could just as well have used the left wedge to define the notion of a general wedge,
+since they are transformed into each other by space inversion.
+
+\begin{lemma}{}{general-wedge-from-right-wedge}
+  If a spacetime region $W$ is a wedge,
+  then there exists an element $g$ of the proper orthochronous Poincaré group
+  such that $W = g \rightwedge$.
+\end{lemma}
+
+\begin{proof}
+  \todo{xxx}
+\end{proof}
+
+In the standard representation of the Lorentz group, the boost (or velocity transformation) along the $x^1$-axis
+with rapidity $2 \pi t$ is given by the matrix\footnote{
+  This matrix depends on the choice of metric signature.
+  Ours is $(+,-,-,-)$.
+  For $(-,+,+,+)$, use
+  \begin{equation*}
+  \Lambda(t) = \begin{pmatrix}
+    \phantom{-}\cosh(2 \pi @ t) & -\sinh(2 \pi @ t) & \; 0 \; & \; 0 \; \\
+    -\sinh(2 \pi @ t) & \phantom{-}\cosh(2 \pi @ t) & 0 & 0 \\
+    0 & 0 & 1 & 0 \\
+    0 & 0 & 0 & 1 \\
+  \end{pmatrix}.
+  \end{equation*}
+  }
+
+\begin{equation*}
+  \Lambda(t) = \begin{pmatrix}
+    \cosh(2 \pi @ t) & \sinh(2 \pi @ t) & \; 0 \; & \; 0 \; \\
+    \sinh(2 \pi @ t) & \cosh(2 \pi @ t) & 0 & 0 \\
+    0 & 0 & 1 & 0 \\
+    0 & 0 & 0 & 1 \\
+  \end{pmatrix}
+\end{equation*}
+
+The following proposition shows that $t \mapsto \Lambda(t)$ is
+a one-parameter subgroup of the stabilizer group of the right wedge
+with respect to the action of the Lorentz group on subsets of Minkowski space.
+
+\begin{proposition}{}{}
+  \begin{enumerate}[label=(\roman*),nosep,leftmargin=*,widest=ii]
+    \item $\Lambda(s + t) = \Lambda(s) \Lambda(t)$ for all $s,t \in \RR$.
+    \item $\Lambda(t) \rightwedge = \rightwedge$ for all $t \in \RR$.
+  \end{enumerate}
+\end{proposition}
+
+\begin{proof}
+  The first property can be verified by direct computation.
+  Let us prove the second.
+  By definition, the image $\Lambda(t) x$ of a vector $x \in M$ lies in $\rightwedge$ if and only if
+  \begin{equation*}
+    x^0 \sinh(2 \pi t) + x^1 \cosh(2 \pi t) 
+    > \abs{x^0 \cosh(2 \pi t) + x^1 \sinh(2 \pi t)},
+  \end{equation*}
+  or equivalently
+  \begin{equation*}
+    x^0 \parens[\big]{\sinh(2 \pi t) \mp \cosh(2 \pi t)}
+    + x^1 \parens[\big]{\cosh(2 \pi t) \mp \sinh(2 \pi t)} > 0
+  \end{equation*}
+  for both sign choices.
+  Using the definitions of the hyperbolic sine and cosine, this may be further simplified to $(x^1 \mp x^0) e^{2 \pi t} > 0$,
+  which holds if and only if $x^1 > \abs{x^0}$,
+  since the exponential is always positive.
+  So we have shown that
+  \begin{equation}
+    \label{equation:image-right-wedge}
+    \Lambda(t) x \in \rightwedge \iff x \in \rightwedge.
+  \end{equation}
+  This implies that $\Lambda(t)\rightwedge \subset \rightwedge$ for all $t \in \RR$.
+  Conversely, given an arbitrary vector $y \in \rightwedge$,
+  we have to find $x \in \rightwedge$ such that $\Lambda(t) x = y$.
+  Consider $x = \Lambda(-t) y$. Clearly, $x \in \rightwedge$, because of $y \in \rightwedge$
+  and~\eqref{equation:image-right-wedge}. Now it follows from $\Lambda(-t) = \Lambda(t)^{-1}$ that in fact $\Lambda(t) x = y$.
+\end{proof}
+
+\begin{theorem}{Bisognano-Wichmann Theorem \textmd{\cite{Bisognano1975}}}{}
+  For the theory of a free scalar field in Minkowski spacetime,
+  let $\spacetimeregion{O} \mapsto \localalg{\spacetimeregion{O}}$ be the net of von Neumann algebras of local observables.
+  If $(J,\Delta)$ is the modular data associated to the algebra $\localalg{\rightwedge}$ of the right wedge and the vacuum $\Omega$, then
+  \begin{equation*}
+    J = \Theta \cdot U\parens[\big]{0, R_{23}(\pi)} \qquad
+    \Delta^{it} = U\parens[\big]{0,\Lambda(t)},
+  \end{equation*}
+  where $U$ is the theory's unitary representation of the proper orthochronous Poincaré group.
+\end{theorem}
+
+\todo{give definition of $\Theta$ and $R_{23}$}
+
+Note that above statement is for the right wedge only.
+Let us investigate how the modular group changes, if we consider another wedge region.
+By \cref{lemma:general-wedge-from-right-wedge} any wedge $W$ can be obtained as $W = g\rightwedge$, where $g$ is a proper orthochronous Poincaré transformation.
+The covariance property~\eqref{equation:poincare-covariance-local-algebras} of $\vNa{R}$ implies
+\begin{equation*}
+  \localalg{W} =
+  U(g) \localalg{\rightwedge} U(g)^*.
+\end{equation*}
+The vacuum $\Omega$ is Poincaré invariant:
+\begin{equation*}
+  U(g) \Omega = \Omega.
+\end{equation*}
+We write $(J_W,\Delta_W)$ for the modular data associated to $(\localalg{W},\Omega)$.
+By \cref{proposition:modular-data-unitary}
+\begin{equation*}
+  J_W = U(g) J U(g)^* \qquad 
+  \Delta_W = U(g) \Delta U(g)^*
+\end{equation*}
+Recall that the modular group $\Delta_W^{it}$ is defined by means of functional calculus.
+This raises the following problem: given a selfadjoint operator $A$, a unitary operator $U$
+and a suitable function $f$ we want to express $f(UAU^*)$ in terms of $f(A)$, if possible.
+Note that two different functional calculi are at play here, the former is for $UAU^*$ and the latter for $A$.
+Simple functions such as polynomials suggest $f(UAU^*) = Uf(A)U^*$.
+That this is generally true is the statement of the following Lemma.
+
+
+\begin{lemma}{}{functional-calclus-unitary-trafo}
+  Suppose that $A$ is a selfadjoint operator on a Hilbert space $\hilb{H}$,
+  with spectral measure $E_A$.
+  Suppose $U$ is an unitary operator on $\hilb{H}$, and
+  let $E_{U\! @AU^*}$ denote the spectral measure of the (selfadjoint) operator $UAU^*$.
+  Then we have $U E_A U^* = E_{U\! @AU^*}$, and
+  \begin{equation*}
+    U f(A) U^* = f(U\! @@AU^*)
+  \end{equation*}
+  for all Borel functions $f : \RR \to \CC$.
+\end{lemma}
+
+\question{Ist diese Aussage korrekt? Ist mein Beweis richtig? Geht der auch einfacher?}
+
+\begin{proof}
+  For each regular value $\lambda \in \rho(A)$ let
+  \begin{equation*}
+    R_A(\lambda) = (A-\lambda)^{-1}
+  \end{equation*}
+  denote the resolvent operator of $A$.
+  This proof is based on Stone's Formula \todo{reference},
+  which relates the resolvent to the spectral projections of $A$:
+  If $E_A$ is the spectral measure of $A$ and $\alpha < \beta$ are real numbers, then
+  \begin{equation*}
+    \stronglim_{\varepsilon \downarrow 0}
+    \frac{1}{\pi i} \int_{\alpha}^{\beta} \bracks{R_A(\lambda + i \varepsilon)  - R_A(\lambda - i \varepsilon)} d\lambda 
+    = E_A \parens[\big]{\bracks{\alpha,\beta}} + E_A \parens[\big]{\parens{\alpha,\beta}}
+  \end{equation*}
+  Recall that a spectral measure is countably additive.
+  As a consequence,
+  \begin{equation*}
+    \stronglim_{\alpha \uparrow a}
+    \stronglim_{\beta \downarrow b}
+    \stronglim_{\varepsilon \downarrow 0}
+    \frac{1}{2\pi i} \int_{\alpha}^{\beta} \bracks{R_A(\lambda + i \varepsilon)  - R_A(\lambda - i \varepsilon)} d\lambda 
+    = E_A \parens[\big]{\bracks{a,b}}
+  \end{equation*}
+  for all $a \in \RR \cup \braces{-\infty}$, $b \in \RR \cup \braces{\infty}$.
+  Observe that $\rho(A) = \rho(U\! @AU^*)$ and that for each (common) regular value $\lambda$ we have
+  \begin{equation*}
+    R_{U\! @AU^*}(\lambda) = U R_A(\lambda) @ U^*\!.
+  \end{equation*}
+  Since conjugation with an unitary commutes with the strong operator limit, we obtain
+  \begin{equation*}
+    E_{U\! @AU^*} \parens[\big]{\bracks{a,b}}
+    = U E_A \parens[\big]{\bracks{a,b}} U^*
+  \end{equation*}
+  for all $a,b \in \RR$.
+  The collection $\mathcal{A}$ of all subsets $S$ of $\RR$ such that
+    $E_{U\! @AU^*} \parens[\big]{S} = U E_A \parens[\big]{S} U^*$
+    is a $\sigma$-algebra on $\RR$.
+    We have shown that all closed intervals belong to $\mathcal{A}$.
+    It is well known that the Borel-$\sigma$-algebra $\mathcal{B}$ of $\RR$
+    is generated by the closed intervals. Hence, $\mathcal{B} \subset \mathcal{A}$.
+    This shows that the spectral measures $U E_A U^*$ and $E_{U\! @AU^*}$ coincide.
+\end{proof}
+
+\begin{equation*}
+  \Delta_W^{it}
+  = U(g) \Delta^{it} U(g)^*
+  = U(g) U\parens[\big]{0,\Lambda(t)} U(g)^*
+  = U\parens[\big]{g(0,\Lambda(t))g^{-1}}
+\end{equation*}
+
+
+Recall that Stones Theorem \todo{add reference} states that
+every strongly continuous one-parameter unitary group
+is of the form $t \mapsto e^{itK}$ with a uniquely determined
+selfadjoint operator $K$, which is called \emph{infinitesimal generator} of the group.
+
+\begin{definition}{Modular Hamiltonian}{}
+  The infinitesimal generator of the modular group associated to a spacetime region $\spacetimeregion{O}$ is called the
+  \emph{modular Hamiltonian}\index{modular!Hamiltonian}\nomenclature{$K_{\spacetimeregion{O}}$}{modular Hamiltonian for $\spacetimeregion{O}$}
+  for said region, and denoted $K_{\spacetimeregion{O}}$.
+\end{definition}
+
+In other words, $K_{\spacetimeregion{O}}$ is the unique selfadjoint operator such that $\Delta_{\spacetimeregion{O}}^{it} = e^{itK_{\spacetimeregion{O}}}$ for all $t \in \RR$.
+
+\begin{proposition}{}{}
+  The modular Hamiltonian for the right wedge is given by $d \Gamma(A)$, where
+  \begin{equation*}
+    A\psi(p) = - \frac{2\pi}{i} \parens[\big]{\partial_0 \psi(p) \, p^1 + \partial_1 \psi(p) \, p^0}
+  \end{equation*}
+\end{proposition}
+
+\section{Complex Lorentz Transformations}
+
+\subsection{Analytic Continuation of the Space-Time Translation Group}
+
+\subsection{Complex Lorentz Boosts}
+
+\begin{lemma}{}{}
+  Suppose $A$ is a selfadjoint operator on some Hilbert space $\hilb{H}$.
+  For all complex numbers $z$ define a closed normal operator $V(z) = e^{izA}$ by means of functional calculus.
+  Let $g$ be a xxx function. Then the range of the bounded operator $g(A)$ is contained in the domain of $V(z)$ for all $z$, and
+  \begin{equation*}
+    V(z) g(A) = \int e^{iz \lambda} g(\lambda) dE_A(\lambda).
+  \end{equation*}
+\end{lemma}
+
+\subsection{A Convolution Theorem for Vector-Valued Tempered Distributions}
+
+\blockcquote{Bisognano1975}{%
+  The extension to vector-valued tempered distributions is trivial.
+}
+
+
+
+\chapterbib
+\cleardoublepage
+
+% vim: syntax=mytex
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