\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.
}



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