path: root/much.tex

\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) \FockVacuum}
\end{equation*}

\section{Misc}

\todo{Put this somewhere 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}} = \Set{b(\varphi(f)) \mid \text{$b$ bounded}, 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 $\BoundedLinearOperators{\hilb{H}}$.

\begin{definition}{Cyclic and Separating Vectors}{}
  Suppose $\hilb{H}$ is a Hilbert space and $\mathcal{A}$ is a $C^*$-subalgebra of $\BoundedLinearOperators{\hilb{H}}$.
  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[A]{$\mathcal{A}'$}{commutant of $\mathcal{A}$}

\begin{proposition}{}{cyclic-separating}
  Let $\hilb{H}$ be a Hilbert space and $\mathcal{A}$ be a $C^*$-subalgebra of $\BoundedLinearOperators{\hilb{H}}$.
  \begin{enumerate}
    \item \label{item:first} A vector is cyclic for $\mathcal{A}$ if and only if it is separating for $\mathcal{A}'$.
    \item \label{item:second} If $\mathcal{A}$ is a von Neumann algebra, then a vector is cyclic and separating for $\mathcal{A}$
  if and only if it is cyclic and separating for $\mathcal{A}'$.
  \end{enumerate}
\end{proposition}

\begin{proof}
  First, suppose that $\Omega \in \hilb{H}$ is cyclic for $\mathcal{A}$.
  If $A'$ is an element of $\mathcal{A}'$ with $A' \Omega = 0$,
  then $A' A \Omega = A A' \Omega = 0$ for all $A \in \mathcal{A}$.
  This means that $A'$ vanishes on the dense subspace $\mathcal{A} \Omega$,
  and thus on all of $\hilb{H}$, i.e.\ $A'=0$.
  This proves that $\Omega$ is separating for $\mathcal{A}'$.

  Conversely, suppose that $\Omega$ is separating for $\mathcal{A}'$.
  We have to show that the closed subspace $\overline{\mathcal{A}\Omega}$ is all of $\hilb{H}$.
  Let $P$ be the orthogonal projection onto $\overline{\mathcal{A}\Omega}$.
  Clearly, any element $A$ of $\mathcal{A}$ maps $\overline{\mathcal{A}\Omega}$ into itself.
  Thus, $PAP=AP$, and the same holds for $A^*$, that is, $PA^*P=A^*P$.
  Taking the adjoint of the second equation, we get $PAP=PA$. Hence, $P \in \mathcal{A}'$.
  Now, $P \Omega = \Omega = I \Omega$, where $I$ is the identity operator on $\hilb{H}$ which obviously also belongs to $\mathcal{A}'$,
  and the assumption that $\Omega$ is separating for $\mathcal{A'}$ implies $P=I$.
  Consequently, $\overline{\mathcal{A} \Omega} = \hilb{H}$.

  Statement~\ref{item:second} directly follows from~\ref{item:first} and
  the fact that $\mathcal{A}'' = \mathcal{A}$ by the Double Commutant Theorem.
\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 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, in analogy 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 B^*\Omega \qquad B \in \mathcal{A'}.
  \end{equation*}
  By definitions 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
  $S_0 A\Omega = A^*\Omega$
  for $A \in \mathcal{A}$
  is called the
  \emph{Tomita operator}\index{Tomita operator}\index{operator!Tomita}\nomenclature[S]{$S$}{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 will be 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 of an operator 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]{$J$}{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 $\hilb{H}$.

\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~\cite[Theorem 18.6]{Zhu1993},
  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 $S = \operatorclosure{S_0}$ be the Tomita operator associated to $(\vNa{M},\Omega)$,
  and let $S' = \operatorclosure{S'_0}$ 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}

Finally, let us outline how modular theory enters into algebraic quantum field theory.

\begin{theorem}{Reeh--Schlieder Theorem}{reeh-schlieder}
  Let $\spacetimeregion{O}$ be any open spacetime region.
  Then the vacuum vector $\Omega$ is cyclic for $\localalg{\spacetimeregion{O}}$.
  If $\spacetimeregion{O}'$ is non-empty, then $\Omega$ is also separating for $\localalg{\spacetimeregion{O}}$.
\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}{wedge}
  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 \Set[\big]{x \in M \given x^1 > \abs{x^0}} 
    \quad \text{and} \quad
    \leftwedge \defequal \Set[\big]{x \in M \given 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 full 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}
  \label{equation:lorentz-boost}
  \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}
    \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}

It can easily be seen that $\rightwedge' = \leftwedge$, and so \cref{theorem:reeh-schlieder} applies.

\begin{theorem}{Bisognano--Wichmann Theorem \textmd{\cite{Bisognano1975}}}{bisognano-wichmann}
  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}

\todo{write down the simpler proof}

\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 $a \in \RR \cup \Set{-\infty}$, $b \in \RR \cup \Set{\infty}$.
  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 Stone's 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[KO]{$K_{\spacetimeregion{O}}$}{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}
\todo{domain, proof}

\bluetext{Maybe this is simpler in Rindler coordinates...}

\section{Complex Lorentz Transformations}

The main result of this section is \cref{proposition:main-result}.

By definition, the \emph{complex Lorentz group}\index{Lorentz group!complex}\nomenclature[LC]{$\ComplexLorentzGroup$}{complex Lorentz group} $\ComplexLorentzGroup$ is the isometry group
of complex Minkowski space $M+iM \cong \CC^4$ with respect to the inner product
\begin{equation*}
  \innerp{z_1}{z_2} = \innerp{x_1}{x_2} - \innerp{y_1}{y_2} + i \parens[\big]{\innerp{x_1}{y_2} + \innerp{x_2}{y_1}}.
\end{equation*}
The \emph{complex Poincaré group}\index{Poincaré group!complex}\nomenclature[PC]{$\ComplexPoincareGroup$}{complex Poincaré group} is the semidirect product $\ComplexPoincareGroup \defequal \CC^4 \ltimes \ComplexLorentzGroup$.
The action of $\ComplexPoincareGroup$ on $M+iM$ is defined in the obvious way.
The complex Poincaré group has just two connected components, the subgroup $\ProperComplexPoincareGroup$ and the subset $\ImproperComplexPoincareTransformations$,
discriminated by the sign of $\det \Lambda \in \Set{\pm 1}$ for its elements $(z,\Lambda)$.
The (real) proper orthochronous Poincaré group $\ProperOrthochronousPoincareGroup$ is a subgroup of $\ProperComplexPoincareGroup$.
Each of the two following sections deals with a subgroup $G$ of $\smash{\ProperOrthochronousPoincareGroup}$,
and the possibility of extending a unitary representation of $G$ to a larger set within $\ProperComplexPoincareGroup$.

\subsection{Analytic Continuation of the Space-Time Translation Group}

%\todo{a short intro}

Let $a \mapsto U(a)$ be a strongly continuous unitary representation of the additive group of $\RR^4$ (on some separable Hilbert space).
By a generalization of Stone's Theorem~\cite[Theorem VIII.12]{ReedSimon1},
there exists a unique projection-valued measure $E$ on $\RR^4$ such that
\begin{equation}
  \label{equation:spectral-resolution-translation}
  U(a) = \int_{\RR^4} \exp(ia \cdot k) \, dE(k) \qquad a \in \RR^4.
\end{equation}
Then one can define a vector $P$ of unbounded selfadjoint operators
\begin{equation*}
  P_i = \int_{\RR^4} k_i \, dE(k) \qquad i=0,\ldots,3
\end{equation*}
which have a common dense domain $D$ and satisfy
\begin{equation*}
  a \cdot P = \int_{\RR^4} a \cdot k \, dE(k) \qquad a \in \RR^4.
\end{equation*}
We are specifically interested in the representation
\begin{equation*}
  U(a) \defequal U(a,I) \qquad a \in \RR^4
\end{equation*}
obtained by restricting the unitary representation of the Poincaré group $\RestrictedPoincareGroup$ on Fock space to the subgroup of spacetime translations.
In this case the vector operator $P$ carries the physical meaning of energy-momentum,
and we impose the so-called \emph{spectrum condition}
\begin{equation*}
  \langle a \cdot P \rangle_{\psi} \ge 0 \qquad
  \forall \psi \in D \;
  \forall a \in \ClosedForwardCone,
\end{equation*}
where $\ClosedForwardCone \defequal \Set{a \in \RR^4 \given a \cdot a \ge 0, a^0 \ge 0}$ is the \emph{closed forward cone}\index{cone!closed forward}\nomenclature[V]{$\ClosedForwardCone$}{closed forward cone}.
It can be shown \cite{Uhlmann1961} that the spectrum condition is equivalent to the statement that
the support of the spectral measure is contained in the closed forward cone, i.e.\ $\supp(E) \subset \ClosedForwardCone$.

Spectral calculus allows us to extend $a \mapsto U(a)$ to complex arguments $z \in \CC^4$ by simply replacing $a$ with $z$ in the spectral resolution~\eqref{equation:spectral-resolution-translation} of $U(a)$. 
However, one obtains, in general, an unbounded operator.
It is a consequence of the spectrum condition that $U(z)$ is bounded whenever $z$ lies in the \emph{closed forward tube}\index{tube!closed}\nomenclature[T]{$\ClosedForwardTube$}{closed forward tube} $\ClosedForwardTube \defequal \RR^4 + i\ClosedForwardCone$.
Observe that the set $\ClosedForwardTube$ is closed under vector addition and thus forms a commutative monoid; it is not a group.

\begin{proposition}{}{}
  For every $z \in \ClosedForwardTube$ the operator
  \nomenclature[U]{$U(z)$}{complex translation}
  \begin{equation}
    \label{equation:definition-complex-translation}
    U(z) \defequal \int_{\ClosedForwardCone} \exp(iz \cdot k) \, dE(k)
  \end{equation}
  is bounded.
  Moreover, $U(w+z) = U(w) U(z)$ for all $w,z \in \ClosedForwardTube$.
\end{proposition}

\begin{proof}
  By a general property of spectral integrals~\cite[Proposition 4.18]{Schmüdgen2012},
  the operator $U(z)$ is bounded if (and only if)
  the function $f(k) = \exp(iz \cdot k)$ is bounded $E$-almost everywhere.
  In view of the fact that $E$ is supported in the closed forward cone $\ClosedForwardCone$,
  it is sufficient to show that $f$ is bounded on $\ClosedForwardCone$.
  %the $E$-essential supremum of the function
  Since $z$ lies in the closed forward tube, $z=x+iy$ with $x \in \RR^4$ and $y \in \ClosedForwardCone$.
  Now $\abs{f(k)} = \exp(-y \cdot k)$, and on $\ClosedForwardCone$ this is bounded by $1$ because $y \cdot k \ge 0$ for all $k \in \ClosedForwardCone$.

  The identity $U(w+z) = U(w) U(z)$ follows from $\exp(i(w+z) \cdot k) = \exp(iw \cdot k)\*\exp(iz \cdot k)$
  and the boundedness of the operators, see~\cite[Proposition 4.16(iii) and (v)]{Schmüdgen2012}.
\end{proof}

\begin{proposition}{}{}
  If $(b,\Lambda) \in \RestrictedPoincareGroup$ and $z \in \ClosedForwardTube$, then $\Lambda z \in \ClosedForwardTube$ and
  \begin{equation*}
    U(b,\Lambda) U(z) U(b,\Lambda)^* = U(\Lambda z).
  \end{equation*}
\end{proposition}

\begin{proof}
  We exploit the uniqueness of the projection-valued measure $E$ satisfying~\eqref{equation:spectral-resolution-translation}.
  Since $\Lambda$ acts continuously on $\RR^4$, $\Lambda^{-1} S$ is a Borel set whenever $S \subset \RR^4$ is, and
  \begin{equation*}
    F(S) \defequal U(b,\Lambda) E(\Lambda^{-1} S) U(b,\Lambda)^*
    \qquad S \in \BorelSigmaAlgebra{\RR^4}
  \end{equation*}
  is a well-defined projection-valued measure on $\RR^4$.
  By the Transformation Formula for Spectral Integrals~\cite[Proposition 4.24]{Schmüdgen2012}, we have
  \begin{align}
    \int_{\RR^4} \exp(iz \cdot k) \, dF(k)
    &= U(b,\Lambda) \bracks[\bigg]{\,\int_{\RR^4} \!\exp(iz \cdot \Lambda k) \, dE(k)} U(b,\Lambda)^* \nonumber\\
    &= U(b,\Lambda) \bracks[\bigg]{\,\int_{\RR^4} \!\exp(i \Lambda^{-1} z \cdot k) \, dE(k)} U(b,\Lambda)^* \nonumber\\
    \label{equation:F-integral}
    &= U(b,\Lambda) U(\Lambda^{-1} z)  U(b,\Lambda)^*
  \end{align}
  for all $z \in \ClosedForwardTube$.
  In particular, for $z=a \in \RR^4$ the last term equals
  \begin{equation*}
    U(b,\Lambda) U(\Lambda^{-1} a)  U(b,\Lambda)^* = U(a) \qquad a \in \RR^4,
  \end{equation*}
  which can be seen by applying $U$ to the identity
  \begin{equation*}
    (b,\Lambda) (\Lambda^{-1} a, I) (b,\Lambda)^{-1} = (a,I).
  \end{equation*}
  Hence, $U(a) = \int \exp(ia \cdot k) dF(k)$ for all $a \in \RR^4$. We conclude $E = F$.
  Now~\eqref{equation:F-integral} asserts that $U(z) = U(b,\Lambda) U(\Lambda^{-1} z) U(b,\Lambda)$ for all $z \in \ClosedForwardTube$
  and a substitution of $z$ by $\Lambda z$ yields the desired identity.
\end{proof}

%\begin{proposition}{Analyticity of the Complex Translation Monoid}{analyticity-complex-translations}
  \begin{proposition}{\textmd{\cite[Theorem 4]{Uhlmann1961}}}{analyticity-complex-translations}
    The operator-valued map $z \mapsto U(z)$ given by~\eqref{equation:definition-complex-translation} is
  strongly continuous on $\ClosedForwardTube$ and
  analytic on $\OpenForwardTube$.
\end{proposition}

\todo{Explain what it means for an operator-valued function of several complex variables to be analytic.}

\begin{lemma}{}{complex-translation}
  Let $g = (b,\Lambda)$ be a proper orthochronous Poincaré transform with $b \in \OpenForwardCone$.
  Then, for all $z \in \OpenForwardTube$
  \begin{equation*}
    gz \in \OpenForwardTube \qquad
    U(g) U(z) = U(gz).
  \end{equation*}
\end{lemma}

\begin{proof}
  \bluetext{Edge of the Wedge}
\end{proof}

Next we consider an operator-valued tempered distribution $u$ that is \emph{covariant}
in the sense that it obeys the relativistic transformation law
\begin{equation}
  \label{equation:covariance-distribution}
  U(g) u(f) U(g)^* = u(f_g) \qquad g \in \RestrictedPoincareGroup, f \in \schwartz{\RR^4},
\end{equation}
where $f_g(x) = f(g^{-1} x)$ for all $x \in M$.
In particular, if $g=(a,I)$ is the translation by a vector $a \in \RR^4$,
then~\eqref{equation:covariance-distribution} and the invariance of the vacuum vector $\FockVacuum$ imply
\begin{equation}
  \label{equation:real-translation-law}
  U(a) u(f) \FockVacuum = u(f_a) \FockVacuum \qquad \forall a \in \RR^4.
\end{equation}
We would like to extend this law to complex translation vectors,
but translating a function defined on $\RR^4$ by a complex vector is not a sensible operation.
Nevertheless, we have $\FT{f_a}(p) = \exp(ia \cdot p) \ft{f}(p)$ in Fourier space,
and $\exp(iz \cdot p) \ft{f}(p)$ is a well defined function of $p \in \RR^4$ even when $z \in \CC^4$.
The obvious idea would be to define $f_z$ as the inverse Fourier transform of this function.
This does not work because $\exp(iz \cdot p) \ft{f}(p)$ is generally not in the Schwartz class.
However, thanks to the spectrum condition we may modify this function outside of the closed forward cone.

\begin{lemma}{}{depends-only-on-restriction}
  Let $u$ be a covariant operator-valued tempered distribution.
  Then the vector $u(f) \FockVacuum$, where $f \in \schwartz{\RR^4}$,
  depends only on the restriction of $\ft{f}$ to $\ClosedForwardCone$.
\end{lemma}

\begin{proof}
  We consider a Schwartz function $g \in \schwartz{\RR^4}$ and
  the operator $G = \int g(k) dE(k)$,
  where $E$ is the unique projection-valued measure on $\RR^4$ such that
  $U(a) = \int \exp(ik \cdot a) dE(k)$ for all $a \in \RR^4$.
  Let $g(k) = (2 \pi)^{-2} \int \ift{g}(a) \exp(ia \cdot k) da$ be the Fourier decomposition of $g$.
  \begin{multline*}
    \hspace{1cm} (2 \pi)^2 G = \int_{\ClosedForwardCone} \!\int_{\RR} \ift{g}(a) \exp(ia \cdot k) \, da \, dE(k) = \\
    = \int_{\RR} \ift{g}(a) \!\int_{\ClosedForwardCone} \exp(ia \cdot k) \, dE(k) \ da
    = \int_{\RR} \ift{g}(a) U(a) \, da \hspace{1cm}
  \end{multline*}
  \question{Darf ich hier wirklich die Integrationsreihenfolge vertauschen?}

  Recall that the Fourier transform of $u$ is defined by $\ft{u}(f) = u(\ft{f}@@)$ for $f \in \schwartz{\RR^4}$.
  We obtain the action of the translation group on $\ft{u}(\ft{f}@@)\FockVacuum$ by definition chasing and~\eqref{equation:real-translation-law}:
  \begin{equation*}
    U(a) \ft{u}(\ft{f}@@)\FockVacuum
    = U(a) u(f)\FockVacuum
    = u(f_a)\FockVacuum
    = \ft{u}(\ft{f}_a)\FockVacuum
    = \ft{u}(\ft{f}e_a)\FockVacuum
  \end{equation*}
  Here $e_a$ stands for the function $e_a(p) = \exp(ia \cdot p)$.
  \begin{equation*}
    G @\ft{u}(\ft{f}@@)\FockVacuum
    = \int \ift{g}(a) @\ft{u}(\ft{f}e_a)\FockVacuum \, da
    = \ft{u} \parens[\bigg]{\ft{f} \int \ift{g}(a) e_a da} \FockVacuum 
    = \ft{u}(\ft{f} g) \FockVacuum
  \end{equation*}
  The second identity is due to the continuity of the vector-valued map $f \mapsto u(f) \FockVacuum$.
  If the support of $g$ does not intersect the support of $E$, i.e.\ the closed forward cone, then $G=0$.
  Thus, $\ft{u}(\ft{f} g) \FockVacuum = 0$.
  This proves that $u(f_1) \FockVacuum = u(f_2) \FockVacuum$ when $\supp(\ft{f_1} - \ft{f_2}) \subset \ClosedForwardCone$.
\end{proof}

This fact inspires the definition
\begin{equation*}
  f_z \defequal d_z * f \qquad z \in \ClosedForwardTube
\end{equation*}
where $d_z$ is any Schwartz function on $\RR^4$ such that $\FT{d_z}(p) = \exp(iz \cdot p)$ for all $p \in \ClosedForwardCone$.
Such a function does exist \todo{elaborate, smooth cutoff}. Then $f_z$ will be Schwartz class as well.
Moreover, $u(f_z) \FockVacuum$ does not depend on the specific choice of $d_z$, by~\cref{lemma:depends-only-on-restriction}.

%\begin{lemma}{}{}
  %For every $z \in \ClosedForwardTube$ there exists a Schwartz function $d_z \in \schwartz{\RR^4}$
  %such that $\ft{e_z} \in \schwartz{\RR^4}$ and $\ft{e_z}(p) = \exp(iz \cdot p)$ for $p \in \ClosedForwardCone$.
%\end{lemma}

\begin{proposition}{}{prp}
  Let $u$ be a covariant operator-valued tempered distribution,
  and let $f \in \schwartz{\RR^4}$ be a test function. Then we have,
  in generalization of~\eqref{equation:real-translation-law},
  \begin{equation*}
    U(z) u(f) \FockVacuum = u(f_z) \FockVacuum \qquad \forall z \in T_+.
  \end{equation*}
\end{proposition}

\begin{proof}
  By \cref{proposition:analyticity-complex-translations},
  the function $z \mapsto U(z) u(f) \FockVacuum$ is analytic on the open forward tube.
  \todo{Zeige, dass $z \mapsto u(f_z) \FockVacuum$ ebenfalls analytisch ist.
  Dann folgt die Behauptung wohl mit Edge of the Wedge~\cite[Theorem 2-17]{Streater1964}}
\end{proof}

\begin{corollary}{}{}
  Let $u$ be a covariant operator-valued tempered distribution,
  and let $f \in \schwartz{\RR^4}$ be a test function. Then we have,
  \begin{equation*}
    U(z) u(f) \FockVacuum = \int dx \, f(x) \, u(d_{z+x}) \FockVacuum \qquad \forall z \in T_+.
  \end{equation*}
\end{corollary}

\begin{proof}
  The convolution formula \cref{proposition:vector-valued-convolution-formula} applied to the vector-valued distribution defined by $f \mapsto \alpha(f) = u(f) \FockVacuum$ yields
  \begin{equation*}
    (\alpha * \tilde{d}_z)(f) = \int dx \, f(x) \, \alpha(\tau_x d_z)
  \end{equation*}
  Using \cref{proposition:prp}, we calculate
  \begin{equation*}
    (\alpha * \tilde{d}_z)(f) = \alpha(d_z * f) = \alpha(f_z) = u(f_z) \FockVacuum = U(z) u(f) \FockVacuum.
  \end{equation*}
  It is easily seen by Fourier transformation that $\tau_x d_z = d_{x+z}$.
  Hence, $\alpha(\tau_x d_z) = u(d_{x+z}) \FockVacuum$.
\end{proof}

\subsection{Complex Lorentz Boosts}

The Lorentz boosts $\Lambda(t)$ given by
the matrices~\eqref{equation:lorentz-boost} in standard representation
have a natural interpretation for complex parameters,
since the hyperbolic functions $\cosh$ and $\sinh$ extend analytically to the whole complex plane.
In view of Lemma xxx it follows immediately that the matrix-valued function $\CC \ni w \mapsto \Lambda(w)$ is entire analytic.
In particular, the vector-valued function $\CC \ni w \mapsto \Lambda(w) z$ is entire analytic for every fixed vector $z \in \CC^4$.

We are particularly interested in the case of a purely imaginary parameter.
The relations $\cosh iz = \cos z$ and $\sinh iz = i \sin z$
between the complex hyperbolic and trigonometric functions imply
\begin{equation*}
  \Lambda(is) = \begin{pmatrix}
    \phantom{i}\cos(2 \pi @ s) & i\sin(2 \pi @ s) & \; 0 \; & \; 0 \; \\
    i\sin(2 \pi @ s) & \phantom{i}\cos(2 \pi @ s) & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
  \end{pmatrix}
  \qquad \forall s \in \RR.
\end{equation*}
For later use, we give the action of $\Lambda(is)$ on a complex four-vector $x+iy$:
\begin{equation}
  \label{equation:pure-imaginary-lorentz-boost}
  \Lambda(is) (x+iy) =
  \begin{pmatrix}
    \cos(2 \pi @ s) x^0 - \sin(2 \pi @ s) y^1 \\
    \cos(2 \pi @ s) x^1 - \sin(2 \pi @ s) y^0 \\
    x^2 \\
    x^3 
  \end{pmatrix}
  +i
  \begin{pmatrix}
    \sin(2 \pi @ s) x^1 + \cos(2 \pi @ s) y^0 \\
    \sin(2 \pi @ s) x^0 + \cos(2 \pi @ s) y^1 \\
    y^2 \\
    y^3 
  \end{pmatrix}
\end{equation}

We

\begin{equation*}
  \mathcal{J} \defequal \Lambda(i/2) = \begin{pmatrix}
    -1 & 0 & \; 0 \; & \; 0 \; \\
    0 & -1 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
  \end{pmatrix}
\end{equation*}

\begin{equation*}
  \mathcal{J}_{\pm} \defequal \Lambda(\pm i/4) = \begin{pmatrix}
    0 & \pm i & \; 0 \; & \; 0 \; \\
    \pm i & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
  \end{pmatrix}
\end{equation*}

We now turn to the unitary representation of (real) Lorentz boosts
\begin{equation*}
  V(t) \defequal U \parens[\big]{0,\Lambda(t)} \qquad t \in \RR
\end{equation*}
on Fock space and aim for an analytic extension similar to the previous section.
By Stone's theorem theorem there exists a unique selfadjoint operator $K$ such that
\begin{equation*}
  V(t) = \exp(itK) = \int_{\RR} \exp(it \lambda) \,dE_K(\lambda),
\end{equation*}
where $E_K$ is the spectral measure on $\RR$ associated to $K$.
Now we define \emph{complex Lorentz boosts} to be the operators
\nomenclature[V]{$V(z)$}{complex Lorentz boost}
\begin{equation*}
  V(z) \defequal \int_{\RR} \exp(iz \lambda) \,dE_K(\lambda) \qquad z \in \CC.
\end{equation*}
In contrast to the previous section, we


\begin{lemma}{}{}
  Suppose $A$ is a selfadjoint unbounded operator on some Hilbert space $\hilb{H}$.
  For each complex number $z$ define the closed normal operator $V(z) = e^{izA}$ by means of functional calculus.
  Let $g \in \schwartz{\RR}$ be a Schwartz function.
  \begin{enumerate}
    \item $V(z) V(w) = V(z + w)$ for all $z,w \in \CC$.
    \item The operator $g(A)$ is bounded, and its range is contained in the domain of $V(z)$ for all $z \in \CC$.
    \item The operator $V(z) g(A)$ is bounded for all $z \in \CC$, and has spectral resolution
      \begin{equation*}
        V(z) g(A) = \int e^{iz \lambda} g(\lambda) dE_A(\lambda).
      \end{equation*}
    \item The function $z \mapsto V(z) g(A)$ is entire analytic.
  \end{enumerate}
\end{lemma}

Remember that a \emph{core}\index{operator!core for an} for a closed densely defined unbounded operator $T$
is, by definition, a linear subspace $\mathcal{D}_0$ of its domain $\Domain{T}$ such that
the closure of the restriction of $T$ to $\mathcal{D}_0$ coincides with $T$.
%symbolically $\overline{T \vert \mathcal{D}_0} = T$.
Each core of $T$ is necessarily a dense subspace of $\Domain{T}$,
but a dense subspace of $\Domain{T}$ need not be a core for $T$.

\begin{lemma}{A Common Core for All Complex Lorentz Boosts}{common-core-for-complex-lorentz-boots}
  Adopt the notation of the foregoing lemma. The linear subspace
  \begin{equation*}
    \mathcal{D}_0 = \Span \Set{\ran g(K) \given g \in \schwartz{\RR}} 
  \end{equation*}
  is a core for $V(z)$ for every $z \in \CC$.
\end{lemma}

\begin{proof}
  $M_n = \bracks{-n,n}$
\end{proof}

\subsection{Application to the Energy Density}

The following three lemmas are variations of the arguments
brought forward by~\citeauthor{Bisognano1975} in their proof of \cref{theorem:bisognano-wichmann}.
The main difference is that we state \cref{lemma:biso1} and \cref{lemma:biso2} as operator identities without reference to a field operator,
and proof \cref{lemma:biso3} for arbitrary Lorentz-covariant operator-valued distributions
rather than products of field operators.
This generalization is necessary for the application to the energy density.
In addition, we provide in \cref{chapter:convolution} a complete proof of the convolution formula for vector-valued distributions.

Roughly speaking, the following lemma asserts that a translation by a complex vector
followed by a suitable imaginary boost is again a complex translation.

\begin{lemma}{}{biso1}
Let $z = x + iy \in \OpenForwardTube$ with $x,y$ real, and suppose $y^2=y^3=0$.
  \begin{enumerate}
    \item If $x \in \rightwedge$, then for all $s \in [0,\tfrac{1}{4}]$ 
      \begin{equation*}
        \Lambda(is) z \in \OpenForwardTube, \qquad
        \ran U(z) \subset \dom V(is), \qquad 
  V(is) U(z) = U \parens[\big]{\Lambda(is) z}.
      \end{equation*}
    \item If $x \in \leftwedge$, then the above holds for all $s \in [0,-\tfrac{1}{4}]$.
  \end{enumerate}
\end{lemma}
\nomenclature[dom]{$\dom T$}{domain of the operator $T$\nomnorefpage}
\nomenclature[ran]{$\ran T$}{range of the operator $T$\nomnorefpage}

\begin{proof}
  The result of applying $\Lambda(is)$ for some $s \in \RR$ to a complex four-vector $x+iy$
  has been given in~\eqref{equation:pure-imaginary-lorentz-boost}.
  To show that this vector lies in the open forward tube,
  we need to verify that its imaginary part lies in the open forward cone $\OpenForwardCone$.
  By definition, a real four-vector $a$ lies in $\OpenForwardCone$ if and only if
  $a^0 > 0$ and $a \cdot a > 0$. If $a^2 = a^3 = 0$, then
  these conditions are easily seen to be equivalent to the conditions
  $a^0 > 0$ and $a^0 \mp a^1 > 0$ (for both sign choices).

  Since the second and third components of the imaginary part of~\eqref{equation:pure-imaginary-lorentz-boost},
  $y^2$ and $y^3$, vanish by assumption, it is sufficient to prove
  \begin{equation}
    \label{equation:inequalities}
    \begin{aligned}
      \sin(2 \pi s) x^1 + \cos(2 \pi s) y^0 &> 0 \ \text{and} \\
      \sin(2 \pi s) \parens[\big]{x^1 \mp x^0} + \cos(2 \pi s) \parens[\big]{y^0 \mp y^1} &> 0.
    \end{aligned}
  \end{equation}
  The assumption $x \in \rightwedge$ implies that
  $x^1 > 0$ and $x^1 \mp x^0 > 0$,
  by \cref{definition:wedge}.
  The assumptions $y \in \OpenForwardCone$ and $y^2 = y^3 = 0$ imply that
  $y^0 > 0$ and $y^0 \mp y^1 > 0$,
  by the argument in the foregoing paragraph.
  So, all we need to do to ensure~\eqref{equation:inequalities} holds,
  is choose $s$ such that both $\sin(2 \pi s)$ and $\cos(2 \pi s)$ are nonnegative.
  (Then, at least one of these will be positive.)
  Clearly, this is true for all $s \in \bracks{0,\tfrac{1}{4}}$.

  \noindent\begin{minipage}{0.5\textwidth}
  \hspace{\parindent} The vector-valued function $\CC \ni s \mapsto \Lambda(is) z$ is entire analytic.
  In particular it is continuous, and we have shown that it maps the compact subset $[0,1/4]$ into the open set $\OpenForwardTube$.
  This implies that there exists a connected open neighborhood $N \subset \CC$ of $[0,1/4]$ such that $\Lambda(is) z \in \OpenForwardTube$ for all $s \in N$.

  \end{minipage} \hfill
  \begin{minipage}{0.45\textwidth}
    \begin{center}
    \begin{tikzpicture}[baseline=10]
      \draw[->] (-1,0) -- (3,0) node[right] {\footnotesize$\Real s$};
      \draw[->] (0,-1) -- (0,2) node[left] {\footnotesize$\Imag s$};
      \draw[faunat,thick,{Parenthesis[]}-{Parenthesis[]}] (0,-0.5) -- (0,0.5);
      \draw[thick,{Bracket[]}-{Bracket[]}] (-0.3pt,0) -- (2,0);
      \draw (1,1) node {$N$};
      \draw (2,0) node[above] {\footnotesize$\tfrac{1}{4}$};
      \draw plot [smooth cycle] coordinates {(2.5,0) (2,1) (1,0.7) (0.3,1) (-0.5,0.7) (-0.3,-0.7) (1.6,-0.7)};
    \end{tikzpicture}
    \end{center}
  \end{minipage}

  Let $\xi \in \hilb{F}$ be arbitrary, and let $\eta$ be in the common dense domain $\mathcal{D}_0$ of the operators $V(is)$ from \cref{lemma:common-core-for-complex-lorentz-boots}.
  Then the function $f_1(s) = \innerp{V(is)^* \eta}{U(z) \xi}$ is well-defined, and entire analytic by Lemma xxx.
  The function $f_2(s) = \innerp{\eta}{U(\Lambda(is) z) \xi}$ is analytic on $N$, by \cref{proposition:analyticity-complex-translations}.
  Since $N$ is an open neighborhood of $0$, there is an $\epsilon >0$ such that $i(-\epsilon,\epsilon) \subset N$.
  By \cref{lemma:complex-translation}, $V(is) U(z) = U(\Lambda(is) z)$ if $is \in \RR$.
  Hence, $f_1$ and $f_2$ agree in an open real neighborhood of $is$.
  Now the Identity Principle implies $f_1 \equiv f_2$ on $N$.
  Since this holds for all $\eta$ in $\mathcal{D}_0$,
  which is a core for $V(is)^{**} = V(is)$,
  we conclude that $U(z) \xi$ lies in the domain of $V(is)$,
  and $V(is) U(z) \xi = U(\Lambda(is) z) \xi$.
  As $\xi$ was arbitary, the proof is complete.
\end{proof}

Remember that $\mathcal{J} = \Lambda(i/2) = \diag(-1,-1,1,1)$.

\begin{lemma}{}{biso2}
  Let $x \in \rightwedge$, and let $e_0 = (1,0,0,0)$ be the forward timelike unit vector. Then
\begin{equation*}
  \stronglim_{\varepsilon \downarrow 0} V(i/4) U(x+i \varepsilon e_0)
  = U \parens[\big]{\Lambda(i/4)x} 
  = \stronglim_{\varepsilon \downarrow 0} V(-i/4) U(\mathcal{J}x+i \varepsilon e_0)
\end{equation*}
\end{lemma}

\begin{proof}
  The first identity follows from \cref{lemma:biso2}(i)
  and the strong continuity of $z \mapsto U(z)$ on $\ClosedForwardTube$ (\cref{proposition:analyticity-complex-translations}).
  For the second identity, note that $\mathcal{J} x \in \leftwedge$ and
  apply \cref{lemma:biso2}(ii), then use $\Lambda(-i/4) \mathcal{J} = \Lambda(i/4)$.
\end{proof}

\begin{lemma}{}{biso3}
  Suppose that $u$ is a covariant operator-valued tempered distribution.
  Let $f \in \schwartz{M}$ with $\supp f \subset \rightwedge$, and
  let $g \in \schwartz{M}$ be arbitrary. Then
  \begin{equation*}
    V(i/2) g(K) u(f) \FockVacuum = g(K) u(f_{\mathcal{J}}) \FockVacuum
  \end{equation*}
\end{lemma}

Here, $K$ is the infinitesimal generator of the group $t \mapsto V(t)$ of real Lorentz boosts,
$\FockVacuum$ is the Fock vacuum, and $\mathcal{J}$ is the Lorentz transformation given by the diagonal matrix $\diag(-1,-1,1,1)$.

\begin{proof}
  a
\end{proof}

\begin{equation*}
  \Delta^{-1/2} g(K) \energydensity(f) \FockVacuum = g(K) \energydensity(f^J) \FockVacuum
\end{equation*}

Die Anwendung auf die Energiedichte $\energydensity$:

\begin{proposition}{}{main-result}
  Suppose $W \subset M$ is any wedge domain, with associated modular operator $\Delta_W$ and modular Hamiltonian $K_W$.
  Let $f \in \schwartz{M}$ with $\supp f \subset W$, and
  let $h \in \schwartz{M}$ be arbitrary. Then
  \begin{equation*}
    \norm{\Delta_W^{-1/2} h(K_W) \energydensity(f) \FockVacuum}
    = \norm{h(K) \energydensity(f_{\mathcal{J}g}) \FockVacuum},
  \end{equation*}
  where $K$ is the modular Hamiltonian of the right wedge $\rightwedge$,
  and $g$ is any element of $\RestrictedPoincareGroup$ such that $W = g \rightwedge$,
  and $\mathcal{J} = \diag(-1,-1,1,1)$.
\end{proposition}
In der Ungleichung aus~\cite{Much2022} ist $h$ eine Gauß-Funktion.

\section{Calculating Gaussians of the Modular Hamiltonian}

coming soon\ldots

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