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diff --git a/much.tex b/much.tex new file mode 100644 index 0000000..58aeab3 --- /dev/null +++ b/much.tex @@ -0,0 +1,502 @@ +\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 |