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@@ -1,23 +1,22 @@ -\chapter{A quantum energy inequality involving local modular data} - +\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} + - \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 somwhere else.} +\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} $\FullLorentzGroup$. + +The \emph{Lorentz group} $\FullLorentzGroup$. \begin{equation*} \FullPoincareGroup = \RR^4 \ltimes \FullLorentzGroup \end{equation*} @@ -44,7 +43,7 @@ Poincaré covariance \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}}'' + \localalg{\spacetimeregion{O}} = \braces{b(\varphi(f)) \mid \text{$b$ bounded}, f \in \realschwartz{M}, \supp f \subset \spacetimeregion{O}}'' \end{equation*} \end{definition} @@ -52,10 +51,10 @@ Poincaré covariance \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 $B(\hilb{H})$. +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 $B(\hilb{H})$. + 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 $\hilb{H}$. @@ -66,19 +65,37 @@ Occasionally, a vector that is both cyclic and separating is called \emph{standa 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 $\mathcal{A}$} +We shall denote the commutant of $\mathcal{S}$ by $\mathcal{S}'$.\nomenclature[A]{$\mathcal{A}'$}{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 $\vNa{M}'$. + 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} - \todo{xxx} - The second assertion directly follows from the first and the fact that $\vNa{M}'' = \vNa{M}$. + 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 $\mathcal{A}$, @@ -87,7 +104,7 @@ 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}. + \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, @@ -100,12 +117,12 @@ the range of $S_0$ coincides with its domain. \begin{proof} By \cref{proposition:cyclic-separating}, $\Omega$ is also cyclic and separating for the commutant $\vNa{A}'$. - Hence we may, analogously to $S_0$, + 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 F_0 B^*\Omega \qquad B \in \mathcal{A'}. + \quad F_0 B\Omega \defequal 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 $B \in \mathcal{A}'$ + 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} = @@ -124,10 +141,10 @@ the range of $S_0$ coincides with its domain. 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 - $S_0 A\Omega = S_0 A^*\Omega$ + $S_0 A\Omega = A^*\Omega$ for $A \in \mathcal{A}$ is called the - \emph{Tomita operator}\index{Tomita operator}\index{operator!Tomita}\nomenclature{$S$}{Tomita operator} + \emph{Tomita operator}\index{Tomita operator}\index{operator!Tomita}\nomenclature[S]{$S$}{Tomita operator} for the pair $(\mathcal{A},\Omega)$. \end{definition} @@ -135,7 +152,7 @@ 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. +as this will be our only use case. \begin{theorem}{Polar Decomposition for Anti-Linear Closed Operators}{polar-decomposition} \index{polar decomposition} @@ -156,7 +173,7 @@ as this is our only use case. \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 +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. @@ -170,7 +187,7 @@ Now we are able to introduce the fundamental objects of modular theory. \end{equation*} be its polar decomposition. The anti-unitary operator $J$ is called - \emph{modular conjugation}\index{modular!conjugation}\nomenclature{$J$}{modular conjugation}. + \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 @@ -187,8 +204,6 @@ Now we are able to introduce the fundamental objects of modular theory. 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 $\hilb{H}$. @@ -202,11 +217,11 @@ The modular group is a strongly continuous one-parameter unitary group on $\hilb \begin{proof} To prove the first assertion, consider any $A \in (U\vNa{M}U^*)''$. - By the double commutant theorem, + 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. + 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 $(U\vNa{M}U^*)'$. @@ -221,15 +236,15 @@ The modular group is a strongly continuous one-parameter unitary group on $\hilb 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 $(U\vNa{M}U^*,U\Omega)$. + 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 + (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 $S' = U S U^*$. @@ -243,12 +258,12 @@ The modular group is a strongly continuous one-parameter unitary group on $\hilb 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} +\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 $\localalg{\spacetimeregion{O}}$. @@ -286,7 +301,7 @@ since they are transformed into each other by space inversion. \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{ +with rapidity $2 \pi t$ is given by the matrix\footnote{% This matrix depends on the choice of metric signature. Ours is $(+,-,-,-)$. For $(-,+,+,+)$, use @@ -300,21 +315,22 @@ with rapidity $2 \pi t$ is given by the matrix\footnote{ \end{equation*} } -\begin{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*} +\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] + \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} @@ -349,7 +365,9 @@ with respect to the action of the Lorentz group on subsets of Minkowski space. 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}}}{} +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 @@ -400,7 +418,7 @@ That this is generally true is the statement of the following Lemma. for all Borel functions $f : \RR \to \CC$. \end{lemma} -\question{Ist diese Aussage korrekt? Ist mein Beweis richtig? Geht der auch einfacher?} +\todo{write down the simpler proof} \begin{proof} For each regular value $\lambda \in \rho(A)$ let @@ -453,14 +471,14 @@ That this is generally true is the statement of the following Lemma. \end{equation*} -Recall that Stones Theorem \todo{add reference} states that +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{$K_{\spacetimeregion{O}}$}{modular Hamiltonian for $\spacetimeregion{O}$} + \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} @@ -472,29 +490,453 @@ In other words, $K_{\spacetimeregion{O}}$ is the unique selfadjoint operator suc 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} \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 $\ImproperComplexPoincareTransformations$, +differentiated by the sign of $\det \Lambda \in \braces{\pm 1}$ 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 $\ProperOrthochronousPoincareGroup$, +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 $\ClosedForwardCone \defequal \braces{a \in \RR^4 \vcentcolon a \cdot a \ge 0, a^0 \ge 0}$ 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.} + +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}{}{} + 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} + \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*} + +\begin{equation*} + \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 \braces{\ran g(K) \vcentcolon g \in \schwartz{\RR}} + \end{equation*} + is a core for $V(z)$ for every $z \in \CC$. +\end{lemma} + +\begin{proof} + xxx +\end{proof} + +\subsection{Application to the Energy Density} + +\bluetext{Achtung: Dieser Abschnitt ist noch roh, lückenhaft und enthält inkonsistente Notation und falsche Aussagen.} +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 xxx and xxx as operator identities without reference to a field operator, +and proof xxx 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}{}{} +Let $z = x + iy \in \OpenForwardTube$ with $x,y$ real, and suppose $y^3=y^4=0$. + \begin{enumerate} + \item If $x \in \rightwedge$, then for all $s \in [0,1/4]$ + \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 $s \in [0,-1/4]$. + \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} + xxx + + \noindent\begin{minipage}{0.5\textwidth} + 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}. + By Lemma xxx, $f_1$ and $f_2$ agree in an open real neighborhood $is$. + Since $N$ is an open neighborhood of $0$, there is an $\epsilon >0$ such that $i(-\epsilon,\epsilon) \subset N$. + It follows that $f_1 \equiv f_2$ on $N$. + core \ldots +\end{proof} + \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 + Let $x \in \rightwedge$ +\begin{equation*} + \stronglim_{\varepsilon \downarrow 0} V(i/4) U(x+i \varepsilon e_0) + = U \parens[\big]{V(i/4)x} + = \stronglim_{\varepsilon \downarrow 0} V(-i/4) U(\mathcal{J}x+i \varepsilon e_0) +\end{equation*} +\end{lemma} + +\begin{proof} + xxx +\end{proof} + +\begin{lemma}{}{} + 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(z) g(A) = \int e^{iz \lambda} g(\lambda) dE_A(\lambda). + V(i/2) g(K) u(f) \FockVacuum = g(K) u(f_{\mathcal{J}}) \FockVacuum \end{equation*} \end{lemma} -\subsection{A Convolution Theorem for Vector-Valued Tempered Distributions} +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 $\diag(-1,-1,1,1)$. -\blockcquote{Bisognano1975}{% - The extension to vector-valued tempered distributions is trivial. -} +\begin{proof} + xxx +\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 \chapterbib \cleardoublepage |