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diff --git a/stresstensor.tex b/stresstensor.tex index 70a79ea..a03b10b 100644 --- a/stresstensor.tex +++ b/stresstensor.tex @@ -161,6 +161,7 @@ into annihilation operators on the left. \end{align*} \section{Quadratic Forms} +\label{section:quadratic-forms} In a typical physics literature treatment of second quantization, the annihilation and creation operators and the quantized field are treated as @@ -169,13 +170,20 @@ disregarding the fact that these may not be operators, in a strict sense, and wi Nonetheless, this notational fiction is useful, and we can uphold it with little effort by giving the pointwise \enquote{operators} rigorous meaning as quadratic forms. +Let us consider a real scalar quantum field with mass parameter $m>0$. +Thus, the single-particle state space is the Hilbert space $\hilb{H} = L^2(X_m^+,\Omega_m)$, +where $X_m^+$ is the upper half of the hyperboloid defined by the condition $p \cdot p = m^2$ in momentum space, +and $\Omega_m$ is the unique normalized Lorentz-invariant measure on it. Given a point $p$ in momentum space, -we define the annihilation operator $a(p)$ with domain TODO by +we define the annihilation operator $a(p)$ to be the operator +whose domain is the finite particle subspace $\BosonFockFinite{\hilb{H}}$ of Bosonic Fock space +and whose action on a vector $\psi = (\psi_n)_{n \ge 0}$ in $\BosonFockFinite{\hilb{H}}$ is given by \begin{equation*} \parens[\big]{a(p) \psi} {}_n (k_1, \ldots, k_n) - = \sqrt{n+1} \, \psi_{n+1} (k_1, \ldots, k_n,p) + = \sqrt{n+1} \, \psi_{n+1} (k_1, \ldots, k_n,p). \end{equation*} -The issue arises when one looks for an adjoint to this operator. +While this is a perfectly well-defined operator, +an issue arises when one looks for an adjoint to this operator. A formal calculation based on the adjoint identity \begin{equation} \label{equation:adjoint-identity} @@ -193,12 +201,12 @@ at least has a chance of being a $n$ Boson state. However, it clearly is not a $L^2$ function. Given any state $\psi'$, we can formally calculate the inner product of $\psi'$ with~\eqref{equation:creation-operator-at-point} -and we use the result to define the $a^\dagger(p)$ +and use the result to define $a^\dagger(p)$ as a mapping that assigns a number to each \emph{pair} of states. That is, we define the creation \enquote{operator} $a^\dagger(p)$ to be the quadratic form \begin{gather*} - a(p)^\dagger : F \times F \longrightarrow \CC \\ + a(p)^\dagger \vcentcolon \BosonFockFinite{\hilb{H}} \times \BosonFockFinite{\hilb{H}} \longrightarrow \CC \\ \innerp[\big]{\psi'}{a(p)^\dagger \psi} \defequal \begin{multlined}[t] @@ -210,13 +218,13 @@ to be the quadratic form \end{gather*} One can verify directly that with this definition the adjoint identity~\eqref{equation:adjoint-identity} -holds for all $\psi,\psi' \in F$. +holds for all $\psi,\psi' \in \BosonFockFinite{\hilb{H}}$. For completeness, we give a precise definition of quadratic form. -\begin{definition}{Quadratic Form}{} +\begin{definition}{Quadratic Form}{quadratic-form} A \emph{quadratic form}\index{quadratic form} $q$ on a complex Hilbert space $\hilb{H}$ is a mapping \begin{equation*} - q : D(q) \times D(q) \to \CC, + q \vcentcolon D(q) \times D(q) \to \CC, \end{equation*} where $D(q)$ is a linear subspace of $\hilb{H}$, called the \emph{form domain}\index{form domain}\index{quadratic form!domain of a}, such that $q$ is conjugate linear in its first argument @@ -225,6 +233,11 @@ For completeness, we give a precise definition of quadratic form. if $D(q)$ is dense in $\hilb{H}$. \end{definition} +The creation \enquote{operator} $a^\dagger(p)$ we considered above +is a densely defined quadratic form on Bosonic Fock space +in the sense of \cref{definition:quadratic-form} +and its form domain is the finite particle subspace. + Any linear operator on a complex Hilbert space $\hilb{H}$ has an obvious interpretation as a quadratic form on $\hilb{H}$, and the form domain agrees with the domain of the operator. @@ -247,8 +260,8 @@ We will use the symbol $\QFequal$ between quadratic forms or operators to indicate their equality as quadratic forms. \todo{statement about domains?} - A natural question is how the smeared operators relate to the pointwise ones. +The answer is simple: \begin{equation*} a(g) \QFequal \int \overline{g(p)} a(p) \, d\Omega_m(p) @@ -258,40 +271,53 @@ A natural question is how the smeared operators relate to the pointwise ones. a^\dagger(g) \QFequal \int g(p) a^\dagger(p) \, d\Omega_m(p) \end{equation*} We have to explain what is meant by the integral on the right hand side. -Suppose $q(p)$ is a quadratic form on $\BosonFock{\hilb{H}}$ for each $q \in \RR^4$, +Suppose $q(p)$ is a quadratic form on $\BosonFock{\hilb{H}}$ for each $p \in \RR^4$, that share a common domain $D \subset D(q(p))$, and $g$ is in $\hilb{H} = L^2(\RR^4,\Omega_m)$. Then we define a quadratic form by \begin{equation*} - \parens{\int g(p) q(p) \, d\Omega_m(p)}(\psi',\psi) - = \int g(p) \parens{q(p)}(\psi',\psi) \, d \Omega_m(p) + \parens[\bigg]{\int g(p) q(p) \, d\Omega_m(p)}(\psi',\psi) + \defequal \int g(p) \parens[\big]{q(p)}(\psi',\psi) \, d \Omega_m(p) \end{equation*} for all $\psi,\psi' \in D$. +A product of the form \begin{equation*} a(p_1)^\dagger \cdots a(p_s)^\dagger a(p_{s+1}) \cdots a(p_r) \end{equation*} +can be rigorously defined as quadratic form by setting \begin{equation*} \innerp[\big]{\psi'}{a(p_1)^\dagger \cdots a(p_s)^\dagger a(p_{s+1}) \cdots a(p_r) \psi} - = \innerp[\big]{a(p_1) \cdots a(p_s) \psi'}{a(p_{s+1}) \cdots a(p_r) \psi} + \defequal \innerp[\big]{a(p_1) \cdots a(p_s) \psi'}{a(p_{s+1}) \cdots a(p_r) \psi}. \end{equation*} \section{Normal Ordering} % The Renormalization Map? -%\blockcquote{Wick1950}{% - %\textelp{} we then proceed to rearrange such a product so as to carry all - %creation operators to the left of all destruction operators \textelp{}. The - %main problem to be solved in carrying out this idea is one of algebraic - %technique \textelp{} -%} - +\blockcquote{Wick1950}{% + \textelp{} we then proceed to rearrange such a product so as to carry all + creation operators to the left of all destruction operators \textelp{}. The + main problem to be solved in carrying out this idea is one of algebraic + technique \textelp{} +} The process of renormalizing a product of field operators has the purpose of discarding infinite constants that occur when calculating the vacuum expectation value. -\todo{present physicists way of introducing normal ordering} -Now let us extract the algebraic essence of the situation. +In physics texts the standard prescription for doing this goes as follows: +(1) Express all field operators by creation and annihilation operators +and expand the expression into a sum of products. +(2) Within each product, move all creation operators to the left of all annihilation operators +while ignoring their commutation relations. + +This ad-hoc way of dealing with infinities is not entirely satisfactory. +For once, it leaves the uniqueness question (whether there is another way to +achieve the desired effect) open. +Moreover, it obscures the facts that we are renormalizing in relation to the vacuum state +(in principle, one could use another state for reference), and that the process +is representation-independent and purely algebraic. + +With this in mind, let us extract the algebraic essence of the situation. The objects of our calculations are the field operators $\Phi(f)$, but it does not matter that these are realized as linear maps on Fock space. Forming the product $\Phi(f)\Phi(g)$ might as well be done purely symbolically, @@ -300,70 +326,128 @@ having the meaning of operator composition; and similar for the other two arithmetic operations, addition and multiplication with a complex scalar. Thus we should calculate with abstract objects $\Phi(f)$ labeled by Hilbert space vectors $f \in \hilb{H}$. -Considering that here $\Phi$ carries no meaning, we can use the label $f$ itself to represent the object. - +Considering that here the symbol $\Phi$ carries no meaning, we can drop it and use the label $f$ itself to represent the object. This leads us to consider formal expressions \begin{equation*} - \alpha^{(0)} e + \sum_{i} \alpha^{(1)}_i z^{(1)}_i + \sum_{j,k} \alpha^{(2)}_{j,k} z^{(2)}_j z^{(2)}_k + \cdots + \alpha^{(0)} e + \sum_{i} \alpha^{(1)}_i z^{(1)}_i + \sum_{j,k} \alpha^{(2)}_{j,k} \, z^{(2)}_j z^{(2)}_k + \cdots \end{equation*} where the $z^{(1)}_i,z^{(2)}_j,z^{(2)}_k,\ldots$ are in $\hilb{H}$, the $\alpha^{(0)},\alpha^{(1)}_i,\alpha^{(2)}_{j,k},\ldots$ are complex numbers, of which only finitely many are nonzero, and $e$ is a special object representing an empty product of $z$'s. To make this mathematically precise: -we are speaking of the non-commutative associative algebra over $\CC$ +we are speaking of the unital non-commutative associative algebra over $\CC$ freely generated by the elements of $\hilb{H}$. The unit of the algebra is $e$. -This in not quite what we want -\todo{explain need for commutation relations} +This is not quite what we want, +as we yet need to account for the commutation relations +$\bracks{\Phi(f),\Phi(g)} = i \Imag \innerp{f}{g}$. By abstract algebra, this is viable by forming the quotient of the free algebra with respect to the two-sided ideal -generated by all elements $zz' - z'z = i \Imag \innerp{z}{z'} \, e$, +generated by all elements $zz' - z'z = i \Imag \innerp{z}{z'} \cdot e$, where $z,z' \in \hilb{H}$. +In addition, we must implement the $\RR$-linear dependence of $\Phi(f)$ on $f$. \begin{definition}{Infinitesimal Weyl Algebra}{} Let $\hilb{H}$ be a complex Hilbert space. - The \emph{infinitesimal Weyl algebra}\index{infinitesimal Weyl algebra} $\WeylAlg(\hilb{H})$ over $\hilb{H}$ - is the non-commutative associative algebra over $\CC$ + The \emph{infinitesimal Weyl algebra}\index{infinitesimal Weyl algebra} $\InfinitesimalWeylAlg(\hilb{H})$ over $\hilb{H}$ + is the unital non-commutative associative algebra over $\CC$ generated by the elements of $\hilb{H}$, with the relations - \begin{equation*} - zz' - z'z = i \Imag \innerp{z}{z'} \, e \qquad z,z' \in \hilb{H}, - \end{equation*} + \begin{align*} + \alpha \cdot z &= 1 \cdot (\alpha z) \qquad \alpha \in \RR, z \in \hilb{H}, \\ + zz' - z'z &= i \Imag \innerp{z}{z'} \, e \qquad z,z' \in \hilb{H}, + \end{align*} where $e$ is the unit of the algebra. + Moreover, the mapping $z \mapsto z^* = z$ + extends to a $*$-operation that turns $\InfinitesimalWeylAlg(\hilb{H})$ into a $*$-algebra. \end{definition} - -\todo{introduce $\Phi$ as representation of $\WeylAlg$} - +This construction is an instance of what is known as a quantization functor. +For a longer discussion of its functorial properties we refer the reader to~\cite{Fewster2012}. + +To avoid any confusion, we emphasize that +$1 \cdot iz$ (coefficient $1$, vector $iz$) and $i \cdot z$ (coefficient $i$, vector $z$) +are always different algebra elements (unless $z=0$), and there no rule that allows one to move imaginary numbers from the coefficient part to the vector part. + +Now let $\Phi = \Phi_{\mathrm{S}}$ be +the Segal quantization for the free Boson field over $\hilb{H}$, +which was discussed in the preceding section. +For each $z \in \hilb{H}$, +we consider $\Phi(z)$ as an everywhere-defined linear operator +on the finite particle space $\BosonFockFinite{\hilb{H}}$ by restriction. +The mapping $z \mapsto \Phi(z)$ thus induces a natural $*$-representation +of the infinitesimal Weyl algebra, +\begin{align*} + \Phi \vcentcolon \InfinitesimalWeylAlg(\hilb{H}) &\longrightarrow \End \parens[\big]{\BosonFockFinite{\hilb{H}}} \\ + w = z_1 \!\cdots z_n &\longmapsto \Phi(w) = \Phi(z_1) \circ \cdots \circ \Phi(z_n), +\end{align*} +since $\InfinitesimalWeylAlg(\hilb{H})$ is generated by elements of $\hilb{H}$ +and $\bracks{\Phi(z),\Phi(z')} = i \Imag \innerp{z}{z'}$ on $\BosonFockFinite{\hilb{H}}$ for each $z,z' \in \hilb{H}$. + +In general, any renormalization scheme for the infinitesimal Weyl algebra $\InfinitesimalWeylAlg$ +will be in relation to a given linear functional, i.e.\ an element $E$ of the dual space $\InfinitesimalWeylAlg'$, +which we interpret to be yielding expectation values. +Each element of $\InfinitesimalWeylAlg$ should then be mapped to an element of $\InfinitesimalWeylAlg$ +that is renormalized in the sense that it has zero expectation value in relation to $E$. +We will see that this can be accomplished by bringing each product into a special order called normal order. +In the case of the free Boson field, +the Fock vacuum state $\FockVacuum$ gives rise to the linear functional +\begin{equation*} + \InfinitesimalWeylAlg(\hilb{H}) \ni w \mapsto E(w) \defequal \innerp[\big]{\FockVacuum}{\Phi(w) \FockVacuum}, +\end{equation*} +which we call the \emph{normal vacuum}. +The normal vacuum has two essential properties: Firstly, +\begin{equation} + \label{equation:normal-vaccum-1} + E(e) = 1 +\end{equation} +since $\Phi(e)$ is the identity operator and the vacuum $\Omega$ is a unit vector. +For stating the second property we need to introduce some notation: \begin{definition}{Annihilator and Creator}{} - Suppose $\WeylAlg$ is the infinitesimal Weyl algebra + Suppose $\InfinitesimalWeylAlg(\hilb{H})$ is the infinitesimal Weyl algebra over some complex Hilbert space $\hilb{H}$. - For all $z \in \hilb{H}$, - we define, as elements of $\WeylAlg$, the \emph{annihilator} + For all $z \in \hilb{H}$ + we define, as elements of $\InfinitesimalWeylAlg(\hilb{H})$, the \emph{annihilator} \begin{equation*} - \weylannihilator(z) = \frac{1}{\sqrt{2}} \parens{z+iz}, + \weylannihilator(z) = \frac{1}{\sqrt{2}} \parens{z+i \cdot iz}, \end{equation*} and the \emph{creator} \begin{equation*} - \weylcreator(z) = \frac{1}{\sqrt{2}} \parens{z-iz}. + \weylcreator(z) = \frac{1}{\sqrt{2}} \parens{z-i \cdot iz}. \end{equation*} \end{definition} +The second important property of the normal vacuum $E$ is +\begin{equation} + \label{equation:normal-vaccum-2} + E\parens[\Big]{\prod_{i=1\vphantom{S}}^{s} \weylcreator(z_i) + \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} \weylannihilator(z_j)} = 0 + \qquad \forall z_1,\ldots,z_r \in \hilb{H}, r \ge 1, 1 \le s \le r. +\end{equation} +This follows from $\Phi(\weylannihilator(z)) = a(z)$, $\Phi(\weylcreator(z)) = a^{\dagger}(z)$. +%XXX more detail -\begin{equation*} - z = \frac{1}{\sqrt{2}} \parens[\big]{\weylannihilator(z) + \weylcreator(z)} -\end{equation*} +It is easy to verify that for all $z,z' \in \hilb{H}$ +\begin{gather*} + z = \frac{1}{\sqrt{2}} \parens[\big]{\weylannihilator(z) + \weylcreator(z)}, \\ + \bracks[\big]{\weylannihilator(z),\weylcreator(z')} = i \Imag \innerp{z}{z'} e. +\end{gather*} +Taken together, these identities show that $\InfinitesimalWeylAlg$ is linearly generated by expressions of the form +$\prod_{i=1}^{s} \weylcreator(z_i) \prod_{j=s+1}^{r} \weylannihilator(z_j)$. +We conclude that the normal vacuum is the \emph{unique} functional on $\InfinitesimalWeylAlg$ +satisfying the conditions~\eqref{equation:normal-vaccum-1} and~\eqref{equation:normal-vaccum-2}, which are physically justified. -A \emph{monomial} in the Weyl algebra $\WeylAlg$ over a complex Hilbert space $\hilb{H}$ is an element of the form -$z_1 \cdots z_r \in \WeylAlg$, where $r \ge 0$ and $z_1,\ldots,z_r$ are in $\hilb{H}$. +A \emph{monomial} in the Weyl algebra $\InfinitesimalWeylAlg$ over a complex Hilbert space $\hilb{H}$ is an element of the form +$z_1 \!\cdots z_r \in \InfinitesimalWeylAlg$, where $r \ge 0$ and $z_1,\ldots,z_r$ are in $\hilb{H}$. We allow $r=0$, meaning that the unit $e$ is a monomial. -The set of all monomials in $\WeylAlg$ is denoted $\Mon(\WeylAlg)$. +The set of all monomials in $\InfinitesimalWeylAlg$ is denoted $\Mon(\InfinitesimalWeylAlg)$. \begin{definition}{Normal Ordering}{} - Let $\hilb{H}$ be a complex Hilbert space and $\WeylAlg$ its associated infinitesimal Weyl algebra. - The mapping $\normord{\,\,}$, defined by + Let $\hilb{H}$ be a complex Hilbert space and $\InfinitesimalWeylAlg$ its associated infinitesimal Weyl algebra. + The mapping $\normord{\,\,}$\nomenclature[:]{$\normord{\,\,}$}{normal ordering}, + defined by linear extension of the mapping \begin{gather} - \Mon(\WeylAlg) \longrightarrow \WeylAlg \nonumber\\ + \Mon(\InfinitesimalWeylAlg) \longrightarrow \InfinitesimalWeylAlg \nonumber\\ \label{equation:normal-ordering} \normord{z_1 \!\cdots z_r} = \frac{1}{\sqrt{2^r}} @@ -372,8 +456,8 @@ The set of all monomials in $\WeylAlg$ is denoted $\Mon(\WeylAlg)$. \prod_{\mathclap{j \in \braces{1,\ldots,r} \setminus I}} \weylannihilator(z_j), \end{gather} is called the \emph{normal} (or \emph{Wick}) \emph{ordering}\index{normal ordering}\index{Wick ordering} on $\hilb{H}$. - A monomial $z_1 \cdots z_r \in \Mon(\WeylAlg)$ is said to be in \emph{normal} (or \emph{Wick}) \emph{order}, - if $\normord{z_1 \cdots z_r} = z_1 \cdots z_r$. + An element $w \in \InfinitesimalWeylAlg$ is said to be in \emph{normal} (or \emph{Wick}) \emph{order}, + if $\normord{w} = w$. \end{definition} The products in~\eqref{equation:normal-ordering} are well defined @@ -395,35 +479,50 @@ if one brings~\eqref{equation:normal-ordering} into the equivalent form \label{equation:normal-ordering-symmetric} \normord{z_1 \!\cdots z_r} = \frac{1}{\sqrt{2^r}} - \sum_{\sigma \in S_r} + \sum_{\sigma \in \SymmetricGroup{r}} \sum_{s=0\vphantom{S}}^{r} \frac{1}{s!(r-s)!} \prod_{i=1\vphantom{S}}^{s} \weylcreator(z_{\sigma(i)}) \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} \weylannihilator(z_{\sigma(j)}) \end{gather} -by basic combinatorial arguments \todo{further explanation?}. -In~\cite{Klein1973}, the factor $\frac{1}{s!(r-s)!}$ is erroneously missing. - +by basic combinatorial arguments: +\nomenclature[S]{$\SymmetricGroup{r}$}{symmetric group on the set $\Set{1,\ldots,r}$} +It is easiest to start from~\eqref{equation:normal-ordering-symmetric} +and observe that the product remains invariant +when we permute creators with creators and annihilators with annihilators. +There are precisely $s!(r-s)!$ such permutations. +Let $\sigma$ be any of these. +The associated product corresponds to that summand of~\eqref{equation:normal-ordering} +for which $I = \Set{\sigma(1),\ldots,\sigma(s)}$. +The term $1 / s!(r-s)!$ cancels out. +In more technical language, we are using the fact that the mapping +\begin{align*} + \SymmetricGroup{r} \times \Set{1,\ldots,r} &\longrightarrow \PowerSet[\big]{\Set{1,\ldots,r}} \\ + (\sigma,s) &\longmapsto \Set{\sigma(1),\ldots,\sigma(s)} +\end{align*} +is surjective and that the preimage of each $s$-element set has cardinality $s!(r-s)!$. +\nomenclature[PX]{$\PowerSet{X}$}{power set of a set $X$} +In~\cite{Klein1973}, the factor $1 / s!(r-s)!$ is erroneously missing +(which makes no difference for $r < 2$). +As a direct consequence of~\eqref{equation:normal-vaccum-2} we obtain \begin{equation*} E(\normord{z_1 \!\cdots z_r}) = 0 \qquad \forall z_1,\ldots,z_r \in \hilb{H}, r \ge 1 \end{equation*} -\begin{equation*} - E\parens[\Big]{\prod_{i=1\vphantom{S}}^{s} \weylcreator(z_i) - \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} \weylannihilator(z_j)} = 0 - \qquad \forall z_1,\ldots,z_r \in \hilb{H}, r \ge 1, 1 \le s \le r -\end{equation*} - +as desired. The normal ordered product is supposed to represent the identical quantity as before ordering, -except that we have adjusted our point of reference, such that measurements yield finite results. -It is therefore \emph{physically reasonable} that the commutation relations -of the normal ordered product with the field are analogous. +except that we have adjusted our point of reference, so that the vacuum expectation value is zero. +It is therefore \emph{physically reasonable} to demand that the commutation relations +of the normal ordered product with the field remain unchanged. As it turns out, this additional property makes the construction of normal ordering \emph{mathematically unique}. +This is the content of the following theorem. -\begin{theorem}{Uniqueness of the Normal Order}{} - Normal ordering is the unique mapping $N : \Mon(\WeylAlg) \to \WeylAlg$ such that +\begin{theorem}{Characterization of the Normal Order}{} + Let $\hilb{H}$ be a complex Hilbert space and + let $E$ be the normal vacuum on $\InfinitesimalWeylAlg(\hilb{H})$. + Then, normal ordering is the unique mapping $N : \Mon(\InfinitesimalWeylAlg) \to \InfinitesimalWeylAlg$ such that \begin{gather*} E\parens[\big]{N(z_1 \!\cdots z_r)} = 0 \\ \bracks{N(z_1 \!\cdots z_r), z'} = @@ -431,6 +530,7 @@ As it turns out, this additional property makes the construction of normal order \end{gather*} for all $z_1,\ldots,z_r,z' \in \hilb{H}$ and all $r \ge 1$. \end{theorem} +A proof of this statement is contained in the proof of~\cite[Theorem 7.1]{Baez1992}. %\begin{theorem}{}{} %The normal ordering is the renormalization with respect to the normal vacuum. @@ -440,10 +540,10 @@ As it turns out, this additional property makes the construction of normal order Before we turn to the problem of defining renormalized products of a quantum field and its derivatives we must clarify what is meant mathematically by the derivative of a field. -For this, we recall that in Wightmans approach to quantum field theory, +For this, we recall that in Wightman’s approach to Quantum Field Theory, a quantum field $\varphi$ on a spacetime manifold $M$ is modeled by an operator valued tempered distribution, -that is a mapping that assigns to each (Schwartz class) test function $f$ on $M$ an unbounded operator $\varphi(f)$ -in the Fock space xxx over some Hilbert space $\hilb{H}$, such that for each fixed pair of states $\psi,\psi'$ +that is, a mapping that assigns to each (Schwartz class) test function $f$ on $M$ an unbounded operator $\varphi(f)$ +in the Fock space $\BosonFock{\hilb{H}}$ over some Hilbert space $\hilb{H}$, such that for each fixed pair of states $\psi,\psi'$ the mapping \begin{equation*} \schwartz{M} \to \CC, \quad @@ -482,32 +582,35 @@ and may be obtained via integration by parts. Naturally, we now define the \emph{distributional derivative} of the field by \begin{equation*} - D \varphi(f) = \varphi(D^{\dagger} f) \qquad \forall f \in \schwartz{\RR^4} + D \varphi(f) = \varphi(D^{\dagger} f) \qquad \forall f \in \schwartz{\RR^4}. \end{equation*} As one expects, $D\varphi$ is an operator-valued tempered distribution on $M=\RR^4$. In terms of creation and annihilation operators we have \begin{equation} - \label{derivative-free-field} + \label{equation:derivative-free-field} D \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(ED^{\dagger}f)^{\dagger} + a(E\overline{D^{\dagger}f})}. \end{equation} In Fourier space the operator $D^\dagger$ corresponds to multiplication with the polynomial -\begin{equation*} +\begin{equation} + \label{equation:d-hat} \ft{D}(p) \defequal \sum_{\alpha} i^{\abs{\alpha}} a_{\alpha} (+p^0)^{\alpha_0} (-p^1)^{\alpha_1} (-p^2)^{\alpha_2} (-p^3)^{\alpha_3} -\end{equation*} +\end{equation} If $D=\partial^{\mu}$, then $\ft{D}(p) = i @ p_{\!\mu}$, were the potential sign is concealed by lowering the index. - Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients. +Plugging~\eqref{equation:derivative-free-field} into the renormalization formula~\eqref{equation:normal-ordering-symmetric} yields \begin{gather} \label{equation:renormalized-product} \normord{D_1 \varphi(f) \cdots D_r \varphi(f)} = \frac{1}{\sqrt{2^r}} - \sum_{\sigma \in S_r} + \sum_{\sigma \in \SymmetricGroup{r}} \sum_{s=0\vphantom{S}}^{r} \frac{1}{s!(r-s)!} \prod_{i=1\vphantom{S}}^{s} a^\dagger(ED^\dagger_{\sigma(i)}f) - \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} a(E\overline{D^\dagger_{\sigma(j)}f}) + \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} a(E\overline{D^\dagger_{\sigma(j)}f}). \end{gather} +%As discussed in \cref{section:quadratic-forms}, +%this is a well-defined quadratic form for each fixed test function $f$. \section{Renormalized Products of the Free Field and~its~Derivatives} @@ -557,7 +660,7 @@ this approach incurs significant technical difficulties. \text{and} \quad P_s(p_1,\ldots,p_r) = \frac{1}{\sqrt{2^r}} \frac{1}{s!(r-s)!} - \sum_{\sigma \in S_r} + \sum_{\sigma \in \SymmetricGroup{r}} \ft{D}_{\sigma(1)}(p_1) \cdots \ft{D}_{\sigma(s)}(p_s) \hspace{1.5cm} \\[-1.5ex] \cdot \overline{\ft{D}_{\sigma(s+1)}(p_{s+1}) \cdots \ft{D}_{\sigma(r)}(p_r)}. \end{multline*} @@ -570,14 +673,15 @@ This is made explicit in the alternative integral representation \label{equation:alternative-integral-representation} \begin{multlined} \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} =\\ - \hspace{1cm} \int dp_1 \!\cdots dp_r + \hspace{1cm} \sum_{s=0}^{r} + \int dp_1 \!\cdots dp_r \, \ft{f}(p_1) \cdots\! \ft{f}(p_s) \, \overline{\ft{f}(p_{s+1}) \cdots\! \ft{f}(p_r)} \, \tilde{K}^s_{\psi'\!,\psi}(p_1,\ldots,p_r) \end{multlined} \end{equation} - where + where the integral kernels are given by \begin{multline*} \tilde{K}^s_{\psi'\!,\psi}(p_1,\ldots,p_r) = P_s(p_1,\ldots,p_r) @@ -588,9 +692,9 @@ This is made explicit in the alternative integral representation \ \overline{\psi'_m(k_1,\ldots,k_{m-s},p_1,\ldots,p_s)} \ \psi_n(k_1,\ldots,k_{n-(r-s)},p_{s+1},\ldots,p_r). \end{multline*} - This will be more convenient for xxx + This representation will be more convenient for xxx -\begin{myproof}[lemma:renormalized-product-integral-representation] +\begin{myproof}{lemma:renormalized-product-integral-representation} From equation~\eqref{equation:renormalized-product}, applying the definition of the Fock space inner product, and moving all creation operators to the left hand side, @@ -598,7 +702,7 @@ This is made explicit in the alternative integral representation \begin{multline*} \innerp{\psi'}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} - \frac{1}{\sqrt{2^r}} \sum_{s=0}^{r} \sum_{\sigma \in S_r} \frac{1}{s!(r-s)!} \\ + \frac{1}{\sqrt{2^r}} \sum_{s=0}^{r} \sum_{\sigma \in \SymmetricGroup{r}} \frac{1}{s!(r-s)!} \\ \cdot \big\langle a(ED_{\sigma(1)}^{\dagger}f) \cdots a(ED_{\sigma(s)}^{\dagger}f) \psi_m, a(ED_{\sigma(s+1)}^{\dagger}f) \cdots a(ED_{\sigma(r)}^{\dagger}f) \psi_n @@ -608,7 +712,7 @@ This is made explicit in the alternative integral representation can only be nonzero if the particle numbers match up after the application of the annihilation operators in each argument, that is if $m-s=n-(r-s)$. - With~\eqref{equation:multiple-annihilation-operators} + By applying~\eqref{equation:multiple-annihilation-operators} twice, this expression may be further expanded into \begin{gather*} \sqrt{m(m-1) \cdots (m-s+1)} @@ -623,16 +727,20 @@ This is made explicit in the alternative integral representation \end{gather*} Now recall that $E$ stands for Fourier transformation (followed by restriction to the mass shell) and that in Fourier space the linear differential operator $D^\dagger$ corresponds to a - multiplication with the function $\hat{D}$, so that + multiplication with the function $\hat{D}$ defined by~\eqref{equation:d-hat}, so that \begin{equation*} ED_{\sigma(i)}^{\dagger}f(p_i) = \hat{D}_{\sigma(i)}(p_i) \cdot \ft{f} (p_i) \qquad \forall i \end{equation*} - By Fubini’s Theorem, the we may interchange the integrals with respect to the variables $p_i$ - with the $k$-integrals. - This allows us to move all factors involving $\ft{f}$ in front of the $k$-integrals. - Finally, we introduce the $\chi$s through the substitution $\overline{\ft{f}} = \chi \ft{f}$, - and combine all terms depending on $\sigma$ into $P_s$. + According to Fubini’s Theorem, we may move the $p$-integrals past the $k$-integrals to the very left of the expression. + This also allows us to move all factors involving $\ft{f}$ or one of the $\ft{D}_{\sigma(i)}$ in front of the $k$-integrals. + Finally, we move the summation over $s$ to the front. + By packaging everything not depending on $\ft{f}$ in new functions $\tilde{K}^{s}_{\psi',\psi}$, + where $s$ runs from $0$ to $r$, + we arrive at the representation~\eqref{equation:alternative-integral-representation}. + + A single-integral representation is achieved by + introducing the functions $\chi$s through the substitution $\overline{\ft{f}} = \chi \ft{f}$, \end{myproof} In the special case that $D_1 = \cdots = D_r = D$ we have @@ -679,7 +787,7 @@ In the following proof it will be convenient to use the abbreviation \omega(p_1,\ldots,p_s) \defequal \omega(p_1) + \cdots + \omega(p_n). \end{equation*} -\begin{myproof}[lemma:integral-kernel-h-bound] +\begin{myproof}{lemma:integral-kernel-h-bound} We have to find an estimate for \begin{equation*} \norm{K_{\psi'\!,\psi}}_1 = @@ -797,7 +905,7 @@ In the following proof it will be convenient to use the abbreviation %$\norm{(1+H)\psi_n}_2 \ge n \epsilon \norm{\psi_n} = \epsilon \norm{N \psi_n}$ In order to determine conditions for the finiteness of the remaining factor involving $F$, - it is desireable to have an estimate of the growth of $P_s$ in terms of $\omega(p_1),\ldots,\omega(p_r)$. + it is desirable to have an estimate of the growth of $P_s$ in terms of $\omega(p_1),\ldots,\omega(p_r)$. Notice that it is sufficient to make an estimate that is valid on the support of the measure $\Omega_m$, that is, the mass shell $X_m^+$, since $F$ appears in an integral with respect to $p_1,\ldots,p_r$. For an arbitrary point $q$ on the mass shell $X_m^+$ we have @@ -875,10 +983,13 @@ In the following proof it will be convenient to use the abbreviation and because $((1+H)^l \psi')_m = (1+H)^l \psi'_m$ for all $m$. \end{myproof} +Now we are in a position to implement the idea of taking the limit $f \to \delta_x$. +Recall that any tempered distribution, and in particular Dirac distributions, may be approximated by Schwarz test functions. + \begin{lemma}{Renormalized Product at a Point}{} In the setting of \cref{lemma:renormalized-product-integral-representation}, assume that $\psi,\psi'$ are in $\Domain{H^l}$. - Let $x$ be any point in $M$ and let $\delta_x \in \tempdistrib{M}$ be the Dirac distribution supported in $x$. + Let $x$ be any point in $M$ and let $\delta_x \in \tempdistrib{M}$ be the Dirac distribution supported at $x$. Then the limit \begin{equation*} \lim_{f \to \delta_x} @@ -888,6 +999,11 @@ In the following proof it will be convenient to use the abbreviation \end{lemma} \begin{proof} + Since the Fourier transformation of tempered distribution + is a continuous mapping $\tempdistribnoarg \to \tempdistribnoarg$, + we have $\ft{f} \to \FT{\delta_x}$ whenever $f \to \delta_x$ in the topology of $\tempdistribnoarg$. + Consequently, $\abs{\ft{f}}$ remains bounded by some constant $C$ while taking the limit. + According to \cref{lemma:renormalized-product-integral-representation} we have \begin{equation*} \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} @@ -895,58 +1011,56 @@ In the following proof it will be convenient to use the abbreviation \, \ft{f}(p_1) \cdots\! \ft{f}(p_r) \, K_{\psi'\!,\psi}(p_1,\ldots,p_r) \end{equation*} - The integrand is dominated by the function $\abs{K_{\psi'\!,\psi}(p_1,\ldots,p_r)}$, - which has finite integral as it is $L^1$ - by \cref{lemma:integral-kernel-h-bound}. + The integrand is dominated by the function $C^r\abs{K_{\psi'\!,\psi}(p_1,\ldots,p_r)}$, + which has finite integral by \cref{lemma:integral-kernel-h-bound}. - Moreover, the integrand converges pointwise to $K_{\psi'\!,\psi}(p_1,\ldots,p_r)$, since $\ft{f} \to 1$ when $f \to \delta_x$. - \todo{With of choice of FT constants, $\ft{f} \to 1/(2\pi)^2$. Change here or change def?} + %Moreover, the integrand converges pointwise to $K_{\psi'\!,\psi}(p_1,\ldots,p_r)$, since $\ft{f} \to 1$ when $f \to \delta_x$. + %\todo{With of choice of FT constants, $\ft{f} \to 1/(2\pi)^2$. Change here or change def?} - Since the Fourier transformation of tempered distribution - is a continuous mapping $\tempdistribnoarg \to \tempdistribnoarg$, - we have $\ft{f} \to \FT{\delta_x}$ whenever $f \to \delta_x$ in the topology of $\tempdistribnoarg$. Recall that $\ft{\delta} = 1$, and thus $\FT{\delta_x}(p) = e^{ix \cdot p}$ for all $p \in M$. - This shows that the integrand converges pointwise to + This shows that the integrand converges pointwise to the function \begin{equation*} + F(p_1,\ldots,p_r) = \sum_{s=0}^r e^{ix \cdot (p_1 + \cdots + p_s)} e^{-ix \cdot (p_{s+1} + \cdots + p_r)} - \tilde{K}_{\psi'\!,\psi}(p_1,\ldots,p_r) + \tilde{K}^s_{\psi'\!,\psi}(p_1,\ldots,p_r) \end{equation*} - - The Dominated Convergence Theorem implies + The Dominated Convergence Theorem implies that $F$ is integrable and that + \begin{equation} + \label{equation:intx} + \lim_{f \to \delta_x} + \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} + = \int dp_1 \!\cdots dp_r \, F(p_1,\ldots,p_r). + \end{equation} + In particular, the limit exists. \end{proof} -\begin{definition}{Renormalized Product at a Point}{} - In the setting of \cref{lemma:renormalized-product-integral-representation}, - the mapping defined by +\begin{definition}{Renormalized Product as a QF-valued distribution}{} + Adopt the assumptions of the foregoing Lemma. + We define two mappings that intentionally share the same name by \begin{gather*} \normord{D_1 \varphi \cdots D_r \varphi} \ \vcentcolon \ - M \to \QF{fock} \\ + M \to \QF{\mathcal{D}(H^l)} \\ \innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(x) \,\psi} = \lim_{f \to \delta_x} \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} \end{gather*} - is called the xxx -\end{definition} - - -\begin{lemma}{Renormalized Product as a QF-valued distribution}{} - In the setting of \cref{lemma:renormalized-product-integral-representation}, + and \begin{equation*} \normord{D_1 \varphi \cdots D_r \varphi} \ \vcentcolon \ - \schwartz{M} \to \QF{fock} + \schwartz{M} \to \QF{\mathcal{D}(H^l)} \end{equation*} \begin{equation*} \innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(f) \,\psi} = \int_M \!dx \ f(x) \ \innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(x) \,\psi} \end{equation*} -\end{lemma} +\end{definition} -\begin{lemma}{TODO}{} +\begin{proposition}{}{} Let $\varphi$ be a free quantum field. Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients. - Suppose that $l$ is a positive integer large enough to satisfy the + Suppose that $l$ is a large enough positive integer. Then we have for all states $\psi,\psi' \in \Domain{H^l}$ \begin{multline*} \innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(f) \,\psi} = \\ @@ -967,10 +1081,10 @@ In the following proof it will be convenient to use the abbreviation \ \psi_n(k_1,\ldots,k_{n-(r-s)},p_{s+1},\ldots,p_r) \end{multline*} and $P_s(p_1,\ldots,p_r)$ is defined as before. -\end{lemma} +\end{proposition} \begin{proof} - a + This follows directly from the definition and~\eqref{equation:intx}. \end{proof} @@ -1005,15 +1119,9 @@ In particular, the energy density is \energydensity = \frac{1}{2} \sum_{\mu=0}^{3} \normord{(\partial^{\mu}\varphi)^2} + \frac{1}{2} m^2 \normord{\varphi^2} \end{equation*} -\begin{multline*} - \innerp{\psi'\!}{\energydensity(f) \,\psi} = \\ - = \int dp_1 dp_2 - \parens{p_1^{\mu} p_2^{\mu} + m^2} - \sum_{s=0}^{r} (-1)^{s+1} - \, \ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r) - \, L_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) -\end{multline*} -where +Now let $f$ be a real-valued Schwarz function, +and let $\psi,\psi' \in \Domain{H^l}$ for a large enough integer $l$. +For convenience we introduce the functions \begin{multline*} L_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} @@ -1023,6 +1131,43 @@ where \ \overline{\psi'_m(k_1,\ldots,k_{m-s},p_1,\ldots,p_s)} \ \psi_n(k_1,\ldots,k_{n-(r-s)},p_{s+1},\ldots,p_r) \end{multline*} +so that +\begin{equation*} + \tilde{K}_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) = + P_s(p_1,\ldots,p_r) \, + L_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r). +\end{equation*} +Since the energy density only contains squares, it suffices to consider $r=2$. + +\begin{multline*} + \innerp{\psi'\!}{\normord{\varphi^2}(f) \,\psi} = \int dp_1 dp_2 \, + \ft{f}(-p_1 - p_2) \, \tfrac{1}{2} L_{\psi'\!,\psi}^{0}(p_1,p_2) \\ + + \ft{f}(p_1 - p_2) \, L_{\psi'\!,\psi}^{1}(p_1,p_2) + + \ft{f}(p_1 + p_2) \, \tfrac{1}{2} L_{\psi'\!,\psi}^{2}(p_1,p_2) \\ + \innerp{\psi'\!}{\normord{(\partial^{\mu}\varphi)^2}(f) \,\psi} = \int dp_1 dp_2 \, + \ft{f}(-p_1 - p_2) \, \parens[\big]{-\tfrac{1}{2} (p_1)_{\mu} (p_2)_{\mu}} \, L_{\psi'\!,\psi}^{0}(p_1,p_2) \\ + + \ft{f}(p_1 - p_2) \, \parens[\big]{+(p_1)_{\mu} (p_2)_{\mu}} \, L_{\psi'\!,\psi}^{1}(p_1,p_2) + + \ft{f}(p_1 + p_2) \, \parens[\big]{-\tfrac{1}{2} (p_1)_{\mu} (p_2)_{\mu}} \, L_{\psi'\!,\psi}^{2}(p_1,p_2) +\end{multline*} + +\begin{equation*} + \bar{p} \defequal \eta p = (p^0,-\Vector{p}) +\end{equation*} + +\begin{equation*} + \sum_{\mu = 0}^{3} p_{\mu} p'_{\mu} = \bar{p} \cdot p' +\end{equation*} + +\begin{proposition}{}{} + \begin{multline*} + \innerp{\psi'\!}{\energydensity(f) \, \psi} = + \frac{1}{4} \int dp \, dp' + (m^2 + \bar{p} \cdot p') + \bracks[\big]{2 \ft{f}(p - p') L^1_{\psi'\!,\psi}(p,p')} \\ + + (m^2 -\bar{p} \cdot p') + \bracks[\big]{\ft{f}(- p - p') L^0_{\psi'\!,\psi}(p,p') + \ft{f}(p + p') L^2_{\psi'\!,\psi}(p,p')} + \end{multline*} +\end{proposition} \begin{theorem}{TODO}{} Let $\varphi$ be a free quantum field. @@ -1038,45 +1183,25 @@ where \quad P_s(p_1,\ldots,p_r) = \frac{1}{\sqrt{2^r}} \frac{1}{s!(r-s)!} - \sum_{\sigma \in S_r} + \sum_{\sigma \in \SymmetricGroup{r}} \ft{D}_{\sigma(1)}(p_1) \cdots \ft{D}_{\sigma(s)}(p_s) \hspace{1.5cm} \\[-1.5ex] \cdot \overline{\ft{D}_{\sigma(s+1)}(p_{s+1}) \cdots \ft{D}_{\sigma(r)}(p_r)}. \end{multline*} \end{theorem} -\begin{definition}{}{} +\begin{proposition}{}{energy-density} \begin{multline*} - \energydensity(f) \QFequal \frac{1}{4} \int dp dp' (p \cdot p' + m^2) - \Big\lbrack \ft{f}(p+p') a(p) a(p') + {}\\ - + 2\ft{f}(p-p') a^\dagger(p) a(p') + \ft{f}(-p-p') a^\dagger(p) a^\dagger(p') \Big\rbrack - \end{multline*} -\end{definition} - -\begin{equation*} - \bar{p} := \eta p = (p^0,-\symbfit{p}) -\end{equation*} - -\begin{proposition}{}{} - \begin{multline*} - \innerp{\psi'}{\energydensity(f) \psi} = - \frac{1}{4} \int dp dp' - (\bar{p} \cdot p' + m^2) - \bracks[\big]{2 \ft{f}(p - p') L^1_{\psi'\!,\psi}(p,p')} \\ - + (-\bar{p} \cdot p' + m^2) - \bracks[\big]{\ft{f}(- p - p') L^0_{\psi'\!,\psi}(p,p') + \ft{f}(p + p') L^2_{\psi'\!,\psi}(p,p')} - \end{multline*} -\end{proposition} - -\begin{proposition}{}{} - \begin{multline*} - \energydensity(f) \QFequal \frac{1}{4} \int dp dp' + \energydensity(f) \QFequal \frac{1}{4} \int dp \, dp' \, (m^2 + \bar{p} \cdot p') - \bracks[\Big]{2\ft{f}(p-p') a^\dagger(p) a(p')} \\ + \bracks[\Big]{2\ft{f}(p-p') a^\dagger(p) a(p')} + {} \\ + (m^2 - \bar{p} \cdot p') \bracks[\Big]{\ft{f}(p+p') a(p) a(p') + \ft{f}(-p-p') a^\dagger(p) a^\dagger(p')} \end{multline*} \end{proposition} +Observe that $\bar{p} \cdot p' = p^0p'^0 + p^1p'^1 + p^2p'^2 + p^3p'^3$ is symmetric in $p$ and $p'$. +Consequently, we could rewrite the first bracketed expression as $2\ft{f}(p-p') a^\dagger(p) a(p')$ + \begin{proposition}{}{} The Fock vacuum $\FockVacuum$ lies in the domain of $\energydensity(f)\QFop{}$ for all test functions $f \in \schwartz{M}$ @@ -1087,6 +1212,29 @@ where and $\psi_n \equiv 0$ for $n \ne 2$. \end{proposition} +\begin{proof} + Since $\ft{f}$ is a Schwarz function, there exists for each $N \in \NN$ a positive constant $C_N$ such that + \begin{equation*} + \abs{\ft{f}(q)}^2 \le \frac{C_N}{1+(q^0)^{2N}} + \end{equation*} + For $p,p'$ on the mass shell we have + \begin{equation*} + m^2 - \bar{p} \cdot p' = + m^2 - \sqrt{m^2 + \norm{\Vector{p}}^2} \sqrt{m^2 + \norm{\Vector{p'}}^2} + \Vector{p} \cdot \Vector{p}', + \end{equation*} + where on the right hand side \enquote*{$\cdot$} stands for the Euclidean scalar product for three-vectors. + The Cauchy-Schwarz Inequality (applied twice) shows + \begin{equation*} + m^2 + \Vector{p} \cdot \Vector{p}' + \le m^2 + \norm{\Vector{p}} \norm{\Vector{p}'} + \le \sqrt{m^2 + \norm{\Vector{p}}^2} \sqrt{m^2 + \norm{\Vector{p'}}^2}. + \end{equation*} + \begin{equation*} + \abs{m^2 - \bar{p} \cdot p'} + \le 2 \sqrt{m^2 + \norm{\Vector{p}}^2} \sqrt{m^2 + \norm{\Vector{p'}}^2} + \end{equation*} +\end{proof} + \section{Essential Selfadjointness of Renormalized Products} \begin{lemma}{H-Bounds for the Renormalized Product}{} |