diff options
Diffstat (limited to 'stresstensor.tex')
| -rw-r--r-- | stresstensor.tex | 302 |
1 files changed, 246 insertions, 56 deletions
diff --git a/stresstensor.tex b/stresstensor.tex index a4bb6fb..4c128c2 100644 --- a/stresstensor.tex +++ b/stresstensor.tex @@ -35,9 +35,13 @@ as a service to the reader. \item Given a complex-valued function $f$ on $M$, we define its \emph{Fourier transform} $\ft{f}\,$ by \begin{equation} \label{fourier-transform} - \hat{f}(p) = \frac{1}{(2 \pi)^2} \int_{M} e^{i p \cdot x} f(x) \, dx + \ft{f}(p) \defequal \int_{M} e^{i p \cdot x} f(x) \, dx \end{equation} - whenever the integral converges. The \emph{inverse Fourier transform} is TODO + whenever the integral converges. The \emph{inverse Fourier transform} is defined by + \begin{equation*} + \label{inverse-fourier-transform} + \ift{f}(p) \defequal \frac{1}{(2 \pi)^2} \int_{M} e^{-i p \cdot x} f(x) \, dx. + \end{equation*} \item To a mathematician $\overline{\phantom{z}}$ usually means complex conjugation and ${}^*$ indicates the Hilbert adjoint of an operator, while a physicist may read ${}^*$ as complex conjugation and denotes the Hilbert adjoint with ${}^{\dagger}$. @@ -85,9 +89,15 @@ as a service to the reader. \begin{equation*} E : \schwartz{M} \to \hilb{H}, \quad f \mapsto Ef = \left.\ft{f}\,\right\vert {X_m^+} \end{equation*} + We define a $\RR$-linear mapping $\phi$ by + \begin{equation*} + \realschwartz{M} \ni f \mapsto \varphi(f) = \Phi_{\mathrm{S}}(Ef) = \frac{1}{\sqrt{2}} \parens*{a(Ef) + a(Ef)^\dagger} + \end{equation*} + This extedns to complex valued test functions $f \in \schwartz{M}$ \begin{equation*} - \realschwartz{M} \ni f \mapsto \Phi(f) = \Phi_{\mathrm{S}}(Ef) = \frac{1}{\sqrt{2}} \parens*{a(Ef) + a(Ef)^\dagger} + \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(E\bar{f}) + a(Ef)^\dagger} \end{equation*} + called the \emph{massive free scalar quantum field} \item annihilation and creation operators, $f \in \schwartz{M}$, $\psi \in \BosonFock{\hilb{H}}$ \begin{align*} @@ -176,7 +186,7 @@ leads to \parens[\big]{a(p)^\dagger \psi} {}_n (k_1, \ldots, k_n) = \frac{1}{\sqrt{n}} \sum_{i=1}^n \delta(p - k_i) \, \psi_{n-1} (k_1, \ldots, \widehat{k_i}, \ldots, k_n), \end{equation} -where the symmetization is necessary to obtain an expression that +where the symmetrization is necessary to obtain an expression that 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 @@ -207,7 +217,7 @@ For completeness, we give a precise definition of quadratic form. q : 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 agrument + such that $q$ is conjugate linear in its first argument and linear in its second argument (i.e.\ sesquilinear). We say that $q$ is \emph{densely defined} if $D(q)$ is dense in $\hilb{H}$. @@ -221,13 +231,13 @@ but one may obtain an operator with trivial domain. \begin{definition}{Operator Associated to a Quadratic Form}{} Suppose $q$ is a densely defined quadratic form on a complex Hilbert space $\hilb{H}$. - The linear \emph{operator associated to}\index{quadratic form!operator associated to a} $q$, denoted $q_{\mathrm{op}}$, + The linear \emph{operator associated to}\index{quadratic form!operator associated to a} $q$, denoted $\QFop{q}$, is defined on the domain \begin{equation*} - D(q_{\mathrm{op}}) = \braces{\psi \in D(q) \mid \text{the map $q(\cdot,\psi) : D(q) \to \CC$ is bounded}}, + D(\QFop{q}) = \braces{\psi \in D(q) \mid \text{the map $q(\cdot,\psi) \vcentcolon D(q) \to \CC$ is bounded}}, \end{equation*} - and maps $\psi \in D(q_{\mathrm{op}})$ to the vector $q_{\mathrm{op}}\psi$ in $\hilb{H}$ satisfying - $q(\psi',\psi) = \innerp{\psi'}{q_{\mathrm{op}}\psi}$, + and maps $\psi \in D(\QFop{q})$ to the vector $\QFop{q}\psi$ in $\hilb{H}$ satisfying + $q(\psi'\!,\psi) = \innerp{\psi'\!}{\QFop{q}\psi}$, which exists and is unique by Riesz’s Representation Theorem. \end{definition} @@ -299,7 +309,7 @@ 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 noncommutative associative algebra over $\CC$ +we are speaking of the non-commutative associative algebra over $\CC$ freely generated by the elements of $\hilb{H}$. The unit of the algebra is $e$. @@ -314,7 +324,7 @@ where $z,z' \in \hilb{H}$. \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 noncommutative associative algebra over $\CC$ + is the 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}, @@ -371,10 +381,10 @@ the formula makes sense even for $r=0$ and asserts that $\normord{e} = e$. The cases $r=1$ and $r=2$ read \begin{align*} \normord{z} &= - \frac{1}{\sqrt{2}} \parens[\big]{\weylannihilator(z) + \weylcreator(z)} = z \\ + \frac{1}{\sqrt{2}} \parens[\big]{\weylannihilator(z) + \weylcreator(z)} = z, \\ \normord{z_1 z_2} &= \frac{1}{2} - \parens[\big]{\weylannihilator(z_1) \weylannihilator(z_2) + \weylannihilator(z_1) \weylcreator(z_2) - + \weylcreator(z_1) \weylannihilator(z_2) + \weylcreator(z_1) \weylcreator(z_2) } + \parens[\big]{ \weylannihilator(z_1) \weylannihilator(z_2) + \weylcreator(z_1) \weylannihilator(z_2) + + \weylcreator(z_2) \weylannihilator(z_1) + \weylcreator(z_1) \weylcreator(z_2) }. \end{align*} This suggests that the normally ordered product $\normord{z_1 \!\cdots z_r}$ is symmetric in $z_1,\ldots,z_n$. This is in fact true, and becomes evident @@ -470,20 +480,19 @@ 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^d} + 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 $M=\RR^d$. -TODO +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} - D \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(ED^{\dagger}f)^{\dagger} + a(ED^{\dagger}f)} + D \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(ED^{\dagger}f)^{\dagger} + a(E\overline{D^{\dagger}f})}. \end{equation} - - -The operator corresponding to $D$ in Fourier space is the multiplication operator +In Fourier space the operator $D^\dagger$ corresponds to muliplication with the polynomial \begin{equation*} - -i \sum_{\alpha} a_{\alpha} p_0^{\alpha_0} (-p_1)^{\alpha_1} (-p_2)^{\alpha_2} (-p_3)^{\alpha_3} + \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*} +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. @@ -494,8 +503,8 @@ Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) \sum_{\sigma \in S_r} \sum_{s=0\vphantom{S}}^{r} \frac{1}{s!(r-s)!} - \prod_{i=1\vphantom{S}}^{s} a^\dagger(D^\dagger_{\sigma(i)}f) - \prod_{\mathclap{j=s+1\vphantom{S}}}^{r} a(D^\dagger_{\sigma(j)}f) + \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}) \end{gather} \section{Renormalized Products of the Free Field and~its~Derivatives} @@ -516,7 +525,7 @@ this approach incurs significant technical difficulties. \begin{lemma}{Integral Representation of the Renormalized Product}{renormalized-product-integral-representation} Let $\varphi$ be the free scalar quantum field with mass parameter $m > 0$. Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients. - Then, for arbitrary Schwartz functions $f \in \schwartz{M}$ and Fock states $\psi,\psi' \in \BosonFock{L^2(X_m^+,d\Omega_m)}$, + Then, for arbitrary Schwartz functions $f \in \realschwartz{M}$ and Fock states $\psi,\psi' \in \BosonFock{L^2(X_m^+,d\Omega_m)}$, we have \begin{equation*} \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} = @@ -552,7 +561,32 @@ this approach incurs significant technical difficulties. \end{multline*} \end{lemma} -TODO(Note about the remaining dependence of $K$ on $f$.) +Note that $K$ has a remaining dependence on $f$ via $\chi$ +even thogh the notation does not indicate this. +This is made explicit in the alternative integral representation + \begin{equation} + \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 + \sum_{s=0}^{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 + \begin{multline*} + \tilde{K}^s_{\psi'\!,\psi}(p_1,\ldots,p_r) = + P_s(p_1,\ldots,p_r) + \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} + \delta_{m-s}^{n-(r-s)} \\ + \cdot \sqrt{m(m-1) \cdots (m-s+1)} \sqrt{n(n-1) \cdots (n-(r-s)+1)} \\ + \cdot \int dk_1 \cdots dk_{m-s} + \ \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 \begin{myproof}[lemma:renormalized-product-integral-representation] From equation~\eqref{equation:renormalized-product}, @@ -617,24 +651,30 @@ In particular, for squares ($r=2$) we have The following assertion is key to realizing the idea of taking the limit $f \to \delta_x$. -\begin{lemma}{}{integral-kernel-h-bound} +\begin{lemma}{H-bounds for the Integral Kernel}{integral-kernel-h-bound} In the setting of \cref{lemma:renormalized-product-integral-representation}, there exist a constant $C$, and a positive integer $l$, - such that for arbitrary states $\psi,\psi' \in \BosonFock{\hilb{H}}$, - and test functions $f \in \schwartz{M}$, + such that for arbitrary test functions $f \in \schwartz{M}$ + and states $\psi,\psi' \in \Domain{H^l}$, the function $K_{\psi'\!,\psi}$ is integrable (that is, $L^1$) and satisfies the $H$-bound \begin{equation*} \norm{K_{\psi'\!,\psi}}_1 \le C \norm{(1+H)^l \psi'} \norm{(1+H)^l \psi}. \end{equation*} + More specifically, it is sufficient to choose $l > rd + r/2$, + where $d$ is the highest order of differentiation occuring in $D_1, \ldots, D_r$. \end{lemma} -The Hamilton operator $H$ acts on $n$-particle states $\psi_n$ -by multiplication with $\omega(p_1)$ -In the following proof it will we convenient to use the abbreviation +The Hamilton operator $H$ acts on $n$-particle states $\psi_n$ as follows: +\begin{gather*} + H \psi_n(p_1,\ldots,p_n) = \parens[\big]{\omega(p_1) + \cdots + \omega(p_n)} \psi_n(p_1,\ldots,p_n) \\ + \shortintertext{where} + \omega(p) = \omega(\symbfit{p}) = \sqrt{m^2 + \abs{\symbfit{p}}^2} = \sqrt{m^2 + (p^1)^2 + (p^2)^2 + (p^3)^2}. +\end{gather*} +In the following proof it will be convenient to use the abbreviation \begin{equation*} - \omega(p_1,\ldots,p_s) = \omega(p_1) + \cdots + \omega(p_s) + \omega(p_1,\ldots,p_s) \defequal \omega(p_1) + \cdots + \omega(p_n). \end{equation*} \begin{myproof}[lemma:integral-kernel-h-bound] @@ -778,28 +818,71 @@ In the following proof it will we convenient to use the abbreviation \le \frac{\omega(p_1) + \cdots + \omega(p_s)}{s} \le \omega(p') \le 1 + \omega(k,p'), \nonumber\\ \shortintertext{hence} - \label{equation:one-plus-omega-estimate} + \label{equation:one-plus-omega-estimate1} \parens[\big]{1+\omega(k,p')} {}^{-a} - \le \parens[\big]{\omega(p_1) \cdots \omega(p_s)} {}^{-a/s}. + \le \parens[\big]{\omega(p_1) \cdots \omega(p_s)} {}^{-a/s}, \\ + \shortintertext{and similarly} + \label{equation:one-plus-omega-estimate2} + \parens[\big]{1+\omega(k,p)} {}^{-a} + \le \parens[\big]{\omega(p_1) \cdots \omega(p_{r-s})} {}^{-a/(r-s)}. \end{gather} - The estimates~\eqref{equation:polynomial-estimate} and~\eqref{equation:one-plus-omega-estimate} entail + The estimates~\eqref{equation:polynomial-estimate},~\eqref{equation:one-plus-omega-estimate1} and~\eqref{equation:one-plus-omega-estimate2} entail \begin{equation*} \abs{F(k,p',p)} \le C_s \prod_{i=1}^{s} \omega(p_i)^{d_i-a/s} - \prod_{j=s+1}^{r} \omega(p_j)^{d_j-a/(r-s)} + \prod_{j=s+1}^{r} \omega(p_j)^{d_j-a/(r-s)}. \end{equation*} + Since the right hand side does not depend on $k$, and the $p$-variables are separated, + the problem reduces to proving that + \begin{equation} + \label{equation:integral-finite} + \int \omega(q)^{-2b} \,d\Omega(q) < \infty + \end{equation} + for $b$ large enough. + Recall that $d \Omega(q) = \omega(q)^{-1} d^3 \symbfit{q}$. + By transformation to spherical coordinates, we find that~\eqref{equation:integral-finite} + is equivalent to + \begin{equation*} + \int \frac{r^2}{(m^2 + r^2)^{b+1/2}} \,dr < \infty + \end{equation*} + It it well known that this holds for $b > 1$. + $a > r d$ + + \begin{equation} + \label{equation:intermediate-result} + \norm{K_{\psi'\!,\psi}}_1 \le + \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} + \sum_{s=0}^{r} \delta_{m-s}^{n-(r-s)} C_s + \underbracket{\norm{(1+H)^l \psi'_m}_2}_{a'_m} + \underbracket{\norm{(1+H)^l \psi_n}_2}_{a_n} + \end{equation} + We introduce auxiliary variables $a'_m, a_n$ as shown above, and for convenience set $a_n = 0$ whenever $n < 0$. + Using this, we rewrite the right hand side of~\eqref{equation:intermediate-result} + and apply the Cauchy-Schwarz Inequality for sequences as follows: + \begin{equation*} + \sum_{s=0}^{r} C_s \sum_{m = 0}^{\infty} a'_m \cdot a_{m+r-2s} \le + \sum_{s=0}^{r} C_s \sqrt{\sum_{m=0}^{\infty} a'^2_m \sum_{n=0}^{\infty} a^2_n} + \end{equation*} + To complete the proof, observe that + \begin{equation*} + \norm{(1+H)^l \psi'}_2 = + \sqrt{\sum_{m=0}^{\infty} \norm{(1+H)^l \psi'_m}_2^2} = + \sqrt{\sum_{m=0}^{\infty} a'^2_m}, + \end{equation*} + and similar for $\psi$, by definition of the inner prouct + and because $((1+H)^l \psi')_m = (1+H)^l \psi'_m$ for all $m$. \end{myproof} \begin{lemma}{Renormalized Product at a Point}{} In the setting of \cref{lemma:renormalized-product-integral-representation}, - assume that $\psi,\psi'$ are in $D^l(H)$. + 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 $x$. Then the limit \begin{equation*} \lim_{f \to \delta_x} \innerp{\psi'\!}{\normord{D_1 \varphi(f) \cdots D_r \varphi(f)} \,\psi} \end{equation*} - exists and depends continously on $x$. + exists and depends continuously on $x$. \end{lemma} \begin{proof} @@ -813,8 +896,22 @@ In the following proof it will we convenient to use the abbreviation 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}. + 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 + \begin{equation*} + \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) + \end{equation*} + The Dominated Convergence Theorem implies \end{proof} @@ -847,20 +944,21 @@ In the following proof it will we convenient to use the abbreviation \begin{lemma}{TODO}{} Let $\varphi$ be a free quantum field. Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients. - Then we have for all states $\psi,\psi' \in D^l(H)$ + Suppose that $l$ is a positive integer large enough to satisfy the + 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} = \\ = \int dp_1 \!\cdots dp_r \sum_{s=0}^{r} \, \ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r) - \, L_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) + \, \tilde{K}_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) \end{multline*} where \begin{multline*} - L_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) = + \tilde{K}_{\psi'\!,\psi}^{s}(p_1,\ldots,p_r) = + P_s(p_1,\ldots,p_r) \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} - \delta_{m-s}^{n-(r-s)} - \ P_s(p_1,\ldots,p_r) \\ + \delta_{m-s}^{n-(r-s)} \\ \cdot \sqrt{m(m-1) \cdots (m-s+1)} \sqrt{n(n-1) \cdots (n-(r-s)+1)} \\ \cdot \int dk_1 \cdots dk_{m-s} \ \overline{\psi'_m(k_1,\ldots,k_{m-s},p_1,\ldots,p_s)} @@ -869,10 +967,12 @@ In the following proof it will we convenient to use the abbreviation and $P_s(p_1,\ldots,p_r)$ is defined as before. \end{lemma} -%\[ - %f(T), f\left( T \right), - %\int_{a}^{b} f\left( x \right) d x, \frac{1}{T}, -%\] +\begin{proof} + a +\end{proof} + + +\section{Definition of the Stress Tensor} In the theory of a real scalar field $\phi$ of mass $m$, the Lagrangian density of the Klein-Gordon action is given by @@ -890,7 +990,7 @@ Raising the index $\nu$ and inserting \cref{lagrangian-density} yields \end{equation*} The \emph{energy density}: \begin{equation*} - \rho = T^{00} = \frac{1}{2} \parens*{\sum_{\mu=0}^{3} (\partial^{\mu}\phi)^2 + m^2 \phi^2} + \energydensity = T^{00} = \frac{1}{2} \parens*{\sum_{\mu=0}^{3} (\partial^{\mu}\phi)^2 + m^2 \phi^2} \end{equation*} The discussion in the previous section enables us to define the \emph{renormalized stress-energy tensor} of a free scalar field $\varphi$ by @@ -900,11 +1000,11 @@ the \emph{renormalized stress-energy tensor} of a free scalar field $\varphi$ by and this is a quadratic form. In particular, the energy density is \begin{equation*} - \rho = \frac{1}{2} \sum_{\mu=0}^{3} \normord{(\partial^{\mu}\phi)^2} + \frac{1}{2} m^2 \normord{\phi^2} + \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'\!}{\rho(f) \,\psi} = \\ + \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} @@ -925,13 +1025,13 @@ where \begin{theorem}{TODO}{} Let $\varphi$ be a free quantum field. Let $D_1, \ldots, D_r$ be linear differential operators with constant (complex) coefficients. - Then we have for all states $\psi,\psi' \in D^l(H)$ + Then we have for all test functions $f \in \schwartz{M}$ \begin{multline*} \normord{D_1 \varphi \cdots D_r \varphi}(f) \QFequal \int dp_1 \!\cdots dp_r \sum_{s=0}^{r} P_s(p_1,\ldots,p_r) \, \ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r) \\ \cdot a^\dagger(p_1) \cdots a^\dagger(p_s) a(p_{s+1}) \cdots a(p_r) \end{multline*} - as quadratic forms on $D^l(H)$, where + as quadratic forms on $\Domain{H^l}$, where \begin{multline*} \quad P_s(p_1,\ldots,p_r) = \frac{1}{\sqrt{2^r}} @@ -944,17 +1044,107 @@ where \begin{definition}{}{} \begin{multline*} - \rho(f) \QFequal \frac{1}{4} \int dp dp' (p \cdot p' + m^2) + \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' + (m^2 + \bar{p} \cdot 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} + +\begin{proposition}{}{} + The Fock vaccum $\fockvaccum$ lies in the domain of $\energydensity(f)\QFop{}$ + for all test functions $f \in \schwartz{M}$ + and $\energydensity(f)\QFop{}\fockvaccum$ is the vector $\psi$ defined by + \begin{equation*} + \psi_2(p,p') = \frac{\sqrt{2}}{4} (m^2 - \bar{p} \cdot p') \ft{f}(-p-p') + \end{equation*} + and $\psi_n \equiv 0$ for $n \ne 2$. +\end{proposition} + +\begin{equation*} + \energydensity(f) \Omega = ? +\end{equation*} + \section{Essential Selfadjointness of Renormalized Products} -a +\begin{lemma}{H-Bounds for the Renormalized Product}{} + \begin{equation*} + \abs{\innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(f)\psi}} \le + C \norm{(I+H)^l \psi} \norm{(I+H)^l \psi'} + \end{equation*} +\end{lemma} -%\nocite{*} +\begin{proof} + This proof is nearly identical to that of \cref{lemma:integral-kernel-h-bound} + and we will only cover the differences. + \begin{equation*} + \begin{multlined}[c] + \abs{\innerp{\psi'\!}{\normord{D_1 \varphi \cdots D_r \varphi}(f)\psi}} \le + \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \sum_{s=0}^{r} + \delta_{m-s}^{n-(r-s)} \\ + \hspace{2.5cm} \cdot + \int \!dk \int \!dp'\! \int \!dp \, \abs*{F(k,p',p) \, G'(k,p') \, G(k,p)}, + \end{multlined} + \end{equation*} + where + \begin{multline} + F(k,p',p) = \parens[\big]{1+\omega(k,p')} {}^{-a} \parens[\big]{1+\omega(k,p)} {}^{-a} P_s(p',p) \\ + \cdot \ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r) + \end{multline} + and $G$ and $G'$ are defined as before. + All we have to do, is verifying that + \begin{equation*} + \sup_k \norm{F(k,\cdot,\cdot)}_2 < \infty + \end{equation*} + for a sufficiently large integer $a$. + Then we obtain the desired $H$-bound with $l=a+r/2$. + + Recall that the Schwartz class is preserved by Fourier transform, translation and multiplication with polynomials. + Moreover, it is well known that Schwartz functions are square-integrable with repect to the Lorentz invariant measure on the mass shell. + Hence, + \begin{equation*} + \int dp_1 \abs{\ft{f}(p_1 + \cdots + p_s - p_{s+1} - \cdots - p_r)}^2 + \end{equation*} + is bounded by a constant independent of $p_2,\ldots,p_r$. +\end{proof} + +\section{Covariance} + +\begin{equation*} + f_g(x) \defequal f(g^{-1} x) \qquad + x \in M, \quad g \in \ProperOrthochronousPoincareGroup. +\end{equation*} + +\begin{theorem}{Covariance}{covariance-renormalized-product} + \begin{equation*} + U(g) \,\normord{D_1 \varphi \cdots D_r \varphi}(f)\, U(g)^{-1} + = \normord{D_1 \varphi \cdots D_r \varphi}(f_g) + \end{equation*} +\end{theorem} \chapterbib \cleardoublepage |