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Diffstat (limited to 'stresstensor.tex')
| -rw-r--r-- | stresstensor.tex | 36 |
1 files changed, 19 insertions, 17 deletions
diff --git a/stresstensor.tex b/stresstensor.tex index 4c128c2..79c0930 100644 --- a/stresstensor.tex +++ b/stresstensor.tex @@ -32,7 +32,9 @@ as a service to the reader. x \cdot y = g_{\mu \nu} x^{\mu} y^{\nu} = x^0y^0 - x^1 y^1 - x^2 y^2 - x^3 y^3 \end{equation*} points $x = (x^0,x^1,x^2,x^3) \in M$ are sometimes written $x = (x^0,\symbfit{x})$ with separated time and space coordinates - \item Given a complex-valued function $f$ on $M$, we define its \emph{Fourier transform} $\ft{f}\,$ by + \item Given a complex-valued function $f$ on $M$, + we define its \emph{Fourier transform}\index{Fourier transform} $\ft{f}\,$ by + \nomenclature[f]{$\ft{f}$}{Fourier transform of $f$} \begin{equation} \label{fourier-transform} \ft{f}(p) \defequal \int_{M} e^{i p \cdot x} f(x) \, dx @@ -93,7 +95,7 @@ as a service to the reader. \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 $f \in \schwartz{M}$ + This extends to complex valued test functions $f \in \schwartz{M}$ \begin{equation*} \varphi(f) = \frac{1}{\sqrt{2}} \parens*{a(E\bar{f}) + a(Ef)^\dagger} \end{equation*} @@ -243,7 +245,7 @@ but one may obtain an operator with trivial domain. We will use the symbol $\QFequal$ between quadratic forms or operators to indicate their equality as quadratic forms. -TODO(statement about domains?) +\todo{statement about domains?} A natural question is how the smeared operators relate to the pointwise ones. @@ -287,7 +289,7 @@ for all $\psi,\psi' \in D$. 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) +\todo{present physicists way of introducing normal ordering} Now let us extract the algebraic essence of the situation. The objects of our calculations are the field operators $\Phi(f)$, @@ -314,7 +316,7 @@ 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) +\todo{explain need for commutation relations} By abstract algebra, this is viable by forming the quotient of the free algebra with respect to the two-sided ideal @@ -332,7 +334,7 @@ where $z,z' \in \hilb{H}$. where $e$ is the unit of the algebra. \end{definition} -TODO(introduce $\Phi$ as representation of $\WeylAlg$) +\todo{introduce $\Phi$ as representation of $\WeylAlg$} \begin{definition}{Annihilator and Creator}{} Suppose $\WeylAlg$ is the infinitesimal Weyl algebra @@ -383,8 +385,8 @@ The cases $r=1$ and $r=2$ read \normord{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) + \weylcreator(z_1) \weylannihilator(z_2) - + \weylcreator(z_2) \weylannihilator(z_1) + \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 @@ -399,7 +401,7 @@ if one brings~\eqref{equation:normal-ordering} into the equivalent form \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?). +by basic combinatorial arguments \todo{further explanation?}. In~\cite{Klein1973}, the factor $\frac{1}{s!(r-s)!}$ is erroneously missing. @@ -488,7 +490,7 @@ In terms of creation and annihilation operators we have \label{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 muliplication with the polynomial +In Fourier space the operator $D^\dagger$ corresponds to multiplication with the polynomial \begin{equation*} \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*} @@ -562,7 +564,7 @@ this approach incurs significant technical difficulties. \end{lemma} Note that $K$ has a remaining dependence on $f$ via $\chi$ -even thogh the notation does not indicate this. +even though the notation does not indicate this. This is made explicit in the alternative integral representation \begin{equation} \label{equation:alternative-integral-representation} @@ -663,7 +665,7 @@ The following assertion is key to realizing the idea of taking the limit $f \to 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 $D_1, \ldots, D_r$. + where $d$ is the highest order of differentiation occurring in $D_1, \ldots, D_r$. \end{lemma} The Hamilton operator $H$ acts on $n$-particle states $\psi_n$ as follows: @@ -869,7 +871,7 @@ In the following proof it will be convenient to use the abbreviation \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 similar for $\psi$, by definition of the inner product and because $((1+H)^l \psi')_m = (1+H)^l \psi'_m$ for all $m$. \end{myproof} @@ -898,7 +900,7 @@ In the following proof it will be convenient to use the abbreviation 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?) + \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 $\tempdistribnoarg \to \tempdistribnoarg$, @@ -1076,9 +1078,9 @@ where \end{proposition} \begin{proposition}{}{} - The Fock vaccum $\fockvaccum$ lies in the domain of $\energydensity(f)\QFop{}$ + The Fock vacuum $\FockVacuum$ 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 + and $\energydensity(f)\QFop{}\FockVacuum$ 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*} @@ -1124,7 +1126,7 @@ where 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. + Moreover, it is well known that Schwartz functions are square-integrable with respect 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 |