\documentclass[12pt, a4paper]{article} \usepackage[left=2cm, right=2cm, top=2cm]{geometry} \makeatletter % ------------------------- % % PACKAGES % % ------------------------- % \usepackage[utf8]{inputenc} % Encoding of the document \usepackage[T1]{fontenc} % Font encoding \usepackage{lmodern} % Latin Modern fonts \usepackage{amsmath, amssymb, amsfonts, bm} % Math packages \usepackage{dsfont} \usepackage{physics} % Simplify writing physics equations \usepackage{graphicx} % Include images \usepackage{tikz} % Drawing package for diagrams \usetikzlibrary{decorations.pathmorphing} \usepackage[compat=1.1.0]{tikz-feynman} \usepackage{hyperref} % Hyperlinks within the document \usepackage{cleveref} % Enhanced cross-referencing \usepackage{geometry} % Page layout \usepackage{fancyhdr} % Custom headers and footers \usepackage{color, xcolor} % Color management \usepackage{sectsty} % Section font customization \usepackage{titlesec} % Title formatting \usepackage{enumitem} % Custom lists \usepackage{marginnote} % Margin notes \usepackage{booktabs} % Enhanced tables \usepackage{caption} % Customize captions \usepackage{subcaption} % Subfigures \usepackage{hyperref} % Hyperlinks \usepackage{siunitx} % SI units formatting \usepackage{thmtools, thm-restate} % Theorem environments \usepackage{mathtools} \usepackage{esvect} \usepackage{gensymb} \usepackage{changepage} \usepackage{tensor} \usepackage{mathrsfs} \usepackage{slashed} \usepackage{simpler-wick} % ------------------------- % % PAGE LAYOUT % % ------------------------- % \geometry{ a4paper, total={170mm,257mm}, left=20mm, right=20mm, top=20mm, bottom=20mm, } % ------------------------- % % HEADER & FOOTER % % ------------------------- % \pagestyle{fancy} \fancyhf{} \fancyhead[L]{WEINBERG SOLUTIONS} \fancyhead[R]{\rightmark} \fancyfoot[C]{\thepage} \renewcommand{\headrulewidth}{0.4pt} \renewcommand{\footrulewidth}{0.4pt} % ------------------------- % % SECTION STYLING % % ------------------------- % \sectionfont{\fontsize{14}{16}\selectfont\bfseries\color{violet}} \subsectionfont{\fontsize{12}{14}\selectfont\bfseries\color{purple}} \subsubsectionfont{\fontsize{11}{13}\selectfont\bfseries\color{magenta}} % ------------------------- % % THEOREM ENVIRONMENTS % % ------------------------- % \declaretheorem[name=Theorem]{theorem} \declaretheorem[name=Lemma, sibling=theorem]{lemma} \declaretheorem[name=Definition, style=definition]{definition} \declaretheorem[name=Proposition, sibling=theorem]{proposition} \declaretheorem[name=Corollary, sibling=theorem]{corollary} % ------------------------- % % CUSTOM COMMANDS % % ------------------------- % \newcommand{\vect}[1]{\mathbf{#1}} % Bold vectors \newcommand{\deriv}[2]{\frac{d #1}{d #2}} % Derivative \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} % Partial derivative \newcommand{\ex}[1]{\langle #1 \rangle} \newcommand{\Var}[1]{\text{Var}\left[ #1 \right]} \newcommand{\Cov}[2]{\text{Cov}\left[ #1 , #2 \right]} \renewcommand*\env@matrix[1][\arraystretch]{% \edef\arraystretch{#1}% \hskip -\arraycolsep \let\@ifnextchar\new@ifnextchar \array{*\c@MaxMatrixCols c}} \makeatother % ------------------------- % % TITLE & AUTHOR % % ------------------------- % \title{ \vspace{2cm} \Huge\textbf{Selected Solutions to Weinberg's TQTF}\\ \vspace{0.5cm} } \author{Cameron Poe, Joey Takach, William Urdahl} \date{\today} % ------------------------- % % DOCUMENT % % ------------------------- % \begin{document} \maketitle \tableofcontents \section{Conventions} We will follow all the same conventions as Weinberg does. Most importantly, a generic four vector is written as $A^{\mu} = (A^1, A^2, A^3, A^0)$, and the metric $\eta^{\mu \nu} = \text{diag}(1,1,1,-1)$. \section{Chapter 2} \subsection{Problem 2.1 -- Quantum Lorentz Transformations for a Massive Particle} We will make use of equation (2.5.23) for how a massive particle state $\Psi_{p,\sigma}$ transforms under a homogenous Lorentz transformation $U(\Lambda)$: \begin{equation} U(\Lambda) \Psi_{p,\sigma} = \sqrt{\frac{(\Lambda p)^0}{p^0}}\sum_{\sigma'}D^{(j)}_{\sigma' \sigma}(W(\Lambda,p))\Psi_{\Lambda p, \sigma'} \end{equation} The most difficult part of this problem is finding what the little group transformation $W$ is. $W$ is given by equation (2.5.10): \begin{equation} W(\Lambda, p) = L^{-1}(\Lambda p) \Lambda L(p) \label{eq:littlegroup-w} \end{equation} Before we compute $W$, we can note two properties it must have. Since the little group for massive particles is $SO(3)$, we know that $W$, a representation of the little group, must be a rotation matrix. The other property is the rotation matrix must be a rotation about the $x$-axis. This is because $\mathbf{p}$ is in the $y$-direction, and therefore the boost $L(p)$ preserves four-vectors' $x$-components. Similarly, the boost $\Lambda$ is in the $z$-direction and preserves $x$-components. The boost $L^{-1}(\Lambda p)$ boosts in the $y$- and $z$-directions, and must also preserve $x$-components. So the rotation $W$ must leave $x$-components invariant, which means the rotation must be about the $x$-axis. The energy of the W-boson in observer $\mathcal{O}$'s frame is $E = \sqrt{p^2 + m^2}$, and therefore the four-momentum is \begin{equation} p^{\mu} = (0,p,0,E) \end{equation} We will use equation (2.5.24) to calculate $L(p)$ and $L^{-1}(\Lambda p)$. The Lorentz factor to go from $k^{\mu}$ to $p^{\mu}$ is $\gamma = \frac{E}{m}$, so $\sqrt{\gamma^2 - 1} = \frac{p}{m}$. The components of the unit three-momentum are $\hat{p}_1 = \hat{p}_3 = 0$ and $\hat{p}_2 = 1$. The boost $L(p)$ is then \begin{equation} L(p) = \begin{bmatrix}[1.2] 1 & 0 & 0 & 0 \\ 0 & \frac{E}{m} & 0 & \frac{p}{m} \\ 0 & 0 & 1 & 0 \\ 0 & \frac{p}{m} & 0 & \frac{E}{m} \\ \end{bmatrix} \end{equation} Since $\mathcal{O}'$ is moving at speed $v$ in the $+z$-direction relative to $\mathcal{O}$, the boost that takes us from $\mathcal{O}$ to $\mathcal{O}'$ is \begin{equation} \Lambda = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \gamma & -v \gamma \\ 0 & 0 & -v \gamma & \gamma \\ \end{bmatrix} \end{equation} where $\gamma = \frac{1}{\sqrt{1-v^2}}$. Note that this $\gamma$ is not referring to the gamma used to previously find $L(p)$, but rather refers to the boost from $\mathcal{O}$ to $\mathcal{O}'$. The four-momenta to $\mathcal{O}'$ is \begin{equation} (\Lambda p)^{\mu} = (0,p,-v \gamma E, \gamma E) \end{equation} The boost $L^{-1}(\Lambda p)$ is the inverse of $L(\Lambda p)$, and therefore boosts a particle with four-momentum $(\Lambda p)^{\mu}$ back into its rest frame. This is equivalent to boosting the particle in the opposite direction it was originally boosted in, so $L^{-1}(\Lambda p) = L(-\Lambda p)$. The Lorentz factor for this boost is $\gamma = \frac{E'}{m} = \frac{\gamma E}{m}$. The expression for $L^i_0$ can be simplified when solving for these components: \begin{equation} L^i_0(p) = \hat{p_i}\sqrt{\gamma^2 - 1} = \frac{p_i}{\abs{\mathbf{p}}}\frac{\abs{\mathbf{p}}}{m} = \frac{p_i}{m} \end{equation} Therefore \begin{align} L^1_0(-\Lambda p) &= 0 \\ L^2_0(-\Lambda p) &= -\frac{p}{m} \\ L^3_0(-\Lambda p) &= \frac{v \gamma E}{m} \end{align} The boost $L^{-1}(\Lambda p)$ then reads \begin{equation} L^{-1}(\Lambda p) = \begin{bmatrix}[1.5] 1 & 0 & 0 & 0 \\ 0 & \frac{\gamma E}{m} \left(\frac{p^2 + v^2 \gamma m E}{p^2 + v^2 \gamma^2 E^2}\right) & \frac{v \gamma p E}{m}\left(\frac{m- \gamma E}{p^2 + v^2 \gamma^2 E^2}\right)& -\frac{p}{m} \\ 0 & \frac{v \gamma p E}{m}\left(\frac{m- \gamma E}{p^2 + v^2 \gamma^2 E^2}\right) & \frac{v^2 \gamma^3 E^3 + m p^2}{m(p^2 + v^2 \gamma^2 E^2)} & \frac{v \gamma E}{m} \\ 0 & -\frac{p}{m} & \frac{v \gamma E}{m} & \frac{\gamma E}{m} \end{bmatrix} \end{equation} Plugging all of this into Equation \ref{eq:littlegroup-w} gives the full little group element \begin{equation} W(\Lambda, p) = \begin{bmatrix}[1.5] 1 & 0 & 0 & 0 \\ 0 & \frac{\gamma m + E}{m + \gamma E} & \frac{v \gamma p}{m + \gamma E} & 0 \\ 0 & -\frac{v \gamma p}{m + \gamma E} & \frac{\gamma m + E}{m + \gamma E} & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{equation} We should note that $W(\Lambda, p)$ has the predicted form of a rotation matrix about the $x$-axis, where we identify $\cos(\theta) = \frac{\gamma m + e}{m + \gamma E}$ and $\sin(\theta) = \frac{v \gamma p}{m + \gamma E}$. Since the W-boson is a spin-1 particle, the representation $D^{(j=1)}_{\sigma' \sigma}$ of $W(\Lambda,p)$ is simply a rotation matrix for 3D vectors, so we can immediately identify \begin{equation} D^{(1)}_{\sigma' \sigma} = \begin{bmatrix}[1.5] 1 & 0 & 0 \\ 0 & \frac{\gamma m + E}{m + \gamma E} & \frac{v \gamma p}{m + \gamma E} \\ 0 & -\frac{v \gamma p}{m + \gamma E} & \frac{\gamma m + E}{m + \gamma E} \\ \end{bmatrix} \end{equation} with subsequent rows and columns numbered -1, 0, and 1. We are now able to write the full transformed state \begin{equation} \begin{split} U(\Lambda)\Psi_{p,+1} &= \sqrt{\frac{\gamma E}{E}} \sum_{\sigma'}D^{(1)}_{\sigma', +1}(W(\Lambda, p))\Psi_{\Lambda p, \sigma'} \\ &= \sqrt{\gamma}\left(D^{(1)}_{-1,+1}\Psi_{\Lambda p, -1} + D^{(1)}_{0,+1}\Psi_{\Lambda p, 0} + D^{(1)}_{+1,+1}\Psi_{\Lambda p, +1}\right) \\ &= \sqrt{\gamma}\left(\frac{v \gamma p }{m+\gamma E}\Psi_{\Lambda p, 0} + \frac{\gamma m + E}{m +\gamma E}\Psi_{\Lambda p, +1}\right) \\ \Aboxed{U(\Lambda)\Psi_{p,+1} &= \frac{\sqrt{\gamma}}{m + \gamma E} \left(v \gamma p \Psi_{\Lambda p, 0} + (\gamma m + E)\Psi_{\Lambda p, +1}\right)} \end{split} \end{equation} We can further check that when $v\rightarrow0$, we get that $U(\Lambda)\Psi_{p,+1} = \Psi_{p,+1}$, as expected. \subsection{Problem 2.2 -- Quantum Lorentz Transformations for a Massless Particle} To solve this problem, we do something similar to 2.1, where we find the little group element $W$, and read off the necessary values from this matrix to find $U(\Lambda)$. The quantum Lorentz transformation for a massless particle is given by equation (2.5.42) in Weinberg: \begin{equation} U(\Lambda) \Psi_{p,\sigma} = \sqrt{\frac{(\Lambda p)^0}{p^0}} e^{i\sigma \theta(\Lambda, p)} \Psi_{\Lambda p, \sigma} \end{equation} Weinberg equation 2.5.43 gives the relation between $W$, $W$ as a Wigner rotation, and the special form of $W$ for massless particles: \begin{equation} W(\Lambda, p) = L^{-1}(\Lambda p) \Lambda L(p) = S(\alpha(\Lambda,p),\beta(\Lambda,p)) R(\theta(\Lambda,p)) \end{equation} where we have the following relations: \begin{equation} S(\alpha, \beta) = \begin{bmatrix} 1 & 0 & -\alpha & \alpha \\ 0 & 1 & -\beta & \beta \\ \alpha & \beta & 1-\xi & \xi \\ \alpha & \beta & -\xi & 1 + \xi \end{bmatrix} \end{equation} \begin{equation} R(\theta) = \begin{bmatrix} \cos\theta & \sin\theta & 0 & 0 \\ -\sin\theta & \cos\theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{equation} \begin{equation} L(p) = R(\boldsymbol{\hat{\mathbf{p}}})B(|\mathbf{p}|/\kappa) \end{equation} \begin{equation} B(u) = \begin{bmatrix}[1.4] 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{u^2 + 1}{2u} & \frac{u^2 - 1}{2u} \\ 0 & 0 & \frac{u^2 - 1}{2u} & \frac{u^2 + 1}{2u} \end{bmatrix} \end{equation} where $L(p)$ is boosting $k^{\mu} = (0,0,\kappa,\kappa)$ to $p^{\mu}$, $R(\boldsymbol{\hat{\mathbf{p}}})$ rotates the three-axis to the direction of the three-momentum $\boldsymbol{\hat{\mathbf{p}}}$. Also, $\xi = \frac{1}{2}(\alpha^2 + \beta^2)$. The four-momentum of the photon to observer $\mathcal{O}$ is $p^{\mu} = (0,p,0,p)$. We boost to observer $\mathcal{O'}$ moving in the $+z$-direction with velocity $\beta = v$. The four-momentum to $\mathcal{O'}$ is then \begin{equation} \Lambda p = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \gamma & -v \gamma \\ 0 & 0 & -v \gamma & \gamma \end{bmatrix} \begin{bmatrix} 0 \\ p \\ 0 \\ p \end{bmatrix} = \begin{bmatrix} 0 \\ p \\ -v \gamma p \\ \gamma p \end{bmatrix} \end{equation} with $\gamma = 1/\sqrt{1-v^2}$. For $L(p)$, we wish to rotate $\mathbf{k}$, the standard massless three-momentum which is along the three-axis, into the $\boldsymbol{\hat{\mathbf{p}}}$ direction. This is the $+y$-direction in our case. This is a rotation of $-\pi/2$ about the one-axis. Weinberg describes other rotation conventions that would give the same result (and also change the phase of one-particle states), but we will go with this as it only requires one rotation.\footnote{See Hagimoto for the other convention.} The rotation matrix then looks like: \begin{equation} R(\boldsymbol{\hat{\mathbf{p}}}) = R_1 (-\pi/2) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -\sin(-\pi/2) & 0 \\ 0 & \sin(-\pi/2) & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{equation} The full $L(p)$ is then \begin{equation} L(p) = \begin{bmatrix}[1.4] 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}[1.4] 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{(p/\kappa)^2 + 1}{2(p/\kappa)} & \frac{(p/\kappa)^2 - 1}{2(p/\kappa)} \\ 0 & 0 & \frac{(p/\kappa)^2 - 1}{2(p/\kappa)} & \frac{(p/\kappa)^2 + 1}{2(p/\kappa)} \end{bmatrix} = \begin{bmatrix}[1.4] 1 & 0 & 0 & 0 \\ 0 & 0 & \frac{(p/\kappa)^2 + 1}{2(p/\kappa)} & \frac{(p/\kappa)^2 - 1}{2(p/\kappa)} \\ 0 & -1 & 0 & 0 \\ 0 & 0 & \frac{(p/\kappa)^2 - 1}{2(p/\kappa)} & \frac{(p/\kappa)^2 + 1}{2(p/\kappa)} \end{bmatrix} \end{equation} To find $L^{-1}(\Lambda p)$, first let $p' = \Lambda p$. We then note \begin{equation} L^{-1}(p') = (R(\boldsymbol{\hat{\mathbf{p}}}')B(|\mathbf{p}'|/\kappa))^{-1} = B^{-1}(|\mathbf{p}'|/\kappa) R^{-1}(\boldsymbol{\hat{\mathbf{p}}}') \end{equation} Solving for $R(\boldsymbol{\hat{\mathbf{p}}}')$, we can once again do a simple rotation about the one-axis, although this time it won't be a nice angle like $\pi/2$. Since this is a massless particle, we must have $|\mathbf{p}'| = E = \gamma p$, so $-\boldsymbol{\hat{\mathbf{p}}}' = (0,-1/\gamma, v)$. Geometrically, we find that $\cos\theta = -v$ and $\sin\theta = -1/\gamma$. Our inverse rotation matrix then just sends $\cos \theta \rightarrow \cos \theta$ and $\sin \theta \rightarrow -\sin \theta$ \begin{equation} R^{-1}(\boldsymbol{\hat{\mathbf{p}}}') = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -v & -1/\gamma & 0 \\ 0 & 1/\gamma & -v & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{equation} The inverse of $B(u)$ is not as simply as letting $u \rightarrow -u$ everywhere, but since the $(3,0)$ and $(0,3)$ elements correspond to $\beta \gamma$, we just put a minus on these since we are boosting in the opposite direction, i.e. in $-\beta$, so \begin{equation} B^{-1}(|\mathbf{p}'|/\kappa) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{(\gamma p/\kappa)^2 + 1}{2(\gamma p/\kappa)} & -\frac{(\gamma p/\kappa)^2 - 1}{2(\gamma p/\kappa)} \\ 0 & 0 & -\frac{(\gamma p/\kappa)^2 - 1}{2(\gamma p/\kappa)} & \frac{(\gamma p/\kappa)^2 + 1}{2(\gamma p/\kappa)} \end{bmatrix} \end{equation} Our full $L(p')$ is: \begin{equation} L^{-1}(p') = \begin{bmatrix}[1.4] 1 & 0 & 0 & 0 \\ 0 & -v & -1/\gamma & 0 \\ 0 & \frac{p}{2\kappa} + \frac{\kappa}{2p\gamma^2} & -\frac{v \kappa^2 + p^2 \gamma^2}{2p\gamma\kappa} & -\frac{p\gamma}{2\kappa} + \frac{\kappa}{2p\gamma} \\ 0 & -\frac{p}{2\kappa} + \frac{\kappa}{2p\gamma^2} & -\frac{p v \gamma}{2\kappa} - \frac{v \kappa}{2p\gamma} & \frac{p\gamma}{2\kappa} + \frac{\kappa}{2p\gamma} \end{bmatrix} \end{equation} The full little group element is then \begin{equation} W = \begin{bmatrix}[1.4] 1 & 0 & 0 & 0 \\ 0 & 1 & \frac{-v \kappa}{p} & \frac{v \kappa}{p} \\ 0 & \frac{v \kappa}{p} & 1 - \frac{v^2 \kappa^2}{2p^2} & \frac{v^2 \kappa^2}{2p^2} \\ 0 & \frac{v \kappa}{p} & -\frac{v^2\kappa^2}{2p^2} & 1+\frac{v^2\kappa^2}{2p^2} \end{bmatrix} \end{equation} Let's now match this $W$ to the one in terms of $\theta$, $\alpha$, and $\beta$. \begin{align} W &= \begin{bmatrix} 1 & 0 & -\alpha & \alpha \\ 0 & 1 & -\beta & \beta \\ \alpha & \beta & 1-\xi & \xi \\ \alpha & \beta & -\xi & 1 + \xi \end{bmatrix} \begin{bmatrix} \cos\theta & \sin\theta & 0 & 0 \\ -\sin\theta & \cos\theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} \cos \theta & \sin \theta & -\alpha & \alpha \\ -\sin \theta & \cos \theta & -\beta & \beta \\ \alpha \cos \theta - \beta \sin \theta & \beta \cos \theta + \alpha \sin \theta & 1 - \xi & \xi \\ \alpha \cos \theta - \beta \sin \theta & \beta \cos \theta + \alpha \sin \theta & -\xi & 1 + \xi \end{bmatrix} \end{align} The top left 2-by-2 block let's us easily read off what $\theta$ is, while the top right 2-by-2 block let's us read $\alpha$ and $\beta$. We see that $\theta = 0$, $\alpha = 0$, and $\beta = v \kappa / p$. Therefore, the transformed photon state is \begin{equation} \boxed{U(\Lambda) \Psi_{p,+1} = \sqrt{\gamma} \Psi_{\Lambda p, +1}} \end{equation} \subsection{Problem 2.3 -- Galilean Group Commutation Relations} To begin, we define our inhomogeneous Galilean transformations as a pair $(G,a)$, where $G$ is a 4x4 matrix acting on 4-vectors, and $a$ is a 4-vector shift representing the translation elements. The group law is the same as the Poincare case: $(\bar G, \bar a)(G,a) = (\bar G G,\bar G a + \bar a)$. The analogues to the Lorentz condition here are two-fold; both the spatial norm of vectors on the same time slice, and the time coordinate of any vector must be preserved under arbitrary (homogeneous) Galilean transformations $G$. These conditions have the form \begin{equation} \sum_i(Gx)_i(Gy)_i = \sum_i x_iy_i \quad \text{and} \quad (Gx)_0=x_0 \ . \end{equation} In component notation, these have the form \begin{equation} G_{ik}G_{jk} = \delta_{ij} \quad \text{and} \quad G_{0\alpha} = \delta_{0\alpha} \ . \end{equation} From here on, we will use latin indices to refer to spatial components (1,2,3) and greek indices for arbitrary components. We will also write the sums implicitly unless the notation is wack. These conditions tell us that the spatial block of the matrix $G$ must be an ordinary 3x3 rotation matrix $R$, and that the bottom row of the matrix is null save for the time component. The boost components ($G_{i0}$) are unconstrained, so we set them as the vector $\vect{v}$. In block-diagonal notation, $G$ then takes the form: \begin{equation} G= \begin{bmatrix} R & \vect{v}\\ 0 & 1 \end{bmatrix} \ . \end{equation} We will be interested in the form of the infinitesimal transformations, eg when we approximately have $G=1+\omega$ (or $G_{\alpha\beta}=\delta_{\alpha\beta}+\omega_{\alpha\beta}$), where $\omega$ is a matrix representing the infinitesimal parameters. Concretely, \begin{equation} G= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} + \begin{bmatrix} r & \vect{b}\\ 0 & 0 \end{bmatrix} = 1+\omega \end{equation} where $r$ is our infinitesimal rotation, and $\vect{b}$ our infinitesimal boost. Applying the Galilean conditions to this infinitesimal form, we obtain \begin{equation} \begin{split} \delta_{ij} = G_{ik}G_{jk} = (\delta_{ik}+\omega_{ik})&(\delta_{jk}+\omega_{jk}) = \delta_{ij} + \omega_{ij} + \omega_{ji} + \mathcal O (\omega^2) \\ &\Rightarrow \omega_{ij} = -\omega_{ji} \end{split} \end{equation} \begin{equation} \delta_{0\alpha} = G_{0\alpha} = \delta_{0\alpha} + \omega_{0\alpha} + \mathcal O(\omega^2) \quad \Rightarrow \quad \omega_{0\alpha}=0 \ . \end{equation} This just confirms what we already knew, namely that the infinitesimal matrix $r$ is antisymmetric, and the bottom row of the infinitesimal matrix $\omega$ is 0. As with the relativistic case, there are 10 independent generators: 3 for rotations, 3 for boosts, and 4 for spacetime translations. When we represent these group elements on the Hilbert space as $U(G,a)$, we can then expand the infinitesimal elements as: \begin{equation} U(1+\omega,\varepsilon) = 1 + \frac{i}{2}r_{ij}J_{ij}+ ib_iK_i - i\varepsilon_i P_i - i\varepsilon_0P_0 \ . \end{equation} The factor of 1/2 again comes from the fact that since $r$ is antisymmetric, we can also choose the generators $J_{ij}$ in our representation to be antisymmetric, so the double sum over $i,j$ doubles the contribution from each independent parameter. Following the procedure in Section 2.4, we can expand the product $U(G,a)U(1+\omega,\varepsilon)U^{-1}(G,a)$ in two different ways; one where we first apply the group law, and one where we first expand. However, there is a key difference to the Galilean case from the Lorentz case. The representation of the Galilean group acting on the Hilbert space is intrinsically projective, and moreover, the central charges cannot be eliminated by a redefinition of the commutators (see Section 2.7). This means that the group law in the representation will have the form: \begin{equation} U(\bar G,\bar a)U(G,a) = \exp[i\phi(\bar G,\bar a ; G, a)]U(\bar G G, \bar G a + \bar a) \ . \end{equation} However, the discussion in Section 2.7 shows that this extra phase is equivalent to the inclusion of central charges in the commutation relations, so we will ignore the phase for now, and manually add in the central charges once we have the basic commutators. We need to expand both sides of the equation: \begin{equation} U(G,a)U(1+\omega,\varepsilon)U^{-1}(G,a) = U(1+G\omega G^{-1},G\varepsilon-G\omega G^{-1}a) \ . \end{equation} Expanding the RHS is made much easier through use of the block diagonal matrices for $G,\omega$ written above. We have: \begin{equation} \begin{split} 1+\frac{i}{2}(RrR^T)_{mn}J_{mn}&+i(R\vect{b}-RrR^T\vect{v})_mK_m\\ &-i\bigl[ (R(\bm{\varepsilon}+a_0\vect{b})+RrR^T(a_0\vect{v}-\vect{a})-\varepsilon_0\vect{v})_mP_m + \varepsilon_0P_0\bigr] \ . \end{split} \end{equation} And for the left side, we have: \begin{equation} U(G,a)\bigl[ 1 + \frac{i}{2}r_{ij}J_{ij}+ ib_iK_i - i\varepsilon_i P_i - i\varepsilon_0P_0 \bigr]U^{-1}(G,a) \ . \end{equation} Expanding out the RHS making sure the indices on the infinitesimal parameters are the same as those on the LHS, we can match the coefficients of those parameters on either side to arrive at the transformation laws for each of the generators: \begin{equation} \begin{split} U(G,a)J_{ij}U^{-1}(G,a)&=R_{mi}R_{nj}\bigl(J_{mn} -2K_mv_n-2P_m(a_0v_n-a_n)\bigr) \\ U(G,a)K_iU^{-1}(G,a) &= R_{mi}(K_m - a_0P_m) \\ U(G,a)P_iU^{-1}(G,a) &= R_{mi}P_m \\ U(G,a)P_0U^{-1}(G,a) &= P_0+v_mP_m \ . \end{split} \end{equation} Now, treating $G,a$ as infinitesimal transformations of the Galilean group, we can expand the $U$'s to first order, eg for the transformation of $P_i$ we get: \begin{equation} \begin{split} (\delta_{mi}+r_{mi})P_m &= \bigl(1+\frac{i}{2}r_{mn}J_{mn}+ ib_mK_m - i\varepsilon _mP_m - i\varepsilon _0P_0\bigr)P_i\bigl(1-\frac{i}{2}r_{mn}J_{mn}- ib_mK_m + i\varepsilon _mP_m + i\varepsilon _0P_0\bigr) \\ \end{split} \end{equation} which, when cancelling out the $P_i$ terms, leads to \begin{equation} \begin{split} r_{mn}\delta_{ni}P_m &= \frac{i}{2}r_{mn}[J_{mn},P_i]+ ib_m[K_m,P_i] - i\varepsilon _m[P_m,P_i] - i\varepsilon _0[P_0,P_i] \ . \end{split} \end{equation} Equating terms with common infinitesimal coefficients (and remembering $r_{mn},J_{mn}$ are antisymmetric), we get for the commutators: \begin{equation} i[J_{mn},P_i] = \delta_{ni}P_m - \delta_{mi}P_n \quad , \quad [K_m,P_i] = [P_m,P_i] = [P_0,P_i] = 0 \ . \end{equation} With the convention $J_{12}\equiv J_3$, etc, we can do the same with the other transformation laws to get all of the commutators: \begin{equation} \begin{split} &[J_i,J_j]=i\epsilon_{ijk}J_k ,\quad [J_i,K_j]=i\epsilon_{ijk}K_k,\quad [K_i,K_j]=0 \\ &[J_i,P_j]=i\epsilon_{ijk}P_k ,\quad [K_i,P_j] = 0,\quad [K_i,P_0] = -iP_i \\ &[P_0,J_i] = [P_0,P_i] = [P_i,P_j] = 0 \ . \end{split} \end{equation} Note that these do not yet match the commutation relations that Weinberg writes on Pg. 62 for the Galilean group. This is because, as discussed before, the work we have done so far has assumed that the Galilean group has an ordinary representation on the Hilbert space, and not a projective one. This does not mean the work we have done is completely invalidated, but it does mean that we have to add to the commutation relations a central charge term proportional to the identity, so that the commutators in general go from: \begin{equation} [t_b,t_c]=iC^a_{\text{ \ }bc}t_a \quad \Rightarrow \quad [t_b,t_c]=iC^a_{\text{ \ }bc}t_a +iC_{bc} \mathds{1} . \end{equation} Thus, we rewrite our commutators above as: \begin{equation} \begin{split} &[J_i,J_j]=i\epsilon_{ijk}J_k + iA_{ij} ,\quad [J_i,K_j]=i\epsilon_{ijk}K_k+ iB_{ij},\quad [K_i,K_j]= iD_{ij} \\ &[J_i,P_j]=i\epsilon_{ijk}P_k+ iE_{ij} ,\quad [K_i,P_j] = iF_{ij},\quad [K_i,P_0] = -iP_i + iL_{i0}\\ &[P_0,J_i] = iN_{0i},\quad [P_0,P_i] =iQ_{0i},\quad [P_i,P_j] = iR_{ij} \ . \end{split} \label{eq:charged_commutators} \end{equation} To determine these constants $C_{bc}$ in more detail, we need to apply the Jacobi identity to each of our commutators, which in general takes the form: \begin{equation} \bigl[t_a,[t_b,t_c]\bigr]+\bigl[t_c,[t_a,t_b]\bigr]+\bigl[t_b,[t_c,t_a]\bigr]=0 \ . \end{equation} As an example of how this works, consider now the Jacobi identity applied to $P_0,K_i,P_j$: \begin{equation} \begin{split} 0&=\bigl[P_0,[K_i,P_j]\bigr] + \bigl[P_j,[P_0,K_i]\bigr] +\bigl[K_i,[P_j,P_0]\bigr] \\ &=[P_j,iP_i-iL_{i0}]=-i[P_i,P_j]=R_{ij} \\ & \qquad \qquad \qquad \quad \Rightarrow R_{ij}=0 \ . \end{split} \end{equation} In the first line, the inner commutators of the first and third terms give only central charges (constants), so the outer commutators vanish. In the second line, we used that the commutator of the two $P_i$ generators gives another charge term, which we could then set to 0 because of the Jacobi identity. We now need to repeat this process for the other commutators. We have a few other similar cases: \ $P_0,J_i,J_j$: \begin{equation} \begin{split} 0&=\bigl[P_0,[J_i,J_j]\bigr] + \bigl[J_j,[P_0,J_i]\bigr] +\bigl[J_i,[J_j,P_0]\bigr] \\ &=i\epsilon_{ijk}[P_0,J_k] = -\epsilon_{ijk}N_{0k} \\ & \qquad \qquad \qquad \quad \Rightarrow N_{0i}=0 \ . \end{split} \end{equation} $P_0,J_i,P_j$: \begin{equation} \begin{split} 0&=\bigl[P_0,[J_i,P_j]\bigr] + \bigl[P_j,[P_0,J_i]\bigr] +\bigl[J_i,[P_j,P_0]\bigr] \\ &=i\epsilon_{ijk}[P_0,P_k] = -\epsilon_{ijk}Q_{0k} \\ & \qquad \qquad \qquad \quad \Rightarrow Q_{0i}=0 \ . \end{split} \end{equation} $J_i,K_j,K_k$: \begin{equation} \begin{split} 0&=\bigl[J_i,[K_j,K_k]\bigr] + \bigl[K_k,[J_i,K_j]\bigr] +\bigl[K_j,[K_k,J_i]\bigr] \\ &=i\epsilon_{ijm}[K_k,K_m] -i\epsilon_{ikm}[K_j,K_m] \\ &= -\epsilon_{ijm}D_{km} + \epsilon_{ikm}D_{jm} \ \end{split} \end{equation} This is the first non-trivial one. First note that from (\ref{eq:charged_commutators}), $D_{ij}$ is antisymmetric by definition, so all diagonal elements vanish. Next, pluging into the above every possibility of $i=j\neq k$ yields $D_{ij}=0$ for $i\neq j$. These combined mean $D_{ij}=0$ uniformly. Now we look at the central charges where the "base" commutation relations make use of the Levi-Civita symbol for their structure constants. Specifically, $A_{ij},B_{ij},E_{ij}$. Note how $A_{ij}$ is antisymmetric by definition, but $B_{ij},E_{ij}$ must be shown to be so. For $E_{ij}$, we consider the Jacobi identity for $P_0,J_i,K_j$: \begin{equation} \begin{split} 0&=\bigl[P_0,[J_i,K_j]\bigr] + \bigl[K_j,[P_0,J_i]\bigr] +\bigl[J_i,[K_j,P_0]\bigr] \\ &=i\epsilon_{ijk}[P_0,K_k] -i[J_i,P_j] \\ &=-\epsilon_{ijk}P_k+ \epsilon_{ijk}L_{i0} + \epsilon_{ijk}P_k +E_{ij} \\ &= -\epsilon_{ijk}P_{k}+E_{ij} \ . \end{split} \label{eq:P0JK} \end{equation} Similarly, we can swap the indices on $J_i$ and $K_j$ to get the result $0= -\epsilon_{jik}P_{k}+E_{ji}= \epsilon_{ijk}P_{k}+E_{ji}$. Adding this and the last line of (\ref{eq:P0JK}) yields $E_{ij}+E_{ji}=0$, so $E_{ij}$ is antisymmetric. To show the same for $B_{ij}$, we look at the Jacobi identity for $J_i,J_j,K_k$: \begin{equation} \begin{split} 0&=\bigl[J_i,[J_j,K_k]\bigr] + \bigl[K_k,[J_i,J_j]\bigr] +\bigl[J_j,[K_k,J_i]\bigr] \\ &=i\epsilon_{jkm}[J_i,K_m] -i\epsilon_{ijm}[J_m,K_k]-i\epsilon_{ikm}[J_j,K_m] \\ &=-\epsilon_{jkm}\epsilon_{imn}K_n +\epsilon_{ijm}\epsilon_{mkn}K_n+\epsilon_{ikm}\epsilon_{jmn}K_n \\ & \quad \ -\epsilon_{jkm}B_{im} +\epsilon_{ijm}B_{mk}+\epsilon_{ikm}B_{jm}\ . \end{split} \label{eq:JJK} \end{equation} The first line of the last equality vanishes from properties of $\epsilon_{ijk}$ (also since the generators $K_n$ are linearly independent to each other and to the identity on the lie algebra vector space, their coefficients must identically vanish if the RHS is equal to 0). The second line, when you plug in all cases of $i=k\neq j$, yields the desired antisymmetry condition, $B_{ij}+B_{ji}=0$ (alternatively, if you're feeling particularly pretentious or feel so inclined as to sacrifice a whole sheet of paper to the algebra gods, you can do some tedious index gymnastics by contracting with various Levi-Civita tensors and Kronecker deltas to arrive at the same thing, but this method is \textit{objectively} dumb and stupid). Now that we know $A_{ij},B_{ij},E_{ij}$ are all antisymmetric, we can move to redefine the corresponding generators to eliminate the central charges. Because these charges are antisymmetric, we can write them in terms of the antisymmetric Levi-Civita tensor, eg: \begin{equation} A_{ij} = \epsilon_{ijk}a_k, \quad B_{ij} = \epsilon_{ijk}b_k, \quad E_{ij} = \epsilon_{ijk}e_k \ , \end{equation} for some sets of undetermined constants $a_k,b_k,e_k$. We then redefine the generators in the following way: \begin{equation} \tilde{J}_i \equiv J_i + a_i , \quad \tilde{K}_i \equiv K_i + b_i , \quad \tilde{P}_i \equiv P_i + e_i \ . \end{equation} With these redefined generators, the commutation relations are rid of the $A,B,E$ charges, eg: \begin{equation} [\tilde{J}_i,\tilde{J}_j] = [J_i,J_j] = i\epsilon_{ijk}J_k+iA_{ij} = i\epsilon_{ijk}(\tilde{J}_k-a_k)+i\epsilon_{ijk}a_k = i\epsilon_{ijk}\tilde{J}_k \ . \end{equation} The same holds with the other commutation relations. From here on, we will work with the $\tilde{J}$ generators, but refer to them without tildes as $J$. We can summarize our results so far by writing the commutation relations: \begin{equation} \begin{split} &[J_i,J_j]=i\epsilon_{ijk}J_k ,\quad [J_i,K_j]=i\epsilon_{ijk}K_k,\quad [K_i,K_j]= 0 \\ &[J_i,P_j]=i\epsilon_{ijk}P_k ,\quad [K_i,P_j] = iF_{ij},\quad [K_i,P_0] = -iP_i + iL_{i0}\\ &[P_0,J_i] = [P_0,P_i] = [P_i,P_j] = 0 \ . \end{split} \label{eq:less_charged_commutators} \end{equation} We still have two charges to deal with, $L_{i0}$ and $F_{ij}$. We can show that $L_{i0}=0$ with the Jacobi identity applied to $P_0,J_i,K_j$: \begin{equation} \begin{split} 0&=\bigl[P_0,[J_i,K_j]\bigr] + \bigl[K_j,[P_0,J_i]\bigr] +\bigl[J_i,[K_j,P_0]\bigr] \\ &=i\epsilon_{ijk}[P_0,K_k] -i[J_i,P_j] \\ &=-\epsilon_{ijk}P_k + \epsilon_{ijk}L_{k0} +\epsilon_{ijk}P_k = \epsilon_{ijk}L_{k0} \\ & \qquad \qquad \qquad \quad \Rightarrow L_{i0}=0 \ . \end{split} \end{equation} Now we consider the final charge, $F_{ij}$. First we consider the Jacobi identity applied to $J_i,P_j,K_k$: \begin{equation} \begin{split} 0&=\bigl[J_i,[P_j,K_k]\bigr] + \bigl[K_k,[J_i,P_j]\bigr] +\bigl[P_j,[K_k,J_i]\bigr] \\ &=i\epsilon_{ijm}[K_k,P_m] -i\epsilon_{ikm}[P_j,K_m] \\ &=-\epsilon_{ijm}F_{km} -\epsilon_{ikm}F_{mj} \ . \end{split} \label{eq:JPK} \end{equation} As before, we can plug in all cases of $i=k\neq j$ to kill the second term and get that $F_{ij}=0$ for all $i\neq j$, meaning all off-diagonal elements are 0. We can then plug in the cases where $ijk$ is some permutation of even permutation of 123, which leads to the condition $F_{11}=F_{22}=F_{33}$, which when combined with the former condition means that $F_{ij}$ is proportional to the identity. In other words, we may choose \begin{equation} F_{ij} = -M\delta_{ij} \ . \end{equation} Because this central charge does not take the form of the structure constants contracted with some set of scalars (equation 2.7.10 of Weinberg), it follows that we cannot eliminate it by redefinition of the generators. This is not explicitly stated by Weinberg, but can be shown by assuming a redefinition of the generators $t_a\rightarrow \tilde{t}_a$ eliminates the central charges, and solving for the condition on the redefinition parameters. To make contact with the Galilean group commutation relations presented by the late and great Steve, we make one final redefinition, namely \begin{equation} H \equiv P_0 + M \ , \end{equation} where our $P_0$ is playing the role of $W$ in the text. This doesn't change the commutation relations because $M$ is proportional to the identity, and therefore commutes with everything. Once we formally enlarge our group by adding in this mass generator $M$, our final commutation relations become: \begin{equation} \boxed{ \begin{split} &[J_i,J_j]=i\epsilon_{ijk}J_k ,\quad [J_i,K_j]=i\epsilon_{ijk}K_k,\quad [K_i,K_j]= 0 \\ &[J_i,P_j]=i\epsilon_{ijk}P_k ,\quad [K_i,P_j] = -iM\delta_{ij},\quad [K_i,P_0] = -iP_i\\ &[P_0,J_i] = [P_0,P_i] = [P_i,P_j] = 0 \\ &[M,t] = 0 \text{ for all generators $t$ in the Lie algebra.} \end{split} } \label{eq:final_commutators} \end{equation} \textbf{Note: }Actually identifying $M$ with the mass of the particles in the theory is another story, and beyond the scope of these solutions. \subsection{Problem 2.4 -- Casimir Elements of O(3,1)} This problem is relatively straightforward, asking us to prove that $P^{\mu} P_{\mu}$ and $W^{\mu} W_{\mu}$, the Casimir elements of $O(3,1)$, do indeed commute with all the generators of the algebra. One could solve this directly by calculating the commutators with $P^{\mu}$ and $J^{\rho \sigma}$, but this is quite involved. An easier way is to show that these Casimir elements transform as scalars under inhomogeneous Lorentz transformation, i.e. $[P^{\mu}P_{\mu}, U(\Lambda, a)] = [W^{\mu}W_{\mu}, U(\Lambda, a)] = 0$, which implies for infinitiessimal parameters that these elements commute with all the generators. This is particularly easy for $P^{\mu}P_{\mu}$ since it transforms as a four-vector under inhomogeneous Lorentz transformations, Weinberg 2.4.9: \begin{equation} U(\Lambda, a) P^{\mu} U^{-1}(\Lambda, a) = \Lambda\indices{_\nu ^\mu} P^{\nu} \end{equation} We therefore have \begin{equation} \begin{split} U(\Lambda, a) P^{\mu} P_{\mu} U^{-1}(\Lambda, a) &= U(\Lambda, a) P^{\mu} U^{-1}(\Lambda, a) U(\Lambda, a)P_{\mu} U^{-1}(\Lambda, a) \\ &= \Lambda\indices{_\nu ^\mu} P^{\nu} \Lambda_{\rho \mu} P^{\rho} \\ &= (\Lambda^{-1})\indices{^\mu _\nu} \Lambda\indices{^\rho _\mu} P^{\nu} P_{\rho} \\ &= \delta\indices{^\rho _\nu} P^{\nu} P_{\rho} \\ &= P^{\rho} P_{\rho} \\ & \implies U(\Lambda, a) P^{\mu} P_{\mu} = P^{\mu} P_{\mu} U(\Lambda, a) \\ & \implies [P^{\mu} P_{\mu}, U(\Lambda, a), ] = 0 \end{split} \end{equation} $W$ is a little harder to work with. We first begin with seeing how $W^{\mu}$ transforms under inhomogeneous Lorentz transformations. A useful reminder is how $J^{\rho \sigma}$ transforms: \begin{equation} U(\Lambda, a) J^{\rho \sigma} U^{-1}(\Lambda, a) = \Lambda\indices{_\mu ^\rho} \Lambda\indices{_\nu ^\sigma} \left(J^{\mu \nu} - a^{\mu}P^{\nu} + a^{\nu} P^{\mu}\right) \end{equation} Applying this to $W^{\mu}$: \begin{equation} \begin{split} U(\Lambda, a) W^{\mu} U^{-1}(\Lambda, a) &= \eta\indices{^\mu ^\nu} \epsilon\indices{_\nu _\alpha _\beta _\gamma} U(\Lambda, a)J\indices{^\alpha ^\beta} P^{\gamma} U^{-1}(\Lambda, a) \\ &= \eta\indices{^\mu ^\nu} \epsilon\indices{_\nu _\alpha _\beta _\gamma} U(\Lambda, a)J\indices{^\alpha ^\beta} U^{-1}(\Lambda, a) U(\Lambda, a) P^{\gamma} U^{-1}(\Lambda, a) \\ &= \eta\indices{^\mu ^\nu} \epsilon\indices{_\nu _\alpha _\beta _\gamma} \Lambda\indices{_\rho ^\alpha} \Lambda\indices{_\sigma ^\beta} \Lambda\indices{_\tau ^\gamma} \left(J^{\rho \sigma} - a^{\rho}P^{\sigma} + a^{\sigma} P^{\rho}\right) P\indices{^\tau} \end{split} \end{equation} Focusing on the second and third terms, \begin{equation} \begin{split} &-\eta\indices{^\mu ^\nu} \epsilon\indices{_\nu _\alpha _\beta _\gamma} \Lambda\indices{_\rho ^\alpha} \Lambda\indices{_\sigma ^\beta} \Lambda\indices{_\tau ^\gamma} a^{\rho}P^{\sigma} + \eta\indices{^\mu ^\nu} \epsilon\indices{_\nu _\alpha _\beta _\gamma} \Lambda\indices{_\rho ^\alpha} \Lambda\indices{_\sigma ^\beta} \Lambda\indices{_\tau ^\gamma} a^{\sigma} P^{\rho} P\indices{^\tau} \\ &= \eta^{\mu \nu} \left( - \epsilon\indices{_\nu _\alpha _\beta _\gamma} (\Lambda^{-1} a)^{\alpha} (\Lambda^{-1} P)^{\beta} (\Lambda^{-1} P)^{\gamma} + \epsilon\indices{_\nu _\alpha _\beta _\gamma} (\Lambda^{-1} a)^{\beta} (\Lambda^{-1} P)^{\alpha} (\Lambda^{-1} P)^{\gamma}\right) \\ &= \eta^{\mu \nu} \left( - \epsilon\indices{_\nu _\alpha _\beta _\gamma} (\Lambda^{-1} a)^{\alpha} (\Lambda^{-1} P)^{\beta} (\Lambda^{-1} P)^{\gamma} + \epsilon\indices{_\nu _\beta _\gamma _\alpha} (\Lambda^{-1} a)^{\beta} (\Lambda^{-1} P)^{\gamma} (\Lambda^{-1} P)^{\alpha}\right) \\ &= 0 \end{split} \end{equation} where to arrive at the third line we did two permutations of the Levi-Cevita symbol $\epsilon\indices{_\nu _\alpha _\beta _\gamma} = \epsilon\indices{_\nu _\beta _\gamma _\alpha}$ and swapped $(\Lambda^{-1} P)^{\alpha}$ and $(\Lambda^{-1} P)^{\gamma}$ since they commute. We are left with \begin{equation} U(\Lambda, a) W^{\mu} U^{-1}(\Lambda, a) = \eta\indices{^\mu ^\nu} \epsilon\indices{_\nu _\alpha _\beta _\gamma} \Lambda\indices{_\rho ^\alpha} \Lambda\indices{_\sigma ^\beta} \Lambda\indices{_\tau ^\gamma} J^{\rho \sigma}P\indices{^\tau} \end{equation} Before we proceed, we will need to recall some identities of the Levi-Cevita symbol. In four dimensions, we have the following identity: \begin{equation} \epsilon_{\mu \alpha \beta \gamma} \epsilon^{\mu \delta \epsilon \zeta} = \delta^{\delta \epsilon \zeta}_{\alpha \beta \gamma} \end{equation} where the $\delta$ on the right is the generalized Kronecker delta. The definition of this symbol is \begin{equation} \delta^{\delta \epsilon \zeta}_{\alpha \beta \gamma} = \sum_{p \in \text{P}} \text{sgn}(p) \delta^{\delta}_{\alpha} \delta^{\epsilon}_{\beta} \delta^{\zeta}_{\gamma} \end{equation} where $\text{P}$ is the set of all possible permutations of either the top or bottom row. Therefore, \begin{equation} \epsilon_{\mu \alpha \beta \gamma} \epsilon^{\mu \delta \epsilon \zeta} = \delta\indices{^\delta _\alpha} \delta\indices{^\epsilon _\beta} \delta\indices{^\zeta _\gamma} + \delta\indices{^\zeta _\alpha} \delta\indices{^\delta _\beta} \delta\indices{^\epsilon _\gamma} + \delta\indices{^\epsilon _\alpha} \delta\indices{^\zeta _\beta} \delta\indices{^\delta _\gamma} - \delta\indices{^\delta _\alpha} \delta\indices{^\zeta _\beta} \delta\indices{^\epsilon _\gamma} - \delta\indices{^\epsilon _\alpha} \delta\indices{^\delta _\beta} \delta\indices{^\zeta _\gamma} - \delta\indices{^\zeta _\alpha} \delta\indices{^\epsilon _\beta} \delta\indices{^\delta _\gamma} \end{equation} We now have \begin{equation} \begin{split} U(\Lambda, a) &W^{\mu} W_{\mu} U^{-1}(\Lambda, a) = U(\Lambda, a) W^{\mu} U^{-1}(\Lambda, a) U(\Lambda, a) W_{\mu} U^{-1}(\Lambda, a) \\ & = \epsilon\indices{^\mu _\alpha _\beta _\gamma} \epsilon\indices{_\mu _\delta _\epsilon _\zeta} \Lambda\indices{_\rho ^\alpha} \Lambda\indices{_\sigma ^\beta} \Lambda\indices{_\tau ^\gamma} \Lambda\indices{_\phi ^\delta} \Lambda\indices{_\chi ^\epsilon} \Lambda\indices{_\psi ^\zeta} J^{\rho \sigma}P\indices{^\tau} J^{\phi \chi}P\indices{^\psi} \\ & = \epsilon\indices{^\mu ^\alpha ^\beta ^\gamma} \epsilon\indices{_\mu _\delta _\epsilon _\zeta} \Lambda\indices{_\rho ^\alpha} \Lambda\indices{_\sigma ^\beta} \Lambda\indices{_\tau ^\gamma} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} J^{\rho \sigma}P\indices{^\tau} J_{\phi \chi}P\indices{_\psi} \end{split} \end{equation} Focusing just on the Levi-Cevita and $\Lambda$ portion \begin{equation} \begin{split} \epsilon\indices{^\mu ^\alpha ^\beta ^\gamma} & \epsilon\indices{_\mu _\delta _\epsilon _\zeta} \Lambda\indices{_\rho ^\alpha} \Lambda\indices{_\sigma ^\beta} \Lambda\indices{_\tau ^\gamma} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} \\ &= (\delta\indices{^\delta _\alpha} \delta\indices{^\epsilon _\beta} \delta\indices{^\zeta _\gamma} + \delta\indices{^\zeta _\alpha} \delta\indices{^\delta _\beta} \delta\indices{^\epsilon _\gamma} + \delta\indices{^\epsilon _\alpha} \delta\indices{^\zeta _\beta} \delta\indices{^\delta _\gamma} - \delta\indices{^\delta _\alpha} \delta\indices{^\zeta _\beta} \delta\indices{^\epsilon _\gamma} - \delta\indices{^\epsilon _\alpha} \delta\indices{^\delta _\beta} \delta\indices{^\zeta _\gamma} \\ & - \delta\indices{^\zeta _\alpha} \delta\indices{^\epsilon _\beta} \delta\indices{^\delta _\gamma}) \Lambda\indices{_\rho ^\alpha} \Lambda\indices{_\sigma ^\beta} \Lambda\indices{_\tau ^\gamma} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} \\ &= \Lambda\indices{_\rho ^\delta} \Lambda\indices{_\sigma ^\epsilon} \Lambda\indices{_\tau ^\zeta} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} + \Lambda\indices{_\rho ^\zeta} \Lambda\indices{_\sigma ^\delta} \Lambda\indices{_\tau ^\epsilon} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} + \Lambda\indices{_\rho ^\epsilon} \Lambda\indices{_\sigma ^\zeta} \Lambda\indices{_\tau ^\delta} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} \\ & - \Lambda\indices{_\rho ^\delta} \Lambda\indices{_\sigma ^\zeta} \Lambda\indices{_\tau ^\epsilon} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} - \Lambda\indices{_\rho ^\epsilon} \Lambda\indices{_\sigma ^\delta} \Lambda\indices{_\tau ^\zeta} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} - \Lambda\indices{_\rho ^\zeta} \Lambda\indices{_\sigma ^\epsilon} \Lambda\indices{_\tau ^\delta} \Lambda\indices{^\phi _\delta} \Lambda\indices{^\chi _\epsilon} \Lambda\indices{^\psi _\zeta} \\ &= \delta\indices{^\phi _ \rho} \delta\indices{^\chi _\sigma} \delta\indices{^\psi _\tau} + \delta\indices{^\psi _ \rho} \delta\indices{^\phi _\sigma} \delta\indices{^\chi _\tau} + \delta\indices{^\chi _ \rho} \delta\indices{^\psi _\sigma} \delta\indices{^\phi _\tau} - \delta\indices{^\phi _ \rho} \delta\indices{^\psi _\sigma} \delta\indices{^\chi _\tau} - \delta\indices{^\chi _ \rho} \delta\indices{^\phi _\sigma} \delta\indices{^\psi _\tau} - \delta\indices{^\psi _ \rho} \delta\indices{^\chi _\sigma} \delta\indices{^\phi _\tau} \\ &= \epsilon\indices{^\mu ^\phi ^\chi ^\psi} \epsilon\indices{_\mu _\rho _\sigma _\tau} \end{split} \end{equation} Therefore \begin{equation} \begin{split} U(\Lambda, a)& W^{\mu} W_{\mu} U^{-1}(\Lambda, a) = \epsilon\indices{^\mu ^\phi ^\chi ^\psi} \epsilon\indices{_\mu _\rho _\sigma _\tau} J^{\rho \sigma}P\indices{^\tau} J_{\phi \chi}P\indices{_\psi} = W^{\mu} W_{\mu} \\ &\implies U(\Lambda, a) W^{\mu} W_{\mu} = W^{\mu} W_{\mu} U(\Lambda, a) \\ &\implies [W^{\mu} W_{\mu}, U(\Lambda, a)] = 0 \end{split} \end{equation} So, since both $P^{\mu}P_{\mu}$ and $W^{\mu}W_{\mu}$ commute with an arbitrary $U(\Lambda, a)$, these operators must also commute with all generators of inhomogeneous Lorentz transformations, and are therefore Casimir elements. \subsection{Problem 2.5 -- Massive Particles in 2+1 Dimensions} When going over to 2+1 dimensions, it's useful to first assess what from Weinberg we can reuse, and what we must derive anew. Weinberg's entire discussion in chapter 2.4 over the Poincare algebra actually makes no reference to the number of dimensions, so we are able to reuse these equations, just making the choice to only let our indices range over $0, 1, 2$. We have \begin{equation} U(1+\omega, \epsilon) = 1 + \frac{i}{2} \omega\indices{_\rho _\sigma} J\indices{^\rho ^\sigma} - i \epsilon_{\rho} P^{\rho} \end{equation} just as before. We can define the following generators in $J\indices{^\rho ^\sigma}$: \begin{equation} J \equiv J\indices{^1 ^2}, \quad K\indices{_1} \equiv J\indices{^0 ^1}, \quad K\indices{_2} \equiv J\indices{^0 ^2} \end{equation} and the momentum generators are the same, just in 2D. In 2+1 dimensions, our standard three-momentum for massive particles is \begin{equation} k^{\mu} = (0, 0, M) \end{equation} Clearly, the little group must be $SO(2)$, since any rotation will preserve the length of zero in the space-like components and not affect the time-like component. The unitary representations of $SO(2)$ are the direct sum of one dimensional phases $e^{i k \theta}$. Let's investigate this more concretely. We can write an element of the little group as \begin{equation} W\indices{^\mu _\nu}(\theta) = \begin{bmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{equation} The little group element for infinitessimal $\theta$ is \begin{equation} (1 + \omega)\indices{^\mu _\nu} = \begin{bmatrix} 1 & \theta & 0 \\ -\theta & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \implies \omega\indices{^\mu _\nu} = \begin{bmatrix} 0 & \theta & 0 \\ -\theta & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \implies \omega\indices{_\mu _\nu} = \begin{bmatrix} 0 & \theta & 0 \\ -\theta & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{equation} Therefore, \begin{equation} U(1+\omega) = 1 + \frac{i}{2}\omega\indices{_\rho _\sigma} J\indices{^\rho ^\sigma} = 1 + i \theta J \end{equation} which generalizes to arbitrary little group element \begin{equation} U(W) = U(\theta) = e^{i \theta J} \end{equation} How does this operator act on a state? Well, $\sigma$ just denotes all other degrees of freedom. Now, since $[J,H] = [J, P^1] = [J, P^2] = 0$ for the state $\Psi_{k, \sigma}$ of the standard momentum, we can take $J$ to belong to our complete set of commuting observables defining the state $\Psi_{k, \sigma}$. Therefore, $\sigma$ is the eigenvalue of $J$ for our states: \begin{equation} U(\theta) \Psi_{k, \sigma} = e^{i \sigma \theta} \Psi_{k, \sigma} \end{equation} The next step is to boost this state to an arbitrary momentum $p$. This is the same as in 3+1 dimensions, so \begin{equation} \Psi_{p, \sigma} = \sqrt{\frac{M}{p^0}} U(L(p)) \Psi_{k, \sigma} \end{equation} with the 2+1 dimensional $L(p)$: \begin{equation} L(p) = \begin{bmatrix} 1 + (\gamma - 1)\hat{p}_1 \hat{p}_1 & (\gamma - 1) \hat{p}_1 \hat{p}_2 & \sqrt{\gamma^2 - 1} \hat{p}_1 \\ (\gamma - 1) \hat{p}_2 \hat{p}_1 & 1 + (\gamma - 1)\hat{p}_2 \hat{p}_2 & \sqrt{\gamma^2 - 1} \hat{p}_2 \\ \sqrt{\gamma^2 - 1} \hat{p}_1 & \sqrt{\gamma^2 - 1} \hat{p}_2 & \gamma \end{bmatrix} \end{equation} where $\gamma = \sqrt{|\mathbf{p}|^2 + M^2}/M$. Knowing the normalization is the same, our next step is to find out what the $D$-matrix elements are. However, before we write these out, we must consider one crucial thing we have not mentioned yet: What are the constraints on the value $\sigma$ can take? We do not have the same ladder-operator algebra as the little group $SO(3)$ in 3+1 dimensions that requires $\sigma$ to take discrete integer or half-integer values. We must therefore investigate the group structure of $SO(2,1)$, specifically its projective representations, since this was let us restrict the value of $\sigma$ for massless particles in 3+1 dimensions, when we lacked the ladder algebra. \subsubsection{An Aside: Projective Representations of SO(2,1)} Since the central charges for $SO(3,1)$ made no reference to the dimensionality of the space, we can likewise conclude that all central charges for $SO(2,1)$ can be eliminated or the generators can be redefined to include the central charges, just like for $SO(3,1)$. However, the topologies are quite different. We proceed like Weinberg. Let's encode an arbitrary three-vector $V^{\mu}$ as a matrix: \begin{equation} v = \begin{bmatrix} V^0 + V^2 & V^1 \\ V^1 & V^0 - V^2 \end{bmatrix} \end{equation} Why do we do this? Well the determinant gives us the invariant length $V^{\mu} V_{\mu}$: \begin{equation} \det v = -(V^0)^2 + (V^1)^2 + (V^2)^2 = V^{\mu} V_{\mu} \end{equation} We notice also that $v$ is symmetric. Therefore, the transformation involving real matrices $\lambda$ \begin{equation} v \rightarrow \lambda v \lambda^T \end{equation} is also symmetric since \begin{equation} (\lambda v \lambda^T)^T = (v \lambda^T) \lambda^T = \lambda v^T \lambda^T = \lambda v \lambda^T \end{equation} How does this transformation affect the determinant? \begin{equation} \det(\lambda v \lambda^T) = \det\lambda \det v \det \lambda^T = (\det \lambda)^2 \det v \end{equation} So, if $\det \lambda = \pm 1$, then $\det v$, the invariant length, is preserved. Let's now restrict our study to $\lambda$ with $\det \lambda = \pm 1$. We can further restrict the $\lambda$ we consider because this arbitrary phase, just like in the 3+1 case, can be chosen such that $\det \lambda = +1$. Further, we can compose two transformations and see they obey the group transformation law: \begin{equation} (\lambda \bar{\lambda})v(\lambda \bar{\lambda})^T = \lambda ( \bar{\lambda}v \bar{\lambda}^T) \lambda^T \end{equation} So, the $\lambda$ form a group, $SL(2,R)$. This group preserves the invariant length of a three-vector, the same effect as elements of $SO(2,1)$. We however note that $\lambda$ and $-\lambda$ are different elements of $SL(2,R)$, but they nevertheless produce the same Lorentz transformation $\lambda v \lambda^T$, and therefore $SO(2,1)$ is not the same as $SL(2,R)$ but rather $SL(2,R)/Z_2$. We now investigate the topology. We apply the polar decomposition theorem to an arbitrary real matrix $\lambda$: \begin{equation} \lambda = o e^s \end{equation} where $o$ is an orthogonal matrix and $s$ is a symmetric matrix: \begin{equation} o^T o = 1, \quad s = s^T \end{equation} From $o$ being orthogonal, we must have $\det o = \pm 1$, however, we know $\det \lambda = +1$, so \begin{equation} \det \lambda = (\det o) (\det e^s) = (\det o) e^{\Tr s} \end{equation} but $e^{\Tr s}$ is a positive number, so $\det o = +1$ for $\det \lambda$ to be positive. But, for $\det \lambda$ to be $+1$ exactly, we also require $\Tr s = 0$, so to summarize \begin{equation} \begin{split} \det o &= 1 \\ \Tr s &= 0 \end{split} \end{equation} The $o$ therefore belong to the group $SO(2)$, and the $s$ are the group of symmetric traceless matrices. In particular, the $o$ correspond to rotations in the spatial components. First note that $V^0 = \frac{1}{2} \Tr v$. If $s = 0$, then $\lambda$ leaves $V^0$ invariant: \begin{equation} V^0 = \frac{1}{2} \Tr v \rightarrow \frac{1}{2}\Tr o v o^T = \frac{1}{2} Tr o^T o v = \frac{1}{2} Tr v = V^0 \end{equation} so $o$ is just a rotation. Our $o$ can be written in general as: \begin{equation} o = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \end{equation} subject to the constraint \begin{equation} \det o = a^2 + b^2 = 1 \end{equation} Which means the topology of $SO(2)$ is the same as $S_1$, the circle. $s$, our 2-by-2 symmetric, traceless matrix can in general be written as: \begin{equation} s = \begin{bmatrix} c & d \\ d & -c \\ \end{bmatrix} \end{equation} with $c, d \in \mathbb{R}$, so the topology of this group is the same as two-dimensional flat space $R_2$. Since the polar decomposition theorem gurantees a unique decomposition, the group $SL(2,R)$ is just $R_2 \cross S_1$. Now, the sign change of $+\lambda$ and $-\lambda$ can only be caused $o$ since $e^{s}$ is always positive, so the topology of $SO(2,1)$ is $R_2 \cross S_1 / Z_2$. It can be shown that the topology of $S_1/Z_2$ is isomorphic to $S_1$, and this topology is not simply connected. The fundamental group of $S_1$ is $Z$. Therefore, unlike with $SO(3,1)$, we have no way to constrict the value $\sigma$ takes. Recall that for $SO(3,1)$, we were able to say that a rotation of $4 \pi$ can be deformed to the identity, since it is composed of two rotations of $2 \pi$ which is a double loop in $S_3/Z_2$. This meant that $e^{i4\pi \sigma} = 1$, which constrained $\sigma$ to be integer or half-integer. However, for $S_1/Z_2$, since the topology is isomorphic to $S_1$, the classes of loops are labeled by winding number. For loops of non-zero winding number, we cannot deform these into the identity, and therefore, for non-zero $\beta$, we have no such identity of $e^{i \beta \sigma} = 1$. Our $\sigma$ is unconstrained. Particles exhibiting this phenomenon have been described as ``anyons,'' and have been studied in condensed matter. The moral of the story is $\sigma$ can take any value. \subsubsection{Continuing on...} The issue with proceeding is the $D$-matrix is in general infinite-dimensional, which is mathematically more unwieldy. We're just going to brush these issues under the rug, and treat $\phi$ as being discrete. This means that our $D$-matrix element is of the form: \begin{equation} D_{\sigma' \sigma}(W) = e^{i \sigma \theta} \delta_{\sigma' \sigma} \end{equation} Therefore, the way an arbitrary state transforms is \begin{equation} \boxed{U(\Lambda) \Psi_{p, \sigma} = \sqrt{\frac{(\Lambda p)^0}{p^0}} e^{i \theta(\Lambda) \sigma} \Psi_{\Lambda p, \sigma}} \end{equation} We now need to know how parity affects our state. There is a bit of an issue with parity in even dimensions. If we try to naively define a parity matrix, it would be \begin{equation} \mathscr{P}\indices{^\mu _\nu} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{equation} since the spatial components are $-1$ and the time-like component is $+1$. However, the determinant of this is $+1$, meaning the parity matrix cannot be represented like this (indeed, this matrix is simply a rotation of $\pi$). We therefore need a matrix that has determinant $-1$ and flips the spatial components. This would be a reflection about the $x-$ or $y-$ axes, so we are left with two options \begin{equation} (\mathscr{P}^{(1)})\indices{^\mu _\nu} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{equation} or \begin{equation} (\mathscr{P}^{(2)})\indices{^\mu _\nu} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{equation} Now, which to choose? It turns out it doesn't matter, so we will show that either choose leads to the same effect on our massive states. However, it is a little less trivial how they are equivalent for massless states, so we will spend extra time on that in the next problem. Weinberg equations (2.6.3) and (2.6.4) will be our starting point: \begin{equation} \begin{split} \mathsf{P} i J\indices{^\rho ^\sigma} \mathsf{P}^{-1} &= i \mathscr{P}\indices{_\rho ^\mu} \mathscr{P}\indices{_\sigma ^\nu} J\indices{^\mu ^\nu} \\ \mathsf{P} i P\indices{^\rho} \mathsf{P}^{-1} &= i \mathscr{P}\indices{_\rho ^\mu} P\indices{^\mu} \end{split} \end{equation} For similar reasons as Weinberg says for 3+1 dimensions, $\mathsf{P}$ must be linear, instead of antilinear, so we can reduce this to \begin{equation} \begin{split} \mathsf{P} J\indices{^\rho ^\sigma} \mathsf{P}^{-1} &= \mathscr{P}\indices{_\rho ^\mu} \mathscr{P}\indices{_\sigma ^\nu} J\indices{^\mu ^\nu} \\ \mathsf{P} P\indices{^\rho} \mathsf{P}^{-1} &= \mathscr{P}\indices{_\rho ^\mu} P\indices{^\mu} \end{split} \end{equation} Now, let's proceed with $\mathscr{P}^{(1)}$. That is, let $\mathsf{P} \equiv U(\mathscr{P}^{(1)})$. The above equations then reduce to \begin{equation} \mathsf{P} J \mathsf{P}^{-1} = -J, \quad \mathsf{P} P^1 \mathsf{P}^{-1} = - P^1, \quad \mathsf{P} P^2 \mathsf{P}^{-1} = P^2, \quad \mathsf{P} H \mathsf{P}^{-1} = H \end{equation} The above operators, of which our state $\Psi_{k,\sigma}$ is an eigenstate of, either commute or anticommute. It follows that $\mathsf{P} \Psi_{k, \sigma}$ is also an eigenstate. And just like Weinberg does, we disregard degeneracies, which means our parity-transformed state differs by just a phase \begin{equation} \mathsf{P} \Psi_{k, \sigma} = \eta_{\sigma} \Psi_{k, \sigma} \label{eq:2-5littlegroup-parity} \end{equation} Unlike in the 3+1 case, $\eta_{\sigma}$ in general depends on $\sigma$, since we do not have the ladder algebra that lets us prove $\eta_{\sigma}$ must only depend on particle species. If we instead use $\mathscr{P}^{(2)}$, we get the same relation as Equation \ref{eq:2-5littlegroup-parity}, since all that changes when $\mathsf{P}$ acts on our generators is: \begin{equation} \mathsf{P} P^1 \mathsf{P}^{-1} = P^1, \quad \mathsf{P} P^2 \mathsf{P}^{-1} = -P^2 \end{equation} The parity transformation on a massive state of arbitrary momentum then reads: \begin{equation} \begin{split} \mathsf{P} \Psi_{p, \sigma} &= \sqrt{\frac{M}{p^0}} \mathsf{P} U(L(p)) \Psi_{k, \sigma} \\ &= \sqrt{\frac{M}{p^0}} U(\mathscr{P}^{(1)}) U(L(p)) U(\mathscr{P}^{(1)})^{-1} U(\mathscr{P}^{(1)})\Psi_{k, \sigma} \\ &= \sqrt{\frac{M}{p^0}} U\left(\mathscr{P}^{(1)}L(p)(\mathscr{P}^{(1)})^{-1}\right) \eta_{\sigma} \Psi_{k, \sigma} \\ &= \eta_{\sigma} \sqrt{\frac{M}{p^0}} U\left(L(\mathscr{P}^{(1)}p)\right) \Psi_{k, \sigma} \\ &= \eta_{\sigma} \Psi_{\mathscr{P}^{(1)}p, \sigma} \end{split} \label{eq:2-5fullstate-parity} \end{equation} where we use the fact that \begin{equation} \begin{split} \mathscr{P}^{(1)}L(p)(\mathscr{P}^{(1)})^{-1} &= \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 + (\gamma - 1)\hat{p}_1 \hat{p}_1 & (\gamma - 1) \hat{p}_1 \hat{p}_2 & \sqrt{\gamma^2 - 1} \hat{p}_1 \\ (\gamma - 1) \hat{p}_2 \hat{p}_1 & 1 + (\gamma - 1)\hat{p}_2 \hat{p}_2 & \sqrt{\gamma^2 - 1} \hat{p}_2 \\ \sqrt{\gamma^2 - 1} \hat{p}_1 & \sqrt{\gamma^2 - 1} \hat{p}_2 & \gamma \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} 1 + (\gamma - 1)\hat{p}_1 \hat{p}_1 & -(\gamma - 1) \hat{p}_1 \hat{p}_2 & -\sqrt{\gamma^2 - 1} \hat{p}_1 \\ -(\gamma - 1) \hat{p}_2 \hat{p}_1 & 1 + (\gamma - 1)\hat{p}_2 \hat{p}_2 & \sqrt{\gamma^2 - 1} \hat{p}_2 \\ -\sqrt{\gamma^2 - 1} \hat{p}_1 & \sqrt{\gamma^2 - 1} \hat{p}_2 & \gamma \end{bmatrix} \\ &= L(\mathscr{P}^{(1)} p) \end{split} \end{equation} If we instead use $\mathscr{P}^{(2)}$, we get the same result since we can also show that \begin{equation} \mathscr{P}^{(2)}L(p)(\mathscr{P}^{(2)})^{-1} = L(\mathscr{P}^{(2)}p) \end{equation} Therefore, regardless of how one chooses the parity matrix $\mathscr{P}$, we have \begin{equation} \boxed{\mathsf{P} \Psi_{p, \sigma} = \eta_{\sigma} \Psi_{\mathscr{P} p, \sigma}} \end{equation} Moving onto time-reversal, the time-reversal matrix is the same in 2+1 dimensions \begin{equation} \mathscr{T}\indices{^\mu _\nu} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \end{equation} We reproduce Weinberg Equations (2.6.5) and (2.6.6) \begin{equation} \begin{split} \mathsf{T} i J\indices{^\rho ^\sigma} \mathsf{T}^{-1} &= i \mathscr{T}\indices{_\rho ^\mu} \mathscr{T}\indices{_\sigma ^\nu} J\indices{^\mu ^\nu} \\ \mathsf{T} i P\indices{^\rho} \mathsf{T}^{-1} &= i \mathscr{T}\indices{_\rho ^\mu} P\indices{^\mu} \end{split} \end{equation} Just like Weinberg mentions, to have no negative energy states, we require $\mathsf{T}$ to be antilinear, so the above equations reduce to \begin{equation} \begin{split} \mathsf{T} J\indices{^\rho ^\sigma} \mathsf{T}^{-1} &= - \mathscr{T}\indices{_\rho ^\mu} \mathscr{T}\indices{_\sigma ^\nu} J\indices{^\mu ^\nu} \\ \mathsf{T} P\indices{^\rho} \mathsf{T}^{-1} &= - \mathscr{T}\indices{_\rho ^\mu} P\indices{^\mu} \end{split} \end{equation} or \begin{equation} \mathsf{T} J \mathsf{T}^{-1} = -J, \quad \mathsf{T} P^1 \mathsf{T}^{-1} = - P^1, \quad \mathsf{T} P^2 \mathsf{T}^{-1} = -P^2, \quad \mathsf{T} H \mathsf{T}^{-1} = H \end{equation} For a massive state with standard three-momentum, time-reversal has no effect on the two-momentum (it's still zero). But there is now an effect on the spin: \begin{equation} J (\mathsf{T} \Psi_{k, \sigma}) = -\mathsf{T} (J \Psi_{k, \sigma}) = -\sigma (\mathsf{T} \Psi_{k, \sigma}) \end{equation} which in turn implies \begin{equation} \mathsf{T} \Psi_{k, \sigma} = \zeta_{\sigma} \Psi_{k, -\sigma} \end{equation} with $\zeta_{\sigma}$ a phase. Again, we have no ladder operator algebra that lets us restrict what the values $\zeta_{\sigma}$ can take. However, just like in the 3+1 dimensions, since $\mathsf{T}$ is an antiunitary operator, we can reabsorb this phase into the definition of $\Psi_{k, \sigma}$, thus showing the phase has no physical effect. Define a new state vector as $\Psi_{k, \sigma}' \equiv \zeta^{1/2}_{\sigma} \Psi_{k, \sigma}$. Therefore \begin{equation} \mathsf{T} \Psi_{k, \sigma}' = \mathsf{T} \zeta^{1/2}_{\sigma} \Psi_{k, \sigma} = \zeta^{\ast 1/2}_{\sigma} \zeta_{\sigma} \Psi_{k, -\sigma} = \zeta^{\ast}_{\sigma} \zeta_{\sigma} \zeta^{1/2}_{\sigma} \Psi_{k, -\sigma} = \Psi_{k, -\sigma}' \end{equation} but we will continue to keep the explict $\zeta_{\sigma}$, just like Weinberg, despite the phases having no physical effect. Following the same procedure as Equation \ref{eq:2-5fullstate-parity}, $\mathsf{T}$ acts on $\Psi_{p, \sigma}$ like \begin{equation} \mathsf{T} \Psi_{p, \sigma} = \sqrt{\frac{M}{p^0}} U\left(\mathscr{T} L(p)\mathscr{T}^{-1}\right) \zeta_{\sigma} \Psi_{k, -\sigma} \end{equation} The time-reversed $L(p)$ is the same as in 3+1 dimensions \begin{equation} \mathscr{T} L(p) \mathscr{T}^{-1} \begin{bmatrix} 1 + (\gamma - 1)\hat{p}_1 \hat{p}_1 & (\gamma - 1) \hat{p}_1 \hat{p}_2 & -\sqrt{\gamma^2 - 1} \hat{p}_1 \\ (\gamma - 1) \hat{p}_2 \hat{p}_1 & 1 + (\gamma - 1)\hat{p}_2 \hat{p}_2 & -\sqrt{\gamma^2 - 1} \hat{p}_2 \\ -\sqrt{\gamma^2 - 1} \hat{p}_1 & -\sqrt{\gamma^2 - 1} \hat{p}_2 & \gamma \end{bmatrix} \end{equation} However, even though the effect is the same, namely $\mathbf{p} \rightarrow -\mathbf{p}$, this is not a parity transformation, but rather a rotation by $\pi$ in 2+1 dimensions. Therefore, \begin{equation} \begin{split} \mathsf{T} \Psi_{p, \sigma} &= \sqrt{\frac{M}{p^0}} U( L(R(\pi)p)) \zeta_{\sigma} \Psi_{k, -\sigma} \\ \implies \Aboxed{\mathsf{T} \Psi_{p, \sigma} &= \zeta_{\sigma} \Psi_{R(\pi)p, -\sigma}} \end{split} \end{equation} \subsection{Problem 2.6 -- Massless Particles in 2+1 Dimensions} We will proceed in a very similar fashion to Weinberg. Even though we are now in 2+1 dimensions, we will still use greek letters to label the three-vector components. We introduce the unit time-like and unit light-like vectors. \begin{equation} k^{\mu} = (0,1,1) ,\quad t^{\mu} = (0,0,1) \end{equation} We introduce the little group elements $W$ that is defined as the group of elements of O(3,1) that leave the light-like vector invariant, $(Wk)^{\mu} = k^{\mu}$. We have the following identities: \begin{equation} \begin{split} &t^{\mu} t_{\mu} = -1 \implies t^{\mu}t_{\mu} = (Wt)^{\mu} (Wt)_{\mu} = -1 \\ t^{\mu} k_{\mu}& = -1 \implies t^{\mu} k_{\mu} = (Wt)^{\mu} (Wk)^{\mu} = (Wt)^{\mu} k_{\mu} = -1 \end{split} \end{equation} If we represent our transformed time-like vector as $(Wt)^{\mu} = (\beta, \zeta, \gamma)$, these two equations give constraints on the transformed time-like vector, and we are left with: \begin{equation} (Wt)^{\mu} = \left(\beta, \frac{\beta^2}{2}, 1+\frac{\beta^2}{2}\right) = \left(\beta, \zeta, 1+\zeta\right) \end{equation} with $\zeta = \beta^2/2$. One possible matrix that maps $t^{\mu}$ to $(Wt)^{\mu}$ is the boost given by \begin{equation} S\indices{^\mu _\nu}(\beta) = \begin{bmatrix} 1 & -\beta & \beta \\ \beta & 1 - \zeta & \zeta \\ \beta & -\zeta & 1 + \zeta \end{bmatrix} \end{equation} One can check that $S(\beta) \in SO^+(3,1)$, but we will not show that here. We now have \begin{equation} (St)^{\mu} = (Wt)^{\mu} \implies (S^{-1} W t)^{\mu} = t^{\mu} \end{equation} Since $t^{\mu}$ is left invariant by the operation $S^{-1} W$, this operation must be a rotation of the spatial coordinates. However, we can also show that \begin{equation} (Sk)^{\mu} = (Wk)^{\mu} \implies (S^{-1} W k)^{\mu} = k^{\mu} \end{equation} which means that if $S^{-1}W$ is the identity, or a rotation about zero degrees, so \begin{equation} W\indices{^\mu _\nu}(\beta) = S\indices{^\mu _\nu}(\beta) = \begin{bmatrix} 1 & -\beta & \beta \\ \beta & 1 - \zeta & \zeta \\ \beta & -\zeta & 1 + \zeta \end{bmatrix} \end{equation} The little group is therefore one dimensional and isomorphic to the the real numbers $\mathbb{R}$ under addition. The little group element for infinitiessimal $\beta$ is \begin{equation} W\indices{^\mu _\nu} = \delta\indices{^\mu _\nu} + \omega\indices{^\mu _\nu} = \delta\indices{^\mu _\nu} + \begin{bmatrix} 0 & -\beta & \beta \\ \beta & 0 & 0 \\ \beta & 0 & 0 \end{bmatrix} \end{equation} where we discard $\zeta$ since it is quadratic in $\beta$. Contracting with the metric gives us \begin{equation} \omega\indices{_\mu _\nu} = \begin{bmatrix} 0 & -\beta & \beta \\ \beta & 0 & 0 \\ -\beta & 0 & 0 \end{bmatrix} \end{equation} The form of the little group element, $U(W)$, for infinitessimal $\beta$ is simple. Weinberg equation (2.4.3) was derived without any explicit reference to the number of dimensions, so we can carry this formula as well as the later commutation relations (2.4.12-2.4.14) over wholesale to 2+1 dimensions. Our indices now range over 0 to 2, and we define the following generators: \begin{equation} J \equiv J\indices{^1 ^2}, \quad K\indices{^1} \equiv J\indices{^0 ^1}, \quad K\indices{^2} \equiv J\indices{^0 ^2} \end{equation} Weinberg equation (2.4.3) then reads \begin{equation} U(1+\omega) = 1 - i \beta J - i \beta K\indices{^1} = 1 + i \beta B \end{equation} where $B \equiv -J - K\indices{^1}$, and the $P^1$, $P^2$ generators are excluded just because their affect on the state $\Psi$ is simply a phase, just like in the 3+1 case. Since $\sigma$ was just our label to denote all other degrees of freedom, $\sigma$ thus labels the eigenvalues of the generator $B$. We then have the following transformation: \begin{equation} U(W) \Psi_{k,\sigma} = e^{i \beta B} \Psi_{k, \sigma} = e^{i\beta \sigma} \Psi_{k, \sigma} \end{equation} which implies a D-matrix of: \begin{equation} D_{\sigma' \sigma}(W) = e^{i \beta \sigma} \delta_{\sigma' \sigma} \end{equation} and a full transformed state as \begin{equation} \boxed{U(\Lambda) \Psi_{p, \sigma} = \sqrt{\frac{(\Lambda p)^0}{p^0}} e^{i \beta(\Lambda) \sigma} \Psi_{\Lambda p, \sigma}} \end{equation} Next we proceed with characterizing how the states change under $P$ and $T$. As mentioned in Problem 2.5, we have two options for our parity matrix $\mathscr{P}$: \begin{equation} (\mathscr{P}^{(1)})\indices{^\mu _\nu} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \quad (\mathscr{P}^{(2)})\indices{^\mu _\nu} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{equation} We will show that both of these lead to the same results, albeit in a slightly more complicated fashion than in the massive case, since our generator $B$ is more complex. First, starting with $\mathscr{P}^{(1)}$, we define $\mathsf{P} \equiv U(\mathscr{P}^{(1)})$. As we saw in Problem 2.5, $\mathsf{P}$ transforms the generators like \begin{equation} \begin{gathered} \mathsf{P} J \mathsf{P}^{-1} = -J, \quad \mathsf{P} P^1 \mathsf{P}^{-1} = - P^1, \quad \mathsf{P} P^2 \mathsf{P}^{-1} = P^2, \\ \quad \mathsf{P} K^1 \mathsf{P}^{-1} = - K^1, \quad \mathsf{P} K^2 \mathsf{P}^{-1} = K^2, \quad \mathsf{P} H \mathsf{P}^{-1} = H \end{gathered} \end{equation} We therefore have \begin{equation} \mathsf{P} B \mathsf{P}^{-1} = - \mathsf{P} J \mathsf{P}^{-1} - \mathsf{P} K^1 \mathsf{P}^{-1} = +J + K^1 = - B \end{equation} We note that the reason we choose to work with $\mathscr{P}^{(1)}$ first is because of the simple effect it has on $B$. We see that \begin{equation} P^{\mu} \mathsf{P} \Psi_{k, \sigma} = k^{\mu} \mathsf{P} \Psi_{k, \sigma} \end{equation} and \begin{equation} B \mathsf{P} \Psi_{k, \sigma} = - \mathsf{P} B \Psi_{k, \sigma} = - \sigma \mathsf{P} \Psi_{k, \sigma} \end{equation} Therefore the standard three-momentum is unaffected by our parity operation and the ``helicity'' changes sign. We can write this as \begin{equation} \mathsf{P} \Psi_{k, \sigma} = \eta_{\sigma} \Psi_{k, -\sigma} \end{equation} with $\eta_{\sigma}$ a phase that depends on the helicity. Like in Problem 2.5, there is no ladder algebra or anything that lets us restrict the values of $\eta_{\sigma}$, so in general it will depend on $\sigma$. We now need to define how our boost $L(p)$ looks for massless particles in 2+1 dimensions and then determine how $\mathsf{P} U(L(p)) \mathsf{P}^{-1}$ behaves. We will use the same boost, just in 2+1 dimensions, followed by a rotation: \begin{equation} L(p) = R(\theta) B(|\mathbf{p}|/\kappa) \end{equation} with \begin{equation} B(u) \equiv \begin{bmatrix} 1 & 0 & 0 \\ 0 & (u^2 + 1)/2u & (u^2 - 1)/2u \\ 0 & (u^2 - 1)/2u & (u^2 + 1)/2u \end{bmatrix} \end{equation} It is clear from this definition of the boost that the parity transformation has no effect \begin{equation} \mathscr{P}^{(1)} B(u) (\mathscr{P}^{(1)})^{-1} = B(u) \end{equation} Our rotation takes the 2-axis into $(\sin\theta, \cos\theta)$, so the corresponding operator\footnote{It should be pointed out here that Weinberg uses the passive point of view when defining his generators of rotations. That is, the $\theta$ in $U(R(\theta)) = \exp(i \mathbf{\theta} \cdot \mathbf{J})$ is the angle by which to rotate the axes, so when we want to rotate a vector by an angle $\theta$, we plug in $-\theta$ into our equation for $U(R(\theta))$.} is \begin{equation} U(R(\theta)) = e^{-i \theta J} \end{equation} Putting this all together, we have \begin{equation} \begin{split} \mathsf{P} \Psi_{p, \sigma} &= \sqrt{\frac{\kappa}{p^0}} \mathsf{P} U(R(\theta) B\left(\frac{|\mathbf{p}|}{\kappa}\right)) \Psi_{k, \sigma} \\ &= \sqrt{\frac{\kappa}{p^0}} P U(R(\theta)) \mathsf{P}^{-1} \mathsf{P} U(B\left(\frac{|\mathbf{p}|}{\kappa}\right)) \mathsf{P}^{-1} \mathsf{P} \Psi_{k, \sigma} \\ &= \eta_{\sigma} \sqrt{\frac{\kappa}{p^0}} e^{+ i \theta J} U(B\left(\frac{|\mathbf{p}|}{\kappa}\right)) \Psi_{k, -\sigma} \\ &= \eta_{\sigma} \sqrt{\frac{\kappa}{p^0}} U(R(-\theta) B\left(\frac{|\mathbf{p}|}{\kappa}\right)) \Psi_{k, -\sigma} \end{split} \end{equation} Since we measure $\theta$ from the 2-axis, rotating our boosted state by $-\theta$ instead of $+\theta$ is the same thing as reflecting our final two-momentum $\mathbf{p}$ across the 2-axis. This is equivalent to the parity matrix acting on our state, so we therefore arrive at \begin{equation} \boxed{\mathsf{P}\Psi_{p, \sigma} = \eta_{\sigma} \Psi_{\mathscr{P}^{(1)} p, -\sigma}} \end{equation} As promised, we now show this outcome is the same if we were to choose the parity matrix $\mathscr{P}^{(2)}$. Defining the parity operator as $\mathsf{P} \equiv U(\mathscr{P}^{(2)})$ gives us the following relations amongst all the generators: \begin{equation} \begin{split} \mathsf{P} J \mathsf{P}^{-1} = -J, \quad \mathsf{P} P^1 \mathsf{P}^{-1} = +P^1, \quad \mathsf{P} P^2 \mathsf{P}^{-1} = -P^2, \\ \quad \mathsf{P} K^1 \mathsf{P}^{-1} = +K^1, \quad \mathsf{P} K^2 \mathsf{P}^{-1} = -K^2, \quad \mathsf{P} H \mathsf{P}^{-1} = H \end{split} \end{equation} There are two issues here. First, our parity operator will now not leave the two-momentum invariant, which is how we defined its effect in the $\mathscr{P}^{(1)}$ case. The second issue, which may be even more troublesome, is that $B$ doesn't tranform simply: \begin{equation} \mathsf{P} B \mathsf{P}^{-1} = - \mathsf{P} J \mathsf{P}^{-1} - \mathsf{P} K^1 \mathsf{P}^{-1} = +J - K^1 \end{equation} We can rectify this by considering an operator consisting of a rotation and $\mathsf{P}$. This rotation would need to bring $k$ to $\mathscr{P}^{(2)}$ so that the combined parity operation and rotation takes $k \rightarrow k$. The specific rotation would simply be $\pi$, so \begin{equation} U(R) = e^{i \pi J} \end{equation} with the effects \begin{equation} \begin{split} U(R^{-1}) P^{\mu} U^{-1}(R^{-1}) &= (\mathscr{P}^{(2)})\indices{^\mu _\nu} P^{\nu} \\ U(R^{-1}) K^{\mu} U^{-1}(R^{-1}) &= (\mathscr{P}^{(2)})\indices{^\mu _\nu} K^{\nu} \end{split} \end{equation} which in turn imply \begin{equation} \begin{split} [U(R^{-1})\mathsf{P}] P^{\mu} [U(R^{-1})\mathsf{P}]^{-1} &= [U(R^{-1})\mathsf{P}] P^{\mu} [\mathsf{P}^{-1} U^{-1}(R^{-1})] \\ &= (\mathscr{P}^{(2)})\indices{^\mu _\nu} U(R^{-1}) P^{\nu} U^{-1}(R^{-1}) \\ &= (\mathscr{P}^{(2)})\indices{^\mu _\nu} (\mathscr{P}^{(2)})\indices{^\nu _\rho} P^{\rho} = P^{\mu} \end{split} \end{equation} and similarly \begin{equation} [U(R^{-1})\mathsf{P}] \mathbf{K} [U(R^{-1})\mathsf{P}]^{-1} = \mathbf{K} \end{equation} with $\mathbf{K} \equiv (K^1, K^2)$. Since the combination operator commutes with the three-momentum operator, the three-momentum $k^{\mu}$ is left invariant: \begin{equation} P^{\mu} [U(R^{-1}) \mathsf{P}] \Psi_{k, \sigma} = [U(R^{-1}) \mathsf{P}] P^{\mu} \Psi_{k, \sigma} = k^{\mu} [U(R^{-1}) \mathsf{P}] \Psi_{k, \sigma} \end{equation} Furthermore, the operator $U(R^{-1}) \mathsf{P}$ transforms $B$ in a simple way: \begin{equation} \begin{split} [U(R^{-1}) \mathsf{P}] B [U(R^{-1}) \mathsf{P}]^{-1} &= [U(R^{-1}) \mathsf{P}] B [\mathsf{P}^{-1} U^{-1}(R^{-1})] \\ &= U(R^{-1}) [+J - K^1] U^{-1}(R^{-1}) \\ &= +J - U(R^{-1}) K^1 U^{-1}(R^{-1}) \\ &= +J - (R^{-1})\indices{^0 _\mu} (R^{-1})\indices{^1 _\nu} J^{\mu \nu} \\ &= +J + K^1 = -B \end{split} \end{equation} where we use the fact that $R^{-1} = \text{diag}(-1,-1,1)$ for a rotation of $\pi$. The ``helicity'' now picks up a minus sign under the combined transformation since our operator anticommutes with $B$: \begin{equation} B [U(R^{-1}) \mathsf{P}] \Psi_{k, \sigma} = -\sigma [U(R^{-1}) \mathsf{P}] \Psi_{k, \sigma} \end{equation} which suggests our state transforms as \begin{equation} U(R^{-1}) \mathsf{P} \Psi_{k, \sigma} = \eta_{\sigma} \Psi_{k, -\sigma} \end{equation} The parity operator acting on a state of arbitrary three-momentum is \begin{equation} \begin{split} \mathsf{P} \Psi_{p, \sigma} &= \sqrt{\frac{\kappa}{p^0}} \mathsf{P} U(R(\theta)B\left(\frac{|\mathbf{p}|}{\kappa}\right)) \Psi_{k,\sigma} \\ &= \sqrt{\frac{\kappa}{p^0}} \mathsf{P} U(R(\theta)B\left(\frac{|\mathbf{p}|}{\kappa}\right)) [P^{-1} U(R)] [U(R^{-1}) \mathsf{P}]\Psi_{k,\sigma} \\ &= \eta_{\sigma} \sqrt{\frac{\kappa}{p^0}} \mathsf{P} U(R(\theta)B\left(\frac{|\mathbf{p}|}{\kappa}\right)) P^{-1} U(R) \Psi_{k,\sigma} \\ &= \eta_{\sigma} \sqrt{\frac{\kappa}{p^0}} \mathsf{P} U(R(\theta)) \mathsf{P}^{-1} U(R) U(B\left(\frac{|\mathbf{p}|}{\kappa}\right)) \Psi_{k,\sigma} \\ &= \eta_{\sigma} \sqrt{\frac{\kappa}{p^0}} U(R(-\theta)) U(R) U(B\left(\frac{|\mathbf{p}|}{\kappa}\right)) \Psi_{k,\sigma} \\ &= \eta_{\sigma} e^{i \pi J} \Psi_{\mathscr{P}^{(1)}, -\sigma} \\ &= \eta_{\sigma} e^{- i \pi \sigma} \Psi_{\mathscr{P}^{(1)}, -\sigma} \end{split} \end{equation} Where in the fourth line we use the fact that $\mathsf{P}^{-1} U(R)$ commutes with $K^2$ and therefore also $U(B(u))$, and in the fifth line that $\mathsf{P} U(R(\theta)) \mathsf{P}^{-1} =U(R(-\theta))$ This is identical to when we used $\mathscr{P}^{(1)}$, except for the phase factor. The phase is just an artifact of using the other partiy matrix. It can be absorbed into the unknown phase factor $\eta_{\sigma}$, which then yields the same parity transformation as with $\mathscr{P}^{(1)}$. \section{Chapter 3} \subsection{Problem 3.1 -- A Separable Interaction} In this problem, we are tasked to construct explicit expressions for the in/out states as well as the S-matrix given the matrix elements of the interaction potential: \begin{equation} (\Phi_\beta,V\Phi_\alpha)=g \ u_\beta \ u^*_\alpha \label{eq:separable_int} \end{equation} where \begin{equation} \sum_\alpha |u_\alpha|^2 = 1 \ . \label{eq:separable_unitarity} \end{equation} To do this, we use the Lippman-Schwinger equations expanded in a basis of free particle states: \begin{equation} \Psi_\alpha^\pm = \Phi_\alpha + \int d\beta \frac{T_{\beta\alpha}^\pm \Phi_\beta}{E_\alpha - E_\beta \pm i\varepsilon} \quad , \quad T_{\beta\alpha}^\pm \equiv (\Phi_\beta, V\Psi_\alpha^\pm) \ . \end{equation} We can recursively expand the expression for the in/out states: \begin{equation} \Psi_\alpha^\pm = \Phi_\alpha + \int d\beta \frac{(\Phi_\beta,V\Phi_\alpha) \Phi_\beta}{E_\alpha - E_\beta \pm i\varepsilon} \ + \int d\beta \int d\gamma\frac{(\Phi_\beta,V\Phi_\gamma)(\Phi_\gamma,V\Phi_\alpha)\Phi_\beta}{(E_\alpha - E_\beta \pm i\varepsilon)(E_\alpha-E_\gamma\pm i\varepsilon)} \ + ... \end{equation} then use (\ref{eq:separable_int}) to rewrite this as: \begin{equation} \begin{split} \Psi_\alpha^\pm &= \Phi_\alpha + g\int d\beta \frac{u_\beta u_\alpha^* \Phi_\beta}{E_\alpha - E_\beta \pm i\varepsilon} \ + g^2\int d\beta \int d\gamma\frac{u_\beta u_\gamma^*u_\gamma u_\alpha^*\Phi_\beta}{(E_\alpha - E_\beta \pm i\varepsilon)(E_\alpha-E_\gamma\pm i\varepsilon)} \ + ... \\ &= \Phi_\alpha + g\int d\beta \frac{u_\beta u_\alpha^* \Phi_\beta}{E_\alpha - E_\beta \pm i\varepsilon} \sum_{n=0}^\infty\left(g\int d\gamma \frac{|u_\gamma|^2}{E_\alpha-E_\gamma \pm i\varepsilon}\right)^n \ . \end{split} \end{equation} The sum is now expressed as a geometric series. We can be sure that this series does indeed converge, because (\ref{eq:separable_unitarity}) guarantees that $|u_\gamma|^2$ dies off sufficiently fast at infinity to use contour integration and pick up poles in the energy from the denominators. We can then define this geometric ratio as $R_\alpha^\pm$, and evaluate the series as $(1-R_\alpha^\pm)^{-1}$. Thus we have an explicit, non-recursive expression for the in/out states: \begin{equation} \Psi_\alpha^\pm = \Phi_\alpha + g\int d\beta \frac{u_\beta u_\alpha^* \Phi_\beta}{(E_\alpha - E_\beta \pm i\varepsilon)(1-R_\alpha^\pm)} \ . \end{equation} We can even remove the resolution of identity at this point to write \begin{equation} \boxed{ \Psi_\alpha^\pm = \biggl[1+(1-R_\alpha^\pm)^{-1}(E_\alpha-H_0\pm i\varepsilon)^{-1}V\biggr]\Phi_\alpha } \end{equation} We can also find an explicit expression for the S-matrix elements $S_{\beta\alpha}=(\Psi_\beta^-,\Psi_\alpha^+)$: \begin{equation} \begin{split} S_{\beta\alpha} &=(\Phi_\beta,\Phi_\alpha) \\ & +(\Phi_\beta,V(E_\beta-H_0 + i\varepsilon)^{-1}(1-R_\beta^+)^{-1}\Phi_\alpha)+(\Phi_\beta,(1-R_\alpha^+)^{-1}(E_\alpha-H_0 + i\varepsilon)^{-1}V\Phi_\alpha) \\ &+(\Phi_\beta,V(E_\beta-H_0 + i\varepsilon)^{-1}(1-R_\beta^+)^{-1}(1-R_\alpha^+)^{-1}(E_\alpha-H_0 + i\varepsilon)^{-1}V\Phi_\alpha) \ . \end{split} \label{eq:sba_expanded} \end{equation} Since $(1-R_\alpha^+)$ is a c-number, we can move it through the inner products with impunity, and then act the $H_0$ terms on the now adjacent eigenstates $\Phi_{\alpha/\beta}$ in the second line. We can then write the second line above as: \begin{equation} \begin{split} &(\Phi_\beta,V\Phi_\alpha)\left( \frac{1}{(1-R^+_\beta)(E_\beta-E_\alpha+i\varepsilon)} +\frac{1}{(1-R^+_\alpha)(E_\alpha-E_\beta+i\varepsilon)}\right) \\ &=g u_\beta u^*_\alpha\left( \frac{1}{(1-R^+_\beta)(E_\beta-E_\alpha+i\varepsilon)} +\frac{1}{(1-R^+_\alpha)(E_\alpha-E_\beta+i\varepsilon)}\right) \ . \end{split} \end{equation} For the third line of (\ref{eq:sba_expanded}), we can insert a complete set of states between the two $H_0$ terms, which gets us \begin{equation} \begin{split} &\frac{1}{(1-R^+_\beta)(1-R^+_\alpha)}\int d\gamma \frac{(\Phi_\beta,V\Phi_\gamma)(\Phi_\gamma,V\Phi_\alpha)}{(E_\beta-E_\gamma + i\varepsilon)(E_\alpha-E_\gamma + i\varepsilon)} \\ &=\frac{g u_\beta u^*_\alpha}{(1-R^+_\beta)(1-R^+_\alpha)}\int d\gamma \frac{g|u_\gamma|^2}{(E_\beta-E_\gamma + i\varepsilon)(E_\alpha-E_\gamma + i\varepsilon)} \ . \end{split} \end{equation} We can then write the entire S-matrix element as \begin{equation} \boxed{ \begin{split} S_{\beta\alpha} = \delta(\beta-\alpha) + \frac{g u_\beta u^*_\alpha}{(1-R^+_\beta)(1-R^+_\alpha)}\Biggl[&\frac{1-R^+_\alpha}{E_\beta-E_\alpha+i\varepsilon} + \frac{1-R^+_\beta}{E_\alpha-E_\beta+i\varepsilon} \\ &+\int d\gamma \frac{g|u_\gamma|^2}{(E_\beta-E_\gamma + i\varepsilon)(E_\alpha-E_\gamma + i\varepsilon)} \Biggr] \ . \end{split}} \end{equation} This can potentially be simplified by combining the first two terms in brackets, but I am weary about the $i\varepsilon$ terms and the pole structure they represent. \subsection{Problem 3.2 -- A Spin-1 Resonance} This question is a simple application of the formulae in chapter 3.7 on resonances. We start with the equation for the cross section of a 2-body channel $n$ going into a 2-body channel $n'$ written in the center-of-mass frame: \begin{equation} \begin{split} \sigma(n\rightarrow n'; E) &= \frac{\pi (2j_R +1)}{k^2 (2s_1 + 1)(2s_2 +1)}\frac{\Gamma_n \Gamma_{n'}}{(E-E_R)^2 + \Gamma^2/4} \\ \end{split} \end{equation} Since we are looking at elastic scattering, $n=n'$. The cross section we are given is also at resonance, so $E=E_R$. Finally, at $\sqrt{s} = 150 \text{ GeV}$, the electron and positron are ultrarelativistic, so $k = 75 \text{ GeV}$. Plugging in the values for the spins and the cross section $\sigma(n\rightarrow n; E_R) = 10^{-34} \text{ cm}^{-2} = 10^{-10}$ bn, we find: \begin{equation} \begin{split} \sigma(n\rightarrow n; E_R) &= \frac{\pi (2j_R +1)}{k^2 (2s_1 + 1)(2s_2 +1)}\frac{\Gamma_n^2}{(E-E_R)^2 + \Gamma^2/4} \\ &= \frac{3 \pi}{k^2} \left(\frac{\Gamma_n}{\Gamma} \right)^2\\ \implies \Aboxed{\Gamma_n / \Gamma &= 0.0124} \end{split} \end{equation} We also have: \begin{equation} \begin{split} \sigma(n \rightarrow n'; E) / \sigma_{\text{total}}(n;E) &= \Gamma_{n'} / \Gamma \\ \implies \Aboxed{\sigma_{\text{total}}(n;E_R) = 8.08 \text{ nb} &= 8.08 \times 10^{-33} \text{ cm}^{-2}} \end{split} \end{equation} \subsection{Problem 3.3 -- Cross Section in Lab Frame} We're not going to attempt this problem because a) the solution is not pretty or insightful and b) this has already been done before multiple times online. For Weinberg's problem specifically see Hagimoto et al. \subsection{Problem 3.7 -- In/Out States For Separable Interactions} Our task here is to find explicit solutions for the in/out states defined by the Lippman-Schwinger equations We are tasked with showing that the states $\Phi_{E \mathbf{p}j \sigma l s n}$ defined by equation (3.7.5) \begin{multline} \left(\Phi_{\mathbf{p}_1 \sigma_1 \mathbf{p}_2 \sigma_2 n'}, \Phi_{E \mathbf{p}j \sigma l s n}\right) = \sqrt{\frac{E}{|\mathbf{p}_1|E_1E_2}} \delta^3(\mathbf{p} - \mathbf{p}_1 - \mathbf{p}_2) \delta(E-E_1-E_2) \delta_{n' , n} \\ \cross \sum_{m,\mu}C_{s_1 s_2} (s,\mu;\sigma_1 \sigma_2) C_{l s}(j,\sigma;m,\mu)Y^m_l(\boldsymbol{\hat{\mathbf{p}}}_1) \end{multline} give the proper normalization in the center of mass frame (Weinberg 3.7.6): \begin{equation} \label{eq:3-7normalization} \left(\Phi_{E' \mathbf{p}' j' \sigma' l' s' n'}, \Phi_{E 0 j \sigma l s n}\right) = \delta^3(\mathbf{p}')\delta(E'-E)\delta_{j',j}\delta_{\sigma',\sigma}\delta_{l',l}\delta_{s',s}\delta_{n',n} \end{equation} To show the validity of Equation \ref{eq:3-7normalization}, we proceed by working with the left hand side. Inserting a resolution of the identity in terms of states $\Phi_{\mathbf{p}_1 \sigma_1 \mathbf{p}_2 \sigma_2 n}$ gives \begin{equation} \begin{split} \label{eq:3-7integrand} \left(\Phi_{E' \mathbf{p}' j' \sigma' l' s' n'}, \Phi_{E 0 j \sigma l s n}\right) &= \int d^3 \mathbf{p}_1 d^3 \mathbf{p}_2 \sum_{\sigma_1 \sigma_2 \bar{n}} \left(\Phi_{E' \mathbf{p}' j' \sigma' l' s' n'}, \Phi_{\mathbf{p}_1 \sigma_1 \mathbf{p}_2 \sigma_2 \bar{n}}\right) \left(\Phi_{\mathbf{p}_1 \sigma_1 \mathbf{p}_2 \sigma_2 \bar{n}}, \Phi_{E 0 j \sigma l s n}\right) \\ &= \int d^3 \mathbf{p}_1 d^3 \mathbf{p}_2 \sum_{\sigma_1 \sigma_2 \bar{n}} \frac{\sqrt{E E'}}{k_1 E_1E_2} \delta^3(\mathbf{p}_1 + \mathbf{p}_2) \delta(E-E_1-E_2) \delta_{\bar{n} , n} \\ & \cross \delta^3(\mathbf{p}' - \mathbf{p}_1 - \mathbf{p}_2) \delta(E'-E_1-E_2) \delta_{\bar{n} , n'} \sum_{m,\mu}C_{s_1 s_2} (s,\mu;\sigma_1 \sigma_2) \\ & \cross C_{l s}(j,\sigma;m,\mu) \sum_{m',\mu'}C_{s_1 s_2} (s',\mu';\sigma_1 \sigma_2) C_{l' s'}(j',\sigma';m',\mu') Y^m_l(\boldsymbol{\hat{\mathbf{p}}}_1) Y^{m'\ast}_{l'}(\boldsymbol{\hat{\mathbf{p}}}_1) \end{split} \end{equation} where $k_i \equiv |\mathbf{p}_i|$ and $E_i \equiv \sqrt{k_i^2 + M_i^2}$. We can rewrite the delta functions as \begin{equation} \begin{split} \delta^3(\mathbf{p}_1 + \mathbf{p}_2) & \delta(E-E_1-E_2) \delta^3(\mathbf{p}' - \mathbf{p}_1 - \mathbf{p}_2) \delta(E'-E_1-E_2) \\ &= \delta^3(\mathbf{p}_1 + \mathbf{p}_2) \delta(E-E_1-E_2) \delta^3(\mathbf{p}') \delta(E'-E) \end{split} \end{equation} Let us now do the integral over $\mathbf{p}_2$. The only parts of Equation \ref{eq:3-7integrand} that depend on $\mathbf{p}_2$ is then \begin{equation} \begin{split} \label{eq:3-7p2int} \int d^3\mathbf{p}_2 & \frac{1}{E_2} \delta^3(\mathbf{p}_1 + \mathbf{p}_2) \delta(E-E_1-E_2) \\ &= \int d^3\mathbf{p}_2 \frac{1}{\sqrt{k_2^2 + M_2^2}} \delta^3(\mathbf{p}_1 + \mathbf{p}_2) \delta(E-E_1-\sqrt{k_2^2 + M_2^2}) \\ & = \frac{1}{\sqrt{k_1^2 + M_2^2}} \delta(E-E_1-\sqrt{k_1^2 + M_2^2}) \end{split} \end{equation} Next, we do the integral over the angular components of $\mathbf{p}_1$ by splitting $d^3\mathbf{p}_1 = k_1^2 dk_1 d\Omega_1$, where $\Omega_1$ is the solid angle of $\boldsymbol{\hat{\mathbf{p}}}_1$. The entire integrand now only depends on $k_1$, except for the spherical harmonics that depend on $\boldsymbol{\hat{\mathbf{p}}}_1$, i.e. $\Omega_1$. \begin{equation} \int d\Omega_1 Y^m_l(\Omega_1) Y^{m'\ast}_{l'}(\Omega_1) = \delta_{l,l'}\delta_{m,m'} \end{equation} Equation \ref{eq:3-7integrand} now reads \begin{equation} \begin{split} \label{eq:3-7integrand2} \delta^3(\mathbf{p}') &\delta(E'-E)\delta_{l,l'}\int dk_1 \frac{Ek_1}{E_1\sqrt{k_1^2+M_2^2}} \delta(E-E_1-\sqrt{k_1^2 + M_2^2}) \\ &\sum_{\mathclap{\substack{\sigma_1 \sigma_2 \bar{n},\\ m,\mu, m',\mu'}}} \delta_{\bar{n} , n} \delta_{\bar{n} , n'} \delta_{m,m'} C_{s_1 s_2} (s,\mu;\sigma_1 \sigma_2) C_{l s}(j,\sigma;m,\mu) C_{s_1 s_2} (s',\mu';\sigma_1 \sigma_2) C_{l' s'}(j',\sigma';m',\mu') \end{split} \end{equation} The sums can be evaluated using the Clebsch-Gordon identities Weinberg gives in the footnote on page 154. The sum over $\sigma_1, \sigma_2$ gives Kronecker deltas $\delta_{s,s'}\delta_{\mu,\mu'}$. Summing over $m',\mu',\bar{n}$ gives \begin{equation} \delta_{s,s'} \delta_{n,n'} \sum_{m,\mu} C_{l s}(j,\sigma;m,\mu) C_{l' s'}(j',\sigma';m,\mu) = \delta_{s,s'} \delta_{n,n'} \delta_{j,j'} \delta_{\sigma,\sigma'} \end{equation} For the $k_1$ integral, one trick to solve this is to convert it into an integral over $E_1$. From $E_1^2 = k_1^2 + M_1^2$, we have $dE_1 = k_1 dk_1/E_1$, so the integral becomes \begin{equation} \int dE_1 \frac{E}{\sqrt{E_1^2-M_1^2+M_2^2}} \delta(E-E_1-\sqrt{E_1^2 - M_1^2 + M_2^2}) \end{equation} Since the delta function is $0$ when $E=E_1 + \sqrt{E_1^2 - M_1^2 +M_2^2}$, we can rewrite the integral as \begin{equation} \int dE_1 \frac{E_1+\sqrt{E_1^2 - M_1^2 + M_2^2}}{\sqrt{E_1^2-M_1^2+M_2^2}} \delta(E-E_1-\sqrt{E_1^2 - M_1^2 + M_2^2}) \end{equation} Let's remind ourselves of the identiy for delta function composition, \begin{equation} \label{eq:3-7deltacomp} \delta(g(x)) = \sum_{x_0} \frac{\delta(x-x_0)}{|g'(x_0)|} \end{equation} where $x_0$ are the zeroes of $g(x)$. The only value of $E_1$ that makes the argument of the delta function zero is \begin{equation} E_0 = \frac{E^2 + M_1^2 - M_2^2}{2E} \end{equation} Therefore $|g'(E_0)|$ is \begin{equation} |g(E_0)| = \frac{E_0 + \sqrt{E_0^2 -M_1^2 +M_2^2}}{\sqrt{E_0^2-M_1^2+M_2^2}} \end{equation} Finally, the full $E_1$ integral evaluates to 1: \begin{equation} \int dE_1 \delta(E_1-E_0) \frac{E_1+\sqrt{E_1^2 - M_1^2 + M_2^2}}{\sqrt{E_1^2-M_1^2+M_2^2}} \frac{\sqrt{E_0^2-M_1^2+M_2^2}}{E_0 + \sqrt{E_0^2 -M_1^2 +M_2^2}} = 1 \end{equation} Equation \ref{eq:3-7integrand2} is therefore reduced to the right hand side of Equation \ref{eq:3-7normalization}, so have we proved \begin{equation} \left(\Phi_{E' \mathbf{p}' j' \sigma' l' s' n'}, \Phi_{E 0 j \sigma l s n}\right) = \delta^3(\mathbf{p}')\delta(E'-E)\delta_{j',j}\delta_{\sigma',\sigma}\delta_{l',l}\delta_{s',s}\delta_{n',n} \end{equation} \section{Chapter 4} \subsection{Problem 4.1 -- Generating Functionals} Important equations for the question \begin{gather} \begin{split} F[v] \equiv 1 + \sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!} \int v^*(q'_1)...v^*(q'_N)v(q_1)...v(q_M)\\ \times S_{q'_1...q'_N,q_1...q_M} dq'_1...dq'_Ndq_1...dq_M \end{split}\\ \begin{split} F^C[v] \equiv \sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!} \int v^*(q'_1)...v^*(q'_N)v(q_1)...v(q_M)\\ \times S^C_{q'_1...q'_N,q_1...q_M} dq'_1...dq'_Ndq_1...dq_M \end{split}\\ S_{\beta\alpha} = \sum_{PART}(\pm)S_{\beta_1\alpha_1}^CS^C_{\beta_2\alpha_2}... \end{gather} Because we only need to consider the bosonic case, we can discard the $\pm$ from now on. In order to make the notation nicer, I'm going to introduce the following notation: \begin{gather} \alpha = q_1...q_M\\ \beta = q'_1 ..q'_N\\ v^*(q'_1)...v^*(q'_N) = \prod_{\beta}^{N} v^*(q'_\beta)\\ v(q_1)...v(q_M) = \prod_{\alpha}^{N} v(q_\alpha)\\ C_{NM} = \int \prod_\beta^N v^*(q_\beta) \prod_\alpha^M v(q_\alpha) S^C_{\alpha \beta} d\beta d\alpha\\ F^C[v] = \sum_{N=1}^{\infty}\sum_{M=1}^{\infty} \frac{C_{NM}}{N!M!}\\ F[v] = 1 + \sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}\int \prod_{\beta}^{N} v^*(q'_\beta)\prod_{\beta}^{N} v^*(q'_\beta) S_{\beta\alpha} d\beta d\alpha \end{gather} We can now replace the $S$ matrix in the expression for $F[v]$ with the definition of the connected $S$ matrices \begin{gather} \begin{split} F[v] = 1 + \sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!} \int \prod_{\beta}^{N} v^*(q'_\beta)\prod_{\alpha}^{M} v(q_\alpha)\\ \times \left(\sum_{PART}S_{\beta_{1}\alpha_{1}}^CS^C_{\beta_{2}\alpha_{2}}...\right) d\beta d\alpha \end{split} \end{gather} factorizing the equation yields: \begin{gather} \begin{split} F[v] = 1 + \sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}\sum_{PART} \\ \times \left(\int \prod_{\beta_1}^{|\beta_1|} v^*(q'_{\beta_1})\prod_{\alpha_1}^{|\alpha_1|}v(q_{\alpha_1})S_{\beta_{1}\alpha_{1}}^C d\alpha_1d\beta_1\right)\\ \times \left(\int \prod_{\beta_2}^{|\beta_2|} v^*(q'_{\beta_2})\prod_{\alpha_2}^{|\alpha_2|}v(q_{\alpha_2})S^C_{\beta_{2}\alpha_{2}}d\alpha_2d\beta_2\right)... \end{split} \end{gather} This can be written much more compressed using the notation introduced earlier because the $q$s are just integration variables: \begin{gather} \begin{split} F[v] = 1 + \sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}\sum_{PART} C_{|\beta_1||\alpha_1|}C_{|\beta_2||\alpha_2|}...\label{equ:genFuncinC} \end{split} \end{gather} Keep in mind \begin{align*} \sum_i |\beta_i| &= N\\ \sum_i |\alpha_i| &= M\\ \end{align*} From Eq. \ref{equ:genFuncinC}, we can pull out the first term of the sum over partitions to get an expression for $F[v]$ in terms of $F^C[v]$ \begin{gather} \begin{split} F[v] = 1 + \sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}\left(C_{NM}+{\sum_{PART}}^\prime C_{|\beta_1||\alpha_1|}C_{|\beta_2||\alpha_2|}...\right) \end{split}\\ \begin{split} F[v] = 1 + F^C[v] +\sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}{\sum_{PART}}^\prime C_{|\beta_1||\alpha_1|}C_{|\beta_2||\alpha_2|}... \end{split} \end{gather} Where ${\sum}^\prime$ is the sum over products of two or more $C$. To get a more general relationship, let us return to the expression including the sum over all partitions: \begin{gather} \begin{split} F[v] = 1 + \sum_{N=1}^{\infty}\sum_{M=1}^{\infty}\frac{1}{N!M!}\sum_{PART} C_{|\beta_1||\alpha_1|}C_{|\beta_2||\alpha_2|}... \end{split} \end{gather} Because the summand does not depend on the the specific partitions, just the lengths of the partitions, I can sum over sets of multiplicities of permutations $\{m_{a,b}\}$, where a corresponds to the number of outgoing particles and b corresponds to the number of incoming particles. Thus, when I sum over the number of incoming and outgoing particles \begin{align*} \sum_{a,b} a\cdot m_{a,b} &= N\\ \sum_{a,b} b\cdot m_{a,b} &= M \end{align*} Thus, I can convert the double sum over all total number of outgoing and incoming particles to a sum over all sets of multiplicities of particles grouped together into all permutations. Now the question is the combinatorial factor for how many ways the particles can be grouped together. There are $N!M!$ different ways of listing the particles, this determines which particles get sorted into what connected piece. Inside each connected piece, the order does not matter, so you must divide by $a!b!$ to remove duplicates. For a certain multiplicity of that cluster appearing, you must divide by this factor each time, so you get a factor of $(a!b!)^m_{a,b}$. Also, each cluster is indistinguishable from each other cluster with the same length, so you must divide by a factor of $m_{a,b}!$. Each set $\{m_{a,b}\}$ is composed of multiple groupings where these factors must be divided out, so you must divide by the product of $m_{a,b}!(a!b!)^{m_{a,b}}$ over all $a$ and $b$. Thus the combinatorial factor becomes: \begin{align*} \frac{N!M!}{\prod_{a,b}m_{a,b}(a!b!)^{m_{a,b}}} \end{align*} Now, the sums over total number of incoming and outgoing particles can be replaced by a single sum over sets of multiplicities of groupings of $|\beta_i|$ outgoing particles and $|\alpha_i|$ incoming particles. \begin{gather} \begin{split} F[v]& = 1 + \frac{1}{N!M!}\\\sum_{\{m_{|\beta_i|,|\alpha_i|}\}}&\frac{M!N!}{\prod_{|\beta_i|,|\alpha_i|}m_{|\beta_i|,|\alpha_i|}!(|\beta_i|!|\alpha_i|!)^{m_{|\beta_i|,|\alpha_i|}}} \prod_{|\beta_i|,|\alpha_i|} C_{|\beta_i||\alpha_i|}^{m_{|\beta_i|,|\alpha_i|}} \end{split} \end{gather} Simplifying \begin{gather} F[v] = \prod_{|\beta_i|,|\alpha_i|}\sum_{m_{|\beta_i|,|\alpha_i|}=0}\frac{1}{m_{|\beta_i|,|\alpha_i|}!(|\beta_i|!|\alpha_i|!)^{m_{|\beta_i|,|\alpha_i|}}} C_{|\beta_i||\alpha_i|}^{m_{|\beta_i|,|\alpha_i|}}\\ F[v] = \prod_{|\beta_i|,|\alpha_i|} e^{C_{|\beta_i||\alpha_i|}/|\beta_i|!|\alpha_i|!}\\ F[v] = \exp\left(\sum_{\beta_i=0}^{\infty}\sum_{\alpha_i=0}^{\infty}\frac{C_{|\beta_i||\alpha_i|}}{|\beta_i|!|\alpha_i|!}\right) \end{gather} Finally, \begin{gather} F[v] = e^{F^C[v]} \end{gather} \subsection{Problem 4.2 -- Spin-0 Particle Interactions} We wish to calculate the $S$-matrix element $S_{\beta \alpha}$ and differential cross section for scattering of spinless bosons with mass $M>0$ in the center-of-mass frame to order $g$ given an interaction: \begin{equation} V = g \int d^3 \mathbf{p}_1 \, d^3 \mathbf{p}_2 \, d^3 \mathbf{p}_3 \, d^3 \mathbf{p}_4 \, \delta^3(\mathbf{p}_1 + \mathbf{p}_2 - \mathbf{p}_3 - \mathbf{p}_4) a^{\dag}(\mathbf{p}_1) a^{\dag}(\mathbf{p}_2) a(\mathbf{p}_3) a(\mathbf{p}_4) \end{equation} We start with the definition of the $S$-matrix in time-dependent pertubation theory to first order in $V$: \begin{equation} S = 1 - i \int_{-\infty}^{\infty} dt \, V(t) = 1 - i \int_{-\infty}^{\infty} dt \, e^{i H_0 t} V e^{-i H_0 t} \end{equation} Taking the inner product with states $\Phi_{\beta}$ and $\Phi_{\alpha}$, we have: \begin{equation} S_{\beta \alpha} = \delta(\beta - \alpha) - i \int_{-\infty}^{\infty} dt \, e^{i (E_{\beta} - E_{\alpha}) t} (\Phi_{\beta}, V \Phi_{\alpha}) \end{equation} We now wish to calculate \begin{equation} (\Phi_{\beta}, V \Phi_{\alpha}) = g \int d^3 \mathbf{p}_1 \, d^3 \mathbf{p}_2 \, d^3 \mathbf{p}_3 \, d^3 \mathbf{p}_4 \, \delta^3(\mathbf{p}_1 + \mathbf{p}_2 - \mathbf{p}_3 - \mathbf{p}_4) (\Phi_{\beta}, a^{\dag}(\mathbf{p}_1) a^{\dag}(\mathbf{p}_2) a(\mathbf{p}_3) a(\mathbf{p}_4) \Phi_{\alpha}) \end{equation} which lets us narrow our focus to just calculating \begin{equation} (\Phi_{\beta}, a^{\dag}(\mathbf{p}_1) a^{\dag}(\mathbf{p}_2) a(\mathbf{p}_3) a(\mathbf{p}_4) \Phi_{\alpha}) = (a(\mathbf{p}_2) a(\mathbf{p}_1) \Phi_{\beta}, a(\mathbf{p}_3) a(\mathbf{p}_4) \Phi_{\alpha}) \label{eq:4-2matrix-elem} \end{equation} Let's now formulate conditions on when $a(\mathbf{p}_3) a(\mathbf{p}_4) \Phi_{\alpha} = 0$ and what happens when both the bra and ket are non-zero in the inner product. In general, we can describe our state $\Phi_{\alpha}$, which let's say consists of $N_{\alpha}$ bosons, as \begin{equation} \Phi_{\alpha} = a^{\dagger}(\mathbf{k}_1) \ldots a^{\dagger}(\mathbf{k}_{N_{\alpha}}) \Phi_0 \end{equation} For generality, let's suppose we have a product of $M$ annihilation operators acting on this state. This would look like \begin{equation} a(\mathbf{p}_1) \ldots a(\mathbf{p}_M) a^{\dagger}(\mathbf{k}_1) \ldots a^{\dagger}(\mathbf{k}_{N_{\alpha}}) \Phi_0 \label{eq:4-2manni-ncreat} \end{equation} If we pass one annihilation operator through all $N_{\alpha}$ creation operators, we are left with a sum where each term is a product of a single delta function and $N_{\alpha}-1$ creation operators acting on the vacuum $\Phi_0$. That is, unless $N_{\alpha}=0$ where we would just get $0$. We can repeat this process for $M$ annihilation operators, where we would get a sum of terms where each has $M$ deltas and $N_{\alpha}-M$ creation operators acting on $\Phi_0$, unless $M>N_{\alpha}$, where we would just get $0$. Since we are operating on $\Phi_{\alpha}$ with $M=2$ annihilation operators, the number of particles in $\Phi_{\alpha}$ must be $\geq 2$ in order to get a non-zero contribution to first-order in pertubation theory. Similarly, we find $N_{\beta} \geq 2$. Now, after we pass all the annihilation operators through Equation \ref{eq:4-2manni-ncreat}, and assuming $N_{\alpha} \geq M$, each of the resulting terms can be expressed as $M$ delta functions acting on a state with $N_{\alpha} - M$ particles. This is due to reapplying the creation operators on the vacuum. This implies that when taking inner products of two such states given by Equation \ref{eq:4-2manni-ncreat}, we require $N_{\beta} = N_{\alpha} \equiv N$. Finally, let's investigate how this interaction look diagramatically. We will have one vertex because we are at first order in pertubation theory. Our interaction has two annihilation operators, so we get two lines going up and meeting at the vertex when we move these operators through the $\Phi_{\alpha}$ creation operators. We also have two creation operators in $V$, so we get two lines exiting the vertex when the $\Phi_{\beta}$ adjoint annihilation operators move past the $V$ creation operators. We are left with $N-2$ annihilation operators from $\Phi_{\beta}$ adjoint and $N-2$ creation operators from $\Phi_{\alpha}$, so we get $N-2$ lines going straight up through the diagram as we pass the annihilation operators through the creation operators. This looks like something like: \begin{equation} (\Phi_{\beta}, V \Phi_{\alpha}) = \; \begin{tikzpicture}[baseline=(a.base)] \begin{feynman} \vertex (i1); \vertex[right=1.5cm of i1] (i2); \vertex[above right=1.06cm of i1] (a); \vertex[above=1.5cm of i1] (f1); \vertex[right=1.5cm of f1] (f2); \vertex[right=0.25cm of i2] (i3); \vertex[right=0.25cm of f2] (f3); \vertex[right=0.25cm of i3] (i4); \vertex[right=0.25cm of f3] (f4); \diagram* { { (i1) -- (a) -- (i2), (f1) -- (a) -- (f2), (i3) -- (f3), (i4) -- (f4), } }; \end{feynman} \end{tikzpicture} \; \ldots \; + \; \begin{tikzpicture}[baseline=(a.base)] \begin{feynman} \vertex (i3); \vertex[above=1.5cm of i3] (f3); \vertex[right=0.25cm of i3] (i1); \vertex[right=1.5cm of i1] (i2); \vertex[above right=1.06cm of i1] (a); \vertex[above=1.5cm of i1] (f1); \vertex[right=1.5cm of f1] (f2); \vertex[right=0.25cm of i2] (i4); \vertex[right=0.25cm of f2] (f4); \diagram* { { (i3) -- (f3), (i1) -- (a) -- (i2), (f1) -- (a) -- (f2), (i4) -- (f4), } }; \end{feynman} \end{tikzpicture} \; \ldots \; + \; \begin{tikzpicture}[baseline=(a.base)] \begin{feynman} \vertex (i3); \vertex[above=1.5cm of i3] (f3); \vertex[right=0.25cm of i3] (i4); \vertex[right=0.25cm of f3] (f4); \vertex[right=0.25cm of i4] (i1); \vertex[right=1.5cm of i1] (i2); \vertex[above right=1.06cm of i1] (a); \vertex[above=1.5cm of i1] (f1); \vertex[right=1.5cm of f1] (f2); \diagram* { { (i3) -- (f3), (i4) -- (f4), (i1) -- (a) -- (i2), (f1) -- (a) -- (f2), } }; \end{feynman} \end{tikzpicture} \; \ldots \; + \; \ldots \end{equation} where the vertical lines occur $N-2$ times. This is equivalent to the Equation (4.4.5) in Weinberg to first order (minus the time dependence): \begin{equation} (\Phi_{\beta}, V \Phi_{\alpha}) = \sum_{\text{clusterings}} \prod_j (\Phi_{\beta_j}, V_j \Phi_{\alpha_j})_{\text{C}} \end{equation} with $V_j = 1$ for clusters that do not involve a vertex. For clusters that do not involve a vertex, we are simply left with $\delta(\beta_j - \alpha_j)$, so we only need to concern ourselves with the diagram that contains a vertex, i.e. the one that corresponds to $N_{\alpha} = N_{\beta} = 2$ scattering. Returning to Equation \ref{eq:4-2matrix-elem}, we want to calculate: \begin{equation} (a(\mathbf{p}_2) a(\mathbf{p}_1) \Phi_{\beta}, a(\mathbf{p}_3) a(\mathbf{p}_4) \Phi_{\alpha}) = (a(\mathbf{p}_2) a(\mathbf{p}_1) a^{\dag}(\mathbf{k}_1') a^{\dag}(\mathbf{k}_2') \Phi_0, a(\mathbf{p}_3) a(\mathbf{p}_4) a^{\dag}(\mathbf{k}_1) a^{\dag}(\mathbf{k}_2) \Phi_0) \label{eq:4-2connected-elem} \end{equation} Focusing on one vector in the inner product: \begin{equation} \begin{split} a(\mathbf{p}_3) a(\mathbf{p}_4) a^{\dag}(\mathbf{k}_1) a^{\dag}(\mathbf{k}_2) \Phi_0 &= a(\mathbf{p}_3) a^{\dag}(\mathbf{k}_1) a(\mathbf{p}_4) a^{\dag}(\mathbf{k}_2) \Phi_0 + \delta^3(\mathbf{p}_4 - \mathbf{k}_1) a(\mathbf{p}_3) a^{\dag}(\mathbf{k}_2) \Phi_0 \\ &= a(\mathbf{p}_3) a^{\dag}(\mathbf{k}_1) a(\mathbf{p}_4) a^{\dag}(\mathbf{k}_2) \Phi_0 + \delta^3(\mathbf{p}_4 - \mathbf{k}_1) \delta^3(\mathbf{p}_3 - \mathbf{k}_2) \Phi_0 \\ &= \delta^3(\mathbf{p}_4 - \mathbf{k}_2) \delta^3(\mathbf{p}_3 - \mathbf{k}_1) \Phi_0 + \delta^3(\mathbf{p}_4 - \mathbf{k}_1) \delta^3(\mathbf{p}_3 - \mathbf{k}_2) \Phi_0 \\ &= \left[ \delta^3(\mathbf{p}_3 - \mathbf{k}_1) \delta^3(\mathbf{p}_4 - \mathbf{k}_2) + \delta^3(\mathbf{p}_3 - \mathbf{k}_2) \delta^3(\mathbf{p}_4 - \mathbf{k}_1) \right] \Phi_0 \end{split} \end{equation} Similarly, we have \begin{equation} a(\mathbf{p}_2) a(\mathbf{p}_1) a^{\dag}(\mathbf{k}_1') a^{\dag}(\mathbf{k}_2') \Phi_0 = \left[ \delta^3(\mathbf{p}_1 - \mathbf{k}_1') \delta^3(\mathbf{p}_2 - \mathbf{k}_2') + \delta^3(\mathbf{p}_1 - \mathbf{k}_2') \delta^3(\mathbf{p}_2 - \mathbf{k}_1') \right] \Phi_0 \end{equation} Equation \ref{eq:4-2connected-elem} reduces to: \begin{equation} \left[ \delta^3(\mathbf{p}_1 - \mathbf{k}_1') \delta^3(\mathbf{p}_2 - \mathbf{k}_2') + \delta^3(\mathbf{p}_1 - \mathbf{k}_2') \delta^3(\mathbf{p}_2 - \mathbf{k}_1') \right] \left[ \delta^3(\mathbf{p}_3 - \mathbf{k}_1) \delta^3(\mathbf{p}_4 - \mathbf{k}_2) + \delta^3(\mathbf{p}_3 - \mathbf{k}_2) \delta^3(\mathbf{p}_4 - \mathbf{k}_1) \right] \end{equation} Therefore, the matrix element of the time-independent interaction potential for $N=2$ is \begin{equation} (\Phi_{\beta}, V \Phi_{\alpha}) = 4 g \, \delta^3(\mathbf{k}_1' + \mathbf{k}_2' - \mathbf{k}_1 - \mathbf{k}_2) \end{equation} The $S$-matrix element for $N=2$ is: \begin{equation} \begin{split} S_{\beta \alpha} &= \delta(\beta - \alpha) - 4 i g \, \delta^3(\mathbf{k}_1' + \mathbf{k}_2' - \mathbf{k}_1 - \mathbf{k}_2) \int_{-\infty}^{\infty} dt \, e^{i(E_1' + E_2' - E_1 - E_2)t} \\ \Aboxed{S_{\beta \alpha} &= \delta(\beta - \alpha)(1 - 8 i \pi g) = \delta^4(k_1' + k_2' - k_1 - k_2)(1-8i\pi g)} \end{split} \end{equation} where we recognize the integral as a representation of the delta function. We may now use Weinberg's result for two particle states, $S^{\text{C}}_{\beta \alpha} = (S - 1)_{\beta \alpha}$, to calculate the $N=2$ connected $S$-matrix element: \begin{equation} \boxed{S^{\text{C}}_{\beta \alpha} = -2 i \pi \delta(\beta - \alpha) (4g) = -2i\pi \, \delta^4(k_1' + k_2' - k_1 - k_2)(4g)} \end{equation} The full $S$-matrix element may then be calculated for any number of particles $N$ by using the fact that the connected $S$-matrix for the one-to-one subdiagrams is simply a momentum conservation delta function, $\delta^4(k_i' - k_i)$. We will omit this since it doesn't give any more interesting information, and it will not affect our cross section. When calculating the cross section, we must restrict ourselves to only states $\Phi_{\alpha}$ and $\Phi_{\beta}$ that have no subset of particles whose momenta are unchanged during the interaction. Therefore, we should only consider the connected $S$-matrix consisting of two particles. The delta function-free matrix element is \begin{equation} M_{\beta \alpha} = 4 g \end{equation} which, using Weinberg Equation (3.4.30), leads to the cross section \begin{equation} \begin{split} \frac{d \sigma(\alpha \rightarrow \beta)}{d \Omega} &= \frac{(2\pi)^4 k' E_1' E_2' E_1 E_2}{E^2 k}|4g|^2 \\ \Aboxed{\frac{d \sigma(\alpha \rightarrow \beta)}{d \Omega} &= g^2 \frac{(4\pi)^4 k' E_1' E_2' E_1 E_2}{E^2 k}} \end{split} \end{equation} \subsection{Problem 4.3 -- Multi-mode Coherent States} In this problem we must construct a state $\Phi_\lambda$ out of the $N$-particle states $\Phi_{q_1,\dots,q_N}$, such that it is an eigenstate of the annihilation operators $a(q)$ with corresponding eigenvalues $\lambda(q)$. To proceed, we first write down a general linear combination of multi-particle states with arbitrary coefficients as: \begin{equation} \Phi_\lambda = C_0\Phi_0+\sum_{N=1}^\infty \int dq_1\dots dq_N C_N(q_1,\dots,q_N)\Phi_{q_1,\dots,q_N} \ , \end{equation} where the functions $C_N$ vary with respect to each of their arguments $q_i$. Our task in this problem is then to determine these coefficients. Starting from this general state, we can immediately determine symmetry conditions on the $C_N$ coefficients based on whether the particles in the states are bosons or fermions. For a given term in the sum, you can't stop me from permuting a few dummy integration variables with the permutation $\mathcal P$ , since they are just labels that we integrate over. We then have \begin{equation}\label{eq:4-3CN_condition} \begin{split} \int dq_1\dots dq_N C_N(q_1,\dots,q_N)\Phi_{q_1,\dots,q_N} &= \int dq_{\mathcal P1}\dots dq_{\mathcal PN} C_N(q_{\mathcal P1},\dots,q_{\mathcal PN})\Phi_{q_{\mathcal P1},\dots,q_{\mathcal PN}} \\ &=\delta_{\mathcal P}\int dq_1\dots dq_N C_N(q_{\mathcal P1},\dots,q_{\mathcal PN})\Phi_{q_1,\dots,q_N} \ , \end{split} \end{equation} where in the second step I put the integration measure back in the regular order at no cost, and I rearranged the state labels in $\Phi_{q_{\mathcal P1},\dots,q_{\mathcal PN}}$ at the cost of a $(\pm)$ permutation factor defined by $\delta_{\mathcal P}$ (defined in Weinberg below equation (4.1.7)). Of course, the two sides must still be equal, so we must conclude that the coefficients $C_N$ must obey the same symmetry rules as the states, that is, \begin{equation} C_N(q_{\mathcal P1},\dots,q_{\mathcal PN}) = \delta_{\mathcal P}C_N(q_1,\dots,q_N)\ . \end{equation} This ensures both sides of (\ref{eq:4-3CN_condition}) remain equal. With these conditions, we can make a recursion relation between $C_N$ and $C_{N+1}$. To enforce $a(q)\Phi_\lambda=\lambda(q)\Phi_\lambda$, we need to compare the $N$th term in the sum with the annihilation operator $a(q)$ acting on the $N+1$th term. Concretely, \begin{equation} \begin{split} \lambda(q)\int dq_1\dots dq_N C_N(q_1,\dots,q_N)\Phi_{q_1,\dots,q_N} &= a(q) \int dq_1\dots dq_{N+1} C_{N+1}(q_1,\dots,q_{N+1})\Phi_{q_1,\dots,q_{N+1}} \ . \end{split} \end{equation} We then act the annihilation operator on the multiparticle state using Weinberg (4.2.3) to get: \begin{equation} \begin{split} \lambda(q)&\int dq_1\dots dq_N C_N(q_1,\dots,q_N)\Phi_{q_1,\dots,q_N} \\ &= \int dq_1\dots dq_{N+1} C_{N+1}(q_1,\dots,q_{N+1}) \sum_{r=1}^{N+1}(\pm)^{r+1} \delta(q-q_r)\Phi_{q_1,\dots,q_{r-1},q_{r+1},\dots,q_{N+1}} \\ &=\sum_{r=1}^{N+1}(\pm)^{r+1} \int dq_1\dots dq_{r-1}dq_{r+1}\dots dq_{N+1} \\ &\qquad \qquad \qquad \times C_{N+1}(q_1,\dots,q_{r-1},q,q_{r+1},\dots,q_{N+1}) \Phi_{q_1,\dots,q_{r-1},q_{r+1},\dots,q_{N+1}} \ . \end{split} \end{equation} From here, we can rewrite \begin{equation} C_{N+1}(q_1,\dots,q_{r-1},q,q_{r+1},\dots,q_{N+1}) = (\pm)^{r-1}C_{N+1}(q,q_1,\dots,q_{r-1},q_{r+1},\dots,q_{N+1}) \end{equation} using the symmetry conditions we determined earlier, which cancels out all $\pm$ signs. Then, for any $q_i$ with $i>r$, we rename the dummy integration variable as $q_i \rightarrow q_{i-1}$. This gets us: \begin{equation} \begin{split} \lambda(q)\int dq_1\dots dq_N C_N(q_1,\dots,q_N)\Phi_{q_1,\dots,q_N} &=\sum_{r=1}^{N+1} \int dq_1\dots dq_{N} C_{N+1}(q,q_1,\dots ,q_{N}) \Phi_{q_1,\dots,q_{N}} \\ &=(N+1) \int dq_1\dots dq_{N} C_{N+1}(q,q_1,\dots ,q_{N}) \Phi_{q_1,\dots,q_{N}} \ . \end{split} \end{equation} Where the sum turned into a prefactor of $(N+1)$ due to the summand being independent of the index $r$. We can now read off the desired recursion relation: \begin{equation} C_{N+1}(q,q_1,\dots ,q_{N}) = \frac{\lambda(q)}{N+1}C_{N}(q_1,\dots ,q_{N}) \ . \end{equation} This also works with the base case, \begin{equation} \begin{split} \lambda(q)C_0\Phi_0 = a(q)\int &dq_1 C_1(q_1)\Phi_{q_1} = C_1(q)\Phi_0 \\ \Rightarrow C_1(q) &= \lambda(q)C_0 \ , \end{split} \end{equation} so we can build up the coefficients $C_N$ by using the recursion relation to get \begin{equation}\label{eq:4-3result} C_{N}(q_1,\dots ,q_{N}) = \frac{C_0}{N!}\prod_{i=1}^N \lambda(q_i) \ , \end{equation} where the constant $C_0$ sets the overall normalization. To find what this normalization is, we first rewrite the state as \begin{equation} \Phi_\lambda = C_0\left[ \Phi_0+\sum_{N=1}^\infty \frac{1}{N!} \int \prod_{i=1}^N \bigl(dq_i \lambda(q_i)\bigr)\Phi_{q_1,\dots,q_N}\right] \ . \end{equation} It sure would be nice if we could normalize this state to unity, so to check if we can let's just calculate the norm divided by a factor of $C_0^2$: \begin{equation} \begin{split} \frac{(\Phi_\lambda,\Phi_\lambda)}{C_0^2} &= 1 + \sum_{N,M}\frac{1}{N!M!}\int \prod_{i=1}^N \bigl(dq_i \lambda^*(q_i)\bigr)\prod_{j=1}^M \bigl(dq'_j \lambda(q'_j)\bigr) \underbrace{(\Phi_{q_1,\dots,q_N},\Phi_{q'_1,\dots,q_M})}_{\delta_{NM}\sum_{\mathcal P}\prod_i\delta(q_i-q'_{\mathcal Pi})} \\ &=1 + \sum_{N}\frac{1}{(N!)^2}\sum_{\mathcal P}\int \prod_{i=1}^N \bigl(dq_i \lambda^*(q_i) dq'_i \lambda(q'_i)\delta(q_i-q'_{\mathcal Pi})\bigr) \ . \end{split} \end{equation} Because we are integrating over all momenta, each element in the permutation sum will end up giving the same exact result, since we can always use the delta functions to evaluate the $q'_{\mathcal Pi}$ integrals, leaving only the same $q_i$ integrals in each term. There are $N!$ elements in the permutation group of $N$ particles, so we get \begin{equation} \begin{split} \frac{(\Phi_\lambda,\Phi_\lambda)}{C_0^2} &= 1 + \sum_{N=1}^\infty\frac{1}{N!}\int \prod_{i=1}^N dq_i |\lambda(q_i)|^2 \\ &=\sum_{N=0}^\infty \frac{1}{N!}\left(\int dq |\lambda(q)|^2\right)^N \\ &=\exp{\int dq |\lambda(q)|^2} \ , \end{split} \end{equation} and we therefore have for the normalization \begin{equation} C_0 = \exp{-\frac{1}{2}\int dq |\lambda(q)|^2} \ . \end{equation} This ensures that our multi-mode coherent state is normalized to unity. All together, we have \begin{equation} \Phi_\lambda = \exp{-\frac{1}{2}\int dq |\lambda(q)|^2}\left[ \Phi_0+\sum_{N=1}^\infty \frac{1}{N!} \int \prod_{i=1}^N \bigl(dq_i \lambda(q_i)\bigr)\Phi_{q_1,\dots,q_N}\right] \ . \end{equation} Additionally, it is worth noting that we can write the multi-particle states in the integrals as a product of creation operators acting on the vacuum state, as in \begin{equation} \Phi_{q_1,\dots,q_N} = \prod_{i=1}^Na^\dag(q_i)\Phi_0 \ , \end{equation} which then allows us to make the integrals separable: \begin{equation} \int \prod_{i=1}^N \bigl(dq_i \lambda(q_i)\bigr)\Phi_{q_1,\dots,q_N} = \int \prod_{i=1}^N \bigl(dq_i \lambda(q_i)a^\dag(q_i)\bigr)\Phi_0 = \left(\int dq \lambda(q)a^\dag(q)\right)^N\Phi_0 \ , \end{equation} and finally write the whole coherent state in terms of a displacement operator acting on the vacuum state: \begin{equation} \Phi_\lambda = D_\lambda\Phi_0 \quad, \quad D_\lambda \equiv \exp{-\frac{1}{2}\int dq |\lambda(q)|^2}\exp{\int dq \lambda(q)a^\dag(q)} \ . \end{equation} \textbf{Note:} This result is easily verified for the bosonic case, where the creation and annihilation operators commute. But for the fermionic case, (\ref{eq:4-3result}) seems to be in contradiction with the symmetry conditions we derived in (\ref{eq:4-3CN_condition}), as a product of numbers $\lambda(q_i)$ seems to be symmetric under interchanges, not antisymmetric as needed. We are therefore driven to one of two conclusions: either there is no such thing as a "coherent state" for fermions, or that the numbers $\lambda(q_i)$ are not real numbers but \textit{Grassmann} numbers for fermionic coherent states, which anticommute rather than commute. \section{Chapter 5} \subsection{Problem 5.1 -- Field Coefficients Satisfying General Transformation Condition} In this problem we want to show that if the zero-momentum field coefficients (suppressing the species index $n$) $u_\ell(\vect{p}=0,\sigma)$ and $v_\ell(\vect{p}=0,\sigma)$ are chosen to satisfy the rotation conditions given by \begin{equation}\label{eq:4.1-rotated_u0} \sum_{\bar{\sigma}}u_{\bar{\ell}}(0,\bar{\sigma})D_{\bar \sigma \sigma}^{(j)}(R) = \sum_\ell D_{\bar \ell \ell}(R)u_{\ell}(0,\sigma) \end{equation} \begin{equation} \sum_{\bar{\sigma}}v_{\bar{\ell}}(0,\bar{\sigma})D_{\bar \sigma \sigma}^{(j)*}(R) = \sum_\ell D_{\bar \ell \ell}(R)v_{\ell}(0,\sigma) , \end{equation} then the non-zero momentum coefficients given by \begin{equation}\label{eq:4.1-boosted_u0} u_{\bar \ell}(\vect q,\sigma)=\sqrt{\frac{m}{q^0}}\sum_\ell D_{\bar \ell \ell}(L(q))u_\ell(0,\sigma) \end{equation} \begin{equation} v_{\bar \ell}(\vect q,\sigma)=\sqrt{\frac{m}{q^0}}\sum_\ell D_{\bar \ell \ell}(L(q))v_\ell(0,\sigma) \end{equation} satisfy the general condition needed for the Lorentz transformations acting on fields, that is \begin{equation}\label{eq:4.1-result} \sum_{\bar \sigma }u_{\bar \ell}(\vect{p}_\Lambda,\bar \sigma)D_{\bar\sigma \sigma}^{(j)}(W(\Lambda,p))=\sqrt{\frac{p^0}{(\Lambda p)^0}}\sum_\ell D_{\bar\ell \ell}(\Lambda)u_\ell(\vect{p},\sigma) \end{equation} \begin{equation} \sum_{\bar \sigma }v_{\bar \ell}(\vect{p}_\Lambda,\bar \sigma)D_{\bar\sigma \sigma}^{(j)*}(W(\Lambda,p))=\sqrt{\frac{p^0}{(\Lambda p)^0}}\sum_\ell D_{\bar\ell \ell}(\Lambda)v_\ell(\vect{p},\sigma) \ . \end{equation} Put briefly, we need to use the first four equations to show the last two. I will focus just on the $u_\ell$ case, as the $v_\ell$ case is completely equivalent. We will start with the LHS of (\ref{eq:4.1-result}), and work our way rightward. First, we will use (\ref{eq:4.1-boosted_u0}) to rewrite it as: \begin{equation} \begin{split} \sum_{\bar \sigma }u_{\bar \ell}(\vect{p}_\Lambda,\bar \sigma)D_{\bar\sigma \sigma}^{(j)}(W(\Lambda,p)) &= \sum_{\bar \sigma }\left(\sqrt{\frac{m}{(\Lambda p)^0}}\sum_\ell D_{\bar \ell \ell}(L(\Lambda p))u_\ell(0,\bar \sigma) \right)D_{\bar\sigma \sigma}^{(j)}(W(\Lambda,p)) \\ &=\sqrt{\frac{m}{(\Lambda p)^0}}\sum_\ell D_{\bar \ell \ell}(L(\Lambda p))\left( \sum_{\bar \sigma }u_\ell(0,\bar \sigma) D_{\bar\sigma \sigma}^{(j)}(W(\Lambda,p))\right) \ . \end{split} \end{equation} Now, since the little group element $W(\Lambda,p)$ is a rotation (the little group of massive particles is $SO(3)$), we can use (\ref{eq:4.1-rotated_u0}) to rewrite the $\bar\sigma$ sum in parentheses as a sum over the field component index $\ell'$. We then have \begin{equation} \begin{split} \sum_{\bar \sigma }u_{\bar \ell}(\vect{p}_\Lambda,\bar \sigma)D_{\bar\sigma \sigma}^{(j)}(W(\Lambda,p)) &= \sqrt{\frac{m}{(\Lambda p)^0}}\sum_\ell D_{\bar \ell \ell}(L(\Lambda p))\left( \sum_{\ell' }D_{\ell \ell'}(W(\Lambda,p))u_{\ell'}(0,\bar \sigma) \right) \ . \end{split} \end{equation} Now we use the definition of the little group element, $W(\Lambda,p)\equiv L^{-1}(\Lambda p)\Lambda L(p)$, as well as the fact that the $D_{\ell\ell'}$ form a representation of the Lorentz group, to make the replacement \begin{equation} \begin{split} D(L(\Lambda p))D(W(\Lambda,p))&= D(L(\Lambda p))D(L^{-1}(\Lambda p)\Lambda L(p))\\ &=D(\Lambda)D(L(p)) \end{split} \end{equation} so we can rewrite the previous equation as \begin{equation} \sum_{\bar \sigma }u_{\bar \ell}(\vect{p}_\Lambda,\bar \sigma)D_{\bar\sigma \sigma}^{(j)}(W(\Lambda,p)) = \sqrt{\frac{m}{(\Lambda p)^0}}\sum_\ell D_{\bar \ell \ell}(\Lambda)\left( \sum_{\ell' }D_{\ell \ell'}(L(p))u_{\ell'}(0,\bar \sigma) \right) \ . \end{equation} Finally, we use (\ref{eq:4.1-boosted_u0}) one more time to rewrite the sum in parenthesis as a boosted coefficient \begin{equation} \sum_{\ell' }D_{\ell \ell'}(L(p))u_{\ell'}(0,\bar \sigma) = \sqrt{\frac{p^0}{m}}u_\ell(\vect{p},\bar\sigma) \ , \end{equation} and we arrive at the desired result, \begin{equation} \boxed{ \sum_{\bar \sigma }u_{\bar \ell}(\vect{p}_\Lambda,\bar \sigma)D_{\bar\sigma \sigma}^{(j)}(W(\Lambda,p)) = \sqrt{\frac{p^0}{(\Lambda p)^0}}\sum_\ell D_{\bar \ell \ell}(\Lambda)u_{\ell}(\vect{p},\bar \sigma) \ . } \end{equation} \subsection{Problem 5.3 -- A Massive Spin 2 Field} Consider a free field $h^{\mu \nu}(x)$ satisfying $h^{\mu \nu}(x) = h^{\nu \mu}(x)$ and $h^{\mu}_{\mu}(x) = 0$, which annihilates and creates a particle of spin two and mass $m \neq 0$. Show how to calculate the coefficient functions $u^{\mu \nu}(\mathbf{p}, \sigma)$, which multiply the annihilation operators $a(\mathbf{p}, \sigma)$ in this field, in such a way that the field transforms under Lorentz transformations like a tensor. What field equations does this field satisfy? Evaluate the function $P^{\mu \nu, \kappa \lambda}(p)$, defined by \begin{equation} \sum_{\sigma} u^{\mu \nu}(\mathbf{p}, \sigma) u^{\kappa \lambda *}(\mathbf{p}, \sigma) \equiv (2p^0)^{-1} P^{\mu \nu, \kappa \lambda}(p) \end{equation} What are the commutation relations of this field? How does the field transform under the inversions $\mathsf{P}, \mathsf{C}, \mathsf{T}$? There are two ways of proceeding. The first way is to calculate the field coefficients for a field $\psi_{a b}(x)$ with $j=2$ in the $(1,1)$ representation of the Lorentz group. This way is a bit trickier in mapping back the 3x3 matrix coefficients $u_{a b}(\mathbf{p}, \sigma)$ to the 4x4 coefficients $u^{\mu \nu}(\mathbf{p}, \sigma)$ of the traceless symmetric tensor $h^{\mu \nu}(x)$. However, one can see that there are the same number of degrees of freedom. For the 3x3 matrix, all matrix elements are free. In the 4x4 case, there are 16 free elements, but then imposing the symmetry conditions yields 10 free elements and the traceless condition lowers the free elements to 9. Instead of calculating $u_{ab}(\mathbf{p}, \sigma)$, we'll go with a second approach that mimics the spin one section of Chapter 5.3. Since our field transforms as a tensor, this means the $D$-matrix takes a very nice form: \begin{equation} D\indices{^\mu ^\nu _\rho _\sigma}(\Lambda) = \Lambda\indices{^\mu _\rho} \Lambda\indices{^\nu _\sigma} \end{equation} Weinberg Equations (5.1.6) and (5.1.7) for the annihilation and creation fields thus read \begin{equation} \begin{split} U_0(\Lambda, a) h^{\mu \nu +}(x) U_0^{-1}(\Lambda, a) &= \Lambda\indices{_\rho ^\mu} \Lambda\indices{_\sigma ^\nu} h^{\rho \sigma +}(x) \\ U_0(\Lambda, a) h^{\mu \nu -}(x) U_0^{-1}(\Lambda, a) &= \Lambda\indices{_\rho ^\mu} \Lambda\indices{_\sigma ^\nu} h^{\rho \sigma -}(x) \end{split} \end{equation} Investigating how this field transforms under translations yields the same result that the fields are Fourier transforms as found in Weinberg Equations (5.1.17) and (5.1.18), so we won't repeat those results here. For boosts, applying our specific form the $D$-matrices to Equations (5.1.21) and (5.1.22) gives us \begin{equation} \begin{split} u^{\mu \nu}(\mathbf{p}, \sigma) &= \sqrt{\frac{m}{p^0}} L(p)\indices{^\mu _\rho} L(p)\indices{^\nu _\sigma} u^{\rho \sigma}(0, \sigma) \\ v^{\mu \nu}(\mathbf{p}, \sigma) &= \sqrt{\frac{m}{p^0}} L(p)\indices{^\mu _\rho} L(p)\indices{^\nu _\sigma} v^{\rho \sigma}(0, \sigma) \end{split} \end{equation} For rotations, Equations (5.1.23) and (5.1.24) become \begin{equation} \begin{split} \sum_{\bar{\sigma}} u^{\mu \nu}(0, \bar{\sigma}) D^{(2)}_{\bar{\sigma} \sigma}(R) &= R\indices{^\mu _\rho} R\indices{^\nu _\sigma} u^{\rho \sigma}(0, \sigma) \\ \sum_{\bar{\sigma}} v^{\mu \nu}(0, \bar{\sigma}) D^{(2)*}_{\bar{\sigma} \sigma}(R) &= R\indices{^\mu _\rho} R\indices{^\nu _\sigma} v^{\rho \sigma}(0, \sigma) \end{split} \end{equation} For an infinitessimal rotation, $D^{(2)} \rightarrow 1 + i \pmb{\theta} \cdot \mathbf{J}^{(2)}$ and $D \rightarrow 1 + i \pmb{\theta} \cdot \pmb{\mathscr{J}}$. Collecting terms linear in $\pmb{\theta}$ gives (5.1.25) and (5.1.26) for tensors: \begin{equation} \begin{split} \sum_{\bar{\sigma}} u^{\mu \nu}(0, \bar{\sigma}) \mathbf{J}_{\bar{\sigma} \sigma} &= u^{\mu \alpha}(0, \sigma) \pmb{\mathscr{J}}\indices{^\nu _\alpha} + u^{\nu \alpha}(0, \sigma) \pmb{\mathscr{J}}\indices{^\mu _\alpha} \\ -\sum_{\bar{\sigma}} v^{\mu \nu}(0, \bar{\sigma}) \mathbf{J}^*_{\bar{\sigma} \sigma} &= v^{\mu \alpha}(0, \sigma) \pmb{\mathscr{J}}\indices{^\nu _\alpha} + v^{\nu \alpha}(0, \sigma) \pmb{\mathscr{J}}\indices{^\mu _\alpha} \label{eq:5-3vectorjeqn} \end{split} \end{equation} Next, let's investigate the following identity: \begin{equation} \begin{split} \sum_{\bar{\sigma}} u^{\mu \nu}(0, \bar{\sigma}) J^2_{\bar{\sigma} \sigma} &= \sum_{\bar{\sigma}, \sigma'} u^{\mu \nu}(0, \bar{\sigma}) \mathbf{J}_{\bar{\sigma} \sigma'} \cdot \mathbf{J}_{\sigma' \sigma} \\ &= \pmb{\mathscr{J}}\indices{^\nu _\alpha} \cdot \sum_{\sigma'} u^{\mu \alpha}(0, \sigma') \mathbf{J}_{\sigma' \sigma} + \pmb{\mathscr{J}}\indices{^\mu _\alpha} \cdot \sum_{\sigma'} u^{\nu \alpha}(0, \sigma') \mathbf{J}_{\sigma' \sigma} \\ &= \pmb{\mathscr{J}}\indices{^\nu _\alpha} \cdot \pmb{\mathscr{J}}\indices{^\alpha _\beta} u^{\mu \beta}(0, \sigma) + \pmb{\mathscr{J}}\indices{^\nu _\alpha} \cdot \pmb{\mathscr{J}}\indices{^\mu _\beta} u^{\alpha \beta}(0, \sigma) \\ &\quad + \pmb{\mathscr{J}}\indices{^\mu _\alpha} \cdot \pmb{\mathscr{J}}\indices{^\alpha _\beta} u^{\nu \beta}(0, \sigma) + \pmb{\mathscr{J}}\indices{^\mu _\alpha} \cdot \pmb{\mathscr{J}}\indices{^\nu _\beta} u^{\alpha \beta}(0, \sigma) \\ &= (\mathscr{J}^2)\indices{^\nu _\beta} u^{\mu \beta}(0, \sigma) + (\mathscr{J}^2)\indices{^\mu _\beta} u^{\nu \beta}(0, \sigma) + \pmb{\mathscr{J}}\indices{^\nu _\alpha} \cdot \pmb{\mathscr{J}}\indices{^\mu _\beta} u^{\alpha \beta}(0, \sigma) + \pmb{\mathscr{J}}\indices{^\mu _\alpha} \cdot \pmb{\mathscr{J}}\indices{^\nu _\beta} u^{\alpha \beta}(0, \sigma) \end{split} \end{equation} We then use the fact that the matrix element for $J^2$ is $J^2_{\bar{\sigma} \sigma} = j(j+1) \delta_{\bar{\sigma} \sigma} = 6 \, \delta_{\bar{\sigma} \sigma}$ to arrive at \begin{equation} 6 u^{\mu \nu}(0, \sigma) = (\mathscr{J}^2)\indices{^\nu _\beta} u^{\mu \beta}(0, \sigma) + (\mathscr{J}^2)\indices{^\mu _\beta} u^{\nu \beta}(0, \sigma) + \pmb{\mathscr{J}}\indices{^\nu _\alpha} \cdot \pmb{\mathscr{J}}\indices{^\mu _\beta} u^{\alpha \beta}(0, \sigma) + \pmb{\mathscr{J}}\indices{^\mu _\alpha} \cdot \pmb{\mathscr{J}}\indices{^\nu _\beta} u^{\alpha \beta}(0, \sigma) \label{eq:5-3bigjsquaredeqn} \end{equation} Next, recall the following relations \begin{equation} \begin{split} &(\mathscr{J}_k)\indices{^0 _0} = (\mathscr{J}_k)\indices{^0 _i} = (\mathscr{J}_k)\indices{^i _0} = 0, \quad (\mathscr{J}_k)\indices{^i _j} = -i \epsilon_{i j k} \\ &(\mathscr{J}^2)\indices{^0 _0} = (\mathscr{J}^2)\indices{^0 _i} = (\mathscr{J}^2)\indices{^i _0} = 0, \quad (\mathscr{J}^2)\indices{^i _j} = 2 \delta\indices{^i _j} \end{split} \end{equation} We therefore immediately see that \begin{equation} u^{0 0}(0, \sigma) = u^{0 i}(0, \sigma) = u^{i 0}(0, \sigma) = 0 \end{equation} For $\mu = i$ and $\nu = j$, it turns out Equation \ref{eq:5-3bigjsquaredeqn} doesn't give us any new constraints. To get further constraints, we must find $u^{\mu \nu}(0, 0)$ and then use the raising and lowering operators to find other spin 3-component coefficients. We also note that the only difference between the top and bottom lines in Equation \ref{eq:5-3vectorjeqn} is a minus sign and the $\mathbf{J}$ being complex conjugated. In our basis, $J^2$ is real, so we find the exact same relation for the $v^{\mu \nu}(0, \sigma)$, or \begin{equation} v^{0 0}(0, \sigma) = v^{0 i}(0, \sigma) = v^{i 0}(0, \sigma) = 0 \end{equation} Working with the three-component of Equation \ref{eq:5-3vectorjeqn}, we first recognize that the matrix element of $J^3$ is $J^3_{\bar{\sigma} \sigma} = \sigma \delta_{\bar{\sigma} \sigma}$, so \begin{equation} \sum_{\bar{\sigma}} u^{i j}(0, \bar{\sigma}) J^3_{\bar{\sigma} \sigma} = \sigma u^{i j}(0, \sigma) \end{equation} which means for the $\sigma=0$ coefficient: \begin{equation} \begin{split} u^{i \alpha}(0,0) (\mathscr{J}_3)\indices{^j _\alpha} + u^{j \alpha}(0,0) (\mathscr{J}_3)\indices{^i _\alpha} &= 0 \\ -i \epsilon_{j m 3} u^{i m}(0,0) -i \epsilon_{i m 3}u^{j m}(0,0) &= 0\\ \epsilon_{j m 3} u^{i m}(0,0) + \epsilon_{i m 3}u^{j m}(0,0) &= 0 \end{split} \end{equation} For $i=3$, we see that \begin{equation} \epsilon_{j m 3} u^{3 m}(0,0) = 0 \end{equation} which for $j=1$ and $j=2$ means that \begin{equation} u^{1 3}(0,0) = u^{3 1}(0,0) = u^{2 3}(0,0) = u^{3 2}(0,0) = 0 \end{equation} For $i = 1$ and $j=1$, the equation reads \begin{equation} \begin{split} \epsilon_{1 2 3}u^{1 2}(0,0) + \epsilon_{1 2 3}u^{1 2}(0,0) &= 0 \\ \implies u^{1 2}(0,0) = - u^{1 2}(0,&0) \end{split} \end{equation} but since $h^{\mu \nu}(x)$ is symmetric, this is only possible if \begin{equation} u^{1 2}(0,0) = u^{2 1}(0, 0) = 0 \end{equation} For $i = 1$ and $j=2$, we have \begin{equation} \begin{split} \epsilon_{2 1 3}u^{1 1}(0,0) + \epsilon_{1 2 3}u^{2 2}(0,0) &= 0 \\ \implies u^{1 1}(0,0) = u^{2 2}(0,&0) \end{split} \end{equation} which together with the traceless condition forces \begin{equation} u^{3 3}(0,0) = -2u^{1 1}(0,0) \end{equation} We can now write the full field coefficient for $\sigma=0$ \begin{equation} u^{\mu \nu}(0,0) = \begin{bmatrix} a & 0 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & -2a & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation} Noting that $J^3$ is real, Equation \ref{eq:5-3vectorjeqn} tells us that the $v^{\mu \nu}(0,0)$ have the same exact form as $u^{\mu \nu}(0,0)$, since the left hand side drops out due to $\sigma=0$. Therefore, we can choose the field normalization $u^{\mu \nu}(0,0) u^{*}_{\mu \nu}(0,0)$ such that \begin{equation} \boxed{u^{\mu \nu}(0,0) = v^{\mu \nu}(0,0) = \frac{1}{\sqrt{6}} \frac{1}{\sqrt{2m}} \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}} \end{equation} From here, we can just use the raising and lowering operators, recalling that the matrix element for the raising and lowering operators in the $j=2$ representation is real and given by \begin{equation} J^{\pm}_{\bar{\sigma} \sigma} = \delta_{\bar{\sigma}, \sigma \pm 1}\sqrt{(2 \mp \sigma)(3 \pm \sigma)} \end{equation} Equation \ref{eq:5-3vectorjeqn} can be rewritten as \begin{equation} \begin{split} \sum_{\bar{\sigma}}J^{\pm}_{\bar{\sigma} \sigma} u^{\mu \nu}(0,\bar{\sigma}) &= u^{\mu \alpha}(0, \sigma)(\mathscr{J}_1)\indices{^\nu _\alpha} + u^{\nu \alpha}(0, \sigma)(\mathscr{J}_1)\indices{^\mu _\alpha} \\ &\quad \pm i u^{\mu \alpha}(0, \sigma)(\mathscr{J}_2)\indices{^\nu _\alpha} \pm i u^{\nu \alpha}(0, \sigma)(\mathscr{J}_2)\indices{^\mu _\alpha} \\ -\sum_{\bar{\sigma}}J^{\mp*}_{\bar{\sigma} \sigma} v^{\mu \nu}(0,\bar{\sigma}) &= v^{\mu \alpha}(0, \sigma)(\mathscr{J}_1)\indices{^\nu _\alpha} + v^{\nu \alpha}(0, \sigma)(\mathscr{J}_1)\indices{^\mu _\alpha} \\ &\quad \pm i v^{\mu \alpha}(0, \sigma)(\mathscr{J}_2)\indices{^\nu _\alpha} \pm i v^{\nu \alpha}(0, \sigma)(\mathscr{J}_2)\indices{^\mu _\alpha} \end{split} \end{equation} where for the second equality, we have a $J^{\mp}*$ since $J^{1*}_{\bar{\sigma} \sigma} \pm i J^{2*}_{\bar{\sigma}\sigma} = (J^1_{\bar{\sigma} \sigma} \mp i J^2_{\bar{\sigma}\sigma})^*$. For $\mu = i$ and $\nu = j$, this reduces to \begin{equation} \begin{split} \sqrt{(2\mp \sigma)(3\pm \sigma)}u^{i j}(0, \sigma \pm 1) &= -i \epsilon_{j a 1} u^{i a}(0, \sigma) - i \epsilon_{i a 1}u^{j a}(0, \sigma) \\ &\quad \pm \epsilon_{j a 2} u^{i a}(0, \sigma) \pm \epsilon_{i a 2}u^{j a}(0, \sigma) \\ -\sqrt{(2\pm \sigma)(3\mp \sigma)}v^{i j}(0, \sigma \mp 1) &= -i \epsilon_{j a 1} v^{i a}(0, \sigma) - i \epsilon_{i a 1}v^{j a}(0, \sigma) \\ &\quad \pm \epsilon_{j a 2} v^{i a}(0, \sigma) \pm \epsilon_{i a 2}v^{j a}(0, \sigma) \label{eq:5-3jpmeqn} \end{split} \end{equation} When we plug in $\sigma=0$, we note that the coefficients vanish for $i=j$ since this requirement forces $a \neq i$ in the right hand side, which is off-diagonal for $u^{\mu \nu}(0,0)$ which is zero. The coefficient also vanishes for $i=1, j=2$ since it either causes the Levi-Cevita symbols to be $0$ or forces $a=3$, which renders the right hand side zero again since $u^{\mu \nu}(0,0)$ is zero off the diagonal. Let's now plug in $i=1, j=3$ \begin{equation} \begin{split} \sqrt{6}u^{1 3}(0, \pm 1) &= -i \epsilon_{3 2 1}u^{1 2}(0, 0) - i\epsilon_{1 a 1}u^{3 a}(0,0) \pm \epsilon_{3 1 2}u^{1 1}(0,0) \pm \epsilon_{1 3 2}u^{3 3}(0,0) \\ &= \pm \frac{3}{\sqrt{6}} \frac{1}{\sqrt{2m}} \\ \implies u^{1 3}(0, \pm 1) &= u^{3 1}(0, \pm 1) = \pm \frac{1}{2} \frac{1}{\sqrt{2m}} \end{split} \end{equation} and for $i=2, j=3$ \begin{equation} \begin{split} \sqrt{6}u^{2 3}(0, \pm 1) &= -i \epsilon_{3 2 1}u^{2 2}(0, 0) - i\epsilon_{2 3 1}u^{3 3}(0,0) \pm \epsilon_{3 1 2}u^{2 1}(0,0) \pm \epsilon_{2 a 2}u^{3 a}(0,0) \\ &= i \frac{3}{\sqrt{6}} \frac{1}{\sqrt{2m}} \\ \implies u^{2 3}(0, \pm 1) &= u^{3 2}(0, \pm 1) = \frac{i}{2} \frac{1}{\sqrt{2m}} \end{split} \end{equation} Now, since $v^{\mu \nu}(0,0)$ and $u^{\mu \nu}(0,0)$ are the same, and since the equations for $u$ and $v$ in \ref{eq:5-3jpmeqn} differ by a sign and describe opposite spins, the coefficients are thus \begin{equation} \boxed{u^{\mu \nu}(0,\pm 1) = -v^{\mu \nu}(0,\mp 1) = \frac{1}{2} \frac{1}{\sqrt{2m}} \begin{bmatrix} 0 & 0 & \pm 1 & 0 \\ 0 & 0 & i & 0 \\ \pm 1 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} } \end{equation} which one can confirm retains the correct normalization $u^{\mu \nu} u^*_{\mu \nu} = 1/2m$. For $\sigma = \pm 2$, we plug $\sigma = \pm 1$ into Equation \ref{eq:5-3jpmeqn}: \begin{equation} \begin{split} 2u^{i j}(0, \pm 2) &= -i \epsilon_{j a 1} u^{i a}(0, \pm 1) - i \epsilon_{i a 1}u^{j a}(0, \pm 1)\pm \epsilon_{j a 2} u^{i a}(0, \pm 1) \pm \epsilon_{i a 2}u^{j a}(0, \pm 1) \\ -2v^{i j}(0, \pm 2) &= -i \epsilon_{j a 1} v^{i a}(0, \pm 1) - i \epsilon_{i a 1}v^{j a}(0, \pm 1) \pm \epsilon_{j a 2} v^{i a}(0, \pm 1) \pm \epsilon_{i a 2}v^{j a}(0, \pm 1) \end{split} \end{equation} Unfortunately, none of the elements of the coefficients are easy to determine outright, so we will have to calculate all six combinations of $i$ and $j$. Starting with diagonal elements, $i=1, j=1$ \begin{equation} \begin{split} 2 u^{1 1}(0, \pm 2) &= - i \epsilon_{1 a 1}u^{1 a}(0, \pm 1) - i \epsilon_{1 a 1}u^{1 a}(0, \pm 1) \pm \epsilon_{1 3 2}u^{1 3}(0, \pm 1) \pm \epsilon_{1 3 2}u^{1 3}(0, \pm 1) \\ &= \mp \frac{\pm 2}{2} \frac{1}{\sqrt{2m}} \\ \implies u^{1 1}(0, \pm 2) &= -\frac{1}{2} \frac{1}{\sqrt{2m}} \end{split} \end{equation} and $i=2, j=2$ \begin{equation} \begin{split} 2 u^{2 2}(0, \pm 2) &= - i \epsilon_{2 3 1}u^{2 3}(0, \pm 1) - i \epsilon_{2 3 1}u^{2 3}(0, \pm 1) \pm \epsilon_{2 a 2}u^{2 a}(0, \pm 1) \pm \epsilon_{2 a 2}u^{2 a}(0, \pm 1) \\ &= -i \frac{2i}{2} \frac{1}{\sqrt{2m}} \\ \implies u^{1 1}(0, \pm 2) &= \frac{1}{2} \frac{1}{\sqrt{2m}} \end{split} \end{equation} and $i=3, j=3$ \begin{equation} \begin{split} 2 u^{3 3}(0, \pm 2) &= - i \epsilon_{3 2 1}u^{3 2}(0, \pm 1) - i \epsilon_{3 2 1}u^{3 2}(0, \pm 1) \pm \epsilon_{3 1 2}u^{3 1}(0, \pm 1) \pm \epsilon_{3 1 2}u^{3 1}(0, \pm 1) \\ &= \frac{2}{2} \frac{1}{\sqrt{2m}} (i^2 \pm(\pm 1)) = 0\\ \implies u^{3 3}(0, \pm 2) &= 0 \end{split} \end{equation} Next, the off-diagonal elements. For $i=1, j=2$ \begin{equation} \begin{split} 2 u^{1 2}(0, \pm 2) &= - i \epsilon_{2 3 1}u^{1 3}(0, \pm 1) - i \epsilon_{1 a 1}u^{2 a}(0, \pm 1) \pm \epsilon_{2 a 2}u^{1 a}(0, \pm 1) \pm \epsilon_{1 3 2}u^{2 3}(0, \pm 1) \\ &= \frac{1}{2} \frac{1}{\sqrt{2m}} (-i(\pm 1) \mp i) = \mp i \frac{1}{\sqrt{2m}}\\ \implies u^{1 2}(0, \pm 2) &= u^{2 1}(0, \pm2) = \mp \frac{i}{2} \frac{1}{\sqrt{2m}} \end{split} \end{equation} and $i=1, j=3$ \begin{equation} \begin{split} 2 u^{1 3}(0, \pm 2) &= - i \epsilon_{3 2 1}u^{1 2}(0, \pm 1) - i \epsilon_{1 a 1}u^{3 a}(0, \pm 1) \pm \epsilon_{3 1 2}u^{1 1}(0, \pm 1) \pm \epsilon_{1 3 2}u^{3 3}(0, \pm 1) \\ &= 0 \\ \implies u^{1 3}(0, \pm 2) &= u^{3 1}(0, \pm2) = 0 \end{split} \end{equation} and $i=2, j=3$ \begin{equation} \begin{split} 2 u^{2 3}(0, \pm 2) &= - i \epsilon_{3 2 1}u^{2 2}(0, \pm 1) - i \epsilon_{2 3 1}u^{3 3}(0, \pm 1) \pm \epsilon_{3 1 2}u^{2 1}(0, \pm 1) \pm \epsilon_{2 a 2}u^{3 a}(0, \pm 1) \\ &= 0 \\ \implies u^{2 3}(0, \pm 2) &= u^{3 2}(0, \pm2) = 0 \end{split} \end{equation} Now, since $v^{\mu \nu}(0,\pm 1)$ only differ from $u^{\mu \nu}(0,\mp 1)$ by a sign, and since the equations for $u$ and $v$ in \ref{eq:5-3jpmeqn} differ by a sign, the coefficients for $\sigma = \pm 2$ must not differ by a sign, and only differ by describing opposite spins, so \begin{equation} \boxed{u^{\mu \nu}(0,\pm 2) = v^{\mu \nu}(0,\mp 2) = \frac{1}{2} \frac{1}{\sqrt{2m}} \begin{bmatrix} -1 & \mp i & 0 & 0 \\ \mp i & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} } \end{equation} which one can check has the proper normalization $u^{\mu \nu}(0, \pm 2)u_{\mu \nu}(0, \pm 2) = 1/2m$. We should also note for all the coefficients, we have the property \begin{equation} u^{\mu \nu}(0, \sigma) = v^{\mu \nu*}(0, \sigma) \end{equation} At this point, we should stop and emphasize that our $u^{\mu \nu}(0, \sigma)$ span a vector space, specifically a five-dimensional complex vector space consisting of symmetric, traceless, 3x3 matrices (3x3 instead of 4x4 since all time-like components are 0). Let's define a new tensor $e^{\mu \nu}(0, \sigma)$ such that \begin{equation} u^{\mu \nu}(0, \sigma) \equiv \frac{1}{\sqrt{2m}} e^{\mu \nu}(0, \sigma) \end{equation} which due to our normalization of the $u^{\mu \nu}(0, \sigma)$, means that the $e^{\mu \nu}(0, \sigma)$ are normalized to $1$. We'll omit the proof, but one can also show that these tensors are orthogonal, so we have the relation \begin{equation} e^{\mu \nu}(0, \sigma) e^*_{\mu \nu}(0, \sigma') = \delta_{\sigma \sigma '} \end{equation} Also, any linear combination of these traceless, symmetric, 3x3 tensors is also a traceless, symmetric, 3x3 tensor. Let $a^{\mu \nu}$ be some linear combination of our $e^{\mu \nu}$: \begin{equation} a^{\mu \nu} = \sum_{\sigma} A_{\sigma} e^{\mu \nu}(0, \sigma) \end{equation} then we have \begin{equation} a^{\mu \nu} = \sum_{\sigma} A_{\sigma} e^{\mu \nu}(0, \sigma) = \sum_{\sigma} A_{\sigma} e^{\nu \mu}(0, \sigma) = a^{\nu \mu} \end{equation} \begin{equation} a\indices{^\mu _\mu} = \sum_{\sigma} A_{\sigma} e\indices{^\mu _\mu} = \sum_{\sigma} A_{\sigma} 0 = 0 \end{equation} \begin{equation} a^{\mu 0} = a^{0 \mu} = \sum_{\sigma} A_{\sigma} e^{\mu 0} = \sum_{\sigma} A_{\sigma} 0 = 0 \end{equation} Therefore, $e^{\mu \nu}(0, \sigma)$ form an orthonormal basis for our complex vector space of traceless, symmetric, 3x3 matrices. Turning next to the polynomial, let's rewrite this in terms of our new basis. \begin{equation} \begin{split} P^{\mu \nu, \lambda \kappa}(p) &= 2p^0 \sum_{\sigma} u^{\mu \nu}(\mathbf{p}, \sigma) u^{\lambda \kappa *}(\mathbf{p}, \sigma) \\ &= 2p^0 \frac{m}{p^0} L(p)\indices{^\mu _\alpha} L(p)\indices{^\nu _\beta} L(p)\indices{^\lambda _\gamma} L(p)\indices{^\kappa _\delta} \sum_{\sigma} u^{\alpha \beta}(0, \sigma) u^{\gamma \delta *}(0, \sigma) \\ &= 2m \frac{1}{2m} L(p)\indices{^\mu _\alpha} L(p)\indices{^\nu _\beta} L(p)\indices{^\lambda _\gamma} L(p)\indices{^\kappa _\delta} \sum_{\sigma} e^{\alpha \beta}(0, \sigma) e^{\gamma \delta *}(0, \sigma) \\ &= L(p)\indices{^\mu _\alpha} L(p)\indices{^\nu _\beta} L(p)\indices{^\lambda _\gamma} L(p)\indices{^\kappa _\delta} P^{\alpha \beta, \gamma \delta}(k) \end{split} \end{equation} with $k^{\mu} = (0, 0, 0, m)$. So, we just need to evaluate $P^{\mu \nu, \lambda, \kappa}(k)$, given by \begin{equation} P^{\mu \nu, \gamma \delta}(k) = \sum_{\sigma} e^{\alpha \beta}(0, \sigma) e^{\gamma \delta *}(0, \sigma) \end{equation} However, this is just a resolution of the identity for a vector space of traceless, symmetric, 3x3 matrices represented in the space of traceless, symmetric, 4x4 matrices. Therefore, our $P^{\mu \nu, \lambda \kappa}(k)$ is just the projection operator from traceless, symmetric, 4x4 matrices onto the space of traceless, symmetric, 3x3 matrices. To proceed, we just need to find an operator that behaves like this. It's useful to recall the projection operator for massive causal vector fields: \begin{equation} \Pi^{\mu \nu}(p) = \eta^{\mu \nu} + \frac{p^{\mu} p^{\nu}}{m^2} \end{equation} which is symmetric \begin{equation} \Pi^{\mu \nu}(p) = \eta^{\mu \nu} + \frac{p^{\mu} p^{\nu}}{m^2} = \eta^{\nu \mu} + \frac{p^{\nu} p^{\mu}}{m^2} = \Pi^{\nu \mu}(p) \end{equation} and idemptotent \begin{equation} \begin{split} \Pi^{\mu \alpha}(p) \Pi\indices{_\alpha ^\nu}(p) &= \left(\eta^{\mu \alpha} + \frac{p^{\mu} p^{\alpha}}{m^2}\right)\left(\delta\indices{^\nu _\alpha} + \frac{p_{\alpha} p^{\nu}}{m^2}\right) \\ &= \eta^{\mu \nu} + \frac{p^{\mu} p^{\nu}}{m^2} + \frac{p^{\mu} p^{\nu}}{m^2} + \frac{p^{\alpha} p_{\alpha} p^{\mu} p^{\nu}}{m^4} \\ &= \eta^{\mu \nu} + 2\frac{p^{\mu} p^{\nu}}{m^2} - \frac{p^{\mu} p^{\nu}}{m^2} \\ &= \eta^{\mu \nu} + \frac{p^{\mu} p^{\nu}}{m^2} \\ &= \Pi^{\mu \nu}(p) \end{split} \end{equation} When evaluated at $p = k$, the projection operator obeys the following properties: \begin{equation} \begin{split} &\Pi^{i 0}(k) = \Pi^{0 i}(k) = \eta^{0 i} + \frac{k^0 k^i}{m^2} = 0 + \frac{m \cdot 0}{m^2} = 0 \\ &\Pi^{0 0}(k) = \eta^{0 0} + \frac{k^0 k^0}{m^2} = -1 + 1 = 0 \end{split} \end{equation} or concisely \begin{equation} \Pi^{0 \mu}(k) = \Pi^{\mu 0}(k) = 0 \label{eq:5-3notimelike} \end{equation} Equation \ref{eq:5-3notimelike} implies that when we contract two $\Pi^{\mu \nu}(k)$ with any 4x4 tensor $a^{\mu \nu}$, we project out the timelike components, or \begin{equation} \bar{a}^{\mu \nu} = \Pi\indices{^\mu _\alpha} \Pi\indices{^\nu _\beta} a^{\alpha \beta} \implies \bar{a}^{0 \nu} = \Pi\indices{^0 _\alpha} \Pi\indices{^\nu _\beta} a^{\alpha \beta} = 0, \quad \bar{a}^{\mu 0} = \Pi\indices{^\mu _\alpha} \Pi\indices{^0 _\beta} a^{\alpha \beta} = 0 \end{equation} So, to construct our $P^{\mu \nu, \lambda \kappa}(k)$, we want to first operate with two projection operators to project out the timelike componenets, then we need to ensure that our result is traceless. To do this, recall that the operator that transforms a 3x3 matrix into a traceless 3x3 matrix is \begin{equation} T^{i j, l k} = \delta^{i l} \delta^{j k} - \frac{1}{3} \delta^{i j} \delta^{l k} \end{equation} which when we promote it to a rank-2 tensor on Minkowski space is \begin{equation} T^{\mu \nu, \lambda, \kappa} = \eta^{\mu \lambda} \eta^{\nu \kappa} - \frac{1}{3} \eta^{\mu \nu} \eta^{\lambda \kappa} \end{equation} Technically, we should change the fraction to $1/4$, but we'll see soon we need to keep the fraction like this since our final product is a 3x3 matrix. Let's call our combined operator so far $\tilde{P}^{\mu \nu, \kappa \lambda}$, so \begin{equation} \begin{split} \tilde{P}^{\mu \nu, \kappa \lambda} &= \left(\delta\indices{^\mu _\alpha} \delta\indices{^\nu _\beta} - \frac{1}{3} \eta^{\mu \nu} \eta_{\alpha \beta}\right)\Pi^{\alpha \lambda}(k) \Pi^{\beta \kappa}(k) \\ &= \Pi^{\mu \lambda}(k) \Pi^{\nu \kappa}(k) - \frac{1}{3} \eta^{\mu \nu} \Pi^{\alpha \lambda}(k) \Pi\indices{_\alpha ^\kappa}(k) \\ &= \Pi^{\mu \lambda}(k) \Pi^{\nu \kappa}(k) - \frac{1}{3} \eta^{\mu \nu} \Pi^{\lambda \kappa}(k) \end{split} \end{equation} which one can verify acts on a tensor to produce something traceless (if we change the fraction to $1/4$). However, there is a bigger issue, regardless of the fraction, which is when we subtract off the part that makes it traceless, we are making the $(0,0)$ component of the resulting tensor non-zero, i.e. \begin{equation} \tilde{a}^{0 0} = \tilde{P}\indices{^0 ^0 _\kappa _\lambda} a^{\kappa \lambda} = -\frac{1}{3} \eta^{0 0} \Pi_{\kappa \lambda}(k) a^{\kappa \lambda} = \frac{1}{3} \Pi_{i j}(k) a^{i j} \neq 0 \end{equation} So, for our full operator, we need to then project out the time-like components again, yielding a full operator \begin{equation} \begin{split} P^{\mu \nu, \lambda \kappa}(k) &= \Pi^{\mu \alpha}(k) \Pi^{\nu \beta}(k) \tilde{P}\indices{_\alpha _\beta ^\lambda ^\kappa} \\ &= \Pi^{\mu \alpha}(k) \Pi^{\nu \beta}(k) \left(\Pi\indices{_\alpha ^\lambda}(k) \Pi\indices{_\beta ^\kappa}(k) - \frac{1}{3} \eta_{\alpha \beta} \Pi^{\lambda \kappa}(k)\right) \\ &= \Pi^{\mu \alpha}(k) \Pi\indices{_\alpha ^\lambda}(k) \Pi^{\nu \beta}(k) \Pi\indices{_\beta ^\kappa}(k) - \frac{1}{3} \Pi^{\mu \alpha}(k) \Pi\indices{_\alpha ^\nu}(k) \\ &= \Pi^{\mu \lambda}(k) \Pi^{\nu \kappa}(k) - \frac{1}{3} \Pi^{\mu \nu}(k) \Pi^{\lambda \kappa}(k) \end{split} \end{equation} which clearly satisfies the no-timelike component condition since it only consists of $\Pi^{\mu \nu}(k)$ operators and also satisfies the traceless condition: \begin{equation} \begin{split} \bar{a}\indices{^\mu _\mu} &= \Pi^{\mu \lambda}(k) \Pi\indices{_\mu ^\kappa}(k)a_{\lambda \kappa} - \frac{1}{3} \Pi\indices{^\mu _\mu}(k) \Pi^{\lambda \kappa}(k) a_{\lambda \kappa} \\ &= \Pi^{\lambda \kappa} a_{\lambda \kappa} - \Pi^{\lambda \kappa} a_{\lambda \kappa} = 0 \end{split} \end{equation} where we used the fact that \begin{equation} \Pi\indices{^\mu _\mu}(k) = \delta\indices{^\mu _\mu} + \frac{k^{\mu} k_{\mu}}{m^2} = 4 - 1 = 3 \end{equation} To get our operator at arbitrary momenta, we just need to boost the operator, and since the $\Pi$ operators obey the condition \begin{equation} \Pi^{\mu \nu}(p) = L(p)\indices{^\mu _\alpha} L(p)\indices{^\nu _\beta} \Pi^{\alpha \beta}(k) \end{equation} we are left with \begin{equation} \boxed{P^{\mu \nu, \lambda \kappa}(p) = \Pi^{\mu \lambda}(p) \Pi^{\nu \kappa}(p) - \frac{1}{3} \Pi^{\mu \nu}(p) \Pi^{\lambda \kappa}(p)} \end{equation} Next, we want to show that these spin-2 massive fields annihilate and create bosons. Let's first find the (anti-)commutator for this field. As a reminder, our annihilation and creation fields are given by \begin{equation} \begin{split} h^{+\mu \nu}(x) &= \sum_{\sigma} \int \frac{d^3 p}{(2\pi)^{3/2}} u^{\mu \nu}(\mathbf{p}, \sigma) e^{i p \cdot x} a(\mathbf{p}, \sigma) \\ h^{-\mu \nu}(x) &= \sum_{\sigma} \int \frac{d^3 p}{(2\pi)^{3/2}} v^{\mu \nu}(\mathbf{p}, \sigma) e^{-i p \cdot x} a^{c \dagger}(\mathbf{p}, \sigma) \end{split} \end{equation} where we are leaving it open whether this particle is its own antiparticle. The (anti-)commutator of these two fields is: \begin{equation} \begin{split} [h^{+\mu \nu}(x), h^{-\lambda \kappa}(y)]_{\mp} &= \sum_{\sigma, \sigma'}\int \frac{d^3p d^3p'}{(2\pi)^3} u^{\mu \nu}(\mathbf{p}, \sigma) v^{\rho \lambda}(\mathbf{p}', \sigma') e^{i p \cdot x - i p' \cdot y} [a(\mathbf{p}, \sigma), a(\mathbf{p}', \sigma')]_{\mp} \\ &= \sum_{\sigma, \sigma'}\int \frac{d^3p d^3p'}{(2\pi)^3} u^{\mu \nu}(\mathbf{p}, \sigma) u^{\rho \lambda*}(\mathbf{p}', \sigma') e^{i p \cdot x - i p' \cdot y} \delta_{\sigma \sigma'} \delta^3(\mathbf{p}' - \mathbf{p}) \\ &= \int \frac{d^3 p}{(2\pi)^3}e^{i p \cdot (x - y)} \frac{1}{2p^0} P^{\mu \nu, \lambda \kappa}(p) \\ &= \left(\eta^{\mu \lambda} - \frac{\partial^{\mu} \partial^{\lambda}}{m^2}\right) \end{split} \end{equation} \section{Chapter 6} \subsection{Problem 6.1} Calculate the connected S-matrix element for scalar-scalar scattering to second order in g for the interaction $V=g \int d^3x \phi(x)^3/3!$. \subsubsection{Feynman Diagrams} To second order in g, the S matrix includes the following Feynman diagrams: \begin{figure}[h] \centering \begin{tikzpicture}[baseline=(current bounding box.center)] \begin{feynman} % Central vertices (bottom and top) \vertex (v1); \vertex[above=2cm of v1] (v2); % Incoming particles (from the bottom) \vertex[below left=0.8cm and 1cm of v1] (i1) {\(\phi\)}; \vertex[below right=0.8cm and 1cm of v1] (i2) {\(\phi\)}; % Outgoing particles (to the top) \vertex[above left=0.8cm and 1cm of v2] (f1) {\(\phi\)}; \vertex[above right=0.8cm and 1cm of v2] (f2) {\(\phi\)}; % Draw the diagram \diagram* { (i1) -- [scalar, momentum=\(p\)] (v1), (i2) -- [scalar,momentum'=\(k\)] (v1), (v1) -- [scalar, edge label=\(\phi\),momentum'=\(q\)] (v2), (v2) -- [scalar,momentum=\(p'\)] (f1), (v2) -- [scalar,momentum'=\(k'\)] (f2), }; \end{feynman} \end{tikzpicture} + \begin{tikzpicture}[baseline=(current bounding box.center)] \begin{feynman} % Central vertices for the propagator \vertex (v1); \vertex[right=2cm of v1] (v2); % Incoming and outgoing particles \vertex[above left=1cm and 0.8cm of v1] (f1) {\(\phi\)}; \vertex[below left=1cm and 0.8cm of v1] (i1) {\(\phi\)}; \vertex[above right=1cm and 0.8cm of v2] (f2) {\(\phi\)}; \vertex[below right=1cm and 0.8cm of v2] (i2) {\(\phi\)}; % Draw the diagram connecting the vertices \diagram* { (i1) -- [scalar,momentum=\(p\)] (v1) -- [scalar,momentum'=\(p'\)] (f1), (v1) -- [scalar, edge label'=\(\phi\),momentum=\(q\)] (v2), (i2)--[scalar,momentum=\(k\)](v2) -- [scalar,momentum=\(k'\)](f2) }; \end{feynman} \end{tikzpicture} + \begin{tikzpicture}[baseline=(current bounding box.center)] \begin{feynman} % Central vertices for the propagator \vertex (v1); \vertex[right=2cm of v1] (v2); % Incoming and outgoing particles \vertex[above left=1cm and 0.8cm of v1] (f1) {\(\phi\)}; \vertex[below left=1cm and 0.8cm of v1] (i1) {\(\phi\)}; \vertex[above right=1cm and 0.8cm of v2] (f2) {\(\phi\)}; \vertex[below right=1cm and 0.8cm of v2] (i2) {\(\phi\)}; % Draw the diagram connecting the vertices \diagram* { (i1) -- [scalar,momentum=\(p\)] (v1) -- [scalar,momentum'=\(k'\)] (f1), (v1) -- [scalar, edge label'=\(\phi\),momentum=\(q\)] (v2), (i2)--[scalar,momentum=\(k\)](v2) -- [scalar,momentum=\(p'\)](f2) }; \end{feynman} \end{tikzpicture} \end{figure} \subsubsection{S-matrix} The contribution to the S-matrix from the first diagram above is \begin{gather} \int d^4 q (2 \pi)^{-6} (16 p^0k^0p'^0k'^0)^{-1/2}(-ig (2\pi)^4 \delta^4(q-p-k)) \frac{-i (2\pi)^{-4}}{q^2+m^2-i\epsilon} (-ig (2\pi)^4 \delta^4(q-p'-k'))\\ (2 \pi)^{-2}(16 p^0k^0p'^0k'^0)^{-1/2} \delta^4(p+k-p'-k') \frac{i g^2}{(p+k)^2+m^2} \end{gather} The contributions from the second and third diagrams are extremely similar, but with just different combinations of momenta in the original delta functions. Therefore, the S-matrix to second order in $g$ is \begin{gather} \begin{split} S_{p',k';p,k} = (2 \pi)^{-2}(16 p^0&k^0p'^0k'^0)^{-1/2} \delta^4(p+k-p'-k') ig^2 \\ &\left( \frac{1}{(p+k)^2+m^2} +\frac{1}{(p-p')^2+m^2}+\frac{1}{(p-k')^2+m^2} \right) \end{split} \end{gather} This expression can be simplified by using the Mandelstam variables: \begin{gather} s=-(p+k)^2\\ t=-(p'-p)^2\\ u=-(k'-p)^2 \end{gather} With these variables, the total S-matrix becomes \begin{gather} S_{p',k';p,k} = -(2 \pi)^{-2} \delta^4(p+k-p'-k') ig^2 \frac{1}{s}\left( \frac{1}{s-m^2} +\frac{1}{t-m^2}+\frac{1}{u-m^2} \right) \end{gather} We can now write the amplitude $M_{p',k';p,k}$ \begin{gather} M_{p',k';p,k} = (2 \pi)^{-3} g^2 \frac{1}{s}\left( \frac{1}{s-m^2} +\frac{1}{t-m^2}+\frac{1}{u-m^2} \right) \end{gather} \subsubsection{Cross-Section} Weinberg gives the differential scattering cross section in terms of the amplitude $M_{p',k';p,k}$ as \begin{equation} \frac{d\sigma}{d\Omega} = \frac{(2\pi)^4 |\mathbf{p}'| p^0 k^0 p'^0 k'^0}{E_{\text{tot}}^2 |\mathbf{p}|}|M_{p',k';p,k}|^2 \end{equation} Therefore, the differential scattering cross section for this interaction to second order in $g$ is \begin{gather} \frac{d\sigma}{d\Omega} = \frac{g^4}{64 \pi^2s}\left( \frac{1}{s-m^2} +\frac{1}{t-m^2}+\frac{1}{u-m^2} \right)^2 \end{gather} \subsection{Problem 6.2} This problem asks for various $S$-matrix elements for a pseudoscalar Yukawa interaction. We will go somewhat further and provide cross-sections for specific scattering processes, both for the pseudoscalar Yukawa interaction and also the scalar Yukawa interaction, just to contrast the form of the cross sections. \subsubsection{Preliminaries} Weinberg gives the spin sums in a somewhat less helpful form, since we normally deal with Dirac adjoints ($\bar{u} = u^{\dagger} \beta$) rather than just normal field adjoints in interaction densities. The spin sums for Dirac adjoints is: \begin{equation} \begin{split} \sum_{\sigma} u_a(p, \sigma) \bar{u}_b (p, \sigma) &= \beta_{c b} \sum_{\sigma} u_a(p, \sigma) u^{\dagger}_c (p, \sigma) \\ &= \frac{1}{2p^0}([-i\slashed{p} + m]\beta)_{ac} \beta_{c b}\\ &= \frac{(-i\slashed{p} + m)_{ab}}{2p^0} \end{split} \end{equation} From here on, we will refer to these spin sums as $\bar{N}(\mathbf{p})$ just to distinguish them from what Weinberg calls $N(\mathbf{p})$. So, we have \begin{equation} \begin{split} \bar{N}(\mathbf{p})_{ab} &= \sum_{\sigma} u_a(p, \sigma) \bar{u}_b (p, \sigma) = \frac{(-i\slashed{p} + m)_{ab}}{2p^0} \\ \bar{M}(\mathbf{p})_{ab} &= \sum_{\sigma} v_a(p, \sigma) \bar{v}_b (p, \sigma) = \frac{(-i\slashed{p} - m)_{ab}}{2p^0} \end{split} \end{equation} We should also find what the complex conjugates of Dirac bilinears are. For a scalar bilinear, composed of $u$ and/or $v$, we have \begin{equation} \begin{aligned} (\bar{u} u)^* &= u^{\dagger} \bar{u}^{\dagger} = u^{\dagger} (u^{\dagger} \beta)^{\dagger} \\ &= u^{\dagger} \beta^{\dagger} u = u^{\dagger} \beta u \\ &= \bar{u}u \end{aligned} \end{equation} For a pseudoscalar bilinear, we have \begin{equation} \begin{aligned} (\bar{u} \gamma_5 u)^* &= u^{\dagger} (u^{\dagger} \beta \gamma_5)^{\dagger} = u^{\dagger} \gamma_5^{\dagger} \beta^{\dagger} u \\ &= u^{\dagger} \gamma_5^{\dagger} \beta u = u^{\dagger} \beta \beta \gamma_5^{\dagger} \beta u \\ &= - u^{\dagger} \beta \gamma_5 u = - \bar{u} \gamma_5 u \end{aligned} \end{equation} \subsubsection{Pseudoscalar Yukawa Interaction} We first consider Bhabha scattering, that is $F + F^c \rightarrow F + F^c$. Since the initial and final particles are distinguishable, we only get $s$- and $t$-channel contributions: \begin{figure}[h] \centering \begin{subfigure}[b]{0.3\textwidth} \centering \feynmandiagram [vertical'= b to a, inline=(b.base)] { i1 [particle=\(p\)] -- [fermion] a -- [fermion] i2 [particle=\(k\)], a -- [scalar] b, f1 [particle=\(p'\)] -- [anti fermion] b -- [anti fermion] f2 [particle=\(k'\)], }; \caption{$s$-channel} \end{subfigure} \begin{subfigure}[b]{0.3\textwidth} \centering \feynmandiagram [horizontal= a to b, inline=(a.base)] { i1 [particle=\(p\)] -- [fermion] a -- [fermion] f1 [particle=\(p'\)], a -- [scalar] b, i2 [particle=\(k\)] -- [fermion] b -- [fermion] f2 [particle=\(k'\)], }; \caption{$t$-channel} \end{subfigure} \end{figure} Using the momentum space rules is not too hard, but we should point out that since we are dealing with fermions, we need to use Weinberg's rule (vi) from chapter 6.1. That is, we need to keep track of the additional minus signs that aries from permuting fermion creation/annihilation operators. We can write out the fermion operators as \begin{equation} a(2'^c)a(1') \bar{\psi}(x) \psi(x) \bar{\psi}(y) \psi(y) a^{\dagger}(1) a^{\dagger}(2^c) \end{equation} which has the following two contractions, corresponding to the $s$- and $t$-channels, respectively: \begin{equation} \wick{ \c1 a(2'^c) \c2{a(1')} \c2{\bar{\psi}}(x) \c1 \psi(x) \c3{\bar{\psi}}(y) \c4 \psi(y) \c4{a^{\dagger}(1)} \c3{a^{\dagger}(2^c)} } \end{equation} and \begin{equation} \wick{ \c1 a(2'^c) \c2{a(1')} \c2{\bar{\psi}}(x) \c4 \psi(x) \c3{\bar{\psi}}(y) \c1 \psi(y) \c4{a^{\dagger}(1)} \c3{a^{\dagger}(2^c)} } \end{equation} to get all the annihilation operators in each pair to the left of the creation operator in each pair thus requires 4 permutations for the $s$-channel and 5 permuations for the $t$-channel. Therefore, the $S$-matrix element for the $t$-channel will have an extra factor of $-1$ due to the odd number of permutations. Now, let's define the mass of the fermions as $m$ and the mass of the bosons as $\mu$. The spins of the incoming fermion and antifermion are labeled $\sigma$ and $\sigma_c$, and the outgoing spins are the same but primed. Then, the full $S$-matrix element, in terms of the Mandelstam variables, is then: \begin{equation} \boxed{\begin{aligned} S_{p' \sigma', k' \sigma_c'; \;p \sigma, k \sigma_c} &= -2\pi i \delta^4(p'+k'-p-k)\frac{g^2}{(2\pi)^3}\left(-\frac{1}{s-\mu^2}\bar{v}(\mathbf{k}, \sigma_c)\gamma_5u(\mathbf{p}, \sigma) \bar{u}(\mathbf{p}', \sigma')\gamma_5 v(\mathbf{k}', \sigma'_c)\right. \\ & \left.\quad + \frac{1}{t-\mu^2}\bar{u}(\mathbf{p}', \sigma')\gamma_5u(\mathbf{p}, \sigma)\bar{v}(\mathbf{k}, \sigma_c)\gamma_5v(\mathbf{k}', \sigma'_c)\right) \\ &\equiv -2\pi i \delta^4(p'+k'-p-k) (\mathcal{M}_s + \mathcal{M}_t) \end{aligned}} \end{equation} Let's now try to compute the spin-averaged cross section, that is \begin{equation} \frac{d\bar{\sigma}}{d\Omega} \equiv \frac{1}{4}\sum_{\text{spins}} \frac{d\sigma}{d\Omega} = \frac{1}{4}\sum_{\text{spins}} \frac{(2\pi)^4 |\mathbf{p}'_1| E_1 E_2 E'_1 E'_2}{|\mathbf{p}_1| E^2_{\text{tot}}} |\mathcal{M}|^2 \equiv \frac{(2\pi)^4 |\mathbf{p}'_1| E_1 E_2 E'_1 E'_2}{|\mathbf{p}_1| E^2_{\text{tot}}} \overline{|\mathcal{M}|^2} \end{equation} where the bar denotes a sum over spins and the proper averaging factor. We also know that since the mass is the same for all of these particles, in the center-of-mass frame, we have \begin{equation} \mathbf{p} = -\mathbf{k}, \quad \mathbf{p}' = - \mathbf{k}', |\mathbf{p}| = |\mathbf{p}'| \implies p^0 = k^0 = p'^0 = k'^0 \equiv E \end{equation} and our cross section simply reduces to \begin{equation} \frac{d\bar{\sigma}}{d\Omega} = \frac{(2\pi)^4 s}{16} \overline{|\mathcal{M}|^2} \end{equation} For now, we just need to calculate the spin-averaged matrix element: \begin{equation} \overline{|\mathcal{M}|^2} = \overline{|\mathcal{M}_s|^2} + \overline{|\mathcal{M}_t|^2} + 2 \Re[\overline{\mathcal{M}_s \mathcal{M}_t^*}] \end{equation} The first term is \begin{equation} \begin{aligned} \overline{|\mathcal{M}_s|^2} &= \frac{1}{4} \sum_{\text{spins} }\frac{g^4}{(2\pi)^6}\frac{1}{(s-\mu^2)^2} \bar{v}(\mathbf{k}, \sigma_c)\gamma_5u(\mathbf{p}, \sigma) \bar{u}(\mathbf{p}', \sigma')\gamma_5 v(\mathbf{k}', \sigma'_c) \\ &\quad\quad \cross \bar{u}(\mathbf{p},\sigma)\gamma_5 v(\mathbf{k}, \sigma_c) \bar{v}(\mathbf{k'},\sigma'_c) \gamma_5 u(\mathbf{p}', \sigma') \\ &= \frac{1}{4} \sum_{\text{spins} }\frac{g^4}{(2\pi)^6}\frac{1}{(s-\mu^2)^2} \bar{v}_a(\mathbf{k}, \sigma_c)(\gamma_5)_{ab}u_b(\mathbf{p}, \sigma) \bar{u}_c(\mathbf{p}', \sigma')(\gamma_5)_{cd} v_d(\mathbf{k}', \sigma'_c) \\ &\quad\quad \cross \bar{u}_e(\mathbf{p},\sigma)(\gamma_5)_{ef} v_f(\mathbf{k}, \sigma_c) \bar{v}_g(\mathbf{k'},\sigma'_c) (\gamma_5)_{gh} u_h(\mathbf{p}', \sigma') \\ &= \frac{1}{4} \frac{g^4}{(2\pi)^6}\frac{1}{(s-\mu^2)^2} \overline{N}(\mathbf{p})_{be} (\gamma_5)_{ef} \overline{M}(\mathbf{k})_{fa} (\gamma_5)_{ab} \overline{N}(\mathbf{p}')_{hc} (\gamma_5)_{cd} \overline{M}(\mathbf{k}')_{dg} (\gamma_5)_{gh} \\ &= \frac{1}{4} \frac{g^4}{(2\pi)^6}\frac{1}{(s-\mu^2)^2} \Tr(\overline{N}(\mathbf{p}) (\gamma_5) \overline{M}(\mathbf{k}) (\gamma_5)) \Tr(\overline{N}(\mathbf{p}') (\gamma_5) \overline{M}(\mathbf{k}') (\gamma_5)) \end{aligned} \end{equation} We can easily do these traces using well-known $\gamma$-matrix trace identities: \begin{equation} \begin{aligned} \Tr(\overline{N}(\mathbf{p}) (\gamma_5) \overline{M}(\mathbf{k}) (\gamma_5)) &= \frac{1}{4p^0 k^0} \Tr[(-i p_{\mu} \gamma^{\mu} + m)\gamma_5(-ik_{\nu}\gamma^{\nu}-m)\gamma_5] \\ &= \frac{1}{4p^0 k^0} \Tr[(-i p_{\mu} \gamma^{\mu} + m)(-ik_{\nu}\gamma_5\gamma^{\nu}\gamma_5-m)] \\ &= \frac{1}{4p^0 k^0} \Tr[(-i p_{\mu} \gamma^{\mu} + m)(+ik_{\nu}\gamma^{\nu}-m)] \\ &= \frac{1}{4p^0 k^0} \Tr[p_{\mu}k_{\nu} \gamma^{\mu} \gamma^{\nu} -m^2] \\ &= \frac{1}{4p^0 k^0} (4 p \cdot k - 4 m^2) = \frac{p \cdot k - m^2}{p^0 k^0} \\ \end{aligned} \end{equation} We can get this in a nicer form by considering: \begin{equation} \begin{aligned} (p+k)^2 &= -s = p^2 + k^2 +2p \cdot k = -2m^2 + 2p \cdot k \\ \implies p \cdot k &= -\frac{s}{2} + m^2 \end{aligned} \end{equation} so, \begin{equation} \Tr(\overline{N}(\mathbf{p}) (\gamma_5) \overline{M}(\mathbf{k}) (\gamma_5)) = -\frac{s}{2p^0 k^0 } \end{equation} The full spin-averaged, squared $s$-channel matrix element is then \begin{equation} \begin{aligned} \overline{|\mathcal{M}_s|^2} &= \frac{1}{4} \frac{g^4}{(2\pi)^6}\frac{1}{(s-\mu^2)^2} \frac{s^2}{4p^0k^0p'^0k'^0} \\ &= \frac{g^4}{(2\pi)^6}\frac{1}{(s-\mu^2)^2}\frac{s^2}{16E^4} \\ &= \frac{g^4}{(2\pi)^6}\frac{1}{(s-\mu^2)^2} \end{aligned} \end{equation} The $t$-channel spin-averaged matrix element squared is \begin{equation} \overline{|\mathcal{M}_t|^2} = \frac{1}{4}\frac{g^4}{(2\pi)^6}\frac{1}{(t-\mu^2)^2} \Tr(\overline{N}(\mathbf{p})\gamma_5\overline{N}(\mathbf{p}')\gamma_5) \Tr(\overline{M}(\mathbf{k})\gamma_5\overline{M}(\mathbf{k}')\gamma_5) \end{equation} and the two traces evaluate to \begin{equation} \begin{aligned} \Tr(\overline{N}(\mathbf{p})\gamma_5\overline{N}(\mathbf{p}')\gamma_5) &= \frac{t}{2p^0 p'^0}\\ \Tr(\overline{M}(\mathbf{k})\gamma_5\overline{M}(\mathbf{k}')\gamma_5) &= \frac{t}{2k^0 k'^0} \end{aligned} \end{equation} which reduces the squared matrix element to \begin{equation} \overline{|\mathcal{M}_t|^2} = \frac{g^4}{(2\pi)^6}\frac{1}{(t-\mu^2)^2} \frac{t^2}{s^2} \end{equation} The interference term looks like \begin{equation} \begin{aligned} \overline{\mathcal{M}_s \mathcal{M}_t^*} &= -\frac{1}{4}\frac{g^4}{(2\pi)^6}\frac{1}{(s-\mu^2)(t-\mu^2)} \Tr[\overline{N}(\mathbf{p}) \gamma_5 \overline{N}(\mathbf{p}') \gamma_5 \overline{M}(\mathbf{k}') \gamma_5 \overline{M}(\mathbf{k}) \gamma_5] \end{aligned} \end{equation} and the trace is pretty complex \begin{equation} \begin{aligned} \Tr[...] &= \frac{1}{s^2}\Tr\left[(p_{\mu}p'_{\nu} \gamma^{\mu} \gamma^{\nu} + i m p'_{\nu} \gamma^{\nu} - i m p_{\mu} \gamma^{\mu} - m^2)\right. \\ & \left.\quad \quad \cross (k'_{\rho} k_{\sigma} \gamma^{\rho} \gamma^{\sigma} + i m k'_{\rho} \gamma^{\rho} - i m k_{\sigma} \gamma^{\sigma} + m^2)\right] \\ &= \frac{1}{s^2}\Tr\left[p_{\mu} p'_{\nu} k'_{\rho} k_{\sigma} \gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma} + m^2 p_{\mu} p'_{\nu} \gamma^{\mu} \gamma^{\nu} - m^2 p'_{\nu} k'_{\rho} \gamma^{\nu} \gamma^{\rho} + m^2 p'_{\nu} k_{\sigma} \gamma^{\nu} \gamma^{\sigma}\right. \\ & \left. \quad \quad + m^2 p_{\mu} k'_{\rho} \gamma^{\mu} \gamma^{\rho} - m^2 p_{\mu} k_{\sigma} \gamma^{\mu} \gamma^{\sigma} + m^2 k'_{\rho} k_{\sigma} \gamma^{\rho} \gamma^{\sigma} + m^4\right] \\ &= \frac{4}{s^2}\left(\left(\frac{s}{2} -m^2\right)^2 + \left(\frac{t}{2} -m^2\right)^2 - \left(\frac{u}{2} -m^2\right)^2 + m^2(s + t + u - 6m^2) + m^4\right) \\ &= \frac{4}{s^2}\left(\left(\frac{s}{2} -m^2\right)^2 + \left(\frac{t}{2} -m^2\right)^2 - \left(\frac{u}{2} -m^2\right)^2 - m^4\right) \\ &= \frac{1}{s^2}\left((s-2m^2)^2 + (t-2m^2)^2 -(u-2m^2)^2 -4m^4 \right) \end{aligned} \end{equation} Therefore, the full interference term is \begin{equation} 2 \Re[\overline{\mathcal{M}_s \mathcal{M}_t^*}] = -\frac{1}{2}\frac{g^4}{(2\pi)^6} \frac{1}{s^2(s-\mu^2)(t-\mu^2)}\left((s-2m^2)^2 + (t-2m^2)^2 -(u-2m^2)^2 -4m^4 \right) \end{equation} and the full Bhabha scattering differential cross section, after simplifying, is \begin{equation} \boxed{\begin{aligned} \frac{d\bar{\sigma}}{d\Omega} = \frac{g^4}{64 \pi^2} \frac{1}{s} \left[\frac{s^2}{(s-\mu^2)^2} + \frac{t^2}{(t-\mu^2)^2} - \frac{s^2 + t^2 - u^2 + 8m^2(u-2m^2)}{2(s-\mu^2)(t-\mu^2)}\right] \end{aligned}} \end{equation} \begin{figure}[h] \centering \begin{subfigure}[b]{0.3\textwidth} \centering \feynmandiagram [vertical'= b to a, inline=(b.base)] { i1 [particle=\(p\)] -- [scalar] a -- [fermion] i2 [particle=\(k\)], a -- [anti fermion] b, f1 [particle=\(p'\)] -- [scalar] b -- [anti fermion] f2 [particle=\(k'\)], }; \caption{$s$-channel} \end{subfigure} \begin{subfigure}[b]{0.3\textwidth} \centering \feynmandiagram [horizontal= a to b, inline=(a.base)] { i1 [particle=\(p\)] -- [scalar] a -- [anti fermion] f1 [particle=\(k'\)], a -- [fermion] b, f2 [particle=\(k\)] -- [anti fermion] b -- [scalar] i2 [particle=\(p'\)], }; \caption{$u$-channel} \end{subfigure} \caption{$B + F^c \rightarrow B + F^c$} \label{fig:6-3_compton} \end{figure} We will now not concern ourselves with the cross sections, and just compute the $S$-matrix elements. Figure \ref{fig:6-3_compton} shows the two diagrams that contribute to $B + F^c \rightarrow B + F^c$ (antiparticle Compton scattering). Both the $s$- and $u$- diagrams require moving the fermion operators past one another an odd number of times, so both get an additional phase of a $-$ sign. The corresponding $S$-matrix element is: \begin{equation} \boxed{\begin{aligned} S_{p', k' \sigma_c'; \;p, k \sigma_c} &= -2\pi i \delta^4(p'+k'-p-k)\frac{g^2}{(2\pi)^3}\frac{1}{2\sqrt{p^0 p'^0}} \left(\frac{\bar{v}(\mathbf{k}, \sigma_c)\gamma_5 (+i(\slashed{p} + \slashed{k}) + m) \gamma_5 v(\mathbf{k'}, \sigma_c')}{s-m^2} \right. \\ & \left.\quad + \frac{\bar{v}(\mathbf{k}, \sigma_c)\gamma_5 (-i(\slashed{p} - \slashed{k}') + m) \gamma_5 v(\mathbf{k'}, \sigma_c')}{u-m^2} \right) \end{aligned}} \end{equation} \begin{figure}[h] \centering \begin{subfigure}[b]{0.3\textwidth} \centering \feynmandiagram [horizontal= a to b, inline=(a.base)] { i1 [particle=\(p\)] -- [fermion] a -- [scalar] f1 [particle=\(p'\)], a -- [fermion] b, f2 [particle=\(k\)] -- [anti fermion] b -- [scalar] i2 [particle=\(k'\)], }; \caption{$t$-channel} \end{subfigure} \begin{subfigure}[b]{0.3\textwidth} \centering \feynmandiagram [horizontal= a to b, inline=(a.base)] { i1 [particle=\(p\)] -- [fermion] a -- [scalar] f1 [particle=\(k'\)], a -- [fermion] b, f2 [particle=\(k\)] -- [anti fermion] b -- [scalar] i2 [particle=\(p'\)], }; \caption{$u$-channel} \end{subfigure} \caption{$F + F^c \rightarrow B + B$} \label{fig:6-3_annihilation} \end{figure} The process $F + F^c \rightarrow B + B$, fermion annihilation, has two contributing diagrams, seen in Figure \ref{fig:6-3_annihilation}. Both the $t$- and $u$-channel diagrams require moving fermion operators past each other an even number of times, so there is no additional phase for either. The $S$-matrix element is: \begin{equation} \boxed{ \begin{aligned} S_{p', k' ; p \sigma, p \sigma_c} &= -2 \pi i \delta^4(p' + k' - p - k) \left(-\frac{g^2}{(2\pi)^3}\frac{1}{2\sqrt{p'^0 k'^0}} \right)\left( \frac{\overline{v}(\mathbf{k}, \sigma_c) \gamma_5 (-i (\slashed{p} - \slashed{p}') + m) \gamma_5 u(\mathbf{p}, \sigma)}{t-m^2} \right. \\ &\quad + \left.\frac{\overline{v}(\mathbf{k}, \sigma_c) \gamma_5 (-i (\slashed{p} - \slashed{k}') + m) \gamma_5 u(\mathbf{p}, \sigma)}{u-m^2} \right) \end{aligned} } \end{equation} \subsubsection{Scalar Yukawa Interaction} Let's now compare the previous results to Bhabha scattering for a scalar Yukawa interaction. The interaction density is now given by \begin{equation} \mathcal{H}(x) = g \bar{\psi}(x)\psi(x) \phi(x) \end{equation} The same diagrams contribute as before, giving an $S$-matrix element that looks like\ \begin{equation} \boxed{\begin{aligned} S_{p' \sigma', k' \sigma_c'; \;p \sigma, k \sigma_c} &= -2\pi i \delta^4(p'+k'-p-k)\frac{g^2}{(2\pi)^3}\left(-\frac{1}{s-\mu^2}\bar{v}(\mathbf{k}, \sigma_c) u(\mathbf{p}, \sigma) \bar{u}(\mathbf{p}', \sigma') v(\mathbf{k}', \sigma'_c)\right. \\ & \left.\quad + \frac{1}{t-\mu^2}\bar{u}(\mathbf{p}', \sigma') u(\mathbf{p}, \sigma)\bar{v}(\mathbf{k}, \sigma_c) v(\mathbf{k}', \sigma'_c)\right) \\ &\equiv -2\pi i \delta^4(p'+k'-p-k) (\mathcal{M}_s + \mathcal{M}_t) \end{aligned}} \end{equation} The spin-averaged matrix element $\mathcal{M}$ squared can likewise be broken up into three terms \begin{equation} \overline{|\mathcal{M}|^2} = \overline{|\mathcal{M}_s|^2} + \overline{|\mathcal{M}_t|^2} + 2 \Re[\overline{\mathcal{M}_s \mathcal{M}_t^*}] \end{equation} The only difference between calculating the above quantity compared to the pseudoscalar Yukawa interaction is that the lack of $\gamma_5$ matrices means we don't pick up an extra minus sign when commuting the $\gamma_5$ matrix through the $\slashed{p}$ matrix in our spin-sums. Sparing the details, one arrives at \begin{equation} \begin{aligned} \overline{|\mathcal{M}_s|^2} &= \frac{g^4}{(2\pi)^6} \frac{1}{(s-\mu^2)^2} \frac{(s-4m^2)^2}{s^2} \\ \overline{|\mathcal{M}_t|^2} &= \frac{g^4}{(2\pi)^6} \frac{1}{(t-\mu^2)^2} \frac{(t-4m^2)^2}{s^2} \\ 2 \Re [\overline{\mathcal{M}_s \mathcal{M}_t^*}] &= - \frac{1}{2} \frac{g^4}{(2\pi)^6} \frac{s^2 + t^2 - u^2 + 8m^2(2u - 2m^2)}{s^2 (s-\mu^2)(t-\mu^2)} \end{aligned} \end{equation} which yields a spin-averaged differential cross section of \begin{equation} \boxed{\begin{aligned} \frac{d\bar{\sigma}}{d\Omega} = \frac{g^4}{64 \pi^2} \frac{1}{s} \left[\frac{(s-4m^2)^2}{(s-\mu^2)^2} + \frac{(t-4m^2)^2}{(t-\mu^2)^2} - \frac{s^2 + t^2 - u^2 + 8m^2(2u-2m^2)}{2(s-\mu^2)(t-\mu^2)}\right] \end{aligned}} \end{equation} \subsection{Problem 6.3} There are four diagrams that contribute to order $g^2$ in the $S$-matrix for $B + B \rightarrow B + B$ scattering. They are: \begin{figure}[h] \centering \begin{subfigure}[c]{0.2\textwidth} \centering \feynmandiagram [vertical= i1 to f2, inline=(b.base)] { i1 [particle=\(p'\)] -- [scalar] a -- [scalar] i2 [particle=\(k\)], f1 [particle=\(k'\)] -- [scalar] a -- [scalar] f2 [particle=\(p\)] }; \end{subfigure} \begin{subfigure}[c]{0.2\textwidth} \centering \feynmandiagram [vertical'= a to b, inline=(a.base)] { i1 [particle=\(k'\)] -- [scalar] a -- [scalar] f1 [particle=\(p'\)], a -- [scalar, half left, momentum=\(q\)] b -- [scalar, half left, momentum=\(q'\)] a, i2 [particle=\(k\)] -- [scalar] b -- [scalar] f2 [particle=\(p\)] }; \end{subfigure} \begin{subfigure}[c]{0.2\textwidth} \centering \feynmandiagram [horizontal= a to b, inline=(a.base)] { i1 [particle=\(p\)] -- [scalar] a -- [scalar] f1 [particle=\(p'\)], a -- [scalar, half left, momentum=\(q\)] b -- [scalar, half left, momentum=\(q'\)] a, i2 [particle=\(k\)] -- [scalar] b -- [scalar] f2 [particle=\(k'\)] }; \end{subfigure} \begin{subfigure}[c]{0.2\textwidth} \centering \tikz[ dash pattern=on 1.8mm off 1.23mm,% adjusted dash pattern line width=.6pt,% adjusted linewidth decoration={ markings, } ]{ \coordinate (A) at (0,0); \coordinate (B) at (.9,1.2); \coordinate (C) at (3,2.4); \coordinate (D) at (3.0,0); \coordinate (E) at (2.1,1.2); \coordinate (F) at (0,2.4); % ~~~ lines, semi circle and dots; arrows-decoration ~~~~~~~~~~~~~~~~ \draw[postaction={decorate}] (A) -- (B) -- (C); \draw[postaction={decorate}] (D) -- (E) -- (F); \draw (B) -- (E) arc[start angle=0, end angle=-180,radius=.6]; % ~~~ overwriting the dash pattern defined above for the circles ~~~ \draw[fill=black,solid] (B) circle[radius=.75mm] (E) circle[radius=.75mm]; } \end{subfigure} \end{figure} Before actually writing down the matrix elements, we need to be careful with prefactors. Since we are dealing with an interaction that is non-linear in a single field and also self-interactions of this interaction type, we may get extra factors that do not cancel with the $1/n!$ factors of the Dyson series. For the order $g$ diagram, there are $4!$ ways of permuting the fields, and so that perfectly cancels with the $1/4!$ from the interaction density. However, for the other three diagrams, we are not so lucky. Their symmetry factors are $2$, which gives a prefactor of $1/2$ in front of the diagrams. Computing symmetry factors can be confusing, so we highly suggest following the following Physics stack exchange post: \url{https://physics.stackexchange.com/a/828871/306033}. Let us now write down the matrix elements for the center-of-mass frame. Since all the particles are the same, and thus their masses are identical, \begin{equation} \mathbf{p} = -\mathbf{k}, \quad \mathbf{p}' = - \mathbf{k}', |\mathbf{p}| = |\mathbf{p}'| \implies p^0 = k^0 = p'^0 = k'^0 \equiv E \end{equation} which means the Mandelstam variable $s$ is simply $s = 4E^2$. The $S$-matrix element to order $g$ is then \begin{equation} \boxed{S_{p', k'; \;p, k} = -2\pi i \delta^4(p' + k' - p - k) \frac{g}{(2\pi)^3}\frac{1}{s}} \end{equation} which implies the matrix element $\mathcal{M}$ is \begin{equation} \mathcal{M} = \frac{g}{(2\pi)^3}\frac{1}{s} \end{equation} Weinberg gives the cross section for 2 particles $\rightarrow$ 2 particles scattering as \begin{equation} \frac{d\sigma}{d\Omega} = \frac{(2\pi)^4 |\mathbf{p}'| p^0 k^0 p'^0 k'^0}{E_{\text{tot}}^2 |\mathbf{p}|}|\mathcal{M}|^2 \end{equation} which reduces in our case to \begin{equation} \boxed{\frac{d\sigma}{d\Omega} = \frac{g^2}{64 \pi^2}\frac{1}{s}} \end{equation} or \begin{equation} \boxed{\sigma = \frac{g^2}{16\pi}\frac{1}{s}} \end{equation} Next, let's compute the order $g^2$ corrections to the $S$-matrix. For the $s$-channel-like interaction, we get: \begin{equation} \begin{split} S_{p', k'; \;p, k} &= \frac{1}{2} \int d^4q d^4q'[-i(2\pi)^4 g \delta^4(q' - q - p - k)][-i(2\pi)^4 g \delta^4(p' + k' + q - q')] \\ &\quad\quad + \left[\frac{1}{(2\pi)^6}\frac{1}{\sqrt{16p^0 k^0 p'^0 k'^0}}\right]\left[\frac{-i}{(2\pi)^4}\frac{1}{q^2 + m^2 - i \epsilon}\right]\left[\frac{-i}{(2\pi)^4}\frac{1}{q'^2 + m^2 - i \epsilon}\right] \\ &= \frac{g^2}{2(2\pi)^6} \frac{1}{s} \delta^4(p' + k' - p - k) \int d^4q \frac{1}{q^2 + m^2} \frac{1}{(q+p+k)^2 + m^2} \\ &= -2\pi i \delta^4(p' + k' - p - k) \frac{g}{(2\pi)^3}\frac{1}{s} \left(\frac{ig}{2(2\pi)^4}\int d^4q \frac{1}{q^2 + m^2} \frac{1}{(q+p+k)^2 + m^2}\right) \end{split} \end{equation} where this final integral is logarithmically divergent. The only difference between this diagram and the other two diagrams is which delta functions are involved. Therefore, the form will be the same except for what $q'$ is constrained to. The entire $S$-matrix to second order in $g$ is thus \begin{equation} \boxed{\begin{aligned} S_{p', k'; \;p, k} &= -2\pi i \delta^4(p' + k' - p - k) \frac{g}{(2\pi)^3}\frac{1}{s} \left(1+\frac{ig}{2(2\pi)^4}\int d^4q \frac{1}{q^2 + m^2} \right. \\ &\quad\quad \cross \left.\left[ \frac{1}{(q+p+k)^2 + m^2} + \frac{1}{(q+p'-p)^2 + m^2} + \frac{1}{(q+k' - p)^2 + m^2}\right]\right) \end{aligned}} \end{equation} \subsection{Problem 6.5 - Heisenberg Picture VEVs} Here we wish to use the theorem given in section 6.4: \begin{equation} \left[ \frac{\delta^r S_{\beta\alpha}[\varepsilon]}{\delta \varepsilon_a(x)\delta \varepsilon_b(y) \dotsb}\right] = (\Psi^-_\beta, T\{-i\mathcal O_a(x),-i\mathcal O_b(y),\dotsb\} \Psi_\alpha^+) \end{equation} to compute the vacuum expectation values $(\Psi_0, \Phi(x)\Psi_0)$ and $(\Psi_0, T\{\Phi(x),\Phi(y)\}\Psi_0)$ to order $g$ and $g^2$ respectively for the interaction \begin{equation} H_\text{int} = \frac{g}{3!}\int d^3x \phi^3(x) \ . \end{equation} To do this, we will need all the diagrams in $S_{00}[\varepsilon]$ that contain at least 1, but at most 2, $\varepsilon(x)$ sources, up to $\mathcal O(g^2)$. \textbf{NB:} Although for scattering processes we usually only use the connected S matrix elements and corresponding Feynman diagrams, this matrix element does not necessarily correspond to a scattering process, and Weinberg says nothing about ignoring the disconnected portions. Therefore, I will keep all terms regardless of their divergences, and you can make your own judgements about which ones Weinberg actually wanted us to compute. There is only one diagram contributing to the first matrix element, namely \begin{figure}[h] \centering \tikz[ % dashed scalar propagators prop/.style ={dashed, line width=.6pt, dash pattern=on 1.8mm off 1.23mm}, % little black dots (interaction vertices) vertex/.style ={fill=black, circle, inner sep=.9pt}, % shaded external blobs blob/.style ={pattern=north east lines, pattern color=black, circle, draw=black, % solid outline line width=.6pt, minimum width=14pt} ]{ % coordinates --------------------------------------------------------- \coordinate (b) at (0,0); % lower blob % upper loop \coordinate (C) at (0,1.3); % loop centre \def\r{.3} % loop radius \path (C) ++(0,-\r) coordinate (v1); % dashed propagators (horizontal legs are shortened by 7 pt) \draw[prop,shorten >=7pt] (b) -- (v1); % dashed loop \draw[prop] (C) circle[radius=\r]; % blobs and vertices \node[blob] at (b) {}; \node[vertex] at (v1) {}; } %\caption{} \label{fig:scalar-loop} \end{figure} (the dashed blob represents the $\varepsilon$ term) whose corresponding position space term is \begin{equation} \frac{g}{2}\int d^4x \int d^4y\ \varepsilon(x) \Delta_F(x-y)\Delta_F(y-y) \ , \end{equation} where the factor of 1/2 comes from the symmetry factor of this diagram. Taking a functional derivative of this with respect to $\varepsilon(x)$ just gets rid of the x integral, as well as the $\varepsilon(x)$, so we are left with \begin{equation} \begin{split} (\Psi_0,\Phi(x)\Psi_0) &= \frac{g}{2}\Delta_F(0)\int d^4y \ \Delta_F(x-y) \\ &= \frac{g}{2m^2}\underbrace{\int \frac{d^4 p}{(2\pi)^4}\frac{1}{p^2 + m^2 -i\epsilon}}_{\Delta_F(0)} \ . \end{split} \end{equation} For the second VEV, we have more diagrams that contribute, given by the sum \begin{figure}[h] \centering %--- global styles that all diagrams share ----------------------------- \tikzset{ prop/.style ={dashed, line width=.6pt, dash pattern=on 1.8mm off 1.23mm}, vertex/.style ={fill=black, circle, inner sep=.9pt}, blob/.style ={pattern=north east lines, pattern color=black, circle, draw=black, line width=.6pt, minimum width=14pt} } %--- little helper macros (each produces one graph) -------------------- \newcommand{\TwoBlobVert}{% \tikz[baseline=(mid.base)]{ \coordinate (top) at (0, .7); \coordinate (bot) at (0,-.7); \coordinate (mid) at (0, 0 ); \draw[prop,shorten >=7pt,shorten <=7pt] (top) -- (bot); \node[blob] at (top) {}; \node[blob] at (bot) {}; }% } \newcommand{\DoubleLoop}{% \tikz[baseline=(c.base)]{ \def\r{.35} \coordinate (c) at (0,0); \coordinate (t) at (0,.8); \coordinate (tp)at (0,.8-\r); \coordinate (b) at (0,-.8); \coordinate (bp)at (0,-.8+\r); \draw[prop] (bp) -- (tp); \draw[prop] (t) circle[radius=\r]; \draw[prop] (b) circle[radius=\r]; \node[vertex] at (tp) {}; \node[vertex] at (bp) {}; }% } \newcommand{\PropLoop}{% \tikz[baseline=(c.base)]{ \def\r{.45} \coordinate (L) at (-\r,0); \coordinate (R) at ( \r,0); \coordinate (c) at (0,0); \draw[prop] (L) -- (R); \draw[prop] (c) circle[radius=\r]; \node[vertex] at (R) {}; \node[vertex] at (L) {}; }% } \newcommand{\VertBlobLoop}{% \tikz[baseline=(mid.base)]{ \def\r{.3} \coordinate (top) at (0,1.0); \coordinate (mid) at (0,0); \coordinate (bot) at (0,-1.0); \coordinate (t) at (0,\r); \coordinate (b) at (0,-\r); \draw[prop,shorten >=7pt] (t) -- (top); \draw[prop,shorten >=7pt] (b) -- (bot); \draw[prop] (mid) circle[radius=\r]; \node[blob] at (top) {}; \node[blob] at (bot) {}; \node[vertex] at (t) {}; \node[vertex] at (b) {}; }% } \newcommand{\BridgeLoop}{% \tikz[baseline=(mid.base)]{ \def\r{.3} \coordinate (L) at (-1.0,-0.5); \coordinate (R) at ( 1.0,-0.5); \coordinate (mid) at ( 0 ,-0.5); \coordinate (up) at ( 0 , 0.3); \draw[prop,shorten >=7pt] (mid) -- (L); \draw[prop,shorten >=7pt] (mid) -- (R); \draw[prop] (mid) -- (up); \draw[prop] (0,0.3+\r) circle[radius=\r]; \node[blob] at (L) {}; \node[blob] at (R) {}; \node[vertex] at (mid) {}; \node[vertex] at (up) {}; }% } %----------------------------------------------------------------------- \[ \TwoBlobVert\; \Biggl( 1 \;+\; \DoubleLoop \;+\; \PropLoop \Biggr) \;+\; \VertBlobLoop \;+\; \BridgeLoop \] \end{figure} For the first term, once we take the functional derivatives of the blob diagram on the outside of the parentheses, it gives a multiplicative factor of $i\Delta_F(x-y)$. The terms in parentheses give a contribution \begin{equation} \begin{split} 1 -ig^2\left[\frac{1}{4}\Delta_F^2(0)\int d^4z_1 \int d^4z_2 \ \Delta_F(z_1-z_2)+\frac{1}{6}\int d^4z_1 \int d^4z_2 \ \Delta_F^3(z_1-z_2)\right] \ , \end{split} \end{equation} where the $1/4,1/6$ are symmetry factors for the diagrams. The first term outside the parenthesis gives a contribution (after stripping off those dirty little epsilons) \begin{equation} \frac{g^2}{2}\int d^4z_1 \int d^4z_2 \ \Delta_F(x-z_1)\Delta_F^2(z_1-z_2)\Delta_F(z_2-y) \end{equation} and the final term gives \begin{equation} \frac{g^2}{2}\Delta_F(0)\int d^4z_1 \int d^4z_2 \ \Delta_F(x-z_1)\Delta_F(z_1-y)\Delta_F(z_1-z_2) \ . \end{equation} \section{Chapter 7} \subsection{Problem 7.1 -- Nonlinear Sigma Model} We are given the Lagrangian density \begin{equation} \mathscr{L} = -\frac{1}{2} \sum_{m n} \partial_{\mu} \Phi^n \partial^{\mu} \Phi^m f_{m n}(\Phi) \end{equation} Let's first define the symmetrized quantity $\tilde{f}$ \begin{equation} \tilde{f}_{m n} (\Phi) \equiv \frac{f_{m n}(\Phi) + f_{m n}(\Phi)}{2} \end{equation} Because we are dealing with scalar fields, derivatives of our fields will commute. We can therefore rewrite our Lagrangian density in terms of this symmetrized matrix since \begin{equation} \begin{aligned} \mathscr{L} &= -\frac{1}{2} \sum_{m n} \partial_{\mu} \Phi^n \partial^{\mu} \Phi^m f_{m n}(\Phi) \\ &= -\frac{1}{2} \sum_{m n} \left( \frac{\partial_{\mu} \Phi^n \partial^{\mu} \Phi^m f_{m n}(\Phi)}{2} + \frac{\partial_{\mu} \Phi^m \partial^{\mu} \Phi^n f_{n m}(\Phi)}{2} \right) \\ &= -\frac{1}{2} \sum_{m n} \left( \frac{\partial_{\mu} \Phi^n \partial^{\mu} \Phi^m f_{m n}(\Phi)}{2} + \frac{\partial_{\mu} \Phi^n \partial^{\mu} \Phi^m f_{n m}(\Phi)}{2} \right) \\ &= -\frac{1}{2} \sum_{m n} \partial_{\mu} \Phi^n \partial^{\mu} \Phi^m \tilde{f}_{m n}(\Phi) \end{aligned} \end{equation} The conjugate momenta are \begin{equation} \begin{aligned} \Pi_n &= \frac{\partial \mathscr{L}}{\partial \dot{\Phi}^n} \\ &= + \frac{1}{2} \frac{\partial}{\partial \dot{\Phi}^n} \left( \sum_{a b} \dot{\Phi}^a \dot{\Phi}^b f_{a b} (\Phi) \right) \\ &= \frac{1}{2} \sum_{a b} \left( \delta^{a}_{n}\dot{\Phi}^b + \delta^{b}_{n} \dot{\Phi}^a \right) f_{a b}(\Phi) \\ &= \sum_m \dot{\Phi}^m \tilde{f}_{m n}(\Phi) \end{aligned} \end{equation} Since $f_{m n}$ is invertible, the sum of it and its transpose $f_{n m}$ is also invertible, so we can solve for the $\dot{\Phi}^n$. \begin{equation} \begin{aligned} & \sum_{m n} \dot{\Phi}^m \tilde{f}_{m n}(\Phi) \tilde{f}^{-1}_{n l}(\Phi) = \sum_n \Pi_{n} \tilde{f}^{-1}_{n l}(\Phi) \\ \implies & \sum_m \dot{\Phi}^m \delta_{m l} = \sum_n \Pi_{n} \tilde{f}^{-1}_{n l}(\Phi) \\ \implies & \dot{\Phi}^n = \sum_m \Pi_{m} \tilde{f}^{-1}_{m n}(\Phi) \end{aligned} \end{equation} We then need to do the Legendre transformation of the Lagrangian density. Unlike Weinberg, let's define (at least for this problem), the density of the full Hamiltonian to be $\mathscr{H}$. Then the Hamiltonian density is \begin{equation} \begin{aligned} \mathscr{H} &= \sum_n \Pi_n \dot{\Phi}^n - \mathscr{L} \\ &= \sum_{m n} \Pi_n \Pi_{m} \tilde{f}^{-1}_{m n}(\Phi) + \frac{1}{2} \sum_{m n} \partial_{\mu} \Phi^n \partial^{\mu} \Phi^m \tilde{f}_{m n}(\Phi) \\ &= \sum_{m n} \Pi_n \Pi_{m} \tilde{f}^{-1}_{m n}(\Phi) - \frac{1}{2} \sum_{m n} \dot{\Phi}^n \dot{\Phi}^m \tilde{f}_{m n}(\Phi) + \frac{1}{2} \sum_{m n} \grad{\Phi^n} \cdot \grad{\Phi^m} \tilde{f}_{m n}(\Phi) \\ &= \sum_{m n} \Pi_n \Pi_{m} \tilde{f}^{-1}_{m n}(\Phi) - \frac{1}{2} \sum_{m n a b} \Pi_a \tilde{f}_{a n}^{-1}(\Phi) \Pi_b \tilde{f}_{b m}^{-1}(\Phi) \tilde{f}_{m n}(\Phi) + \frac{1}{2} \sum_{m n} \grad{\Phi^n} \cdot \grad{\Phi^m} \tilde{f}_{m n}(\Phi) \\ &= \frac{1}{2} \sum_{m n} \Pi_n \Pi_{m} \tilde{f}^{-1}_{m n}(\Phi) + \frac{1}{2} \sum_{m n} \grad{\Phi^n} \cdot \grad{\Phi^m} \tilde{f}_{m n}(\Phi) \end{aligned} \end{equation} To make further progress, we need to say something more about how the function $f_{m n}(\Phi)$ behaves. In our original Lagrangian, it is necessary that we can decompose $f_{m n}(\Phi)$ into the identity plus some other function of $\Phi$ so that we retain the required form of $H_0$. Therefore, we define (like Weinberg in Ch 9.3): \begin{equation} \tilde{f}_{m n}(\Phi) = \left(1 + U(\Phi)\right)_{m n} \end{equation} We know the inverse of the matrix above must exist, but we need to get it in the form of $1 + B$ where $B$ is some other term, that way we can recover $H_0$. Using the Woodbury formula we have \begin{equation} \tilde{f}^{-1}_{m n}(\Phi) = \left(1 + U(\Phi)\right)^{-1}_{m n} = \left[ 1 - \left(1 + U^{-1}(\Phi)\right)^{-1}\right]_{m n} \end{equation} The justification that $U(\Phi)$ allows such a procedure is that we require $H$ contain $H_0$ of the free theory, and this identity ensures we are able to write such a term down. Therefore, our full Hamiltonian density is \begin{equation} \begin{aligned} \mathscr{H} &= \mathscr{H}_0 + \mathscr{H}_{int} \\ \mathscr{H}_0 &= \frac{1}{2} \sum_{m} \Pi_{m} \Pi_{m} + \frac{1}{2} \sum_{m} \grad{\Phi}^m \cdot \grad{\Phi}m \\ \mathscr{H}_{int} &= \frac{1}{2} \sum_{m n} \Pi_m \Pi_n \left(1 + U^{-1}(\Phi) \right)^{-1}_{m n} + \frac{1}{2} \sum_{m n} \grad{\Phi}^m \cdot \grad{\Phi}^{n} U_{m n}(\Phi) \end{aligned} \end{equation} Passing into the interaction picture, we can then write our interaction as \begin{equation} \boxed{V(t) = \int d^3 x \left(\frac{1}{2} \sum_{m n} \pi_m \pi_n \left(1 + U^{-1}(\phi) \right)^{-1}_{m n} + \frac{1}{2} \sum_{m n} \grad{\phi}^m \cdot \grad{\phi}^{n} U_{m n}(\phi)\right)} \end{equation} \subsection{Problem 7.2} Consider a theory of scalar fields $\Phi^n$ and Dirac fields $\Psi^i$, with Lagrangian density $\mathcal{L}=\mathcal{L}_0 + \mathcal{L}_1$, where $\mathcal{L}_0$ is the usual free-field Lagrangian density, and $\mathcal{L}_1$ is an interaction term involving $\Phi^n$ and $\Psi^i$, but not their derivatives. Derive an explicit expression for the symmetric energy-momentum tensor $\Theta^{\mu\nu}$. First, we find the non-symmetric energy momentum tensor $T^{\mu\nu}$: \begin{gather} T^{\mu\nu} = \eta^{\mu\nu} \mathcal{L} - \frac{\partial \mathcal{L}}{\partial(\partial_\mu \Phi^n)} \partial^\nu \Phi^n - \frac{\partial \mathcal{L}}{\partial(\partial_\mu \Psi^i)} \partial^\nu \Psi^i \end{gather} Only the free part of the Lagrangian contains derivative factors, so we only need to consider that part of the Lagrangian to get the form of the derivative terms above. The free-field Lagrangian for the combined scalar and Dirac fields is the following: \begin{gather} \label{equ:Prob7.2_freefield} \mathcal{L}_0 = \sum_{n,i}-\frac{1}{2}\partial_\mu \Phi^n \partial^\mu \Phi^n - \frac{1}{2}m^2\left(\Phi^n\right)^2 - \bar{\Psi}^i(\gamma^\mu\partial_\mu+m)\Psi^i \end{gather} Combining this expression for the free Lagrangian and the stress energy tensor, \begin{gather} T^{\mu\nu} = \eta^{\mu\nu} \mathcal{L} + \frac{1}{2}\frac{\partial (\partial_\alpha \Phi^b \partial^\alpha \Phi^b)}{\partial(\partial_\mu \Phi^n)} \partial^\nu \Phi^n + \frac{\partial (\bar{\Psi}^b(\gamma^\alpha\partial_\alpha)\Psi^b)}{\partial(\partial_\mu \Psi^i)} \partial^\nu \Psi^i\\ T^{\mu\nu} = \eta^{\mu\nu} \mathcal{L} + \partial^\mu \Phi^n \partial^\nu \Phi^n + \bar{\Psi}^i \gamma^\mu \partial^\nu \Psi^i \end{gather} Now, we can use the definition of the Belinfante symmetric stress energy tensor: \begin{gather} \Theta^{\mu\nu} = T^{\mu\nu} -\frac{i}{2}\partial_\kappa\left[\frac{\partial \mathcal{L}}{\partial(\partial_\kappa \Psi^l)} \mathcal{J}^{\mu\nu}\Psi^l - \frac{\partial \mathcal{L}}{\partial(\partial_\mu \Psi^l)} \mathcal{J}^{\kappa\nu} \Psi^l - \frac{\partial \mathcal{L}}{\partial(\partial_\nu \Psi^l)} \mathcal{J}^{\kappa\mu} \Psi^l\right] \end{gather} Only the Dirac fields $\Psi^l$ appear in this equation because the generators for the Lorentz group for scalars are all zero. For Dirac fields, the generators of Lorentz group can be expressed in terms of the commutator of the gamma matrices: \begin{gather} \mathcal{J}^{\mu\nu} = -\frac{i}{4}\left[\gamma^\mu,\gamma^\nu\right] \end{gather} Using this expression of the generators in the definition of the Belinfante tensor, \begin{gather} \Theta^{\mu\nu} = T^{\mu\nu} -\frac{1}{8}\partial_\kappa\left[\frac{\partial \mathcal{L}}{\partial(\partial_\kappa \Psi^l)} \left[\gamma^\mu,\gamma^\nu\right]\Psi^l - \frac{\partial \mathcal{L}}{\partial(\partial_\mu \Psi^l)} \left[\gamma^\kappa,\gamma^\nu\right] \Psi^l - \frac{\partial \mathcal{L}}{\partial(\partial_\nu \Psi^l)} \left[\gamma^\kappa,\gamma^\mu\right] \Psi^l\right] \end{gather} From Eq. \ref{equ:Prob7.2_freefield}, we can express the derivatives of the Lagrangian in terms of the Dirac fields, \begin{gather} \Theta^{\mu\nu} = T^{\mu\nu} -\frac{1}{8}\partial_\kappa\left[\bar{\Psi}^l\gamma^\kappa \left[\gamma^\mu,\gamma^\nu\right]\Psi^l - \bar{\Psi}^l\gamma^\mu \left[\gamma^\kappa,\gamma^\nu\right] \Psi^l - \bar{\Psi}^l\gamma^\nu \left[\gamma^\kappa,\gamma^\mu\right] \Psi^l\right]\\ \Theta^{\mu\nu} = T^{\mu\nu} -\frac{1}{8}\partial_\kappa \left( \bar{\Psi}^l\left[\gamma^\kappa \left[\gamma^\mu,\gamma^\nu\right] - \gamma^\mu \left[\gamma^\kappa,\gamma^\nu\right] - \gamma^\nu \left[\gamma^\kappa,\gamma^\mu\right] \right]\Psi^l\right) \end{gather} This can be simplified somewhat by using the totally antisymmetric tensor defined in Eq. 5.4.11 of Weinberg: \begin{gather} \mathcal{A}^{\rho\sigma\tau} \equiv \gamma^{\left[\rho\right.} \gamma^\sigma \gamma^{\left.\tau \right]}\\ \mathcal{A}^{\rho\sigma\tau} = \gamma^\rho \gamma^\sigma \gamma^\tau - \gamma^\rho \gamma^\tau \gamma^\sigma - \gamma^\sigma \gamma^\rho \gamma^\tau + \gamma^\tau \gamma^\rho \gamma^\sigma + \gamma^\sigma \gamma^\tau \gamma^\rho - \gamma^\tau \gamma^\sigma \gamma^\rho\\ \mathcal{A}^{\rho\sigma\tau} = \gamma^\rho \left[\gamma^\sigma, \gamma^\tau \right] - \gamma^\sigma \left[\gamma^\rho, \gamma^\tau\right] + \gamma^\tau \left[\gamma^\rho, \gamma^\sigma\right] \end{gather} Using the totally antisymmetric tensor, the Belinfante tensor can be written as \begin{gather} \Theta^{\mu\nu} = T^{\mu\nu} -\frac{1}{8}\partial_\kappa\left(\bar{\Psi}^l\left[\mathcal{A}^{\kappa \mu \nu} - 2\gamma^\nu \left[\gamma^\kappa,\gamma^\mu\right] \right]\Psi^l\right) \end{gather} Putting everything together, the symmetric stress energy tensor is \begin{gather} \Theta^{\mu\nu} = \eta^{\mu\nu} \mathcal{L} + \partial^\mu \Phi^n \partial^\nu \Phi^n + \bar{\Psi}^i \gamma^\mu \partial^\nu \Psi^i -\frac{1}{8}\partial_\kappa\left(\bar{\Psi}^l\left[\mathcal{A}^{\kappa \mu \nu} - 2\gamma^\nu \left[\gamma^\kappa,\gamma^\mu\right] \right]\Psi^l\right) \end{gather} Another possible way to simplify the expression is as follows: \begin{gather} \Theta^{\mu\nu} = T^{\mu\nu} -\frac{1}{8}\partial_\kappa \left( \bar{\Psi}^l\left[ \left[\gamma^\kappa, \gamma^\mu \right] \gamma^\nu +\{\gamma^\mu,\gamma^\nu\}\gamma^\kappa - \{\gamma^\kappa,\gamma^\nu\}\gamma^\mu \right]\Psi^l\right)\\ \Theta^{\mu\nu} = T^{\mu\nu} -\frac{1}{4}\partial_\kappa \left( \bar{\Psi}^l\left[ \gamma^\kappa\gamma^\mu \gamma^\nu +\eta^{\mu\nu}\gamma^\kappa - \eta^{\kappa \nu}\gamma^\mu -\eta^{\kappa \mu}\gamma^\nu \right]\Psi^l\right)\\ \begin{split} \Theta^{\mu\nu} = \eta^{\mu\nu} \mathcal{L} +& \partial^\mu \Phi^n \partial^\nu \Phi^n + \bar{\Psi}^i \gamma^\mu \partial^\nu \Psi^i\\ &-\frac{1}{4}\left(\partial_\kappa\left(\bar{\Psi}^l \left(\gamma^\kappa\gamma^\mu \gamma^\nu+ \eta^{\mu\nu}\gamma^\kappa\right) \Psi^l\right) -\partial^\nu\left(\bar{\Psi}^l \kappa^\mu \Psi^l\right)-\partial^\mu\left(\bar{\Psi}^l \gamma^\nu \Psi^l\right) \right) \end{split} \end{gather} \subsection{Problem 7.4 -- Scalar QED with a Massive Photon} \textbf{DISCLAIMER: Because this is all based on classical field theory, I will be extremely loose with operator ordering, as I am not sure how to perfectly avoid the quantum-mechanical ordering ambiguities when computing conserved currents, Legendre transformations, etc. All orderings are exclusively vibes-based.} Here we must canonically quantize the theory given by the Lagrangian: \begin{equation} \mathscr L = -(D_\mu \Phi)^\dag (D^\mu \Phi) -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}m^2V_\mu V^\mu - \mathscr H (\Phi^\dag \Phi) \end{equation} where $D_\mu \equiv \partial_\mu -igV_\mu$. First, it will be useful to note that this Lagrangian has a global phase symmetry under the transformation $\Phi \rightarrow \Phi + i\varepsilon \Phi$, which corresponds to a conserved current (equation 7.3.15 in the text) \begin{equation} \begin{split} J^\mu &= -i \frac{\partial \mathcal L}{\partial(\partial_\mu \Phi)}\Phi + i\Phi^\dag \frac{\partial \mathcal L}{\partial(\partial_\mu \Phi^\dag)} \\ &= i \Bigl[ (D^\mu \Phi)^\dag \Phi - \Phi^\dag D^\mu \Phi \Bigr] \\ \end{split} \end{equation} such that $\partial_\mu J^\mu = 0$, which in turn implies the conservation of the charge $\int d^3x J^0$. Going forward, I will rescale this current so that it contains the charge, that is, I will take \begin{equation}\label{eq:7-4covariant_scalar_current} J^\mu \equiv ig \Bigl[ (D^\mu \Phi)^\dag \Phi - \Phi^\dag D^\mu \Phi \Bigr] \ . \end{equation} Now we need to find the canonical momenta given this Lagrangian. Recall that, like the case in section 7.5 of the text, $V^0$ is an auxiliary field, as its conjugate is always equal to zero (a primary constraint). The non-zero canonical momenta are equal to: \begin{equation}\label{eq:7-4canonical_momenta} \begin{split} \Pi^i \equiv \frac{\partial \mathscr L}{\partial (\partial_0 V^i)}=& F^{i0}=\dot V^i +\partial_i V^0\\ P \equiv \frac{\partial \mathscr L}{\partial (\partial_0 \Phi)} = (D_0\Phi)^\dag = \dot \Phi^\dag -igV^0 \Phi^\dag \quad &, \quad P^\dag \equiv \frac{\partial \mathscr L}{\partial (\partial_0 \Phi^\dag)} = D_0\Phi =\dot \Phi + igV^0 \Phi \ . \end{split} \end{equation} To find our Hamiltonian, we need to apply the Legendre transformation \begin{equation} \mathscr H = \vect{\Pi}\cdot \dot{\vect{V}} + P\dot\Phi + \dot{\Phi}^\dag P^\dag - \mathscr L \end{equation} and we also need to write it in terms of the canonical variables $\vect V, \vect \Pi, \Phi, P, \Phi^\dag, P^\dag$. From (\ref{eq:7-4canonical_momenta}), we can eliminate all of the time derivatives in favor of the canonical momenta, but we still will have explicit dependence on $V^0$, which is not a canonical variable. To eliminate it, we use the Euler-Lagrange equations for the vector field $V^\mu$. We have as usual that \begin{equation} \frac{\partial \mathscr L}{\partial(\partial_\mu V_\nu)}=-F^{\mu\nu} \ , \end{equation} but for the other side of Euler-Lagrange, we get \begin{equation} \begin{split} \pd{\mathscr L}{V_\nu}&=-\eta^{\mu\rho}\pd{ }{V_\nu}\biggl[ \bigl[(\partial_\mu + igV_\mu)\Phi^\dag\bigr]\bigl[(\partial_\rho - igV_\rho)\Phi\bigr] +\frac{1}{2}m^2 V_\mu V_\rho\biggr] \\ &=-\eta^{\mu\rho}\biggl[ ig \delta^{\nu}_\mu \Phi^\dag (D_\rho\Phi)-ig\delta^\nu_\rho (D_\mu\Phi)^\dag \Phi + \frac{1}{2}m^2(\delta^\nu_\mu V_\rho + V_\mu\delta^\nu_\rho)\biggr]\\ &=-ig\Phi^\dag(D^\nu\Phi) + ig(D^\nu\Phi)^\dag\Phi -m^2 V^\nu \\ &=gJ^\nu-m^2V^\nu \ . \end{split} \end{equation} So the full Euler-Lagrange equations are \begin{equation}\label{eq:7.4Vec_EL} \partial_\mu F^{\mu\nu} = m^2V^\nu - J^\nu \ , \end{equation} and for $\nu=0$, we use $\Pi^i = F^{i0}$ to get \begin{equation} \partial_i \Pi^i = m^2 V^0 - J^0 \ . \end{equation} Solving for $V^0$ yields \begin{equation} V^0 = \frac{1}{m^2}\bigl[ \partial_i\Pi^i + J^0\bigr] \ . \end{equation} Now, this may seem like a cheat, since $J^0$ seems to implicitly depend on $V^0$ through the time component of the covariant derivative $D^0 = \partial^0 - igV^0$. However, this is actually fine, since we can just as easily write $J_0$ in terms of the canonical variables \textit{only} by using (\ref{eq:7-4canonical_momenta}): \begin{equation} \begin{split} J^0 &= ig \Bigl[ (D^0 \Phi)^\dag \Phi - \Phi^\dag D^0 \Phi \Bigr] \\ &= ig \Bigl[ -P \Phi + \Phi^\dag P^\dag \Bigr] \\ \end{split} \end{equation} so our expression for $V^0$ really is purely in terms of the canonical variables. We can now express everything in our Hamiltonian density fully in terms of the canonical variables. For the time derivatives, we have (starting to use vector notation instead of component notation) \begin{equation} \dot{\vect V}=\vect \Pi - \frac{1}{m^2}\nabla(\nabla\cdot \vect\Pi + J^0) \quad , \quad \dot \Phi = P^\dag - i\frac{g}{m^2}\Phi(\nabla \cdot \vect \Pi + J^0) \ . \end{equation} We can then express the first part of the Legendre transformation as \begin{equation} \vect{\Pi}\cdot \dot{\vect{V}} + P\dot\Phi + \dot{\Phi}^\dag P^\dag = \vect \Pi^2 +2|P|^2 + \frac{1}{m^2}\bigl[\nabla \cdot \vect\Pi + J^0\bigr]^2 \ . \end{equation} We must also write the Lagrangian density in terms of the canonical variables, which we can do term by term: \begin{equation} -(D_\mu \Phi)^\dag (D^\mu \Phi) = |P|^2 - |\nabla\Phi|^2-g^2 \vect V^2 |\Phi|^2 + igV_i\bigl[(\partial^i\Phi^\dag)\Phi - \Phi^\dag(\partial^i\Phi)\bigr] \end{equation} \begin{equation} -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} = \frac{1}{2}\vect\Pi^2 - \frac{1}{2}(\nabla \times \vect V)^2 \end{equation} \begin{equation} -\frac{1}{2}m^2V_\mu V^\mu = \frac{1}{2m^2}\bigl[\nabla\cdot \vect\Pi+J^0\bigr]^2 -\frac{1}{2}m^2 \vect V^2 \ . \end{equation} Putting all of this together, we can write out the full Hamiltonian, separating out the free part: \begin{equation} H_0 = \int d^3\vect x \Bigl\{\frac{1}{2}\vect\Pi^2 +\frac{1}{2m^2}(\nabla\cdot\vect\Pi)^2 + \frac{1}{2}(\nabla \times \vect V)^2 +\frac{1}{2}m^2 \vect V^2 + |P|^2 +|\nabla\Phi|^2 + \mu^2|\Phi|^2 \Bigr\} \end{equation} \begin{equation} V(t=0) = \int d^3\vect x \Bigl\{\frac{1}{m^2}(\nabla\cdot\vect\Pi)J^0 + \frac{1}{2m^2}\bigl[J^0\bigr]^2+ g^2 \vect V^2 |\Phi|^2 -ig\vect V\cdot \bigl[(\nabla\Phi^\dag)\Phi - \Phi^\dag (\nabla\Phi)\bigr] + \tilde{\mathscr H}(\Phi^\dag\Phi) \Bigr\} \end{equation} where $\mu^2$ is the coefficient appearing in front of the $\Phi^\dag\Phi$ term in $\mathscr H(\Phi^\dag \Phi)$ (in general, $\mu^2$ could be zero), and $\tilde{\mathscr H} = \mathscr H(\Phi^\dag\Phi) - \mu^2 \Phi^\dag \Phi$. Now we pass to the interaction picture, and make the replacements $\vect V\rightarrow\vect v$, $\vect \Pi \rightarrow \bm{\pi}$, $\Phi\rightarrow\phi$, $P\rightarrow \pi_\phi$. We then follow Weinberg, and invent the quantity \begin{equation} v^0 =\frac{1}{m^2}\nabla \cdot \bm{\pi} \end{equation} so that we can write our field equations for $\vect v$ in a covariant manner: \begin{equation} \partial_\mu v^\mu =0 \quad , \quad (\Box -m^2)v^\mu = 0 \ . \end{equation} One can also verify that the interaction picture scalar field satisfies the expected field equation $(\Box-\mu^2)\phi =0$. With our choice of $v^0$, the interaction now looks like \begin{equation} V(t)= \int d^3\vect x \Bigl\{v^0J^0 + \frac{1}{2m^2}\bigl[J^0\bigr]^2+ g^2 \vect v^2 |\phi|^2 -ig\vect v\cdot \bigl[(\nabla\phi^\dag)\phi - \phi^\dag (\nabla\phi)\bigr] + \tilde{\mathscr H}(\phi^\dag\phi) \Bigr\} \ . \end{equation} To clean this up a bit, we can define the "non-covariant" current (the usual conserved current for a free charged scalar field with a $U(1)$ global symmetry) \begin{equation} J^\mu_\phi \equiv ig\bigl[(\partial^\mu \phi^\dag)\phi - \phi^\dag (\partial^\mu\phi)\bigr] \end{equation} and note that the time component is actually the same as that of $J^\mu$ in the interaction picture: \begin{equation} \begin{split} J^0&=ig \Bigl[ -P \Phi + \Phi^\dag P^\dag \Bigr] \rightarrow ig \Bigl[ -\pi_\phi \phi + \phi^\dag \pi_\phi^\dag \Bigr] = J^0_\phi \end{split} \end{equation} where we used the Hamilton equations in the interaction picture \begin{equation} \dot \phi = \partial_0 \phi = \frac{\delta H_0}{\delta \pi_\phi} = \pi_\phi^\dag \quad \text{(and the conjugate equation)} \ . \end{equation} This allows us to condense our expression for the interaction to \begin{equation} V(t) = \int d^3\vect x \Bigl\{ -v_\mu J^\mu_\phi + \frac{1}{2m^2}\bigl[J_\phi^0\bigr]^2+ g^2 \vect v^2 |\phi|^2 + \tilde{\mathscr H}(|\phi|^2) \Bigr\} \ . \end{equation} The non-covariant term $\propto (J^0_\phi)^2$ looks strange, but it is this term that cancels the non-local piece of the propagator we encountered in section 6.2 of the text. In fact, we need to apply this trick again to the other non-covariant piece of the interaction, by adding and subtracting a term $g^2(v^0)^2|\phi|^2$. This gives \begin{equation} \boxed{V(t) = \int d^3\vect x \Bigl\{ -v_\mu J^\mu_\phi + g^2 v_\mu v^\mu |\phi|^2 + \tilde{\mathscr H}(|\phi|^2) + \frac{1}{2m^2}\bigl[J_\phi^0\bigr]^2 + \left|J^0_\text{eff}\right|^2\Bigr\}} \ , \end{equation} where we defined $J_\text{eff}^\mu = g v^\mu \phi^\dag$ as the effective "current" that the derivative $\partial_\mu \phi$ couples to. This extra non-covariant piece, like the other one, will cancel problematic terms in the propagator resulting from contracting a $\partial_\mu \phi$ field with a $\partial_\nu \phi^\dag$ field (see Weinberg section 6.2). \subsection{Problem 7.5 -- Scalar QED Belinfante Tensor} Here we need to construct the symmetric Belinfante tensor $\Theta^{\mu\nu}$ for the theory in the previous problem. We also need to find the current associated with the symmetry $\delta \Phi = i\varepsilon\Phi, \delta V^\mu=0$, but we already found this in the previous problem, \begin{equation} \begin{split} J^\mu = i g\Bigl[ (D^\mu \Phi)^\dag \Phi - \Phi^\dag D^\mu \Phi \Bigr] \ . \end{split} \end{equation} To find the Belinfante tensor, we use the formula (7.4.11) in the text, which in our case reduces to \begin{equation} \Theta^{\mu\nu} = T^{\mu\nu}-\frac{i}{2}\partial_\kappa\left(\left[ \pd{\mathscr L}{(\partial_\kappa V^\rho)}(\mathscr J^{\mu\nu})^\rho_{\ \sigma} -\pd{\mathscr L}{(\partial_\mu V^\rho)}(\mathscr J^{\kappa\nu})^\rho_{\ \sigma} -\pd{\mathscr L}{(\partial_\nu V^\rho)}(\mathscr J^{\kappa\mu})^\rho_{\ \sigma} \right]V^\sigma\right) \end{equation} where $\mathscr J^{\mu\nu}$ are the generators of the Lorentz group in the fundamental representation. We then first need to find the energy-momentum tensor $T^{\mu\nu}$, which is defined through \begin{equation} T^{\mu\nu} = \eta^{\mu\nu}\mathscr L - \pd{\mathscr L}{(\partial_\mu \Psi^\ell)}\partial^\nu \Psi^\ell = \eta^{\mu\nu}\mathscr L + \Bigl[ (D^\mu\Phi)^\dag \partial^\nu \Phi + \text{h.c.}\Bigr] + F^{\mu\rho}\partial^\nu V_\rho \ . \end{equation} Our complete Belinfante tensor is then given by the gross expression \begin{equation}\label{eq:7.5gross_belinfante} \begin{split} \Theta^{\mu\nu} &= -\eta^{\mu\nu}\Bigl[(D_\rho \Phi)^\dag (D^\rho \Phi) +\frac{1}{4}F_{\rho\sigma}F^{\rho\sigma} + \frac{1}{2}m^2V_\rho V^\rho +\mathscr H (\Phi^\dag \Phi)\Bigr] \\ & \quad + \Bigl[ (D^\mu\Phi)^\dag \partial^\nu \Phi + \text{h.c.}\Bigr] + F^{\mu\rho}\partial^\nu V_\rho \\ & \quad +\frac{i}{2}\partial_\kappa\left(\Bigl[ F^{\kappa\rho}(\mathscr J^{\mu\nu})_{\rho\sigma} -F^{\mu\rho}(\mathscr J^{\kappa\nu})_{\rho \sigma} -F^{\nu\rho}(\mathscr J^{\kappa\mu})_{\rho \sigma} \Bigr]V^\sigma\right) \ . \end{split} \end{equation} We can simplify this through use of the explicit form of the $\mathscr J$ generators (given component-wise in equations 5.3.8-9 in the text) \begin{equation} (\mathscr J^{\mu\nu})_{\rho\sigma} = -i(\eta^\mu_{\ \rho}\eta^\nu_{\ \sigma} - \eta^{\mu}_{\ \sigma}\eta^\nu_{\ \rho}) \ . \end{equation} With this we can substantially simplify the last term in $\Theta^{\mu\nu}$ down to \begin{equation}\label{eq:7.5gross_final_term} \begin{split} \frac{i}{2}\partial_\kappa\left( \dotsb \right) &= -\partial_\kappa (F^{\mu\kappa}V^\nu) \\ &=(\partial_\kappa F^{\kappa\mu})V^\nu - F^{\mu\rho}\partial_\rho V^\nu \\ &= (m^2V^\mu - J^\mu)V^\nu - F^{\mu\rho}\partial_\rho V^\nu \ , \end{split} \end{equation} where we used the Euler-Lagrange equations (\ref{eq:7.4Vec_EL}) from the previous problem. Inserting this back into $\Theta^{\mu\nu}$ gives us the slightly cleaner expression \begin{equation} \boxed{ \begin{split} \Theta^{\mu\nu} &= -\eta^{\mu\nu}\Bigl[(D_\rho \Phi)^\dag (D^\rho \Phi) +\frac{1}{4}F_{\rho\sigma}F^{\rho\sigma} + \frac{1}{2}m^2V_\rho V^\rho +\mathscr H (\Phi^\dag \Phi)\Bigr] \\ & \quad + \Bigl[ (D^\mu\Phi)^\dag (D^\nu \Phi) + \text{h.c.}\Bigr] + F^{\mu\rho}F^\nu_{\ \rho} + m^2V^\mu V^\nu \ . \end{split} } \end{equation} where we combined the the last two terms of (\ref{eq:7.5gross_final_term}) with the similar ones in the second line of (\ref{eq:7.5gross_belinfante}). \end{document}