├── mandl_shaw_solution.pdf ├── README.md ├── mandl_shaw_solution.synctex.gz ├── LICENSE ├── mandl_shaw_solution.out ├── mandl_shaw_solution.aux ├── mandl_shaw_solution.tex └── mandl_shaw_solution.log /mandl_shaw_solution.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sanhacheong/mandl_shaw_qft_sol/HEAD/mandl_shaw_solution.pdf -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # mandl_shaw_qft_sol 2 | Solution to Problems in Quantum Field Theory by Franz Mandl & Graham Shaw 3 | -------------------------------------------------------------------------------- /mandl_shaw_solution.synctex.gz: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/sanhacheong/mandl_shaw_qft_sol/HEAD/mandl_shaw_solution.synctex.gz -------------------------------------------------------------------------------- /LICENSE: -------------------------------------------------------------------------------- 1 | MIT License 2 | 3 | Copyright (c) 2017 Sanha Cheong 4 | 5 | Permission is hereby granted, free of charge, to any person obtaining a copy 6 | of this software and associated documentation files (the "Software"), to deal 7 | in the Software without restriction, including without limitation the rights 8 | to use, copy, modify, merge, publish, distribute, sublicense, and/or sell 9 | copies of the Software, and to permit persons to whom the Software is 10 | furnished to do so, subject to the following conditions: 11 | 12 | The above copyright notice and this permission notice shall be included in all 13 | copies or substantial portions of the Software. 14 | 15 | THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR 16 | IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, 17 | FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE 18 | AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER 19 | LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, 20 | OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE 21 | SOFTWARE. 22 | -------------------------------------------------------------------------------- /mandl_shaw_solution.out: -------------------------------------------------------------------------------- 1 | \BOOKMARK [1][-]{section.1}{Photons and the Electromagnetic Field}{}% 1 2 | \BOOKMARK [2][-]{subsection.1.1}{}{section.1}% 2 3 | \BOOKMARK [3][-]{subsubsection.1.1.1}{}{subsection.1.1}% 3 4 | \BOOKMARK [3][-]{subsubsection.1.1.2}{}{subsection.1.1}% 4 5 | \BOOKMARK [3][-]{subsubsection.1.1.3}{}{subsection.1.1}% 5 6 | \BOOKMARK [3][-]{subsubsection.1.1.4}{}{subsection.1.1}% 6 7 | \BOOKMARK [3][-]{subsubsection.1.1.5}{}{subsection.1.1}% 7 8 | \BOOKMARK [3][-]{subsubsection.1.1.6}{}{subsection.1.1}% 8 9 | \BOOKMARK [2][-]{subsection.1.2}{}{section.1}% 9 10 | \BOOKMARK [3][-]{subsubsection.1.2.1}{}{subsection.1.2}% 10 11 | \BOOKMARK [3][-]{subsubsection.1.2.2}{}{subsection.1.2}% 11 12 | \BOOKMARK [2][-]{subsection.1.3}{}{section.1}% 12 13 | \BOOKMARK [1][-]{section.2}{Lagrangian Field Theory}{}% 13 14 | \BOOKMARK [2][-]{subsection.2.1}{}{section.2}% 14 15 | \BOOKMARK [2][-]{subsection.2.2}{}{section.2}% 15 16 | \BOOKMARK [2][-]{subsection.2.3}{}{section.2}% 16 17 | \BOOKMARK [2][-]{subsection.2.4}{}{section.2}% 17 18 | \BOOKMARK [2][-]{subsection.2.5}{}{section.2}% 18 19 | \BOOKMARK [1][-]{section.3}{The Klein-Gordon Field}{}% 19 20 | \BOOKMARK [2][-]{subsection.3.1}{}{section.3}% 20 21 | \BOOKMARK [2][-]{subsection.3.2}{}{section.3}% 21 22 | \BOOKMARK [2][-]{subsection.3.3}{}{section.3}% 22 23 | \BOOKMARK [2][-]{subsection.3.4}{}{section.3}% 23 24 | \BOOKMARK [2][-]{subsection.3.5}{}{section.3}% 24 25 | \BOOKMARK [2][-]{subsection.3.6}{}{section.3}% 25 26 | \BOOKMARK [3][-]{subsubsection.3.6.1}{}{subsection.3.6}% 26 27 | \BOOKMARK [3][-]{subsubsection.3.6.2}{}{subsection.3.6}% 27 28 | \BOOKMARK [1][-]{section.4}{The Dirac Field}{}% 28 29 | \BOOKMARK [2][-]{subsection.4.1}{}{section.4}% 29 30 | \BOOKMARK [2][-]{subsection.4.2}{}{section.4}% 30 31 | \BOOKMARK [2][-]{subsection.4.3}{}{section.4}% 31 32 | \BOOKMARK [2][-]{subsection.4.4}{}{section.4}% 32 33 | \BOOKMARK [2][-]{subsection.4.5}{}{section.4}% 33 34 | \BOOKMARK [1][-]{section.5}{Photons: Covariant Theory}{}% 34 35 | -------------------------------------------------------------------------------- /mandl_shaw_solution.aux: -------------------------------------------------------------------------------- 1 | \relax 2 | \providecommand\hyper@newdestlabel[2]{} 3 | \providecommand\HyperFirstAtBeginDocument{\AtBeginDocument} 4 | \HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined 5 | \global\let\oldcontentsline\contentsline 6 | \gdef\contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}} 7 | 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\documentclass[letterpaper,12pt]{article} 2 | \usepackage[letterpaper, left=0.8in, right=0.8in, top=1in, bottom=1in]{geometry} 3 | \usepackage[titletoc]{appendix} 4 | \usepackage[utf8]{inputenc} 5 | \usepackage{xcolor} 6 | \usepackage{enumerate} 7 | \usepackage{multicol} 8 | \usepackage{tikz} 9 | \usepackage{amsmath,amssymb} 10 | \usepackage{setspace} 11 | \usepackage{natbib} 12 | \usepackage{graphicx} 13 | \usepackage[justification=centering]{caption} 14 | \usepackage{subcaption} 15 | \usepackage{hyperref} 16 | \usepackage{mathrsfs} 17 | \usepackage{calligra} 18 | \usepackage{float} 19 | \usepackage{physics} 20 | \usepackage{slashed} 21 | \usepackage{tikz-feynman} 22 | \usepackage{cancel} 23 | \usepackage{tgtermes} 24 | \usepackage{titlesec} 25 | \usepackage{changepage} 26 | \usepackage{tabto} 27 | 28 | \setlength{\parindent}{0.4in} 29 | 30 | \newcommand{\linia}{\rule{\linewidth}{0.5pt}} 31 | 32 | % my own titles 33 | \title{ 34 | \huge{Solution to Problems in\\Quantum Field Theory}\\ 35 | \vspace{0.1in} 36 | \Large{by Franz Mandl \& Graham Shaw} 37 | } 38 | 39 | \author{Sanha Cheong} 40 | \def\email{\texttt{\href{mailto:sanha@stanford.edu}{sanha@stanford.edu}}} 41 | \def\institution{Stanford University} 42 | 43 | \date{\today} 44 | 45 | \makeatletter 46 | \renewcommand{\maketitle}{ 47 | \begin{center} 48 | \textsc{\@title} 49 | \end{center} 50 | \vspace{2ex} 51 | \Large{\@author} \hfill \Large{\@date} \\ 52 | \large{\email} \hfill $ $ \\ 53 | \large{\institution} 54 | \\ 55 | \linia 56 | } 57 | \makeatother 58 | %%% 59 | 60 | % custom footers and headers 61 | \usepackage{fancyhdr,lastpage} 62 | \pagestyle{fancy} 63 | \lhead{} 64 | \chead{} 65 | \rhead{} 66 | \cfoot{} 67 | \rfoot{Page \thepage\ /\ \pageref*{LastPage}} 68 | \renewcommand{\headrulewidth}{0pt} 69 | \renewcommand{\footrulewidth}{0pt} 70 | % 71 | 72 | % custom section/problems 73 | \titleformat{\section}[block] 74 | {\vspace{0.3in}\LARGE\bfseries\centering} 75 | {\thesection}{1em}{} 76 | 77 | \renewcommand{\thesubsection}{\arabic{subsection}} 78 | \titleformat{\subsection}[runin] 79 | {\vspace{1ex}} 80 | {\thesubsection.}{0em}{} 81 | 82 | \renewcommand{\thesubsubsection}{(\roman{subsubsection})} 83 | \titleformat{\subsubsection}[runin] 84 | {\vspace{-1ex}} 85 | {\thesubsubsection}{0em}{} 86 | 87 | \newenvironment{problem}{\subsection{}\begin{adjustwidth}{0.25in}{}\vspace{-\baselineskip}}{\end{adjustwidth}} 88 | 89 | \newenvironment{subproblem}{\subsubsection{}\begin{adjustwidth}{0.4in}{}\vspace{-\baselineskip}}{\end{adjustwidth}} 90 | % 91 | 92 | % mathematical & typeset commands 93 | \setlength{\fboxsep}{0pt} 94 | \setlength{\fboxrule}{.1pt} 95 | \makeatletter 96 | \newcommand*\dotp{\mathpalette\bigcdot@{.5}} 97 | \newcommand*\bigcdot@[2]{\mathbin{\vcenter{\hbox{\scalebox{#2}{$\m@th#1\bullet$}}}}} 98 | \makeatother 99 | \newcommand{\fig}[2]{ 100 | \begin{figure}[bth] 101 | \centering 102 | \includegraphics[width = 13cm]{#1} 103 | \caption {#2} 104 | \end{figure}} 105 | \newcommand{\der}[2]{\frac{\diff{#1}}{\diff{#2}}} 106 | \newcommand{\nder}[3]{\frac{\diff[#1]{#2}}{\ndiff[#1]{#3}}} 107 | \newcommand{\pder}[2]{\frac{\partial #1}{\partial #2}} 108 | \newcommand{\npder}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} 109 | \newcommand{\lagr}{\mathscr{L}} 110 | \newcommand{\hamil}{\mathscr{H}} 111 | \DeclareMathOperator{\dalem}{\Box} 112 | \makeatletter 113 | \def\diff{\@ifnextchar[{\@with}{\@without}} 114 | \def\@with[#1]#2{\mathrm{d}^#1#2} 115 | \def\@without#1{\mathrm{d}#1} 116 | \makeatother 117 | \newcommand\colcancel[2][black]{\renewcommand\CancelColor{\color{#1}}\cancel{#2}} 118 | \newcommand{\define}{\equiv} 119 | \newcommand{\done}{\tag*{$\blacksquare$}} 120 | \newcommand{\overbar}[1]{ 121 | \mkern 1.5mu \overline{\mkern-1.5mu\raisebox{0pt}[\dimexpr\height+0.5mm\relax]{$#1$}\mkern-1.5mu}\mkern 1.5mu 122 | } 123 | \newcommand{\timep}[1]{\mathrm{T}\left\{#1\right\}} 124 | % 125 | 126 | \bibliographystyle{abbrv} 127 | 128 | %%%----------%%%----------%%%----------%%%----------%%% 129 | 130 | \begin{document} 131 | 132 | \maketitle 133 | 134 | \section{Photons and the Electromagnetic Field} 135 | 136 | \begin{problem} 137 | The free-radiation field inside a cubic enclosure is given by the state: 138 | \begin{equation*} 139 | \ket{c} = \exp\left(-\frac{1}{2}\abs{c}^2\right) 140 | \sum_{n=0}^{\infty} \frac{c^n}{\sqrt{n!}} \ket{n} 141 | \end{equation*} 142 | where $c=\abs{c}e^{i\delta}$ is any complex number and $\ket{n}$ is the state with $n$ photons with some specific wavevector $\vb{k}$ and transverse polarization $r$, omitted in the expression. 143 | 144 | (Such a state $\ket{c}$ is called \emph{a coherent state} and is extremely useful in quantum optics and other bosonic quantum field theory.) 145 | 146 | \begin{subproblem} 147 | $\ket{c}$ is normalized: $\braket{c}{c}=1$. 148 | \begin{align*} 149 | \braket{c}{c} 150 | &= \exp\left(-\abs{c}^2\right) 151 | \left(\sum_{m=0}^{\infty}\bra{m}\frac{{c^*}^m}{\sqrt{m!}}\right) 152 | \left(\sum_{n=0}^{\infty}\frac{c^n}{\sqrt{n!}}\ket{n}\right) \\ 153 | &= e^{-\abs{c}^2} 154 | \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} 155 | \frac{\abs{c}^2 e^{i(n-m)\delta}}{\sqrt{m!}\sqrt{n!}}\braket{m}{n} \\ 156 | \intertext{At this point, we invoke the orthonormality $\braket{m}{n}=\delta_{mn}$ to obtain: } 157 | &= e^{-\abs{c}^2} 158 | \sum_{m=0}^{\infty}\sum_{n=0}^{\infty} 159 | \frac{\abs{c}^2 e^{i(n-m)\delta}}{\sqrt{m!}\sqrt{n!}} \delta_{mn} \\ 160 | &= e^{-\abs{c}^2} \sum_{n=0}^{\infty} \frac{\abs{c}^2}{n!} \\ 161 | &= e^{-\abs{c}^2} e^{\abs{c}^2} \\ 162 | &=1 163 | \end{align*} 164 | \end{subproblem} 165 | 166 | \begin{subproblem} 167 | $\ket{c}$ is an eigenstate of $a_r(\vb{k})$ with the complex eigenvalue $c$: $a_r(\vb{k}) \ket{c} = c\ket{c}$. 168 | \begin{align*} 169 | a_r(\vb{k}) \ket{c} &= 170 | \exp\left(-\frac{1}{2}\abs{c}^2\right) 171 | \sum_{n=0}^{\infty} \frac{c^n}{\sqrt{n!}} a_r(\vb{k})\ket{n} \\ 172 | &= \exp\left(-\frac{1}{2}\abs{c}^2\right) 173 | \left[0 + \sum_{n=1}^{\infty} \frac{c^n}{\sqrt{n!}} \sqrt{n}\ket{n-1}\right] \\ 174 | \intertext{If we redefine the index by $m=n-1$, we get:} 175 | &= \exp\left(-\frac{1}{2}\abs{c}^2\right) 176 | \sum_{m=0}^{\infty} \frac{c^{m+1}}{\sqrt{m!}} \ket{m} \\ 177 | &= c\ket{c} 178 | \end{align*} 179 | \end{subproblem} 180 | 181 | \begin{subproblem} 182 | The expected number of photons in the enclosure $\overbar{N}\define\ev{N}{c}$ is equal to $\abs{c}^2$. 183 | \begin{align*} 184 | \overbar{N} &\define \ev{N}{c} \\ 185 | &= \ev{a^\dagger a}{c} \\ 186 | &= \abs\Big{a\ket{c}}^2 \\ 187 | \intertext{Since $a\ket{c}=c\ket{c}$ as shown in (ii), we get:} 188 | &= \abs\Big{c\ket{c}}^2 \\ 189 | &= \abs{c}^2 190 | \end{align*} 191 | \end{subproblem} 192 | 193 | \begin{subproblem} 194 | The RMS fluctuation of the photon number is given by: 195 | \begin{equation*} 196 | \left(\Delta N\right)^2 \define \ev{N^2}{c} - \overbar{N}^2 = \abs{c}^2 197 | \end{equation*} 198 | Recall that $\left[a_r(\vb{k}),a^\dagger_s(\vb{k}')\right]=\delta_{rs}\delta_{\vb{k}\vb{k}'}$. Hence, in our case with some specific wavevector $\vb{k}$ and polarization $r$, $\left[a, a^\dagger\right]=1$. Then, 199 | \begin{align*} 200 | N^2 &= \left(a^\dagger a\right) \left(a^\dagger a\right) \\ 201 | &= a^\dagger (aa^\dagger) a \\ 202 | &= a^\dagger (N + 1) a \\ 203 | &= a^\dagger N a + N 204 | \end{align*} 205 | Then, the RMS fluctuation becomes: 206 | \begin{align*} 207 | \left(\Delta N\right)^2 &\define \ev{N^2}{c} - \overbar{N}^2 \\ 208 | &= \ev{a^\dagger N a + N}{c} - \overbar{N}^2 \\ 209 | &= \ev{a^\dagger N a}{c} + \overbar{N} - \overbar{N}^2 \\ 210 | &= \abs{c}^2 \ev{N}{c} + \overbar{N} - \overbar{N}^2 \\ 211 | &= \abs{c}^4 + \abs{c}^2 - \abs{c}^4 \\ 212 | &= \abs{c}^2 213 | \end{align*} 214 | \end{subproblem} 215 | 216 | \begin{subproblem} 217 | The expected value of the electric field $\vb{E}$ is given by: 218 | \begin{equation*} 219 | \overbar{\vb{E}} \define \ev{\vb{E}}{c} = -\vb*{\varepsilon}_r(\vb{k}) 2\sqrt{\frac{\hbar \omega_{\vb{k}}}{2V}} \abs{c} \sin(\vb{k} \dotp \vb{x} - \omega_{\vb{k}} t + \delta) 220 | \end{equation*} 221 | where $V$ is the volume of the cubic enclosure. 222 | 223 | Since we are only dealing with free radiation (i.e. no charge distribution, $\phi=0$), we can let the longitudinal field $\vb{E}_\text{L}$ be \emph{zero} in the Coulomb gauge. Then, the remaining electric field is purely transverse, which is given by: 224 | \begin{align*} 225 | \vb{E}\left(\vb{x},t\right) 226 | &= -\frac{1}{c} \pder{\vb{A}\left(\vb{x},t\right)}{t} \\ 227 | &= i \vb*{\varepsilon}_r(\vb{k}) \sqrt{\frac{\hbar \omega_{\vb{k}}}{2V}} 228 | \left[a_r(\vb{k})e^{i\left(\vb{k}\dotp\vb{x} - \omega_{\vb{k}}t\right)} - a^\dagger_r(\vb{k})e^{-i\left(\vb{k}\dotp\vb{x} - \omega_{\vb{k}}t\right)}\right] 229 | \end{align*} 230 | 231 | At this point, we define the phase of the electric field: $\theta \define \vb{k}\dotp\vb{x} - \omega_{\vb{k}}t$. 232 | 233 | The mean electric field $\overbar{\vb{E}}$ is then: 234 | \begin{align*} 235 | \overbar{E} &\define \ev{\vb{E}}{c} \\ 236 | &= i\vb*{\varepsilon}_r(\vb{k}) \sqrt{\frac{\hbar\omega_{\vb{k}}}{2V}} 237 | \left[\ev{a_r(\vb{k})}{c}e^{i\theta} - \ev{a^\dagger_r(\vb{k})}{c}e^{-i\theta}\right] \\ 238 | \intertext{From (ii), $a_r(\vb{k})\ket{c} = c\ket{c}$ and, similarly, $\bra{c}a^\dagger_r(\vb{k}) = \bra{c}c^*$. Also, recall that $c=\abs{c}e^{i\delta}$. Hence, we get:} 239 | &= i\vb*{\varepsilon}_r(\vb{k}) \sqrt{\frac{\hbar\omega_{\vb{k}}}{2V}} 240 | \left[ce^{i\theta} - c^*e^{-i\theta}\right] \\ 241 | &= i\vb*{\varepsilon}_r(\vb{k}) \sqrt{\frac{\hbar\omega_{\vb{k}}}{2V}} 242 | \left[2i\Im{ce^{i\theta}}\right] \\ 243 | &= -\vb*{\varepsilon}_r(\vb{k}) 2\sqrt{\frac{\hbar \omega_{\vb{k}}}{2V}} \abs{c} \sin(\vb{k} \dotp \vb{x} - \omega_{\vb{k}} t + \delta) 244 | \end{align*} 245 | \end{subproblem} 246 | 247 | \begin{subproblem} 248 | The RMS fluctuation $\Delta E$ in the electric field is given by: 249 | \begin{equation*} 250 | \left(\Delta E\right)^2 \define \ev{E^2}{c} - \ev{E}{c}^2 = \frac{\hbar\omega_{\vb{k}}}{2V} 251 | \end{equation*} 252 | Note that the polarization vector $\vb*{\varepsilon}_r$ has been omitted and the electric field here is treated as a scalar quantity, since we are dealing with a definite pure polarization. 253 | 254 | First, we compute the operator $E^2$: 255 | \begin{align*} 256 | E^2 &\define EE \\ 257 | &= \left[i\sqrt{\frac{\hbar\omega_{\vb{k}}}{2V}} 258 | \left(ae^{i\theta} - a^\dagger e^{-i\theta}\right)\right]^2 \\ 259 | &= -\frac{\hbar\omega_{\vb{k}}}{2V}\left[a^2 e^{i2\theta} + a^{\dagger2}e^{-i2\theta} - aa^\dagger - a^\dagger a\right] \\ 260 | &= -\frac{\hbar\omega_{\vb{k}}}{2V}\left[a^2 e^{i2\theta} + a^{\dagger2}e^{-i2\theta} - 2a^\dagger a - 1\right] 261 | \end{align*} 262 | 263 | Then, the RMS fluctuation $\Delta E$ is: 264 | \begin{align*} 265 | \left(\Delta E\right)^2 &\define 266 | \ev{E^2}{c} - \ev{E}{c}^2 \\ 267 | &= -\frac{\hbar\omega_{\vb{k}}}{2V}\left[e^{i2\theta}\ev{a^2}{c} + e^{-i2\theta}\ev{a^{\dagger2}}{c} - 2\ev{a^\dagger a}{c} - \braket{c}\right] \\ 268 | &\qquad - 4\frac{\hbar\omega_{\vb{k}}}{2V}\abs{c}^2\sin^2\left(\theta + \delta\right) \\ 269 | &= -\frac{\hbar\omega_{\vb{k}}}{2V}\left[c^2e^{i2\theta} + c^{*2}e^{-i2\theta} - 2\abs{c}^2 - 1 + 4\abs{c}^2\sin^2\left(\theta+\delta\right) \right] \\ 270 | &= \frac{\hbar\omega_{\vb{k}}}{2V} 271 | \left[1 - c^2e^{i2\theta} - \left(c^2e^{i2\theta}\right)^* + 2\abs{c}^2 - 4\abs{c}^2\sin^2\left(\theta+\delta\right)\right] \\ 272 | &= \frac{\hbar\omega_{\vb{k}}}{2V} 273 | \left[1 - \abs{c}^2\left(2\Re{e^{i2(\theta+\delta)}} - 2 + 4\sin^2(\theta+\delta)\right) \right] \\ 274 | &= \frac{\hbar\omega_{\vb{k}}}{2V} 275 | \left[1 - \abs{c}^2\left(2\cos^2(\theta+\delta)-2\sin^2(\theta+\delta) - 2 + 4\sin^2(\theta+\delta)\right)\right] \\ 276 | &= \frac{\hbar\omega_{\vb{k}}}{2V} 277 | \end{align*} 278 | \end{subproblem} 279 | 280 | From the above calculations, we observe that, in the coherent state $\ket{c}$ which is a superposition of states with all possible number of photons (we sum from $n=0$ to $n=\infty$), the relative fluctuations in both the mean number of photons and the mean electric field strength vanish: 281 | \begin{gather*} 282 | \frac{\Delta N}{\overbar{N}} = \abs{c}^{-1} = \overbar{N}^{-1/2} \\[0.5ex] 283 | \frac{\Delta E}{\overbar{E}} \propto \abs{c}^{-1} = \overbar{N}^{-1/2} 284 | \end{gather*} 285 | as $\overbar{N} \rightarrow \infty$. Hence, in this limit, the state $\ket{c}$ behaves like a classical electromagnetic field. In other words, classical beam of electromagnetic radiation is retained from quantum electrodynamics by considering a \emph{coherent state} in the limit of infinite mean number of photons. 286 | \end{problem} 287 | 288 | 289 | 290 | \begin{problem} 291 | The Lagrangian of a point particle of mass $m$ and electric charge $q$ moving in an electromagnetic potential $(\phi,\vb{A})$ is given by: 292 | \begin{equation*} 293 | L\left(\vb{x},\dot{\vb{x}}\right) = \frac{1}{2}m\dot{\vb{x}}^2 + \frac{q}{c}\vb{A}\dotp\dot{\vb{x}} - q\phi 294 | \end{equation*} 295 | 296 | \begin{subproblem} 297 | The momentum $\vb{p}$ conjugate to $\vb{x}$ is: 298 | \begin{equation*} 299 | \vb{p} \define \pder{L}{\dot{\vb{x}}} = m\dot{\vb{x}} + \frac{q}{c}\vb{A} 300 | \end{equation*} 301 | 302 | The Euler-Lagrange equations become: 303 | \begin{align*} 304 | \der{}{t}\left(\pder{L}{\dot{\vb{x}}}\right) - \pder{L}{\vb{x}} &= \der{}{t}\left(m\dot{x_i} + \frac{q}{c}A_i\right) - \frac{q}{c}\pder{(A_j\dot{x}_j)}{x_i} + q\pder{\phi}{x_i} \\ 305 | &= m\der{}{t}\dot{x_i} + \frac{q}{c}\left[\pder{A_i}{t} + \dot{x}_j\pder{A_i}{x_j}\right] - \frac{q}{c}\pder{A_j}{x_i}\dot{x_j} + q\pder{\phi}{x_i} \\ 306 | &= m\der{}{t}\dot{x_i} + q\left(-\frac{1}{c}\pder{A_i}{t} - \pder{\phi}{x_i}\right) + \frac{q}{c}\dot{x}_j\left(\pder{A_i}{x_j} - \pder{A_j}{x_i}\right) \\ 307 | &= m\der{}{t}\dot{\vb{x}} + q\left(-\frac{1}{c}\pder{\vb{A}}{t}-\pder{\phi}{\vb{x}}\right) + \frac{q}{c}\dot{\vb{x}}\dotp\left(\curl\vb{A}\right) 308 | \end{align*} 309 | where we have used the index notation; repeated indices imply summation. 310 | 311 | The electromagnetic fields $\vb{E}, \vb{B}$ are described in terms of the potentials as: 312 | \begin{equation*} 313 | \vb{E} = -\pder{\phi}{\vb{x}} - \frac{1}{c}\pder{\vb{A}}{t} \,, \, 314 | \vb{B} = \curl \vb{A} 315 | \end{equation*} 316 | and therefore the Euler-Lagrange equations give: 317 | \begin{equation*} 318 | m\der{}{t}\dot{\vb{x}} = q\left(\vb{E} + \frac{1}{c}\dot{\vb{x}}\dotp\vb{B}\right) 319 | \end{equation*} 320 | \end{subproblem} 321 | 322 | \begin{subproblem} 323 | Given a Lagrangian $L\left(\vb{x},\dot{\vb{x}}\right)$, the Hamiltonian $H(\vb{x},\vb{p})$ is: 324 | \begin{equation*} 325 | H(\vb{x},\vb{p}) = \dot{\vb{x}}\dotp\vb{p} - L 326 | \end{equation*} 327 | 328 | Hence, in our case, the Hamiltonian becomes: 329 | \begin{align*} 330 | H(\vb{x},\vb{p}) &= \dot{\vb{x}}\dotp\left(m\dot{\vb{x}}+\frac{q}{c}\vb{A}\right) - \left(\frac{1}{2}m\dot{\vb{x}}^2 + \frac{q}{c}\vb{A}\dotp\dot{\vb{x}} - q\phi\right) \\ 331 | &= \frac{1}{2}m\dot{\vb{x}}^2 + q\phi \\ 332 | \intertext{and since $m\dot{\vb{x}} = \vb{p} - \frac{q}{c}\vb{A}$, we get the desired Hamiltonian:} 333 | &= \frac{1}{2m}\left(\vb{p} - \frac{q}{c}\vb{A}\right)^2 + q\phi 334 | \end{align*} 335 | 336 | With this Hamiltonian, the Hamiltonian equations of motion give: 337 | \begin{align*} 338 | \dot{\vb{x}} &= \pder{H}{\vb{p}} \\ 339 | &= \frac{1}{m}\left(\vb{p} - \frac{q}{c}\vb{A}\right) 340 | \end{align*} 341 | from which we retain: 342 | \begin{equation*} 343 | \vb{p} = m\dot{\vb{x}} + \frac{q}{c}\vb{A} 344 | \end{equation*} 345 | and also 346 | \begin{align*} 347 | \dot{\vb{p}} &= m\der{}{t}\dot{x_i} + \frac{q}{c}\left(\pder{A_i}{t} + \dot{x_j}\pder{A_i}{x_j}\right) \\ 348 | &= -\pder{H}{\vb{x}} \\ 349 | &= -\frac{1}{m}\left(p_j-\frac{q}{c}A_j\right) 350 | \pder{}{x_i}\left(-\frac{q}{c}A_j\right) 351 | - q\pder{\phi}{x_i} \\ 352 | &= +\frac{q}{c}\dot{x}_j\pder{A_j}{x_i} - q\pder{\phi}{x_i} 353 | \end{align*} 354 | from which we retain: 355 | \begin{equation*} 356 | m\der{}{t}\dot{\vb{x}} = q\left(\vb{E}+\frac{1}{c}\dot{\vb{x}}\dotp\vb{B}\right) 357 | \end{equation*} 358 | \end{subproblem} 359 | \end{problem} 360 | 361 | 362 | 363 | \begin{problem} 364 | Consider \textit{a Thomson Scattering} process such that an unpolarized photon with a wavevector $\vb{k}$ \textit{collides} with an electron, and a new photon with a wavevector $\vb{k}'$ linearly polarized in a specific direction is emitted at an angle $\theta$ relative to the incoming photon ($\hat{\vb{k}}\dotp\hat{\vb{k}}'=\cos\theta$). 365 | 366 | The differential cross-section for a Thomson scattering of photons with definite initial and final polarizations $\alpha, \beta$ is given by: 367 | \begin{equation*} 368 | \sigma_{\alpha\rightarrow\beta} \, \diff{\vb*{\Omega}} 369 | = r_0^2 \left[\vb*{\varepsilon}_\alpha\dotp\vb*{\varepsilon}'_\beta\right]^2 \, \diff{\vb*{\Omega}} 370 | \end{equation*} 371 | where $\vb*{\varepsilon}_\alpha, \vb*{\varepsilon}'_\beta$ are the initial and final polarization vectors respectively, and $r_0 \define \frac{e^2}{4\pi mc^2}$ is the classical electron radius. 372 | 373 | To apply this result to our case, we average over the initial polarization $\alpha$ and fix a definite final polarization, say $\beta=1$. That is, 374 | \begin{align*} 375 | \sigma_{\text{unp.}\rightarrow 1} \, \diff{\vb*{\Omega}} 376 | &= \frac{1}{2}\sum_{\alpha=1,2} \sigma_{\alpha\rightarrow 1} \, \diff{\vb*{\Omega}} 377 | \end{align*} 378 | 379 | Since $\vb*{\varepsilon}_1, \vb*{\varepsilon}_2$, and $\hat{\vb{k}}$ form an orthonormal coordinate system, 380 | \begin{equation*} 381 | \sum_{\alpha=1,2} \left(\vb*{\varepsilon}_\alpha \dotp \vb*{\varepsilon}'_\beta\right)^2 382 | = 1 - \left(\hat{\vb{k}} \dotp \vb*{\varepsilon}'_\beta\right)^2 383 | \end{equation*} 384 | Similarly, $\vb*{\varepsilon}'_1, \vb*{\varepsilon}'_2$, and $\hat{\vb{k}}'$ also form an orthonormal coordinate system. In this coordinate system, we can write: 385 | \begin{equation*} 386 | \hat{\vb{k}} = \left(\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta\right) 387 | \end{equation*} 388 | 389 | Therefore, we obtain: 390 | \begin{align*} 391 | \sigma_{\text{unp.}\rightarrow 1} \, \diff{\vb*{\Omega}} 392 | &= \frac{1}{2}\sum_{\alpha=1,2} r_0^2 \left[\vb*{\varepsilon}_\alpha \dotp \vb*{\varepsilon}'_1\right]^2 \, \diff{\vb*{\Omega}} \\ 393 | &= \frac{1}{2}r_0^2\left(1 - \sin^2\theta\cos^2\phi\right) \, \diff{\vb*{\Omega}} 394 | \end{align*} 395 | 396 | If we consider both of the two perpendicular final polarizations, we obtain the unpolarized differential cross-section: 397 | \begin{align*} 398 | \sigma_{\text{unp.}} \, \diff{\vb*{\Omega}} &= 399 | \left(\sigma_{\text{unp.}\rightarrow 1} + \sigma_{\text{unp.}\rightarrow 2}\right) \, \diff{\vb*{\Omega}} \\ 400 | &= \frac{1}{2}r_0^2 \left(2-\sin^2\theta\cos^2\phi-\sin^2\theta\sin^2\phi\right) \, \diff{\vb*{\Omega}} \\ 401 | &= \frac{1}{2}r_0^2\left(1+\cos^2\theta\right) \, \diff{\vb*{\Omega}} 402 | \end{align*} 403 | which is identical Eq.~(1.69a). 404 | 405 | Now, consider the case when the scattering angle $\theta = 90^\circ$. Since the choice of the final state polarization vectors $\vb*{\varepsilon}'_1, \vb*{\varepsilon}'_2$ are arbitrary within a plane perpendicular to $\hat{\vb*{k}}'$, we choose them such that: 406 | \begin{equation*} 407 | \vb*{\varepsilon}'_1 = \frac{\hat{\vb{k}}\cross\hat{\vb{k}}'}{\abs{\hat{\vb{k}}\cross\hat{\vb{k}}'}} 408 | \end{equation*} 409 | and naturally $\vb*{\varepsilon}'_2 = \hat{\vb{k}}' \cross \vb*{\varepsilon}'_1$ for an orthonormal basis. By construction, $\hat{\vb{k}} = \left(\cos\phi, \sin\phi, 0\right)$ is perpendicular to $\vb*{\varepsilon}'_1$ and therefore $\phi=\pi/2$. Then, 410 | \begin{align*} 411 | \sigma_{\alpha\rightarrow 2} = 0 412 | \end{align*} 413 | which implies that the final state polarization must be 100\% in the $\vb*{\varepsilon}'_1$ direction, which is the direction normal to the plane of scattering. 414 | 415 | \end{problem} 416 | 417 | 418 | 419 | \section{Lagrangian Field Theory} 420 | 421 | \begin{problem} 422 | Show the the transformation 423 | \begin{equation*} 424 | \lagr'\left(\phi_r, \phi_{r,\alpha}\right) = \lagr\left(\phi_r,\phi_{r,\alpha}\right) + \partial_\alpha \Lambda^\alpha(x) 425 | \end{equation*} 426 | where $\Lambda^\alpha(x), \alpha=0, ... \,, 3$ are arbitrary functions of the fields $\phi_r(x)$ does not alter the equations of motions. 427 | 428 | First, since $\Lambda^\alpha$ are functions of $\phi_r(x)$ only, we have: 429 | \begin{equation*} 430 | \pder{\Lambda^\alpha}{x^\alpha} 431 | = \pder{\phi_r}{x^\alpha} 432 | \pder{\Lambda^\alpha}{\phi_r} 433 | = \phi_{r,\alpha} \pder{\Lambda^\alpha}{\phi_r} 434 | \end{equation*} 435 | 436 | Then, the Euler-Lagrange equations with the new Lagrangian density $\lagr'$ become: 437 | \begin{align*} 438 | \pder{}{x^\alpha}\left(\pder{\lagr'}{\phi_{r,\alpha}}\right) - \pder{\lagr'}{\phi_r} 439 | &= \pder{}{x^\alpha}\left(\pder{\lagr}{\phi_{r,\alpha}} + \pder{}{\phi_{r,\alpha}}\left(\phi_{s,\alpha}\pder{\Lambda^\alpha}{\phi_s}\right)\right) \\ 440 | &\qquad - \pder{}{\phi_r}\left(\lagr + \phi_{s,\alpha}\pder{\Lambda^\alpha}{\phi_s}\right) \\ 441 | &= \pder{}{x^\alpha}\left(\pder{\lagr}{\phi_{r,\alpha}} + \delta_{rs}\pder{\Lambda^\alpha}{\phi_s}\right) - \pder{\lagr}{\phi_r} - \phi_{s,\alpha}\frac{\partial^2 \Lambda^\alpha}{\partial\phi_r \partial\phi_s} \\ 442 | &= \pder{}{x^\alpha}\left(\pder{\lagr}{\phi_{r,\alpha}}\right) -\pder{\lagr}{\phi_r} \\ 443 | &\qquad + \pder{}{x^\alpha}\left(\pder{\Lambda^\alpha}{\phi_r}\right) - \phi_{s,\alpha}\frac{\partial^2 \Lambda^\alpha}{\partial\phi_r \partial\phi_s} \\ 444 | &= \pder{}{x^\alpha}\left(\pder{\lagr}{\phi_{r,\alpha}}\right) -\pder{\lagr}{\phi_r} \\ 445 | &\qquad + \pder{\phi_s}{x^\alpha}\frac{\partial^2 \Lambda^\alpha}{\partial\phi_s \partial\phi_r} - \phi_{s,\alpha}\frac{\partial^2 \Lambda^\alpha}{\partial\phi_r \partial\phi_s} \\ 446 | &= \pder{}{x^\alpha}\left(\pder{\lagr}{\phi_{r,\alpha}}\right) -\pder{\lagr}{\phi_r} \\ 447 | &= 0 448 | \end{align*} 449 | \end{problem} 450 | 451 | 452 | 453 | \begin{problem} 454 | The real Klein-Gordon field is described by the Hamiltonian density: 455 | \begin{equation*} 456 | \hamil(x) = \frac{1}{2}\left(c^2\pi^2(x) + \left(\grad{\phi}\right)^2 + \mu^2\phi^2\right) 457 | \end{equation*} 458 | and the Hamiltonian of the field is: 459 | \begin{equation*} 460 | H \define \int \diff[3]{\vb{x}} \, \hamil(x) 461 | \end{equation*} 462 | 463 | The commutation relations of the fields are: 464 | \begin{gather*} 465 | \left[\phi(\vb{x},t), \pi(\vb{x}',t)\right] = i\hbar\delta\left(\vb{x}-\vb{x}'\right) \,,\\ 466 | \left[\phi(\vb{x},t), \phi(\vb{x}',t)\right] 467 | = \left[\pi(\vb{x},t), \pi(\vb{x}',t)\right] = 0 468 | \end{gather*} 469 | 470 | Before computing the commutation relations involving the Hamiltonian $H$, let us first compute some simpler commutators. 471 | \begin{align*} 472 | \left[\pi^2(\vb{x}',t), \phi(\vb{x},t)\right] 473 | &= \pi(\vb{x}',t)\pi(\vb{x}',t)\phi(\vb{x},t) 474 | - \phi(\vb{x},t)\pi(\vb{x}',t)\pi(\vb{x}',t) \\ 475 | &= \pi(\vb{x}',t)\pi(\vb{x}',t)\phi(\vb{x},t) \\ 476 | &\qquad - \Big(\pi(\vb{x}',t)\phi(\vb{x},t) + \left[\phi(\vb{x},t), \pi(\vb{x}',t)\right]\Big)\pi(\vb{x}',t) \\ 477 | &= \pi(\vb{x}',t)\left[\pi(\vb{x}',t), \phi(\vb{x},t)\right] 478 | - \left[\phi(\vb{x},t), \pi(\vb{x}',t)\right]\pi(\vb{x}',t) \\ 479 | &= -2i\hbar \delta(\vb{x}-\vb{x}')\pi(\vb{x}',t) 480 | \end{align*} 481 | and 482 | \begin{align*} 483 | \left[\phi^2(\vb{x'},t), \pi(\vb{x},t)\right] 484 | &= \phi(\vb{x}',t)\phi(\vb{x}',t)\pi(\vb{x},t) 485 | - \pi(\vb{x},t)\phi(\vb{x}',t)\phi(\vb{x}',t) \\ 486 | &= \phi(\vb{x}',t)\phi(\vb{x}',t)\pi(\vb{x},t) \\ 487 | &\qquad - \Big(\phi(\vb{x}',t)\pi(\vb{x},t) + \left[\pi(\vb{x},t), \phi(\vb{x}',t)\right]\Big)\phi(\vb{x}',t) \\ 488 | &= \phi(\vb{x}',t)\left[\phi(\vb{x}',t), \pi(\vb{x},t)\right] - \left[\pi(\vb{x},t), \phi(\vb{x}',t)\right]\phi(\vb{x}',t) \\ 489 | &= 2i\hbar \delta(\vb{x}'-\vb{x})\phi(\vb{x}',t) 490 | \end{align*} 491 | 492 | Also, define $\grad' \define \left(\pder{}{x'}, \pder{}{y'}, \pder{}{z'}\right)$. (This is to emphasize that the partial derivatives are taken with respect to $\vb{x}'$, and $\grad'$ therefore does not act on functions of $\vb{x}$, e.g. $\grad{'\phi(\vb{x},t)}=0$.) Then, we have: 493 | \begin{equation*} 494 | \left[\left(\grad'{\phi(\vb{x}',t)}\right)^2, \phi(\vb{x},t)\right] = 0 495 | \end{equation*} 496 | and 497 | \begin{align*} 498 | \left[\left(\grad'{\phi(\vb{x}',t)}\right)^2, \pi(\vb{x},t)\right] &= 499 | \Big(\grad'{\phi(\vb{x}',t)}\dotp\grad'{\phi(\vb{x}',t)}\Big)\pi(\vb{x},t) \\ 500 | &\qquad - \pi(\vb{x},t)\Big(\grad'{\phi(\vb{x}',t)}\dotp\grad'{\phi(\vb{x}',t)}\Big) \\ 501 | &= \grad'{\phi(\vb{x}',t)}\dotp\Big(\grad'{\phi(\vb{x}',t)\pi(\vb{x},t)}\Big) \\ 502 | &\qquad - \grad'{\Big(\pi(\vb{x},t)\phi(\vb{x}',t)\Big)} \dotp \grad'{\phi(\vb{x}',t)} \\ 503 | &= \grad'{\phi(\vb{x}',t)} \dotp \Big(\pi(\vb{x},t)\grad'{\phi(\vb{x}',t)} + \grad'{[\phi(\vb{x}',t), \pi(\vb{x},t)]}\Big) \\ 504 | &\qquad - \Big(\grad'{\phi(\vb{x}',t)}\pi(\vb{x},t) - \grad'{[\phi(\vb{x}',t),\pi(\vb{x},t)]}\Big) \dotp \grad'{\phi(\vb{x}',t)} \\ 505 | &= \grad'{\phi(\vb{x}',t)} \dotp i\hbar\grad'{\delta(\vb{x'}-\vb{x})} + i\hbar\grad{\delta(\vb{x}-\vb{x}')} \dotp \grad'{\phi(\vb{x}',t)} \\ 506 | &= 2i\hbar \grad'{\delta(\vb{x}-\vb{x}')} \dotp \grad'{\phi(\vb{x}',t)} 507 | \end{align*} 508 | 509 | Now, we simply combine the above results to compute the commutator relations with the Hamiltonian: 510 | \begin{align*} 511 | \left[H, \phi(x)\right] &= \left(\int \diff[3]{\vb{x}'}\,\hamil(\vb{x}',t)\right)\phi(\vb{x},t) - \phi(\vb{x},t)\left(\int \diff[3]{\vb{x}'}\,\hamil(\vb{x}',t)\right) \\ 512 | &= \int \diff[3]{\vb{x}'} \, 513 | \frac{1}{2}\Big(c^2\pi^2(\vb{x}',t) + (\grad'{\phi}(\vb{x}',t))^2 + \mu^2\phi^2(\vb{x}',t)\Big)\phi(\vb{x},t) \\ 514 | &\qquad - \int \diff[3]{\vb{x}'} \, 515 | \phi(\vb{x},t)\frac{1}{2}\Big(c^2\pi^2(\vb{x}',t) + (\grad'{\phi}(\vb{x}',t))^2 + \mu^2\phi^2(\vb{x}',t)\Big) \\ 516 | &= \int \diff[3]{\vb{x}'}\, 517 | \frac{1}{2}c^2\left[\pi^2(\vb{x}',t), \phi(\vb{x},t)\right] \\ 518 | &= \int \diff[3]{\vb{x}'}\, 519 | \frac{1}{2}c^2 \left(-2i\hbar\delta(\vb{x}-\vb{x}')\pi(\vb{x}',t)\right) 520 | \\ 521 | &= -i\hbar c^2 \pi(x) 522 | \end{align*} 523 | and 524 | \begin{align*} 525 | [H, \pi(x)] 526 | &= \int \diff[3]{\vb{x}'} \, 527 | \frac{1}{2}\Big(\left[\left(\grad'{\phi(\vb{x}',t)}\right)^2,\pi(\vb{x},t)\right] + \mu^2\left[\phi^2(\vb{x}',t), \pi(\vb{x},t)\right]\Big) \\ 528 | &= \int \diff[3]{\vb{x}'} \, 529 | \frac{1}{2}\Big(2i\hbar\grad'{\delta(\vb{x}-\vb{x}')}\dotp\grad'{\phi(\vb{x}',t)} + 2i\hbar\mu^2\delta(\vb{x}'-\vb{x})\phi(\vb{x}',t)\Big) \\ 530 | &= i\hbar(\mu^2 - \laplacian)\phi(x) 531 | \end{align*} 532 | \end{problem} 533 | 534 | \pagebreak 535 | 536 | \begin{problem} 537 | Consider the Lagrangian density given by: 538 | \begin{equation*} 539 | \lagr = -\frac{1}{2}\big[\partial_\alpha\phi_\beta(x)\big]\big[\partial^\alpha\phi^\beta(x)\big] + 540 | \frac{1}{2}\big[\partial_\alpha\phi^\alpha(x)\big]\big[\partial_\beta\phi^\beta(x)\big] + 541 | \frac{\mu^2}{2}\phi_\alpha(x)\phi^\alpha(x) 542 | \end{equation*} 543 | where $\phi^\alpha(x)$ is a real vector field. 544 | 545 | The corresponding Euler-Lagrange equations are: 546 | \begin{align*} 547 | \pder{}{x^\beta}\left(\pder{\lagr}{\phi^\alpha_{\;,\beta}}\right) - \pder{\lagr}{\phi^\alpha} 548 | &= \pder{}{x^\beta}\Big(-\phi_\alpha^{\;,\beta} + \delta^\beta_\alpha \partial_\gamma\phi^\gamma\Big) - \mu^2\phi_\alpha \\ 549 | &= -\partial_\beta \partial^\beta \phi_\alpha + \partial_\alpha \partial_\gamma \phi^\gamma - \mu^2\phi_\alpha \\ 550 | &= -\dalem\phi_\alpha + \partial_\alpha\partial_\beta \phi^\beta - \mu^2\phi_\alpha \\ 551 | &= 0 552 | \end{align*} 553 | Rewriting $\phi_\alpha = g_{\alpha\beta} \phi^\beta$ and rearranging, we get: 554 | \begin{equation*} 555 | \left[g_{\alpha\beta}\left(\dalem +\mu^2\right) - \partial_\alpha\partial_\beta\right]\phi^\beta(x) = 0 556 | \end{equation*} 557 | 558 | Now, if we take apply an additional derivative $\partial^\alpha$ to the field equations, we get: 559 | \begin{align*} 560 | \partial^\alpha\left[g_{\alpha\beta}\left(\dalem+\mu^2\right) - \partial_\alpha\partial_\beta\right]\phi^\beta(x) 561 | &= \partial_\beta \dalem \phi^\beta + \mu^2 \partial_\beta\phi^\beta - \dalem \partial_\beta \phi^\beta \\ 562 | &= \mu^2 \partial_\beta \phi^\beta \\ 563 | &= 0 564 | \end{align*} 565 | For non-zero $\mu$ (i.e. the fields are \emph{massive}), this implies \emph{the Lorenz Gauge} condition: 566 | \begin{equation*} 567 | \partial_\alpha \phi^\alpha (x) = 0 568 | \end{equation*} 569 | \end{problem} 570 | 571 | 572 | 573 | \begin{problem} 574 | The 3-momentum operator of the fields is defined as: 575 | \begin{equation*} 576 | P^j \define \int \diff[3]{\vb{x}} \, \pi_r(x) \pder{\phi_r(x)}{x_j} 577 | \end{equation*} 578 | 579 | This operator satisfies the following commutation relations: 580 | \begin{align*} 581 | \left[P^j, \phi_r(x)\right] &\define P^j\phi_r(x) - \phi_r(x)P^j \\ 582 | &= \int \diff[3]{\vb{x}'} \, \pi_r(\vb{x}',t)\pder{\phi_r(\vb{x}',t)}{x'_j}\phi_r(\vb{x},t) - \phi_r(\vb{x,t})\pi_r(\vb{x}',t)\pder{\phi_r(\vb{x}',t)}{x'_j} \\ 583 | \intertext{because the derivatives are with respect to $\vb{x}'$ and does not affect $\phi_r(\vb{x},t)$,} 584 | &= \int \diff[3]{\vb{x}'} \, \left[\pi_r(\vb{x}',t), \phi_r(\vb{x},t)\right]\pder{\phi_r(\vb{x}',t)}{x'_j} \\ 585 | &= -\int \diff[3]{\vb{x}'} \, i\hbar\delta\left(\vb{x}-\vb{x}'\right)\pder{\phi_r(\vb{x}',t)}{x'_j} \\ 586 | &= -i\hbar \pder{\phi_r(x)}{x_j} 587 | \end{align*} 588 | and, similarly, 589 | \begin{align*} 590 | \left[P^j, \pi_r(x)\right] 591 | &= \int \diff[3]{\vb{x}'} \, \pi_r(\vb{x}',t)\pder{\phi_r(\vb{x}',t)}{x'_j}\pi_r(\vb{x},t) - \pi_r(\vb{x},t)\pi_r(\vb{x}',t)\pder{\phi_r(\vb{x}',t)}{x'_j} \\ 592 | &= \int \diff[3]{\vb{x}'} \, 593 | \pi_r(\vb{x}',t)\left[\pder{\phi_r(\vb{x}',t)}{x'_j}, \pi_r(\vb{x},t)\right] \\ 594 | &= \int \diff[3]{\vb{x}'} \, 595 | \pi_r(\vb{x}',t) \pder{}{x'_j}\left[\phi_r(\vb{x}',t), \pi_r(\vb{x},t)\right] \\ 596 | &= i\hbar \int \diff[3]{\vb{x}'} \, 597 | \pi_r(\vb{x}',t) \pder{}{x'_j}\delta\left(\vb{x}'-\vb{x}\right) \\ 598 | &= -i\hbar \pder{\pi_r(x)}{x_j} 599 | \end{align*} 600 | 601 | Consider an arbitrary operator $F(x)=F(\phi_r(x),\pi_r(x))$ that can be written as a power series of $\phi_r(x)$ and $\pi_r(x)$: 602 | \begin{equation*} 603 | F(x) = \sum_{n=0}^{\infty} a_n \phi^n_r(x) + b_n \pi^n_r(x) 604 | \end{equation*} 605 | where $a_n$'s and $b_n$'s are arbitrary complex numbers. 606 | 607 | Using the commutator identity 608 | \begin{equation*} 609 | \left[A, B^n\right] = nB^{n-1}\left[A,B\right] 610 | \end{equation*} 611 | we se that $F(x)$ satisfies: 612 | \begin{align*} 613 | \left[P^j, F(x)\right] &= \sum_{m,n=0}^{\infty} a_m\left[P^j, \phi^m_r(x)\right] + b_n\left[P^j, \pi^n_r(x)\right] \\ 614 | &= \sum_{m,n=0}^{\infty} a_m m\phi^{m-1}_r(x) \left[P^j, \phi_r(x)\right] + b_n n\pi^{n-1}_r(x) \left[P^j, \pi_r(x)\right] \\ 615 | &= -i\hbar\sum_{m,n=0}^{\infty} a_m m\phi^{m-1}_r(x)\pder{\phi_r(x)}{x_j} + b_n n\pi^{n-1}_r(x)\pder{\pi_r(x)}{x_j} \\ 616 | &= -i\hbar \pder{F(x)}{x_j} 617 | \end{align*} 618 | 619 | Define the 4-momentum of the fields as: $P^\alpha = (H/c, P^j), \alpha = 0, ... \,, 3$ where $H$ is the Hamiltonian of the fields. Then, combining the above result with the Heisenberg equation of motion: 620 | \begin{equation*} 621 | [H, F(x)] = -i\hbar c \pder{F(x)}{x_0} 622 | \end{equation*} 623 | gives the covariant equations of motion: 624 | \begin{equation*} 625 | \left[P^\alpha, F(x)\right] = -i\hbar \pder{F(x)}{x_\alpha} 626 | \end{equation*} 627 | \end{problem} 628 | 629 | 630 | 631 | \begin{problem} 632 | Consider a scalar field $\phi_r(x_\alpha)$ invariant under translations $x_\alpha \longrightarrow x'_\alpha = x_\alpha + \delta_\alpha$ where $\delta_\alpha$ is a constant 4-vector. i.e., 633 | \begin{align*} 634 | \phi'(x'_\alpha) &= \phi(x_\alpha) \\ 635 | \intertext{or, equivalently,} 636 | \phi'(x_\alpha) &= \phi(x_\alpha - \delta_\alpha) 637 | \end{align*} 638 | 639 | In the limit of an infinitesimal transformation ($\delta_\alpha \ll 1$), we can write the transformation as: 640 | \begin{align*} 641 | \phi'(x_\alpha) &= \phi(x_\alpha - \delta_\alpha) \\ 642 | &= \phi(x_\alpha) - \delta_\alpha \pder{\phi}{x_\alpha} + \mathcal{O}\left(\delta^2\right) \\ 643 | &\approx \phi(x_\alpha) + \delta_\alpha\frac{1}{i\hbar}\left[P^\alpha, \phi(x)\right] 644 | \end{align*} 645 | where we have used the result from the previous problem: $\left[P^\alpha, F(x)\right] = -i\hbar\pder{F(x)}{x_\alpha}$. 646 | 647 | Let us write the corresponding unitary transformation $U$ as: 648 | \begin{equation*} 649 | U = e^{i\delta_\alpha T^\alpha} 650 | \end{equation*} 651 | where $T^\alpha$ is a Hermitian operator ($T^{\alpha\dagger}=T^\alpha$). In terms of $U$, the same translation transformation can also be written as: 652 | \begin{align*} 653 | \phi'(x) &= U\phi(x)U^\dagger \\ 654 | &\approx (1+i\delta_\alpha T^\alpha) \phi(x) (1-i\delta_\alpha T^\alpha) \\ 655 | &= \phi(x) + i\delta_\alpha\left[T^\alpha, \phi(x)\right] + \mathcal{O}(\delta^2) 656 | \end{align*} 657 | 658 | Comparing the two results, we see that $T^\alpha = -P^\alpha / \hbar$ and therefore: 659 | \begin{equation*} 660 | U = e^{-i\delta_\alpha P^\alpha / \hbar} 661 | \end{equation*} 662 | 663 | \end{problem} 664 | 665 | 666 | 667 | \section{The Klein-Gordon Field} 668 | 669 | \begin{problem} 670 | A real Klein-Gordon field can be written as: 671 | \begin{align*} 672 | \phi(x) &= \phi^+(x) + \phi^-(x) \\ 673 | &= \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V\omega_{\vb{k}}}} \left[a(\vb{k})e^{-ikx} + a^\dagger(\vb{k})e^{ikx}\right] 674 | \end{align*} 675 | where $\vb{k}$ is the wavenumber of the field and $k^0 = \omega_{\vb{k}}/c$. 676 | 677 | Then, we have: 678 | \begin{align*} 679 | \dot{\phi}(x) &\define \pder{\phi(x)}{t} \\ 680 | &= \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V\omega_{\vb{k}}}} 681 | \left[\left(-i\frac{\omega_{\vb{k}}}{c}c\right)a(\vb{k})e^{-ikx} + \left(i\frac{\omega_{\vb{k}}}{c}c\right) a^\dagger(\vb{k})e^{ikx}\right] \\ 682 | &= -i\sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V \omega_{\vb{k}}}} \omega_{\vb{k}} 683 | \left[a(\vb{k})e^{-ikx} - a^\dagger(\vb{k})e^{-ikx}\right] 684 | \end{align*} 685 | and therefore: 686 | \begin{align*} 687 | i\dot{\phi}(x) + \omega_{\vb{k}}\phi(x) 688 | &= \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V\omega_{\vb{k}}}} \omega_{\vb{k}} 2a(\vb{k})e^{-ikx} 689 | \end{align*} 690 | 691 | To re-write the expression in the $\vb{k}$-space, use Fourier transform, i.e. apply $\int_{V} \diff[3]{\vb{x}} \, e^{ik'x}$ on both sides. Then, we get: 692 | \begin{align*} 693 | \int_{V} \diff[3]{\vb{x}} \, e^{ik'x} \left(i\dot{\phi}(x) + \omega_{\vb{k}}\phi(x)\right) 694 | &= \int_{V} \diff[3]{\vb{x}} \, \sum_{\vb{k}} \sqrt{\frac{2\hbar c^2 \omega_{\vb{k}}}{V}} a(\vb{k})e^{-ikx} e^{ik'x} \\ 695 | &= \int_{V} \diff[3]{\vb{x}} \, \sum_{\vb{k}} \sqrt{\frac{2\hbar c^2 \omega_{\vb{k}}}{V}} a(\vb{k}) \delta_{\vb{k}\vb{k}'} \\ 696 | &= \sqrt{2\hbar c^2 V \omega_{\vb{k}'}} a(\vb{k}') 697 | \end{align*} 698 | Rewriting this, we get: 699 | \begin{equation*} 700 | a(\vb{k}) = \left(2\hbar c^2 V \omega_{\vb{k}}\right)^{-1/2} \int_{V} \diff[3]{\vb{x}} \, e^{ikx}\left(i\dot{\phi}(x) + \omega_{\vb{k}}\phi(x)\right) 701 | \end{equation*} 702 | 703 | 704 | 705 | We define $x=(\vb{x},t)$ and $x'=(\vb{x}',t)$ to denote two different points in space. Then, we have the following commutator relationship: 706 | \begin{align*} 707 | \left[a(\vb{k}), a^\dagger(\vb{k}')\right] 708 | &= \frac{1}{2\hbar c^2 V \sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \int_V \diff[3]{\vb{x}} \int_V \diff[3]{\vb{x}'} \, e^{i\left(kx-k'x'\right)} \\ 709 | &\qquad \times \Big[ \left(i\dot{\phi}(x) + \omega_{\vb{k}}\phi(x)\right)\left(-i\dot{\phi}(x') + \omega_{\vb{k}'}\phi(x')\right) \\ 710 | &\qquad \hspace{5em} - \left(-i\dot{\phi}(x') + \omega_{\vb{k}'}\phi(x')\right)\left(i\dot{\phi}(x) + \omega_{\vb{k}}\phi(x)\right)\Big] 711 | \end{align*} 712 | since $\phi(x)$ is a real field and thus a Hermitian operator after quantization. 713 | 714 | Given the commutator relationship $[\phi(\vb{x},t), \dot{\phi}(\vb{x}',t)]$, this then becomes: 715 | \begin{align*} 716 | \left[a(\vb{k}, a^\dagger(\vb{k}'))\right] &= \frac{1}{2\hbar c^2 V \sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \int_V \diff[3]{\vb{x}} \int_V \diff[3]{\vb{x}'} \, e^{i\left(kx-k'x'\right)} \\ 717 | &\qquad \times \Big[i\omega_{\vb{k}'}[\dot{\phi}(x), \phi(x')] - i\omega_{\vb{k}}[\phi(x), \dot{\phi}(x')]\Big] \\ 718 | &= \frac{1}{2\hbar c^2 V \sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \int_V \diff[3]{\vb{x}} \int_V \diff[3]{\vb{x}'} \, e^{i\left(kx-k'x'\right)} \\ 719 | &\qquad \times \Big[\omega_{\vb{k}'}\hbar c^2 \delta\left(\vb{x}'-\vb{x}\right) + \omega_{\vb{k}}\hbar c^2 \delta\left(\vb{x} - \vb{x}'\right)\Big] \\ 720 | &= \frac{ \hbar c^2 \left(\omega_{\vb{k}}+\omega_{\vb{k}'}\right)}{2\hbar c^2 V \sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \int_V \diff[3]\vb{x} \, 721 | e^{i\left(k-k'\right)x} \\ 722 | &= \frac{\omega_{\vb{k}} + \omega_{\vb{k}'}}{2V\sqrt{\omega_{\vb{k}}\omega_{\vb{k}'}}} V \delta_{\vb{k}\vb{k}'} e^{i\left(\omega_{\vb{k}} - \omega_{\vb{k}'}\right)t} \\ 723 | &= \delta_{\vb{k}\vb{k}'} 724 | \end{align*} 725 | 726 | Also, we see that: 727 | \begin{align*} 728 | \left[a(\vb{k}), a(\vb{k}')\right] 729 | &= \frac{1}{2\hbar c^2 V \sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \int_V \diff[3]{\vb{x}} \int_V \diff[3]{\vb{x}'} \, e^{i\left(kx+k'x'\right)} \\ 730 | &\qquad \times \Big[ \left(i\dot{\phi}(x) + \omega_{\vb{k}}\phi(x)\right)\left(i\dot{\phi}(x') + \omega_{\vb{k}'}\phi(x')\right) \\ 731 | &\qquad \hspace{5em} - \left(i\dot{\phi}(x') + \omega_{\vb{k}'}\phi(x')\right)\left(i\dot{\phi}(x) + \omega_{\vb{k}}\phi(x)\right)\Big] \\ 732 | &= \frac{1}{2\hbar c^2 V \sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \int_V \diff[3]{\vb{x}} \int_V \diff[3]{\vb{x}'} \, e^{i\left(kx+k'x'\right)} \\ 733 | &\qquad \times \Big[i\omega_{\vb{k}'}[\dot{\phi}(x),\phi(x')] + i\omega_{\vb{k}}[\phi(x), \dot{\phi}(x')]\Big] \\ 734 | &= \frac{1}{2\hbar c^2 V \sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \int_V \diff[3]{\vb{x}} \int_V \diff[3]{\vb{x}'} \, e^{i\left(kx+k'x'\right)} \\ 735 | &\qquad \times \Big[+\omega_{\vb{k}'}\hbar c^2\delta\left(\vb{x}'-\vb{x}\right) - \omega_{\vb{k}}\hbar c^2 \delta(\vb{x}-\vb{x}')\Big] \\ 736 | &= \frac{1}{2 V \sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \int_V \diff[3]{\vb{x}} \, e^{i\left(k+k'\right)x} \Big[\omega_{\vb{k}'} - \omega_{\vb{k}}\Big] \\ 737 | &= \frac{1}{2\sqrt{\omega_{\vb{k}}\omega_{\vb{k}'}}} \delta_{\vb{k},-\vb{k}'} (\omega_{\vb{k}'} - \omega_{\vb{k}}) 738 | \end{align*} 739 | 740 | However, $\omega_{\vb{k}} = \omega_{-\vb{k}}$. Hence, when the delta function is non-zero (i.e. $\vb{k} = -\vb{k}'$), $\omega_{\vb{k}}-\omega_{\vb{k}'} = 0$. Therefore, 741 | \begin{gather*} 742 | \left[a(\vb{k}), a(\vb{k}')\right] = 0 \\ 743 | \intertext{and, similarly,} 744 | \left[a^\dagger(\vb{k}), a^\dagger(\vb{k}')\right] = 0 745 | \end{gather*} 746 | \end{problem} 747 | 748 | 749 | 750 | \begin{problem} 751 | A pair of complex Klein-Gordon fields $\phi(x)$ and $\phi^\dagger(x)$ can be described by two independent real fields $\phi_r(x), r=1,2$: 752 | \begin{equation*} 753 | \phi = \frac{1}{\sqrt{2}}\left(\phi_1 + i\phi_2\right) \,, \, 754 | \phi^\dagger = \frac{1}{\sqrt{2}}\left(\phi_1 - i\phi_2\right) 755 | \end{equation*} 756 | and each of the real fields can be expanded as: 757 | \begin{equation*} 758 | \phi_r(x) = \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V \omega_{\vb{k}}}} \left[a_r(\vb{k})e^{-ikx} + a^\dagger_r(\vb{k})e^{ikx}\right] 759 | \end{equation*} 760 | 761 | Now, suppose we wish to expand the complex field $\phi(x)$ as: 762 | \begin{equation*} 763 | \phi(x) = \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V \omega_{\vb{k}}}} \left[a(\vb{k})e^{-ikx} + b^\dagger(\vb{k})e^{ikx}\right] 764 | \end{equation*} 765 | This expansion, in terms of the real fields $\phi_r(x)$, requires: 766 | \begin{align*} 767 | &= \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V \omega_{\vb{k}}}} \frac{1}{\sqrt{2}}\left[a_1(\vb{k})e^{-ikx} + a^\dagger_1(\vb{k})e^{ikx} + ia_2(\vb{k})e^{-ikx} + ia^\dagger_2(\vb{k})e^{ikx}\right] \\ 768 | &= \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V \omega_{\vb{k}}}} 769 | \left[\frac{1}{\sqrt{2}}\big(a_1(\vb{k}) + ia_2(\vb{k})\big)e^{-ikx} + \frac{1}{\sqrt{2}}\big(a^\dagger_1(\vb{k}) + ia^\dagger_2(\vb{k})\big)e^{ikx}\right] 770 | \end{align*} 771 | 772 | Therefore, this expansion of the complex field $\phi(x)$ requires: 773 | \begin{align*} 774 | a(\vb{k}) = \frac{1}{\sqrt{2}}\left(a_1(\vb{k}) + ia_2(\vb{k})\right) 775 | \end{align*} 776 | and 777 | \begin{align*} 778 | b(\vb{k}) = \frac{1}{\sqrt{2}}\left(a_1(\vb{k}) - ia_2(\vb{k})\right) 779 | \end{align*} 780 | 781 | From here on, let $x=(\vb{x},t)$ and $x'=(\vb{x}',t)$. In other words, we will not consider commutators of operators at different times in this problem. 782 | 783 | Now, from the commutator relationships of the real field operators, we can compute those of the complex field operators. 784 | \begin{align*} 785 | \left[\phi(x), \dot{\phi}^\dagger(x')\right] 786 | &= \left[\frac{1}{\sqrt{2}}\left(\phi_1(x)+i\phi_2(x)\right), \frac{1}{\sqrt{2}}\left(\dot{\phi}_1(x') - i\dot{\phi}_2(x')\right)\right] \\ 787 | &= \frac{1}{2}\left(\left[\phi_1(x), \dot{\phi}_1(x')\right] + \left[\phi_2(x), \dot{\phi}_2(x')\right]\right) \\ 788 | &= i\hbar c^2 \delta\left(\vb{x}-\vb{x}'\right) 789 | \end{align*} 790 | since $\left[\phi_r(x), \phi_r(x')\right] = i \hbar c^2 \delta\left(\vb{x}-\vb{x}'\right)$ for any real field $\phi_r$. 791 | 792 | As one would expect based on the commutators of real fields, we see that: 793 | \begin{align*} 794 | \big[\phi(x), \phi(x')\big] &= \big[\dot{\phi}(x), \dot{\phi}(x')\big] \\ 795 | = \big[\phi^\dagger(x), \phi^\dagger(x')\big] &= \big[\dot{\phi}^\dagger(x), \dot{\phi}^\dagger(x')\big] = 0 796 | \end{align*} 797 | Also, because $\phi_1, \phi_2$ commute (at a given time) and $\phi, \phi^\dagger$ are merely the linear combinations, we also see that: 798 | \begin{equation*} 799 | \left[\phi(x), \phi^\dagger(x')\right] = \left[\dot{\phi}(x), \dot{\phi}^\dagger(x')\right] = 0 800 | \end{equation*} 801 | (In this sense, $\phi$ and $\phi^\dagger$ are independent fields.) 802 | 803 | One less intuitive (and different from the properties of real fields) result is that: 804 | \begin{align*} 805 | \left[\phi(x), \dot{\phi}(x')\right] 806 | &= \left[\frac{1}{\sqrt{2}}\left(\phi_1(x)+i\phi_2(x)\right), \frac{1}{\sqrt{2}}\left(\dot{\phi}_1(x') + i\dot{\phi}_2(x')\right)\right] \\ 807 | &= \frac{1}{2} \left(\left[\phi_1(x), \dot{\phi}_1(x')\right] - \left[\phi_2(x), \dot{\phi}_2(x')\right]\right) \\ 808 | &= 0 809 | \end{align*} 810 | and, similarly, $\left[\phi^\dagger(x), \dot{\phi}^\dagger(x')\right]=0$. 811 | 812 | Furthermore, we see that: 813 | \begin{align*} 814 | \left[a(\vb{k}), a^\dagger(\vb{k}')\right] 815 | &= \left[\frac{1}{\sqrt{2}}\big(a_1(\vb{k}) + ia_2(\vb{k})\big), \frac{1}{\sqrt{2}}\big(a^\dagger_1(\vb{k}')-ia^\dagger_2(\vb{k}')\big)\right] \\ 816 | &= \frac{1}{2}\Big(\big[a_1(\vb{k}), a^\dagger_1(\vb{k}')\big] + \big[a_2(\vb{k}), a^\dagger_2(\vb{k}')\big]\Big) \\ 817 | &= \delta_{\vb{k}\vb{k}'} 818 | \end{align*} 819 | and, similarly, $\left[b(\vb{k}), b^\dagger(\vb{k}')\right] = \delta_{\vb{k}, \vb{k}'}$. 820 | 821 | It is easy to see that 822 | \begin{equation*} 823 | \big[a(\vb{k}), a(\vb{k}')\big] = \big[b(\vb{k}), b(\vb{k}')\big] = \big[a(\vb{k}), b(\vb{k}')\big] = 0 824 | \end{equation*} 825 | since these will involve annihilation operators \emph{only}, and the two real fields are independent. The same holds for their Hermitian conjugates (creation operators only). 826 | 827 | Lastly, we also can compute: 828 | \begin{align*} 829 | \big[a(\vb{k}), b^\dagger(\vb{k}')\big] 830 | &= \left[\frac{1}{\sqrt{2}}\big(a_1(\vb{k}) + ia_2(\vb{k})\big), \frac{1}{\sqrt{2}}\big(a^\dagger_1(\vb{k}') + ia^\dagger_2(\vb{k}')\big)\right] \\ 831 | &= \frac{1}{2}\Big(\big[a_1(\vb{k}), a^\dagger_1(\vb{k}')\big] - \big[a_2(\vb{k}), a^\dagger_2(\vb{k}')\big]\Big) \\ 832 | &= 0 833 | \end{align*} 834 | \end{problem} 835 | 836 | 837 | 838 | \begin{problem} 839 | The Feynman $\Delta$-function can be written as: 840 | \begin{equation*} 841 | \Delta_\text{F}(x) = \frac{1}{(2\pi)^4} \int\diff^4{k} \, \frac{e^{-ikx}}{k^2-\mu^2+i\epsilon} 842 | \end{equation*} 843 | where we let $\epsilon$ tend to zero after the integration. 844 | 845 | Applying the operator $\left(\dalem+\mu^2\right)$ to this function, we get: 846 | \begin{align*} 847 | \left(\dalem + \mu^2\right) \Delta_\text{F}(x) 848 | &= \left(\dalem + \mu^2\right) \frac{1}{(2\pi)^4} \int\diff[4]{k} \, \frac{e^{-ikx}}{k^2-\mu^2+i\epsilon} \\ 849 | &= \frac{1}{(2\pi)^4} \int\diff[4]{k}\, \frac{1}{k^2-\mu^2+i\epsilon}\left(\dalem + \mu^2\right)e^{-ikx} \\ 850 | &= \frac{1}{(2\pi)^4} \int\diff[4]{k} \, \frac{1}{k^2-\mu^2+i\epsilon} \left(-k^2+\mu^2\right)e^{-ikx} \\ 851 | &= \frac{1}{(2\pi)^4} \int\diff[4]{k} \, \frac{-e^{-ikx}}{1+i\kappa} \\ 852 | &= -\delta^{(4)}(x) 853 | \end{align*} 854 | where we have defined $\kappa = \epsilon/(k^2-\mu^2)$ and let it tend to zero. 855 | 856 | Hence, the Feynman $\Delta$-function satisfies the inhomogeneous Klein-Gordon equation. 857 | \end{problem} 858 | 859 | 860 | 861 | \begin{problem} 862 | Let $\phi$ be a complex Klein-Gordon field, representing a charged meson. It can be written as: 863 | \begin{equation*} 864 | \phi(x) = \phi^+(x) + \phi^-(x) = \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V\omega_{\vb{k}}}} \big[a(\vb{k})e^{-ikx} + b^\dagger(\vb{k})e^{ikx}\big] 865 | \end{equation*} 866 | where the operators $a$ and $b$ follow the commutator relationships derived earlier. 867 | 868 | Recall from Problem~2 that: 869 | \begin{equation*} 870 | \big[a(\vb{k}), a(\vb{k}')\big] = \big[b^\dagger(\vb{k}), b^\dagger(\vb{k}')\big] = \big[a(\vb{k}), b^\dagger(\vb{k}')\big] = 0 871 | \end{equation*} 872 | and therefore, unlike the commutator of a real field, we now have: 873 | \begin{equation*} 874 | \left[\phi(x), \phi(x')\right] = 0 875 | \end{equation*} 876 | for any two different points $x,x'$ in spacetime. 877 | 878 | Instead, for complex fields, we see that: 879 | \begin{align*} 880 | \big[\phi(x), \phi^\dagger(x')\big] 881 | &= \big[\phi^+(x), \phi^{\dagger-}(x')\big] + \big[\phi^-(x), \phi^{\dagger+}(x')\big] 882 | \end{align*} 883 | is non-zero. 884 | 885 | The first commutator is: 886 | \begin{align*} 887 | \big[\phi^+(x), \phi^{\dagger-}(x')\big] 888 | &= \frac{\hbar c^2}{2V} \sum_{\vb{k}, \vb{k}'} \frac{1}{\sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \big[a(\vb{k}), a^\dagger(\vb{k}')\big]e^{-i\left(kx-k'x\right)} \\ 889 | &= \frac{\hbar c^2}{2V} \sum_{\vb{k}, \vb{k}'} \frac{1}{\sqrt{\omega_{\vb{k}} \omega_{\vb{k}'}}} \delta_{\vb{k}\vb{k}'} e^{-i\left(kx-k'x\right)} \\ 890 | &= \frac{\hbar c^2}{2V} \sum_{\vb{k}} \frac{e^{-ik(x-x')}}{\omega_{\vb{k}}} 891 | \intertext{which, upon taking the limit $V\rightarrow\infty$, becomes:} 892 | &= \frac{\hbar c^2}{2(2\pi)^3} \int \diff[3]{\vb{k}} \, \frac{e^{-ik(x-x')}}{\omega_{\vb{k}}} \\ 893 | &\define i\hbar c \Delta^+(x-x') 894 | \end{align*} 895 | where 896 | \begin{equation*} 897 | \Delta^+(x-x') \define \frac{-ic}{2(2\pi)^3}\int\frac{\diff[3]{\vb{k}}}{\omega_{\vb{k}}} e^{-ik(x-x')} 898 | \end{equation*} 899 | is defined the same way as it was defined for real field Klein-Gordon fields (neutral mesons). Similarly, the second commutator is: 900 | \begin{align*} 901 | \big[\phi^-(x), \phi^{\dagger+}(x')\big] 902 | &= i\hbar c \Delta^-(x-x') 903 | \end{align*} 904 | 905 | Therefore, we have: 906 | \begin{equation*} 907 | \big[\phi(x), \phi^\dagger(x')\big] = i\hbar c \left(\Delta^+(x-x') + \Delta^-(x-x')\right) \define i\hbar c \Delta(x-x') 908 | \end{equation*} 909 | where $\Delta(x) \define \Delta^+(x) + \Delta^-(x)$. 910 | 911 | Now, define the Feynman $\Delta$-function of a complex Klein-Gordon field $\phi$ as: 912 | \begin{equation*} 913 | i\hbar c \Delta_\text{F}(x-x') \define \ev{\timep{\phi(x)\phi^\dagger(x')}}{0} 914 | \end{equation*} 915 | where $\timep{\hspace{1em}}$ is the time-ordered product. 916 | 917 | Then, this Feynman $\Delta$-function, or \emph{the Feynman propagator}, turns out to be of the same form as that of a real Klein-Gordon field: 918 | \begin{align*} 919 | \Delta_\text{F}(x) &= \theta(t)\Delta^+(x) - \theta(-t)\Delta^-(x) \\ 920 | &= \frac{1}{(2\pi)^4} \int \diff[4]{k} \frac{e^{-ikx}}{k^2-\mu^2+i\epsilon} 921 | \end{align*} 922 | where $\theta(t)$ is the Heaviside step function. 923 | 924 | If $t't$, then $\timep{\phi(x)\phi^\dagger(x')} = \phi^\dagger(x')\phi(x)$, and therefore: 932 | \begin{align*} 933 | i\hbar c \Delta_\text{F}(x-x') &= \ev{\phi^\dagger(x')\phi(x)}{0} \\ 934 | &= \ev{\phi^{\dagger+}(x')\phi^-(x)}{0} 935 | \end{align*} 936 | which is interpreted as: an anti-particle is created by $b^\dagger(\vb{k})$ at $x$, then propagates to $x'$, and is absorbed (annihilated) by $b(\vb{k})$ at $x$. 937 | \end{problem} 938 | 939 | 940 | 941 | \begin{problem} 942 | The charge conjugation operator $\mathscr{C}$ is defined by: 943 | \begin{equation*} 944 | \phi(x) \longrightarrow \mathscr{C}\phi(x)\mathscr{C}^{-1} = \eta_\text{c} \, \phi^\dagger(x) 945 | \end{equation*} 946 | where $\mathscr{C}$ is a unitary operator which leaves the vacuum invariant (i.e., $\mathscr{C}\ket{0} = \ket{0}$), and $\eta_\text{c}$ is a phase factor. 947 | 948 | First, note that: 949 | \begin{align*} 950 | \phi^\dagger(x) &\rightarrow \mathscr{C}\phi^\dagger(x)\mathscr{C}^{-1} \\ 951 | \intertext{and, since $\mathscr{C}$ is unitary, we can rewrite this as:} 952 | &= (\mathscr{C}^{-1})^\dagger \phi^\dagger(x) C^\dagger \\ 953 | &= \left(\mathscr{C} \phi(x) \mathscr{C}^{-1}\right)^\dagger \\ 954 | &= \left(\eta_\text{c} \phi^\dagger(x)\right)^\dagger \\ 955 | &= \eta_\text{c}^* \, \phi(x) 956 | \end{align*} 957 | 958 | Then, under the charge conjugation, the complex Klein-Gordon Lagrangian density $\lagr = \mathrm{N}\Big((\partial_\alpha \phi^\dagger) (\partial^\alpha \phi) - \mu^2 \phi^\dagger\phi\Big)$ becomes: 959 | \begin{align*} 960 | \lagr' &= \mathrm{N}\Big((\eta^*_\text{c}\partial_\alpha \phi) (\eta_\text{c}\partial^\alpha \phi^\dagger) - \mu^2 (\eta^*_\text{c}\phi)(\eta_\text{c}\phi^\dagger) \Big) \\ 961 | &= \left(\eta^*_\text{c} \eta_\text{c}\right) 962 | \mathrm{N}\Big((\partial_\alpha \phi)(\partial^\alpha \phi^\dagger) - \mu^2 \phi\phi^\dagger \Big) \\ 963 | \intertext{Since all commutators inside the normal product vanish, this simplifies to:} 964 | &= \mathrm{N}\Big((\partial_\alpha \phi^\dagger) (\partial^\alpha \phi) - \mu^2 \phi^\dagger\phi\Big) \\ 965 | &= \lagr 966 | \end{align*} 967 | Thus, the Lagrangian density is invariant under the charge conjugation. 968 | 969 | 970 | On the other hand, the charge-current density 971 | \begin{equation*} 972 | s^\alpha = -i\frac{q}{\hbar} \mathrm{N}\Big(\phi \partial^\alpha\phi^\dagger - \phi^\dagger \partial^\alpha\phi\Big) 973 | \end{equation*} 974 | becomes: 975 | \begin{align*} 976 | s'^\alpha &= -i\frac{q}{\hbar} \mathrm{N}\Big((\eta_\text{c}\phi^\dagger)\partial^\alpha(\eta^*_\text{c}\phi) - (\eta^*_\text{c}\phi)\partial^\alpha(\eta_\text{c}\phi^\dagger)\Big) \\ 977 | &= -i\frac{q}{\hbar} (\eta^*_\text{c}\eta_\text{c}) \mathrm{N}\Big(\phi^\dagger\partial^\alpha \phi - \phi \partial^\alpha \phi^\dagger\Big) \\ 978 | &= +i \frac{q}{\hbar} \mathrm{N}\Big(\phi \partial^\alpha\phi^\dagger - \phi^\dagger \partial^\alpha\phi\Big) \\ 979 | &= -s^\alpha 980 | \end{align*} 981 | Thus, the charge-current density changes sign under the charge conjugation. 982 | 983 | Writing $\phi(x), \phi^\dagger(x)$ out in their expansion forms in terms of the absorption and creation operators, we see that: 984 | \begin{align*} 985 | \mathscr{C} \phi(x) \mathscr{C}^{-1} 986 | &= \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V\omega_{\vb{k}}}} \big[\mathscr{C} a(\vb{k}) \mathscr{C}^{-1} e^{-ikx} + \mathscr{C} b^\dagger(\vb{k}) \mathscr{C}^{-1} e^{ikx}\big] \\ 987 | &= \eta_\text{c} \, \phi^\dagger(x) \\ 988 | &= \eta_\text{c} \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V \omega_{\vb{k}}}} \big[a^\dagger(\vb{k})e^{ikx} + b(\vb{k})e^{-ikx}\big] 989 | \end{align*} 990 | which implies that: 991 | \begin{equation*} 992 | \mathscr{C} a(\vb{k}) \mathscr{C}^{-1} = \eta_\text{c}\, b(\vb{k}) \,, 993 | \, \mathscr{C} b(\vb{k}) \mathscr{C}^{-1} = \eta^*_\text{c} \, a(\vb{k}) 994 | \end{equation*} 995 | 996 | Then, we also see that, under the charge conjugation, single-particle states transform as: 997 | \begin{align*} 998 | \mathscr{C} \ket{a,\vb{k}} &= \mathscr{C} a^\dagger(\vb{k}) \ket{0} \\ 999 | &= \mathscr{C} a^\dagger(\vb{k}) \mathscr{C}^{-1} \ket{0} \\ 1000 | &= \left(\mathscr{C}a(\vb{k})\mathscr{C}^{-1}\right)^\dagger \ket{0} \\ 1001 | &= \eta^*_\text{c} \, b^\dagger(\vb{k})\ket{0} \\ 1002 | &= \eta^*_\text{c} \ket{b,\vb{k}} 1003 | \intertext{and, similarly,} 1004 | \mathscr{C} \ket{b,\vb{k}} &= \eta_\text{c} \ket{a,\vb{k}} 1005 | \end{align*} 1006 | where we have used the fact that the vacuum state $\ket{0}$ is unchanged by $\mathscr{C}$ or $\mathscr{C}^{-1}$. 1007 | 1008 | Thus, the charge conjugation $\mathscr{C}$ \emph{interchanges} the particles ($a$-particles) and the anti-particles ($b$-particles). 1009 | \end{problem} 1010 | 1011 | 1012 | 1013 | \begin{problem} 1014 | The parity transformation (i.e. spatial inversion) of the Hermitian (real) Klein-Gordon field $\phi(x)$ is defined by: 1015 | \begin{equation*} 1016 | \phi(\vb{x},t) \longrightarrow \mathscr{P}\phi(\vb{x},t)\mathscr{P}^{-1} = \eta_\text{p} \, \phi(-\vb{x},t) 1017 | \end{equation*} 1018 | where the parity operator $\mathscr{P}$ is unitary which leaves the vacuum invariant (i.e., $\mathscr{P}\ket{0} = \ket{0}$), and $\eta_\text{p} = \pm1$ is called \emph{the intrinsic parity} of the field. Note that $\mathscr{P}$ \emph{only} influences the spatial argument of the field, but not the time $t$. 1019 | 1020 | Under the given parity transformation, the Lagrangian density of the real Klein-Gordon field $\lagr = \frac{1}{2} \left((\partial_\alpha \phi)(\partial^\alpha \phi) - \mu^2\phi^2\right)$ transform as: 1021 | \begin{align*} 1022 | \lagr'(\vb{x},t) &= \frac{1}{2} \left[\left(\eta_\text{p}\pder{\phi(-\vb{x},t)}{x^\alpha}\right)\left(\eta_\text{p}\pder{\phi(-\vb{x},t)}{x_\alpha}\right) - \mu^2 \eta_\text{p}^2 \phi^2(-\vb{x},t) \right] \\ 1023 | &= \frac{\eta_\text{p}^2}{2}\left[\frac{1}{c^2}\npder{2}{\phi(-\vb{x},t)}{t} - \left(\grad{\phi(-\vb{x},t)}\right)^2 - \mu^2\phi^2(-\vb{x},t)\right] \\ 1024 | &= \frac{1}{2} \left[\frac{1}{c^2}\npder{2}{\phi(\vb{x},t)}{t} - (-1)^2(\grad{\phi(\vb{x},t)})^2 - \mu^2\phi^2(-\vb{x},t)\right] \\ 1025 | &= \frac{1}{2} \left[(\partial_\alpha \phi(\vb{x},t))(\partial^\alpha \phi(\vb{x,}t)) - \mu^2\phi^2(-\vb{x},t)\right] 1026 | \end{align*} 1027 | where we have used the chain rule for $\grad$ and the fact that $\eta_\text{p}^2=+1$. 1028 | 1029 | Also, $\phi^2$ is necessarily an even function of each component of $x$. i.e., 1030 | \begin{equation*} 1031 | \phi^2(\vb{x},t) = \phi^2(-\vb{x},t) 1032 | \end{equation*} 1033 | 1034 | Thus, the Lagrangian density $\lagr$ is invariant under the parity transformation: 1035 | \begin{equation*} 1036 | \lagr'(\vb{x},t) = \frac{1}{2}\left[(\partial_\alpha \phi(\vb{x},t))(\partial^\alpha \phi(\vb{x},t)) - \mu^2\phi^2(\vb{x},t)\right] = \lagr(\vb{x},t) 1037 | \end{equation*} 1038 | 1039 | Next, using the field operator $\phi(x)$ in the expansion form in terms of the absorption and creation operators, we can write: 1040 | \begin{align*} 1041 | \phi(-\vb{x},t) &= \sum_{\vb{k}'} \sqrt{\frac{\hbar c^2}{2V \omega_{\vb{k}'}}} \big[a(\vb{k}')e^{-i(\omega_{\vb{k}'}t + \vb{k}'\dotp\vb{x})} + a^\dagger(\vb{k}')e^{i(\omega_{\vb{k}'}t + \vb{k}'\dotp\vb{x})}\big] 1042 | \intertext{Define $\vb{k} = -\vb{k}'$. Note that $\omega_{\vb{k}'} = \omega_{-\vb{k}} =\omega_{\vb{k}}$. Then, we can write:} 1043 | &= \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V\omega_{\vb{k}}}} 1044 | \big[a(-\vb{k})e^{-i(\omega_{\vb{k}}t - \vb{k}\dotp\vb{x})} + a^\dagger(-\vb{k})e^{i(\omega_{\vb{k}}t - \vb{k}\dotp\vb{x})}\big] 1045 | \end{align*} 1046 | 1047 | Then, the parity transformation of $\phi(x)$ can be written as: 1048 | \begin{align*} 1049 | \mathscr{P} \phi(\vb{x},t) \mathscr{P}^{-1} 1050 | &= \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V \omega_{\vb{k}}}} 1051 | \big[\mathscr{P}a(\vb{k})\mathscr{P}^{-1}e^{-i(\omega_{\vb{k}}t - \vb{k}\dotp\vb{x})} 1052 | + \mathscr{P}a^\dagger(\vb{k})\mathscr{P}^{-1}e^{i(\omega_{\vb{k}}t - \vb{k\dotp\vb{x}})}\big] \\ 1053 | &= \eta_\text{p} \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V\omega_{\vb{k}}}} 1054 | \big[a(-\vb{k})e^{-i(\omega_{\vb{k}}t - \vb{k}\dotp\vb{x})} + a^\dagger(-\vb{k})e^{i(\omega_{\vb{k}}t - \vb{k}\dotp\vb{x})}\big] 1055 | \end{align*} 1056 | 1057 | Therefore, we derive the transformation laws for the absorption and creation operators under $\mathscr{P}$: 1058 | \begin{equation*} 1059 | \mathscr{P} a(\vb{k}) \mathscr{P}^{-1} = \eta_\text{p} \, a(-\vb{k}) \,, \, \mathscr{P} a^\dagger(\vb{k}) \mathscr{P}^{-1} = \eta_\text{p} \, a^\dagger(-\vb{k}) 1060 | \end{equation*} 1061 | 1062 | Then, we can derive how an arbitrary $n$-particle state transforms: 1063 | \begin{align*} 1064 | \mathscr{P} \ket{\vb{k}_1, ... \,, \vb{k}_n} 1065 | &= \mathscr{P} a^\dagger(\vb{k}_1) \,... \, a^\dagger(\vb{k}_n) \ket{0} \\ 1066 | &= \mathscr{P} a^\dagger(\vb{k}_1) \,... \, a^\dagger(\vb{k}_n) \mathscr{P}^{-1} \ket{0} \\ 1067 | &= \left(\mathscr{P}a^\dagger(\vb{k}_1)\mathscr{P}^{-1}\right)\,...\, 1068 | \left(\mathscr{P}a^\dagger(\vb{k}_n)\mathscr{P}^{-1}\right) \ket{0} \\ 1069 | &= \eta_\text{p}^n a^\dagger(-\vb{k}_1) \, ... \, a^\dagger(-\vb{k}_n) \ket{0} \\ 1070 | &= \eta_\text{p}^n \ket{-\vb{k}_1, ... \,, -\vb{k}_n} 1071 | \end{align*} 1072 | 1073 | Now, let $A$ and $B$ arbitrary operators, and define: 1074 | \begin{equation*} 1075 | B_0=B , \quad B_n =\left[A, B_{n-1}\right] , \quad \text{for } n=1, 2, ... 1076 | \end{equation*} 1077 | Then, the following holds identically: 1078 | \begin{equation*} 1079 | e^{i\alpha A} B e^{-i\alpha A} = \sum_{n=0}^{\infty} \frac{(i\alpha)^n}{n!} B_n 1080 | \end{equation*} 1081 | 1082 | We choose $B = B_0 = a(\vb{k})$. 1083 | 1084 | \begin{subproblem} 1085 | First, we define: 1086 | \begin{gather*} 1087 | \mathscr{P}_1 \define e^{i\alpha_1 A_1} \\ 1088 | \intertext{where} 1089 | \alpha_1 = -\frac{\pi}{2} , \quad A_1 = \sum_{\vb{k}}a^\dagger(\vb{k})a(\vb{k}) 1090 | \end{gather*} 1091 | Then, we can calculate the commutators $[A_1, B_{n-1}]$: 1092 | \begin{align*} 1093 | [A_1, B_0] &= \left[\sum_{\vb{k}'} a^\dagger(\vb{k}')a(\vb{k}'), a(\vb{k})\right] \\ 1094 | &= \sum_{\vb{k}'} \left[a^\dagger(\vb{k}'), a(\vb{k})\right]a(\vb{k}') \\ 1095 | &= -\sum_{\vb{k}'} \delta_{\vb{k}'\vb{k}} a(\vb{k}') \\ 1096 | &= -a(\vb{k}) 1097 | \end{align*} 1098 | and, by induction, 1099 | \begin{equation*} 1100 | \left[A_1, B_{n-1}\right] = (-1)^n \, a(\vb{k}) 1101 | \end{equation*} 1102 | 1103 | Therefore, we get: 1104 | \begin{align*} 1105 | \mathscr{P}_1 a(\vb{k}) \mathscr{P}_1^{-1} 1106 | &= e^{i\alpha_1A_1} a(\vb{k}) e^{-i\alpha_1A_1} \\ 1107 | &= \sum_{n=0}^{\infty} \frac{1}{n!} \left(+i\frac{\pi}{2}\right)^n \, a(\vb{k}) \\ 1108 | &= e^{i\pi/2} a(\vb{k}) \\ 1109 | &= i a(\vb{k}) 1110 | \end{align*} 1111 | \end{subproblem} 1112 | 1113 | \begin{subproblem} 1114 | Next, we define: 1115 | \begin{gather*} 1116 | \mathscr{P}_2 \define e^{i\alpha_2 A_2} \\ 1117 | \intertext{where} 1118 | \alpha_2 = \frac{\pi}{2}\eta_\text{p} , \quad 1119 | A_2 = \sum_{\vb{k}} a^\dagger(\vb{k}) a(-\vb{k}) 1120 | \end{gather*} 1121 | Then, the commutators $[A_2, B_{n-1}]$ are: 1122 | \begin{align*} 1123 | [A_2, B_{n_1}] &= (-1)^n \, a(-\vb{k}) 1124 | \end{align*} 1125 | 1126 | Therefore, we get: 1127 | \begin{align*} 1128 | \mathscr{P}_2 a(\vb{k}) \mathscr{P}_2^{-1} 1129 | &= e^{i\alpha_2A_2} a(\vb{k}) e^{-i\alpha_2A_2} \\ 1130 | &= \sum_{n=0}^{\infty} \frac{1}{n!} \left(-i\frac{\pi}{2}\eta_\text{p}\right)^n a(-\vb{k}) \\ 1131 | &= e^{-i\eta_\text{p}\pi/2} a(-\vb{k}) \\ 1132 | &= -i \eta_\text{p} a(-\vb{k}) 1133 | \end{align*} 1134 | \end{subproblem} 1135 | 1136 | Since $\alpha_r$'s and $A_r$'s are real for both $r=1,2$, $\mathscr{P}_r = e^{i\alpha_r A_r}$ are unitary. Therefore, the product $\mathscr{P}_1 \mathscr{P}_2$ is unitary as well. Note that the product operator satisfies the following identity: 1137 | \begin{align*} 1138 | \left(\mathscr{P}_1\mathscr{P}_2\right) a(\vb{k}) \left(\mathscr{P}_1\mathscr{P}_2\right)^{-1} 1139 | &= \mathscr{P}_1 \mathscr{P}_2 a(\vb{k}) \mathscr{P}_2^{-1} \mathscr{P}_1^{-1} \\ 1140 | &= \mathscr{P}_1 \left(-i\eta_\text{p}a(-\vb{k})\right) \mathscr{P}_1^{-1} \\ 1141 | &= +\eta_\text{p} a(-\vb{k}) 1142 | \end{align*} 1143 | which was derived earlier in this problem as the parity transformation law for the absorption operator $a(\vb{k})$. 1144 | 1145 | Thus, $\mathscr{P}_1\mathscr{P}_2$ gives an explicit form for the parity operator $\mathscr{P}$. 1146 | \end{problem} 1147 | 1148 | \pagebreak 1149 | 1150 | 1151 | 1152 | \section{The Dirac Field} 1153 | 1154 | \begin{problem} 1155 | For any given Dirac field $\psi$, we have the anti-commutator relationship: 1156 | \begin{equation*} 1157 | \big\{\psi(x), \overbar{\psi}(y)\big\} \define \big[\psi(x), \overbar{\psi}(y)\big]_+ = i S(x-y) 1158 | \end{equation*} 1159 | where 1160 | \begin{gather*} 1161 | S(x) = S^+(x) + S^-(x) = \left(i\gamma^\mu\pder{}{x^\mu} + \frac{mc}{\hbar}\right) \Delta(x) \\ 1162 | \Delta(x) = \Delta^+(x) + \Delta^-(x) 1163 | = \frac{-c}{(2\pi)^3} \int \frac{\diff[3]{\vb{k}}}{\omega_{\vb{k}}} \sin(kx) 1164 | \end{gather*} 1165 | 1166 | First, note that we have: 1167 | \begin{align*} 1168 | \pder{}{x^\mu} \Delta(x) &= \frac{-c}{(2\pi)^3} \int \frac{\diff[3]{\vb{k}}}{\omega_{\vb{k}}} \pder{}{x^\mu} \sin(k_\nu x^\nu) \\ 1169 | &= \frac{-c}{(2\pi)^3} \int \frac{\diff[3]{\vb{k}}}{\omega_{\vb{k}}} k_\mu \cos\left(\omega_{\vb{k}}t - \vb{k}\dotp\vb{x}\right) 1170 | \end{align*} 1171 | 1172 | Now, consider two points $x,y$ at a same given time. That is, 1173 | \begin{equation*} 1174 | x=\left(ct, \vb{x}\right) \text{ and } y=\left(ct,\vb{y}\right) 1175 | \end{equation*} 1176 | Then, the anti-commutator above becomes: 1177 | \begin{align*} 1178 | \left.\big\{\psi(x), \overbar{\psi}(y) \big\}\right|_{x^0=y^0} 1179 | &= iS(0, \vb{x}-\vb{y}) \\ 1180 | &= \frac{-ic}{(2\pi)^3} \int \frac{\diff[3]{\vb{k}}}{\omega_{\vb{k}}} 1181 | \Big[i\gamma^\mu k_\mu \cos\left(-\vb{k}\dotp(\vb{x}-\vb{y})\right) + \sin\left(-\vb{k}\dotp(\vb{x}-\vb{y})\right)\Big] 1182 | \end{align*} 1183 | Note that, if $\vb{x}-\vb{y} \neq \vb{0}$, the integrand is an odd function of $\vb{k}$. Therefore, the anti-commutator vanishes if $x^0=y^0$ and $x^i \neq y^i$. However, if $x=y$, the above expression becomes: 1184 | \begin{align*} 1185 | \frac{-ic}{(2\pi)^3} \int \frac{\diff[3]{\vb{k}}}{\omega_{\vb{k}}} 1186 | \Big[i\gamma^\mu k_\mu \cos0 + \sin0\Big] 1187 | &= \frac{-ic}{(2\pi)^3} \int \frac{\diff[3]{\vb{k}}}{\omega_{\vb{k}}} 1188 | i\Big[\gamma^0\frac{\omega_{\vb{k}}}{c} - \gamma^i k_i\Big] \\ 1189 | &= +\gamma^0 \frac{1}{(2\pi)^3} \int \diff[3]{\vb{k}} 1190 | \end{align*} 1191 | where the odd part of the integrand $\gamma^i k_i$ vanishes again. 1192 | 1193 | Therefore, the equal-time commutator for the Dirac field $\psi$ is: 1194 | \begin{equation*} 1195 | \big\{\psi(x), \overbar{\psi}(y)\big\} \define \big[\psi(x), \overbar{\psi}(y)\big]_+ = \gamma^0 \delta(\vb{x}-\vb{y}) 1196 | \end{equation*} 1197 | \end{problem} 1198 | 1199 | 1200 | 1201 | \begin{problem} 1202 | \begin{align*} 1203 | \left(i\gamma^\mu\pder{}{x^\mu} - \frac{mc}{\hbar}\right) S(x) 1204 | &= \left(i\gamma^\mu\pder{}{x^\mu} - \frac{mc}{\hbar}\right) \left(i\gamma^\nu\pder{}{x^\nu} + \frac{mc}{\hbar}\right) \Delta(x) \\ 1205 | &= \left(-\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu + \left(\frac{mc}{\hbar}\right)^2\right) \Delta(x) \\ 1206 | \intertext{Recall that $\gamma^\mu\gamma^\nu = g^{\mu\nu}$. Defining $\mu\define mc/\hbar$, we get:} 1207 | &= -\left(\dalem + \mu^2 \right)\Delta(x) \\ 1208 | &= 0 1209 | \end{align*} 1210 | because $\Delta(x)$ satisfies the Klein-Gordon equation, which was mentioned in Chapter~3. 1211 | 1212 | Similarly, we also have: 1213 | \begin{align*} 1214 | \left(i\gamma^\mu\pder{}{x^\mu} - \frac{mc}{\hbar}\right) S_\text{F}(x) 1215 | &= \left(i\gamma^\mu\pder{}{x^\mu} - \frac{mc}{\hbar}\right) \left(i\gamma^\nu\pder{}{x^\nu} + \frac{mc}{\hbar}\right) \Delta_\text{F}(x) \\ 1216 | &= -\left(\dalem + \mu^2 \right)\Delta_\text{F}(x) \\ 1217 | &= +\delta^{(4)}(x) 1218 | \end{align*} 1219 | where the last step is the result of Chapter~3 Problem~3. 1220 | 1221 | Hence, the functions $S(x)$ and $S_\text{F}(x)$ satisfy the homogeneous and inhomogeneous Dirac equations respectively (because $\Delta(x)$ and $\Delta_\text{F}(x)$ satisfy the homogeneous and inhomogeneous Klein-Gordon equations). 1222 | \end{problem} 1223 | 1224 | 1225 | 1226 | \begin{problem} 1227 | The charge-current density operator of the Dirac field $\psi(x)$ is given by: 1228 | \begin{equation*} 1229 | s^\mu (x) = -ec \overbar{\psi}(x) \gamma^\mu \psi(x) 1230 | \end{equation*} 1231 | 1232 | Let $x,y$ be two different points at equal times. In other words, we have $x^0=y^0=ct$, but $\vb{x} \neq \vb{y}$. Then, thanks to the results from Problem~1, we have: 1233 | \begin{equation*} 1234 | \big\{\psi(x), \overbar{\psi}(y)\big\} = 0 1235 | \end{equation*} 1236 | Also, we have the general anti-commutator relationships: 1237 | \begin{equation*} 1238 | \big\{\psi(x), \psi(y)\big\} = \big\{\overbar{\psi}(x), \overbar{\psi}(y)\big\} = 0 1239 | \end{equation*} 1240 | 1241 | \Large{SKIPPING FOR NOW. PHYSICALLY INTUITIVE, COMPUTATIONALLY ANNOYING} 1242 | 1243 | %The following identities will be useful for this problem: 1244 | %\begin{gather*} 1245 | %\big[AB, C\big] = A\big[B, C\big] + \big[A, C\big]B = A\big\{B,C\big\} - \big\{A, C\big\} B \\ 1246 | %\big[A, BC\big] = \big\{A, B\}C - B\big\{A, C\big\} 1247 | %\end{gather*} 1248 | % 1249 | %Now, we can write: 1250 | %\begin{align*} 1251 | % \big[s^\mu(x), s^\nu(y)\big] &= 1252 | % (ec)^2 \Big[\overbar{\psi}(x)\gamma^\mu\psi(x), \overbar{\psi}(y)\gamma^\nu\psi(y)\Big] \\ 1253 | % &= (ec)^2 \Big[\overbar{\psi}(x)\gamma^\mu\big[\psi(x), \overbar{\psi}(y)\gamma^\nu\psi(y)\big] + \big[\overbar{\psi}(x)\gamma^\mu, \overbar{\psi}(y)\gamma^\nu\psi(y)\big] \psi(x) \Big] \\ 1254 | % &= (ec)^2 \Big[\overbar{\psi}(x)\gamma^\mu\Big( \cancelto{0}{\big\{\psi(x), \overbar{\psi}(y)\big\}}\gamma^\nu\psi(y) - \overbar{\psi}(y)\big\{\psi(x), \gamma^\nu\psi(y)\big\}\Big) \\ 1255 | % & \hspace{5em} + \Big(\big\{\overbar{\psi}(x)\gamma^\mu, \overbar{\psi}(y)\big\}\gamma^\nu\psi(y) - \overbar{\psi}(y)\big\{\overbar{\psi}(x)\gamma^\mu, \gamma^\nu\psi(y)\big\}\Big)\psi(x) \Big] \\ 1256 | % &= (ec)^2 \Big[ -\overbar{\psi}(x) \gamma^\mu\overbar{\psi}(y) \Big(\gamma^\nu \cancelto{0}{\big\{\psi(x), \psi(y)\big\}} + \big[\psi(x), \gamma^\nu \big] \psi(y)\Big) \\ 1257 | % & \hspace{5em} + \Big(\cancelto{0}{\big\{\overbar{\psi}(x), \overbar{\psi}(y)\big\}}\gamma^\mu + \overbar{\psi}(x)\big[\gamma^\mu, \overbar{\psi}(y)\big]\Big) \gamma^\nu\psi(y)\psi(x) \\ 1258 | % & \hspace{5em} - \overbar{\psi}(y) \Big(\big\{\overbar{\psi}(x), \gamma^\nu\psi(y)\big\}\gamma^\mu + \overbar{\psi}(x)\big[\gamma^\mu, \gamma^\nu\psi(y)\big]\Big)\psi(x) \Big] \\ 1259 | % &= (ec)^2 \Big[ -\overbar{\psi}(x)\gamma^\mu\overbar{\psi}(y) \big[\psi(x), \gamma^\nu\big]\psi(y) + \overbar{\psi}(x) \big[\gamma^\mu, \overbar{\psi}(y)\big]\gamma^\nu\psi(y)\psi(x) \\ 1260 | % & \hspace{5em} -\overbar{\psi}(y) \Big( \gamma^\nu \cancelto{0}{\big\{\overbar{\psi}(x), \psi(y)\big\}} + \big[\overbar{\psi}(x), \gamma^\nu\big]\psi(y)\Big)\gamma^\mu\psi(x) \\ 1261 | % & \hspace{5em} -\overbar{\psi}(y)\overbar{\psi}(x) \big[\gamma^\mu, \gamma^\nu\psi(y)\big]\psi(x) \Big] \\ 1262 | % &= (ec)^2 \Big[ -\overbar{\psi}(x)\gamma^\mu\overbar{\psi}(y)\psi(x)\gamma^\nu\psi(y) + \colcancel[red]{\overbar{\psi}(x)\gamma^\mu\overbar{\psi}(y)\gamma^\nu\psi(x)\psi(y)} \\ 1263 | % &\hspace{5em} + \colcancel[red]{\overbar{\psi}(x)\gamma^\mu\overbar{\psi}(y)\gamma^\nu\psi(y)\psi(x)} - \colcancel[orange]{\overbar{\psi}(x)\overbar{\psi}(y)\gamma^\mu\gamma^\nu\psi(y)\psi(x)} \\ 1264 | % &\hspace{5em} - \colcancel[blue]{\overbar{\psi}(y)\overbar{\psi}(x)\gamma^\nu\psi(y)\gamma^\mu\psi(x)} + \overbar{\psi}(y)\gamma^\nu\overbar{\psi}(x)\psi(y)\gamma^\mu\psi(x) \\ 1265 | % &\hspace{5em} - \colcancel[orange]{\overbar{\psi}(y)\overbar{\psi}(x)\gamma^\mu\gamma^\nu\psi(y)\psi(x)} + \colcancel[blue]{\overbar{\psi}(y)\overbar{\psi}(x)\gamma^\nu\psi(y)\gamma^\mu\psi(x)} 1266 | % \Big] 1267 | %\end{align*} 1268 | % 1269 | % 1270 | %Now, let $\vb{x} \neq \vb{y}$. Then, we have: 1271 | %\begin{align*} 1272 | % \big[s^\mu(x), s^\nu(y)\big] &= 1273 | % (ec)^2 \Big[\overbar{\psi}(x)\gamma^\mu\psi(x)\overbar{\psi}(y)\gamma^\nu\psi(y) \\ 1274 | % &\hspace{4em} - \overbar{\psi}(y)\gamma^\nu\psi(y)\overbar{\psi}(x)\gamma^\mu\psi(x) \Big] 1275 | %\end{align*} 1276 | \end{problem} 1277 | 1278 | 1279 | 1280 | \begin{problem} 1281 | Recall the expansion of the real Klein-Gordon field $\phi(x)$: 1282 | \begin{equation*} 1283 | \phi(x) = \phi^+(x) + \phi^-(x) = \sum_{\vb{k}} \sqrt{\frac{\hbar c^2}{2V\omega_{\vb{k}}}} \left[a(\vb{k})e^{-ikx} + a^\dagger(\vb{k})e^{ikx}\right] 1284 | \end{equation*} 1285 | 1286 | Now, supposes that we impose the \emph{anti}-commutator relationships, instead of the normal commutator relationships, on the absorption and creation operators: 1287 | \begin{gather*} 1288 | \left\{a(\vb{k}), a^\dagger(\vb{k}')\right\} = \delta_{\vb{k}\vb{k}'} \\ 1289 | \left\{a(\vb{k}), a(\vb{k}') \right\} = \left\{a^\dagger(\vb{k}), a^\dagger(\vb{k}') \right\} = 0 1290 | \end{gather*} 1291 | This is equivalent to imposing \emph{Fermi-Dirac statistics}, instead of Bose-Einstein statistics. 1292 | 1293 | The following identity will be useful in this problem. 1294 | \begin{equation*} 1295 | \big[A, B\big] = AB - BA = 2AB - \big\{A, B\} 1296 | \end{equation*} 1297 | 1298 | Now, consider two points $x,y$ in spacetime with a space-like separation $(x-y)$. First, we have: 1299 | \begin{align*} 1300 | \big\{\phi(x), \phi(y)\big\} 1301 | &= \frac{\hbar c^2}{2V}\sum_{\vb{k},\vb{k}'} \frac{1}{\sqrt{\omega_{\vb{k}}\omega_{\vb{k}'}}} 1302 | \Big[-\big\{a(\vb{k}), a^\dagger(\vb{k}')\big\}e^{-i(kx-k'y)} - \big\{a^\dagger(\vb{k}), a(\vb{k}')\big\}e^{i(kx-k'y)}\Big] \\ 1303 | &= -\frac{\hbar c^2}{2V} \sum_{\vb{k},\vb{k}'} \frac{1}{\sqrt{\omega_{\vb{k}}\omega_{\vb{k}'}}} \delta_{\vb{k}\vb{k}'}\left(e^{-i(kx-k'y)} + e^{i(kx-k'y)}\right) \\ 1304 | &= -\frac{\hbar c^2}{V} \sum_{\vb{k}} \frac{1}{\omega_{\vb{k}}} \cos\left(k(x-y)\right) 1305 | \end{align*} 1306 | 1307 | Since this quantity is Lorentz-invariant, it suffices to show its behavior for any $x,y$ with a space-like separation. Hence, we choose $x,y$ such that they are at equal time, but with non-zeros spatial separation $\vb{x}-\vb{y}$. Then, each cosine term is a non-zero even function of $\vb{k}$. Hence, the sum is non-zero. 1308 | 1309 | Similarly, we can see that: 1310 | \begin{align*} 1311 | \big[\phi(x), \phi(y)\big] 1312 | &= 2\phi(x)\phi(y) - \big\{\phi(x), \phi(y)\big\} \\ 1313 | &= 2\phi(x)\phi(y) + \frac{\hbar c^2}{2V} \sum_{\vb{k},\vb{k}'} \frac{1}{\sqrt{\omega_{\vb{k}}\omega_{\vb{k}'}}} \delta_{\vb{k}\vb{k}'}\left(e^{-i(kx-k'y)} + e^{i(kx-k'y)}\right) \\ 1314 | &= 2\phi(x)\phi(y) + \frac{\hbar c^2}{V} \sum_{\vb{k}} \frac{1}{\omega_{\vb{k}}} \cos\left(k(x-y)\right) 1315 | \end{align*} 1316 | is also non-zero. 1317 | 1318 | Therefore, having imposed the anti-commutator relationships on $a(\vb{k}), a^\dagger(\vb{k})$, we see that neither the commutator nor the anti-commutator of the Klein-Gordon field $\phi$ can be zero for $x,y$ with a space-like separation. 1319 | 1320 | This will prevent the commutator of any bilinear physical observable of $\phi$ at $x,y$ to be zero, which is a violation of \emph{microcausality}. Thus, we observe that the absorption and creation operators of the Klein-Gordon field must be quantized using the commutator relationships (i.e. Bose-Einstein statistics). 1321 | \end{problem} 1322 | 1323 | 1324 | 1325 | \begin{problem} 1326 | For a Dirac field $\psi$, we define the chiral phase transformations as: 1327 | \begin{equation*} 1328 | \psi(x) \rightarrow \psi'(x) = \exp(i\alpha\gamma_5)\psi(x) \,, \, 1329 | \psi^\dagger(x) \rightarrow {\psi^\dagger}'(x) = \psi^\dagger(x)\exp(-i\alpha\gamma_5) 1330 | \end{equation*} 1331 | where $\gamma_5 \define i\gamma_0\gamma_1\gamma_2\gamma_3$ and $\alpha$ is an arbitrary real parameter. 1332 | 1333 | Note that $\gamma_5$ anti-commutes with all $\gamma_\mu$'s ($\mu=0, 1, 2, 3$). That is, each time the order of product between $\gamma_5$ and a $\gamma_\mu$ is exchanged, the term gains a negative sign. 1334 | 1335 | Under this transformation, the Lagrangian density 1336 | \begin{equation*} 1337 | \lagr = c\overbar{\psi}(x)\left[i\hbar\gamma^\mu\pder{}{x^\mu} - mc\right]\psi(x) 1338 | \end{equation*} 1339 | transforms to: 1340 | \begin{align*} 1341 | \lagr' &= c {\bar{\psi}}'(x) \gamma^0 \left[i\hbar\gamma^\mu\pder{}{x^\mu} - mc\right] \psi'(x) \\ 1342 | &= c \psi^\dagger(x) \exp(-i\alpha\gamma_5) \gamma^0 \left[i\hbar\gamma^\mu\pder{}{x^\mu} - mc\right] \exp(i\alpha\gamma_5) \psi(x) 1343 | \end{align*} 1344 | The first term has two powers of the Dirac $\gamma$-matrix: the $\gamma^0$ originating from the adjoint spinor $\overbar{\psi}'$ and the $\gamma^\mu$ in front of the derivative. Hence, the first term can be re-ordered such that the two exponential factors due to the chiral phase transformation cancel each other. However, the second term (i.e. the mass term) has only one $\gamma^0$, and thus gains a negative sign after such re-ordering. Hence, 1345 | \begin{align*} 1346 | \lagr' &= c \overbar{\psi}(x)\left[i\hbar\gamma^\mu\pder{}{x^\mu} + mc\right]\psi(x) 1347 | \end{align*} 1348 | Therefore, the Lagrangian density is invariant only if the mass is zero. 1349 | 1350 | The infinitesimal chiral phase transformation is obtained in the limit $\alpha \ll 1$: 1351 | \begin{align*} 1352 | \psi'(x) &= \psi(x) + \delta\psi(x) = \psi(x) + i\alpha\gamma_5\psi(x) \\ 1353 | {\psi^\dagger}'(x) &= \psi^\dagger(x) + \delta\psi^\dagger(x) = \psi^\dagger(x) - i\alpha \psi^\dagger(x) \gamma_5 1354 | \end{align*} 1355 | Because the massless Lagrangian density is invariant under this transformation, we have the corresponding conserved current: 1356 | \begin{align*} 1357 | f^\mu (x) &= \left(\pder{\lagr}{\psi_{,\mu}}\right)\delta\psi(x) \\ 1358 | &= \big(c\overbar{\psi}(x)i\hbar\gamma^\mu\big) i\alpha\gamma_5\psi(x) \\ 1359 | &= -\alpha \hbar c \overbar{\psi}(x)\gamma^\mu \gamma_5 \psi(x) 1360 | \end{align*} 1361 | 1362 | Re-writing without the constants, we see that the axial vector current 1363 | \begin{align*} 1364 | j^\mu_\text{A}(x) = \overbar{\psi}(x) \gamma^\mu \gamma_5 \psi(x) 1365 | \end{align*} 1366 | is conserved under the chiral phase transformation of a massless Dirac field. 1367 | 1368 | Now, for a given massive Dirac field $\psi(x)$, consider two new fields: 1369 | \begin{equation*} 1370 | \psi_\text{L}(x) \define \frac{1}{2}\left(1-\gamma_5\right)\psi(x) \,,\, \psi_\text{R}(x) \define \frac{1}{2}\left(1+\gamma_5\right)\psi(x) 1371 | \end{equation*} 1372 | The subscripts $_{\text{L},\text{R}}$ stand for ``left-handed" and ``right-handed" components of the Dirac field $\psi$. They are also called ``negative/positvie helicity" components, respectively. 1373 | 1374 | Note that: 1375 | \begin{align*} 1376 | \psi_\text{L} + \psi_\text{R} = \psi \,,\, 1377 | \psi_\text{R} - \psi_\text{L} = \gamma_5 \psi 1378 | \end{align*} 1379 | 1380 | Applying the operator $\left(i\hbar\slashed{\partial}-mc\right)$ to the left of $\psi_{\text{L},\text{R}}$ gives their equations of motion: 1381 | \begin{align*} 1382 | \left(i\hbar\slashed{\partial}-mc\right)\psi_\text{L}(x) 1383 | &= \left(i\hbar\gamma^\mu\pder{}{x^\mu}-mc\right) \frac{1}{2}\left(1-\gamma_5\right)\psi(x) \\ 1384 | &= \frac{1}{2} \left[\left(i\hbar\gamma^\mu\pder{}{x^\mu}-mc\right)\psi(x) 1385 | - \left(i\hbar\gamma^\mu\pder{}{x^\mu}-mc\right)\gamma_5\psi(x)\right] \\ 1386 | &= \frac{1}{2} \left[0 + \gamma_5 \left(i\hbar\gamma^\mu\pder{}{x^\mu} + mc\right)\psi(x)\right] \\ 1387 | &= \frac{1}{2} \left[\gamma_5\left(i\hbar\gamma^\mu\pder{}{x^\mu}-mc\right)\psi(x) + 2mc\gamma_5 \psi(x)\right] \\ 1388 | &= mc\gamma_5\psi(x) \\ 1389 | &= mc\left(\psi_\text{R}(x) - \psi_\text{L}(x)\right) 1390 | \end{align*} 1391 | where we have used that $\psi(x)$ satisfies the Dirac equation: $\left(i\hbar\slashed{\partial}-mc\right)\psi(x)=0$. 1392 | 1393 | Similarly, we also have: 1394 | \begin{align*} 1395 | \left(i\hbar\slashed{\partial} - mc\right) \psi_\text{R}(x) 1396 | &= \frac{1}{2}\left(i\hbar\slashed{\partial} - mc\right)(1+\gamma_5) \psi(x) \\ 1397 | &= \frac{1}{2}\gamma_5\left(-i\hbar\slashed{\partial}-mc\right)\psi(x) \\ 1398 | &= -mc\gamma_5\psi(x) \\ 1399 | &= -mc\left(\psi_\text{R}(x) - \psi_\text{L}(x)\right) 1400 | \end{align*} 1401 | 1402 | If the mass $m$ of the Dirac field $\psi(x)$ was zero, then the equations of motion for $\psi_\text{L}, \psi_\text{R}$ would be decoupled: 1403 | \begin{equation*} 1404 | i\hbar\gamma^\mu\pder{}{x^\mu}\psi_{\text{L},\text{R}} = 0 1405 | \end{equation*} 1406 | 1407 | Now, consider the Lagrangian density: 1408 | \begin{equation*} 1409 | \lagr_\text{L} = i\hbar c \overbar{\psi}_\text{L}(x) \gamma^\mu \partial_\mu \psi_\text{L}(x) 1410 | \end{equation*} 1411 | where 1412 | \begin{equation*} 1413 | \overbar{\psi}_\text{L} = \overbar{\psi} \frac{1-\gamma_5}{2} 1414 | \end{equation*} 1415 | 1416 | The corresponding Euler-Lagrange equations are: 1417 | \begin{align*} 1418 | \pder{}{x^\mu}\left(\pder{\lagr_\text{L}}{\psi_{\text{L},\mu}}\right) - \pder{\lagr_\text{L}}{\psi_\text{L}} 1419 | &= \pder{}{x^\mu} \left(i\hbar c \overbar{\psi}_\text{L} \gamma^\mu\right) \\ 1420 | &= i\hbar c \pder{}{x^\mu} \left(\psi^\dagger \gamma^0 \frac{1-\gamma_5}{2}\right)\gamma^\mu \\ 1421 | &= i\hbar c \pder{}{x^\mu} \left(\psi^\dagger \frac{1+\gamma_5}{2}\gamma^0\right) \gamma^\mu \\ 1422 | &= i\hbar c \pder{}{x^\mu} \overbar{\psi_\text{R}} \gamma^\mu \\ 1423 | &= 0 1424 | \end{align*} 1425 | and 1426 | \begin{align*} 1427 | \pder{}{x^\mu}\left(\pder{\lagr_\text{L}}{\overbar{\psi}_{\text{L},\mu}}\right) - \pder{\lagr_\text{L}}{\overbar{\psi}_\text{L}} 1428 | &= - i\hbar c \gamma^\mu \pder{}{x^\mu} \psi_\text{L} \\ 1429 | &= 0 1430 | \end{align*} 1431 | 1432 | Therefore, the given Lagrangian $\lagr_\text{L}$ describes the right-handed massless anti-fermions and the left-handed massless fermions only. 1433 | \end{problem} 1434 | 1435 | 1436 | 1437 | \section{Photons: Covariant Theory} 1438 | 1439 | 1440 | 1441 | \end{document} 1442 | -------------------------------------------------------------------------------- /mandl_shaw_solution.log: -------------------------------------------------------------------------------- 1 | This is pdfTeX, Version 3.14159265-2.6-1.40.16 (TeX Live 2015/Debian) (preloaded format=pdflatex 2017.9.6) 4 OCT 2017 07:20 2 | entering extended mode 3 | restricted \write18 enabled. 4 | %&-line parsing enabled. 5 | **mandl_shaw_solution.tex 6 | (./mandl_shaw_solution.tex 7 | LaTeX2e <2016/02/01> 8 | Babel <3.9q> and hyphenation patterns for 81 language(s) loaded. 9 | 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' s m g ' on line 565. 1813 | ................................................. 1814 | ................................................. 1815 | . LaTeX info: "xparse/define-command" 1816 | . 1817 | . Defining command \dyad with sig. '' on line 566. 1818 | ................................................. 1819 | ................................................. 1820 | . LaTeX info: "xparse/define-command" 1821 | . 1822 | . Defining command \op with sig. '' on line 567. 1823 | ................................................. 1824 | ................................................. 1825 | . LaTeX info: "xparse/define-command" 1826 | . 1827 | . Defining command \ketbra with sig. '' on line 568. 1828 | ................................................. 1829 | ................................................. 1830 | . LaTeX info: "xparse/define-command" 1831 | . 1832 | . Defining command \expectationvalue with sig. 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'' on line 663. 1893 | ................................................. 1894 | ................................................. 1895 | . LaTeX info: "xparse/define-command" 1896 | . 1897 | . Defining command \paulixmatrix with sig. '' on line 665. 1898 | ................................................. 1899 | ................................................. 1900 | . LaTeX info: "xparse/define-command" 1901 | . 1902 | . Defining command \pauliymatrix with sig. '' on line 666. 1903 | ................................................. 1904 | ................................................. 1905 | . LaTeX info: "xparse/define-command" 1906 | . 1907 | . Defining command \paulizmatrix with sig. '' on line 667. 1908 | ................................................. 1909 | ................................................. 1910 | . LaTeX info: "xparse/define-command" 1911 | . 1912 | . Defining command \paulimatrix with sig. 'm' on line 678. 1913 | ................................................. 1914 | ................................................. 1915 | . LaTeX info: "xparse/define-command" 1916 | . 1917 | . Defining command \pmat with sig. '' on line 679. 1918 | ................................................. 1919 | ................................................. 1920 | . LaTeX info: "xparse/define-command" 1921 | . 1922 | . Defining command \diagonalmatrix with sig. 'O{} >{\SplitList {,}}m ' on line 1923 | . 681. 1924 | ................................................. 1925 | ................................................. 1926 | . LaTeX info: "xparse/define-command" 1927 | . 1928 | . Defining command \@dmat with sig. 'mmggggggg' on line 714. 1929 | ................................................. 1930 | ................................................. 1931 | . LaTeX info: "xparse/define-command" 1932 | . 1933 | . Defining command \dmat with sig. 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