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1 \documentclass[a4paper]{article} 2 3 \usepackage[T1]{fontenc} 4 \usepackage[utf8]{inputenc} 5 6 \usepackage{mathptmx} 7 8 \usepackage[a4paper, total={6in, 8in}]{geometry} 9 \usepackage{subcaption} 10 \usepackage[shortlabels]{enumitem} 11 \usepackage{amssymb} 12 \usepackage{amsthm} 13 \usepackage{mathtools} 14 \usepackage{braket} 15 \usepackage{bbm} 16 \usepackage{graphicx} 17 \usepackage{float} 18 \usepackage{yhmath} 19 \usepackage{tikz} 20 \usetikzlibrary{calc,decorations.markings} 21 \usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} 22 \usepackage[parfill]{parskip} 23 \usepackage[backend=biber, sorting=none]{biblatex} 24 \newcommand{\hbbar}{{\raisebox{0.05ex}{$\mathchar '26$}\mkern -9mu\raisebox{-0.15ex}{$\mathchar '26$}\mkern -9muh}} 25 \addbibresource{uni.bib} 26 \pagestyle{myheadings} 27 \markright{Popovic, Vogel\hfill Dispertion relations \hfill} 28 29 30 \title{Universität Wien\\ Fakultät für Physik\\ 31 \vspace{1.25cm}Laborpraktikum Theoretische Physik 2021S \\ Dispertion relations 32 } 33 \author{Milutin Popovic \& Tim Vogel \vspace{1cm}\\ Betreuer: Peter Stoffer} 34 \date{30. Juni, 2021} 35 36 \begin{document} 37 \maketitle 38 \noindent\rule[0.5ex]{\linewidth}{1pt} 39 \begin{abstract} 40 \end{abstract} 41 \noindent\rule[0.5ex]{\linewidth}{1pt} 42 \newcommand{\PV}{\mathop{\mathrlap{\pushR}}\!\int} 43 \newcommand{\pushR}{\mathchoice 44 {\mkern2.5mu P} 45 {\scriptstyle P} 46 {\scriptscriptstyle P} 47 {\scriptscriptstyle P} 48 } 49 50 \tableofcontents 51 52 \section{Introduction} 53 Within the tools of complex analysis, there exists the possibility, to form 54 relations between observable quantities of physicals systems, for example 55 dispersion in a dielectric medium. This can be taken even further, by using the 56 same methods within particle-physical problems, where the now more popular 57 methods of qauntum chromodynamics do not apply, which would be low energy 58 hadronic processes. We will firstly apply these concepts to the simple example 59 of the harmonic oscillator and finally work out more complex problems, 60 regarding the pion vector form factor. The reader is expected to be familiar 61 with the subject of complex analysis, especially analyticity/holomorphicity of 62 a function, integration of complex functions, the residue theorem and the 63 Schwartz reflection principle. 64 65 66 67 68 69 70 71 72 73 74 \section{Damped harmonic oscillator} 75 Considering a free harmonic oscillator, the equation of motion accounts to: 76 \begin{equation} 77 \Ddot{x}(t)+\gamma\Dot{x}(t)+\omega_0^2x(t)=0 78 \end{equation} 79 where $\gamma > 0$ is the damping coefficient, and $\omega_0$ the angular frequency 80 of the oscillator. Using the exponential ansatz we can arrive at an general 81 solution to this ordinary differential equation 82 \begin{align} 83 &x(t) = a e^{-i\omega_1 t} + b e^{-i\omega_2 t} \\ 84 &\nonumber \\ 85 \text{with:} \nonumber\\ 86 &\omega_{1/2} = \pm \sqrt{\omega_0^2 - (\frac{\gamma}{2})^2} - 87 i\frac{\gamma}{2} 88 \end{align} 89 where $a$ and $b$ are calculated based on the Couchy boundary conditions. 90 91 For the case $\omega_0 > \frac{\gamma}{2}$ we can rewrite the solution 92 \begin{align} 93 &x(t) = \left(a e^{-i\tilde{\omega}_0 t} + b e^{-i\tilde{\omega}_0 94 t}\right) e^{-\frac{\gamma}{2}t}\\ 95 &\nonumber \\ 96 \text{with:} \nonumber\\ 97 &\tilde{\omega}_{0} = \sqrt{\omega_0^2 - \left(\frac{\gamma}{2}\right)^2} 98 \end{align} 99 \subsection{External Force} 100 101 Now consider a harmonic oscillator with an external force $F(t)$ driving it 102 \begin{align}\label{eq:force} 103 \Ddot{x}(t)+\gamma\Dot{x}(t)+\omega_0^2x(t)=\frac{F(t)}{m} =: f(t). 104 \end{align} 105 By Fourier transforming the equation we can arrive at an equation for the 106 greens function in Fourier space. Note that the Fourier transform of 107 $x(t)$ is 108 \begin{align} 109 \hat{x}(t) &= \frac{1}{2\pi}\int_\infty^\infty d\omega 110 X(\omega) e^{-i\omega t} 111 \end{align} 112 so the Fourier transforms of $\dot{x}$ and $\Ddot{x}$ are 113 \begin{align} 114 \mathcal{F}(\dot{x}) &= -i\omega X(\omega)\\ 115 \mathcal{F}(\Ddot{x}) &= -\omega^2 X(\omega)\\ 116 \end{align} 117 and the equation \ref{eq:force} turns into 118 \begin{align} 119 (-\omega^2 - i\gamma \omega + \omega_0^2) X(\omega) = F(\omega) 120 \end{align} 121 122 The Green's function can be represented in Fourier space like the following 123 \begin{align} 124 G(\omega)= \frac{1}{-\omega^2 - i\gamma \omega + \omega_0^2} 125 \end{align} 126 127 The Maximum of the squared modulus $|G(\omega)|^2$ for $\gamma \ll \omega_0$ is 128 roughly at $\omega_0$, thus the width at half maximum can be calculated by 129 looking for two $\omega$'s that satisfy 130 \begin{align} 131 \frac{1}{2}|G(\omega_0)|^2 &= |G(\omega)|^2\\ 132 \frac{1}{2} \frac{1}{\omega_0^2\gamma^2} &= |G(\omega)|^2 133 \end{align} 134 135 The exact solutions are 136 \begin{align} 137 \tilde{\omega}_1 &= \omega_0\sqrt{-0.5\left(\frac{\gamma}{\omega_0}\right)^2 138 - 1.0\frac{\gamma}{\omega_0}(0.25\left(\frac{\gamma}{\omega_0}\right)^2 139 + 1)^{\frac{1}{2}} + 1}\\ 140 \tilde{\omega}_2 &= \omega_0\sqrt{-0.5\left(\frac{\gamma}{\omega_0}\right)^2 141 + 1.0\frac{\gamma}{\omega_0}(0.25\left(\frac{\gamma}{\omega_0}\right)^2 142 + 1)^{\frac{1}{2}} + 1} 143 \end{align} 144 With help of Taylor expansion in the linear order in $\frac{\gamma}{\omega_0}$ 145 gives us the approximation for the with at half maximum 146 \begin{align} 147 \tilde{\omega}_2 - \tilde{\omega}_1 \simeq \gamma 148 \end{align} 149 150 In the figure below we plotted the squared modulus of $|G(\omega)|^2$ 151 \begin{figure}[H] 152 \centering 153 \includegraphics[width=\textwidth]{plots_sec2.png} 154 \caption{On the left the squared modulus $|G(\omega)|^2$ in $\omega 155 \in [\omega_0 - 2\gamma, w_0 + w\gamma]$ for $\gamma \ll \omega_0$, precisely 156 $\gamma = \omega_0/20$ for $\omega_0 = 1$ and on the right $arg(G(\omega))$} 157 \end{figure} 158 159 160 Next we want calculate the Green's function in terms of time 161 \begin{align} 162 g(t) = \frac{1}{2\pi} \int^\infty_\infty d\omega G(\omega)e^{-i\omega t}. 163 \end{align} 164 Furthermore we can transform this to the complex integral where we have two singularities 165 at We have two singularities at $z_{1/2} = - \frac{i\gamma}{2} \pm 166 \tilde{\omega}_{0}$. We have the following integral 167 path 168 169 \begin{center} 170 \begin{tikzpicture}[decoration={markings, 171 mark=at position 13cm with {\arrow[line width=2pt]{>}} 172 } 173 ] 174 % The axes 175 \draw[help lines,->] (-4,0) -- (4,0) coordinate (xaxis); 176 \draw[help lines,->] (0,-3.5) -- (0,1) coordinate (yaxis); 177 178 % The path 179 \path[draw,line width=0.8pt,postaction=decorate] 180 (3,0) node[above right] {} arc (0:-180:3) -- (-3, 0) 181 node[above left] {} -- (-3, 0) -- (3, 0); 182 183 % The labels 184 \draw[thick, ->] (0,0) -- (2.1, -2.1) node[midway, fill=white] {$R$}; 185 \node[below] at (xaxis) {$\text{Re}$}; 186 \node[left] at (yaxis) {$\text{Im}$}; 187 \node at (0.2,-1.5) {$\nu$}; 188 \node at (1, -3.2) {$C(R)$}; 189 \end{tikzpicture} 190 \end{center} 191 192 The complex integral representation is 193 \begin{align} 194 \oint_\nu dz G(z) e^{-izt} &= 195 \lim_{R \rightarrow \infty } 196 \bigg( 197 \int_{C(R)} + \int_{-R}^{R} 198 \bigg) 199 dz\ G(z) e^{-izt}\\ 200 &= 2\pi i\sum_j \text{I}(C_j, z_j) \text{Res}_j 201 \end{align} 202 Keep in mind that the integral from $R$ to $-R$ is the integral we are 203 trying to solve, that is pulling the limit we have one integral over the real 204 axis. Because of Jordan's lemma, the integral over the complex curve vanishes 205 \begin{align} 206 \big| \int_{C(R)} dz G(z) e^{-izt}\big| \leq \frac{\pi}{t}M_R 207 \end{align} 208 where $M_R:= \max_{C(R)}\{G(Re^{i\varphi})\}$.It can easily be seen that $M_R$ 209 converges to $0$ as $R$ goes to infinity. Thus the only value the integral can 210 take is $0$ and we can calculate the real integral with the residues 211 \begin{align} 212 \text{Res}_1 &= \frac{e^{i z t}}{(z - z_1)(z - z_2)} (z - z_1)\bigg|_{z=z_1} 213 \\ 214 &= -\frac{e^{-iz_1t}}{z_1 - z_2} = \frac{e^{-\frac{\gamma}{2}t} 215 e^{i\tilde{\omega}_0 t}}{2\tilde{\omega}_0}\\ 216 \nonumber \\ 217 \text{Res}_2 &= \frac{e^{i z t}}{(z - z_1)(z - z_2)} (z - z_2)\bigg|_{z=z_2} 218 \\ 219 &= -\frac{e^{-iz_1t}}{z_2 - z_1} = -\frac{e^{-\frac{\gamma}{2}t} 220 e^{-i\tilde{\omega}_0 t}}{2\tilde{\omega}_0}\\ 221 \end{align} 222 with the index $\text{I}(C_R, z_i)$ being $1$, because the curve goes around the 223 singularities once. 224 225 The integral evolves to 226 \begin{align} 227 \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega G(\omega) e^{-i\omega t}= 228 \frac{\sin(\tilde{\omega}_0t)}{\tilde{\omega}_0} e^{-\frac{\gamma}{2}t}. 229 \end{align} 230 Treating the cases $t<0$ and $t>0$ separately we can join them with the 231 Heaviside-theta function $\theta(t)$, the Green's function for the damped 232 harmonic oscillator is 233 \begin{align} 234 g(t) = \frac{\sin(\tilde{\omega}_0t}{\tilde{\omega}_0} 235 e^{-\frac{\gamma}{2}t} \theta(t) 236 \end{align} 237 With convolution we can arrive at a solution for the damped harmonic oscillator 238 for an arbitrary driving force $f(t)$ 239 \begin{align} 240 x(t) = \int_{-\infty}^{t} dt' 241 \frac{\sin(\tilde{\omega}_0 (t-t'))}{\tilde{\omega}_0} 242 e^{-\frac{\gamma}{2}(t-t')} f(t'). 243 \end{align} 244 245 \subsection{Green's Function and dispersion relations} 246 Next we want to compute the following integral 247 \begin{align} 248 0 = \oint_C d\omega' \frac{G(\omega')}{\omega - \omega'}, \;\;\;\; 249 \text{with} \;\;\; \omega' = \omega_r + i\omega_i 250 \end{align} 251 along the following contour 252 253 \begin{center} 254 \begin{tikzpicture}[decoration={markings, 255 mark=at position 0.5cm with {\arrow[line width=2pt]{>}}, 256 mark=at position 5cm with {\arrow[line width=2pt]{>}}, 257 mark=at position 13cm with {\arrow[line width=2pt]{>}}, 258 mark=at position 15cm with {\arrow[line width=2pt]{>}} 259 } 260 ] 261 % The axes 262 \draw[help lines,->] (-3.5,0) -- (3.5,0) coordinate (xaxis); 263 \draw[help lines,->] (0,-0.5) -- (0,3.5) coordinate (yaxis); 264 265 % The path 266 \path[draw,line width=0.8pt,postaction=decorate] 267 (2,0) -- (3, 0) node[below right] {} arc (0:180:3) -- (0.5, 0) 268 arc (180:0:0.75); 269 270 % The labels 271 \draw[thick, ->] (0,0) -- (-2.1, 2.1) node[midway, fill=white] {$R$}; 272 \draw[thick, ->] (1.25,0) -- (1.75, 0.5) node[midway, above] {$\varrho$}; 273 \node[below] at (xaxis) {$\text{Re}$}; 274 \node[left] at (yaxis) {$\text{Im}$}; 275 \node[circle,inner sep=1pt,label=below:{$\omega$}, fill=black] at (1.25,0) {}; 276 \end{tikzpicture} 277 \end{center} 278 so the integral representation is 279 \begin{align} 280 \oint_C d\omega' \frac{G(\omega')}{\omega - \omega'} = 281 \lim_{\substack{R\rightarrow \infty \\ \varrho \rightarrow 0^+}} 282 \bigg( \int_{C(R)} + \int_{(C(\rho)} + \int_{-R}^{\omega - 283 \varrho} + \int_{\omega +\varrho}^R 284 \bigg) 285 \;d\omega'\ \frac{G(\omega')}{\omega - \omega'} 286 \end{align} 287 We need show that the integral over the big circle goes to $0$. We know that for 288 $\omega' \neq \omega$ we have 289 \begin{align} 290 \big|\frac{G(\omega')}{\omega' - \omega}\big| &= 291 \big|\frac{1}{\omega'^3}\frac{1}{(1-\frac{\omega_1}{\omega'})(1-\frac{\omega_2}{\omega'}) 292 (1 - \frac{\omega}{\omega'})}\big| 293 \leq \frac{1}{R^3} 294 \end{align} 295 thus 296 \begin{align} 297 \bigg| 298 \int_{C(R)} d\omega' \frac{G(\omega')}{\omega - \omega'} 299 \bigg| \leq \frac{2\pi R}{R^3} = \frac{2\pi}{R^2} 300 \xrightarrow[R\rightarrow \infty]{} 0. 301 \end{align} 302 The small circle can be calculated with the Residue theorem with the pole at 303 $\omega$ 304 \begin{align} 305 \int_{C(\varrho)}d\omega' \frac{G(\omega')}{\omega - \omega'} = 2\pi i 306 \text{I}(C(\varrho), \omega) \text{Res}(\frac{G(\omega')}{\omega - \omega'}, 307 \omega) = i\pi G(\omega). 308 \end{align} 309 Note that we go around $\omega$ only $1/2$ times. Reconstructing the integral 310 equation we get 311 \begin{align} 312 -i\pi G(\omega) = \lim_{\varrho \rightarrow 0^+} 313 \big( 314 \int_{-R}^{\omega -\varrho} + \int_{\omega +\varrho}^R 315 \big) \;d\omega'\ \frac{G(\omega')}{\omega' - \omega} 316 \end{align} 317 which is exactly the Cauchy Principal Value. Furthermore we can rewrite 318 $G(\omega)$ into real and imaginary parts 319 \begin{align} 320 \text{Re} (G(\omega)) = \frac{1}{\pi} \PV d\omega' \frac{\text{Im} 321 (G(\omega'))}{\omega' - \omega}\\ 322 \text{Im} (G(\omega)) = \frac{1}{\pi} \PV d\omega' \frac{\text{Re} 323 (G(\omega'))}{\omega' - \omega}\\ 324 \end{align} 325 which are Hilbert transforms of each other, the equations are also known 326 as ``dispersion relations''. It should be noted that these equations also allow 327 negative frequencies. Let us derrive an representation for only positive 328 frequencies. We start off by a simple statement 329 \begin{align} 330 G(-\omega^*) = G(\omega)^*. 331 \end{align} 332 In our case this is obviously true 333 \begin{align} 334 &G(-\omega^*) = \frac{1}{-(\omega^*)^2 + i\gamma \omega^* + \omega_0^2}\\ 335 \nonumber \\ 336 &G(\omega)^* = \frac{1}{-(\omega^2)^* + i\gamma \omega^* + \omega_0^2} = 337 G(-\omega^*) 338 \end{align} 339 Now we choose $\omega \in \mathbb{R}^+$, our relation then becomes 340 $G(-\omega) = G(\omega)^*$. Then we get 341 \begin{align} 342 \text{Re} (G(\omega)) &= \frac{1}{\pi} \PV_0^\infty d\omega' \frac{2\omega'\text{Im} 343 (G(\omega'))}{\omega'^2 - \omega^2}\\ 344 \text{Im} (G(\omega)) &= -\frac{1}{\pi} \PV_0^\infty d\omega' \frac{2\omega'\text{Re} 345 (G(\omega'))}{\omega'^2 - \omega^2} 346 \end{align} 347 348 To round this chapter up we would like to show one last identity in the sense 349 of distributions 350 \begin{align} 351 \lim_{\varepsilon \rightarrow 0^+} \frac{1}{\omega' -\omega \mp 352 i\varepsilon} = \text{P}(\frac{1}{\omega' - \omega}) \pm i\pi\delta(\omega' 353 - \omega). 354 \end{align} 355 Let us extend the fraction with $\omega' - \omega \pm i\varepsilon$. 356 \begin{align} 357 \frac{\omega' - \omega \pm i\varepsilon}{(\omega' -\omega \mp 358 i\varepsilon)(\omega' - \omega \pm i\varepsilon)} = \frac{\omega' - \omega 359 \pm i\varepsilon}{(\omega' - \omega)^2 + \varepsilon^2}. 360 \end{align} 361 That means for a test function $f(\omega')$ we have 362 \begin{align} 363 \lim_{\varepsilon \rightarrow 0^+} \int_{-\infty}^\infty d\omega' 364 \frac{f(\omega')}{\omega' - \omega \mp i\varepsilon} &= 365 \lim_{\varepsilon \rightarrow 0^+} 366 \int_{-\infty}^\infty d\omega' 367 \frac{(\omega' - \omega \pm i\varepsilon) f(\omega')}{(\omega' - \omega)^2 368 + \varepsilon^2}\\ 369 & = 370 \lim_{\varepsilon \rightarrow 0^+}\bigg( 371 \int_{-\infty}^\infty d\omega' 372 \frac{f(\omega')(\omega' - \omega )}{(\omega' - \omega)^2 + \varepsilon^2} 373 \pm i\varepsilon 374 \int_{-\infty}^\infty d\omega'\frac{f(\omega')}{(\omega' - \omega)^2 + \varepsilon^2} 375 \bigg)\label{eq:id}. 376 \end{align} 377 Let us look into the first integral in equation \ref{eq:id}, we can rewrite it 378 \begin{align} 379 \lim_{\varepsilon \rightarrow 0^+}\int_{-\infty}^\infty 380 d\omega'\frac{f(\omega')(\omega' - \omega )}{(\omega' - \omega)^2 + \varepsilon^2} 381 &= \lim_{\varepsilon,\varrho \rightarrow 0^+} 382 \bigg( 383 \int_{-\infty}^{\omega - \varrho}d\omega'\frac{f(\omega')(\omega' - 384 \omega )}{(\omega' - \omega)^2 + \varepsilon^2} 385 + \int_{\omega 386 \varrho}^{\infty}d\omega'\frac{f(\omega')(\omega' - \omega )}{(\omega' 387 - \omega)^2 + \varepsilon^2} + \\ 388 &+\int_{\omega -\varrho}^{\omega+ 389 \varrho}d\omega'\frac{f(\omega')(\omega' - \omega )}{(\omega' - 390 \omega)^2 + \varepsilon^2} 391 \bigg) 392 \\ 393 &= \PV_{-\infty}^{\infty}d\omega' \frac{f(\omega')}{(\omega' - \omega)} 394 \end{align} 395 The integral from $\omega - \varrho$ to $\omega + \varrho$ can be calculated by 396 pulling out $f(\omega)$ out of the integral and directly computing it, which 397 gives then vanishes. In second integral we approximate $f(\omega')$ to 398 $f(\omega)$ in the region $\omega' \simeq \omega$ 399 \begin{align} 400 \varepsilon \int_{-\infty}^{\infty}d\omega' \frac{f(\omega')}{(\omega' - 401 \omega)^2 + \varepsilon^2} &\simeq \varepsilon f(\omega) \int_{-\infty}^{\infty} 402 \frac{1}{(\omega' -\omega)^2 + \varepsilon^2}\\ 403 &= \pi f(\omega). 404 \end{align} 405 Which means the identity is 406 \begin{align} 407 \lim_{\varepsilon \rightarrow 0^+} \int_{-\infty}^{\infty} d\omega' 408 \frac{f(\omega')}{\omega' -\omega \mp i\varepsilon} = 409 \PV_{-\infty}^{\infty} d\omega' \frac{f(\omega')}{\omega' -\omega} \pm 410 i\pi f(\omega)\label{eq:pv} 411 \end{align} 412 \section{Potential scattering in quantum mechanics} 413 If we consider elastic scattering of a spinless particle off a 414 time-independent, spherically symmetric potential of finite range, we look for 415 stationary solutions $\psi$ of the Schrödinger equation 416 \begin{equation} 417 -\frac{\hbbar^2}{2m}\Vec{\nabla^2}\psi(\Vec{x})+V(r)\psi(\Vec{x})=E\psi(\Vec{x}) 418 \end{equation} 419 Since the potential is spherically symmetric, it only depends on $r$ and for 420 large values of $r$, it can be shown, that the asymptotic form of $\psi$ looks 421 like: 422 \begin{equation} 423 \psi(r,\theta)\approx A[e^{ikz}+f(E,\theta)\frac{e^{ikr}}{r}] 424 \end{equation} 425 Where $kr\gg 1$, and k given by 426 \begin{equation} 427 k=\frac{\sqrt{2mE}}{\hbbar} 428 \end{equation} 429 We also define the scattering angle $\theta$ by $z=r\cos{\theta}$, and since 430 there is no dependence on the azimuthal angle $\phi$. we can define the 431 incoming and outgoing parts of the wave function as follows: 432 \begin{equation} 433 \psi_{in}=Ae^{ikz} 434 \end{equation} 435 and 436 \begin{equation} 437 \psi_{out}=Af(E,\theta)\frac{e^{ikr}}{r} 438 \end{equation} 439 Where the factor $\frac{1}{r}$ is carried, to conserve probability. The complex 440 function $f(E,\theta)$ is the so called scattering amplitude. We are now 441 interested in the differential cross section $\frac{d\sigma}{d\Omega}$, which 442 is defined as the ratio of number of particles per unit time, that are 443 scattered into the surface element $dS=r^2d\Omega(\theta,\phi)$ and the number 444 of incoming particles per unit time, per are orthogonal to the beam direction. 445 Expressed via probability currents, we thus obtain: 446 \begin{equation} 447 \frac{d\sigma}{d\Omega}=\frac{\Vec{j}_{out}\cdot \Vec{e}_rr^2}{|\Vec{j}_{in}|} 448 \end{equation} 449 With $\Vec{e}_r$ being a unit vector in direction of the radius, and the currents given as: 450 \begin{equation} 451 \Vec{j}=\frac{i\hbbar}{2m}(\psi\Vec{\nabla}\psi^*-\psi^*\Vec{\nabla}\psi) 452 \end{equation} 453 By applying these equations, we obtain the differential crosssection as: 454 \begin{equation} 455 \frac{d\sigma}{d\Omega}=|f(E,\theta)|^2 456 \end{equation} 457 With the scattering amplitude, being given as: 458 \begin{equation} 459 f(E,\theta)=\sum_{l=0}^\infty(2l+1)f_l(E)P_l(cos\theta) 460 \end{equation} 461 where $l$ denotes the magnitude of orbital angular momentum, and 462 $P_l(cos\theta)$ are the Legendre polynomials. 463 We can now work out the total crosssection $\sigma$ via the integral: 464 \begin{equation} 465 \sigma=\int{d\Omega\frac{d\sigma}{d\Omega}} 466 \end{equation} 467 We do this, by applying the orthogonality relation 468 \begin{equation} 469 \int{d\Omega P_l(cos\theta)P_{l'}(cos\theta)}=\frac{4\pi}{(2l+1)}\delta_{ll'} 470 \end{equation} 471 Thus, we obtain: 472 \begin{equation} 473 \sigma=\sum_{l,l'}(2l+1)(2l'+1)f^*_l(E)f_{l'}(E)\int{d\Omega P_l(cos\theta)P_{l'}(cos\theta)} 474 \end{equation} 475 Which, finally leads to: 476 \begin{equation} 477 \sigma=4\pi\sum_l(2l+1)|f(E)|^2 478 \end{equation} 479 480 481 482 \section{The pion vector from factor and the Omn\`es Problem} 483 Insert Text 484 \subsection{Unitarity of the scattering matrix} 485 Insert Text 486 \begin{align} 487 \label{eq:rec} 488 \text{Im}(F^V_\pi(s) = F^V_\pi(s) e^{-i\delta_{\pi\pi}(s)}\sin\delta_{\pi\pi}(s) 489 \end{align} 490 \subsection{The Omn\`es Problem} 491 The equations \ref{eq:rec} allow us to carefully reconstruct the pion Vector Form 492 Factor, based on strictly formulated conditions. This is known as the Omn\`es 493 Problem. First of all the equation tells us that $F_\pi^V(s)$ is a complex 494 valued function, as $s$ is an analytic variable in the complex plane, apart 495 from a cut complex s-plane $\Gamma = [s_0, \infty) \subset \mathbb{R}$, where 496 $s = 4M_\pi^2 > 0$. To summerize the conditions are 497 \begin{itemize} 498 \item $F_\pi^V(s)$ is analytic on the cut complex s-plane 499 $\mathbb{C}\backslash \Gamma$ 500 \item $F_\pi^V(s)\in \mathbb{R} \;\;\; \forall\; s \in 501 \mathbb{R}\backslash \Gamma$ 502 \item $\lim_{\varepsilon \rightarrow 0}(F_\pi^V(s+i\varepsilon) 503 e^{-i\delta_{\pi\pi}(s)}) \in \mathbb{R}$ on $\Gamma$ for a real 504 bounded fucntion $\delta_{\pi\pi}(s)$ 505 \item We assume $F_\pi^V(0) = 1$ and $F_\pi^V(s)$ has no zeros. 506 \end{itemize} 507 We start off with the Couchy Integral 508 \begin{align} 509 \ln(F(s)) = \frac{1}{2\pi i}\oint_C ds' \frac{\ln(F(s'))}{s'-s} 510 \end{align} 511 over the following contour 512 \begin{center} 513 \begin{tikzpicture}[decoration={markings, 514 mark=at position 0.5cm with {\arrow[line width=2pt]{>}}, 515 mark=at position 5cm with {\arrow[line width=2pt]{>}}, 516 mark=at position 13cm with {\arrow[line width=2pt]{>}}, 517 mark=at position 21cm with {\arrow[line width=2pt]{>}}, 518 mark=at position 23cm with {\arrow[line width=2pt]{>}} 519 } 520 ] 521 % The axes 522 \draw[help lines,->] (-3.5,0) -- (3.5,0) coordinate (xaxis); 523 \draw[help lines,->] (0,-3.5) -- (0,3.5) coordinate (yaxis); 524 525 % The path 526 \path[draw,line width=0.8pt,postaction=decorate] 527 (1.5,0.1) -- (3, 0.1) node[below right] {} arc (4:360:3) -- (1.5, -0.1) 528 node[below right] {} arc(345:15:0.4); 529 530 % The labels 531 \draw[thick, ->] (0,0) -- (-2, 2) node[midway, fill=white] {$R$}; 532 \draw[thick, ->] (1.1,0) -- (1.1, 0.4) node[above] {$\varrho$}; 533 \node[below] at (xaxis) {$\text{Re}$}; 534 \node[left] at (yaxis) {$\text{Im}$}; 535 \node[circle,inner sep=1pt,label=below:{$s_0$}, fill=black] at (1.1,0) {}; 536 \end{tikzpicture} 537 \end{center} 538 This means the integral can be separated into 539 \begin{align} 540 \oint_C ds' \frac{\ln(F(s'))}{s'-s} = 541 \lim_{\substack{\varepsilon \rightarrow 0}} 542 \bigg( 543 \int_{C(R)} + \int_{C(\varrho)} + \int_{s_0+i\varepsilon}^{\infty 544 +i\varepsilon} + \int^{s_0-i\varepsilon}_{\infty 545 -i\varepsilon} 546 \bigg) ds' \frac{\ln(F(s'))}{s'-s} 547 \end{align} 548 The integrals over $C(R)$ and $C(\varrho)$ dissapear. For the last two 549 integrals we can use the Schwarz reflection principle and then we get 550 \begin{align} 551 \oint_C ds' \frac{\ln(F(s'))}{s'-s} = \frac{1}{\pi} \int_{s_0}^{\infty}ds' 552 \text{Im}\left( 553 \frac{\ln(F(s'))}{s'-s}\right) 554 \end{align} 555 where we used $\text{Im}(z) = \frac{z - z^*}{2i}$ to write the imaginary part 556 here. We look now at equation \ref{eq:rec} and refactor it 557 \begin{align} 558 &\frac{F_\pi^V(s)-F_\pi^V(s)^*}{2i} = F_\pi^V e^{i\delta_{\pi\pi}}(s) 559 \sin(\delta_{\pi\pi}(s)) \\ 560 &F_\pi^V(s) = F_\pi^V(s)^* e^{2i \delta_{\pi\pi}(s)}\\ 561 &\ln(F_\pi^V(s)) = \ln((F_\pi^V e^{-i\delta_{\pi\pi}})^*)+ 562 i\delta_{\pi\pi}(s).\label{eq:use} 563 \end{align} 564 Now we use this equation to compute the integral with variation in $s 565 \rightarrow s+i\varepsilon$ as $\varepsilon$ goes to infinity. 566 \begin{align} 567 \ln(F_\pi^V(s)) &= \lim_{\substack{\varepsilon \rightarrow \infty}} 568 \ln(F_\pi^V(s+i\varepsilon)) = \\ 569 &= \lim_{\substack{\varepsilon \rightarrow \infty}}\frac{1}{\pi} 570 \int_{s_0}^{\infty} 571 \text{Im}\left(\frac{F_\pi^V(s')}{s'-s-i\varepsilon}\right)=\\ 572 &=\lim_{\substack{\varepsilon \rightarrow \infty}}\frac{1}{\pi} 573 \int_{s_0}^{\infty} 574 \text{Im}\left(\frac{\ln((F_\pi^V e^{-i\delta_{\pi\pi}})^*)+ 575 i\delta_{\pi\pi}}{s'-s-i\varepsilon}\right) 576 \end{align} 577 the part $F_\pi^V e^{-i\delta_{\pi\pi}}$ needs to be real that means 578 \begin{align} 579 \ln(F_\pi^V(s)) &= \lim_{\substack{\varepsilon \rightarrow \infty}}\frac{1}{\pi} 580 \int_{s_0}^{\infty} 581 \frac{\delta_{\pi\pi}(s')}{s'-s-i\varepsilon} =\\ 582 &= \ln(F_\pi^V(0)) + \lim_{\substack{\varepsilon \rightarrow \infty}} 583 \frac{s}{\pi} 584 \int_{s_0}^{\infty} 585 \frac{\delta_{\pi\pi}(s')}{s'(s'-s-i\varepsilon)} 586 \end{align} 587 with the condition $F_\pi^V(0) = 1$ and the relation from \ref{eq:pv} we get 588 \begin{align} 589 F_\pi^V(s) = \exp 590 \bigg( 591 \frac{s}{\pi}\PV_{s_0}^\infty ds' \frac{\delta_{\pi\pi}(s')}{s'(s'-s)} 592 + i\delta_{\pi\pi}(s) 593 \bigg). 594 \end{align} 595 To compute the principal value integral we use the following trick 596 \begin{align} 597 \frac{s}{\pi}\PV_{s_0}^\infty ds' \frac{\delta_{\pi\pi}(s')}{s'(s'-s)} = 598 \frac{2}{\pi} \int_{s_0}^\infty ds' \frac{\delta_{\pi\pi}(s') 599 -\delta_{\pi\pi}(s)}{s'(s'-s)}+ 600 \delta_{\pi\pi}\frac{s}{\pi}\PV_{s_0}^\infty \frac{1}{s'(s'-s)}. 601 \end{align} 602 The first integral can be computed numerically, the second one has an analytic 603 solution for $s > s_0$. We use the definition of the principal value and circle 604 around $s$ in a small half circle with the radius $r$. 605 \begin{align} 606 \PV_{s_0}^\infty \frac{1}{s'(s'-s)} = \lim_{\substack{r\rightarrow0}} 607 \bigg( 608 \int_{s_0}^{s-r}ds' + \int_{s+r}^{\infty} ds' 609 \bigg)\frac{1}{s'(s'-s)} 610 \end{align} 611 then we simply integrate and plug in 612 \begin{align} 613 \PV_{s_0}^\infty \frac{1}{s'(s'-s)} &= \lim_{\substack{r\rightarrow 0}} 614 \bigg( 615 \frac{\ln(s'-s) - \ln(s')}{s}\big|_{s'= s_0}^{s'= s-r} 616 \frac{\ln(s'-s)-\ln(s')}{s}\big|_{s'=s+r}^{s'=\infty} 617 \bigg) =\\ 618 &=\frac{1}{s}\ln\left(\frac{s_0}{s_0-s}\right) 619 \end{align} 620 that means the second integral is 621 \begin{align} 622 \delta_{\pi\pi}\frac{s}{\pi}\PV_{s_0}^\infty \frac{1}{s'(s'-s)} 623 =\delta_{\pi\pi}(s)\frac{1}{\pi} \ln\left(\frac{s_0}{s_0 -s }\right) 624 \end{align} 625 626 Lastly we would like to plot the modulus of the Omn\`es representation of the 627 pion vector form 628 factor. For the phase we would usually use experimentall value, but in our case 629 we will use the Breit-Wigner representation of the pion VVF to compute the 630 phase $\delta_{\pi\pi}$. A reminder the Breit-Wigner representation is the 631 following 632 \begin{align} 633 F^V_\pi(s)_{BW} = \frac{M_\varrho^2}{M_\varrho^2 - s - iM_\varrho 634 \Gamma_\varrho(s)} 635 \end{align} 636 where 637 \begin{align} 638 \Gamma_\varrho(s) := \Gamma_\varrho\frac{s}{M_\varrho^2} \left( 639 \frac{\sigma_\pi(s)}{\sigma_\pi(M_\varrho^2)} 640 \right)^3 \theta(s-4M_\pi^2), \;\;\;\;\;\; \sigma_\pi(s) := 641 \sqrt{1-\frac{4M_\pi^2}{s}}. 642 \end{align} 643 Thus our phase shift will be 644 \begin{align} 645 \delta_{\pi\pi}(s) := \arg\left(F_\pi^V(s)_{BW}\right) 646 \end{align} 647 where we will use numerical values for $M_\varrho = 0.77\ \text{GeV}$, 648 $\Gamma_\varrho = 0.15\ \text{GeV}$, $M_\pi = 0.14 \text{GeV}$. 649 \begin{figure}[H] 650 \centering 651 \includegraphics[width=0.9\textwidth]{./omnes_bw.png} 652 \caption{Plot of the modulus of the Breit-Wigner(red) and the Omn\`es 653 representation(black) of the pion Vector From Factor for $s \in [0, 1]$ in $GeV^2$} 654 \end{figure} 655 \nocite{mathe} 656 \nocite{stoffer} 657 \nocite{omnes} 658 \printbibliography 659 \end{document}