英文版本 (pdfLaTeX)
使用 Malmoe + dolphin 主題與 mathpazo 數學字型套件(這個最好直接從 miktex package manager 更新,否則一邊 compile 一邊裝字型會一直跳出確認視窗,很麻煩),mathpazo 可以讓方程式的粗體看起來比較自然,你可以自己切換一下比較它的效果,通常我們以 pdfLaTeX 來編譯 beamer 的檔案,會想用 beamer 來作投影片的人大多以英語作為內文。preamble 區主要放入的指令為:
\usetheme{Malmoe}
\usecolortheme{dolphin}
\useoutertheme{miniframes}
\usepackage{mathpazo}
Click here to edit.
另外我還常用一種與 a4size 幾乎相同的字型主題(主要是要克服粗體方程式字體問題),看起來比較尖銳,不過效果也很好,適合排版字數較多版面比較密集的投影片,preamble 區主要放入的指令為:
\documentclass[8pt]{beamer}
\setbeamersize{text margin left=8mm, text margin right=8mm}
\usetheme{Warsaw}
\setbeamertemplate{blocks}[rounded][shadow=true]
\usefonttheme{professionalfonts}
\usefonttheme{serif}
\setbeamertemplate{navigation symbols}{}
\usepackage[absolute,overlay]{textpos}
Click here to edit.
Malmoe + dolphin 主題與 mathpazo 數學字型套件範本
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Warsaw 主題 + professionalfonts 字體範本
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中文版本
要使用中文版本的 beamer 可以使用 XeLaTeX 來編譯,必需在 preamble 加入以下設定
\usepackage{fontspec}
%有這個套件才能自由設定字體
\setsansfont[BoldFont=cwTeXQHei-Bold]{cwTeX Q Ming}
%cwTeX 預設無粗體,所以必須特別為英文字體設定粗體字體
%若不會用到 \textbf 指令可以去掉[BoldFont=cwTeXQHei-Bold]
\newfontfamily\mysm{CMU Typewriter Text}
%打程式碼的時候會另外用到 type writer font
%取用時打 {\mysm YOURCODE} 即可
%沒裝 CMU 自行可以用 Courier New 來取代(較細)
\XeTeXlinebreaklocale "zh"
\XeTeXlinebreakskip = 0pt plus 1pt
%使用 XeLaTeX 的主要斷行設定
很重要的一點在於: beamer 預設取用的是 sans 字體,所以要把 setmainfont 改成 setsansfont 才行。
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英文版 pre-1 原始碼, mathpazo 數學字型:
\documentclass[12pt]{beamer}
\usetheme{Malmoe}
\usecolortheme{dolphin}
%\usefonttheme{professionalfonts}
\useoutertheme{miniframes}
%{infolines}
%\usepackage{xmpmulti}
%\usepackage{fouriernc}
%\usepackage[T1]{fontenc}
%\usepackage{fourier}
%\usepackage[scaled=0.9]{helvet}
\usepackage{bm}
\usepackage{titlesec}
\usepackage{graphicx,subfigure}
\usepackage{amsmath}
\usepackage{amsfonts}
%\usepackage{mathptmx}
%\usepackage[scaled=.90]{helvet}
%\usepackage{courier}
%\usepackage[T1]{fontenc}
\usepackage{mathpazo}
%\usepackage[bitstream-charter]{mathdesign}
\usepackage{amssymb}
\usepackage{url}
\usepackage{appendix}
\usepackage{indentfirst}
\usepackage{makeidx}
\usepackage{array}
\newcommand{\proposition}[1]{\bigskip\noindent\fbox{\begin{minipage}{0.95\textwidth}~\begin{minipage}{0.95\textwidth}#1\end{minipage}\\\smallskip\end{minipage}}\bigskip\vline\vline\vline}
\newcommand{\excercise}{\addtocounter{subsection}{+1}
\subsubsection{Ex~\underline{\thesubsection}}}
\newcommand{\stag}[1]{\tag{\arabic{equation}#1}}
\newcommand{\qdo}[0]{\quad}
\newcommand{\qdt}[0]{\quad\quad}
\newcommand{\qdthr}[0]{\quad\quad\quad}
\newcommand{\qdf}[0]{\quad\quad\quad\quad}
\newcommand{\qdfiv}[0]{\quad\quad\quad\quad\quad}
\newcommand{\norm}[1]{\|#1\|}
\newcommand{\sdelta}{{\scriptstyle\Delta}}
\newcommand{\bra}[1]{\langle#1|}
\newcommand{\ket}[1]{|#1\rangle }
\newcommand{\braket}[2]{\langle#1|#2\rangle }
\newcommand{\proj}[2]{|#1\rangle\langle#2|}
\newcommand{\expect}[2]{\langle #1\rangle_{#2}}
\newcommand{\Kappa}[0]{\bm{\mathcal{K}}}
\newcommand{\parity}[0]{\bm{\mathcal{P}}}
\newcommand{\parityl}[0]{\bm{p}}
\newcommand{\charge}[0]{\bm{\mathcal{C}}}
\newcommand{\ptplus}[0]{\\[+5pt]}
%%%%%%
\title{Lorentz Symmetry, Weyl Spinors, Chirality and Dirac Equation}
\author{Kow Lung Chang}
\institute{Physics Department, National Taiwan University}
%%%%%%
\begin{document}
\begin{frame}
\titlepage
\end{frame}
%%%%%page1
\begin{frame}
\frametitle{Contents}
$\blacklozenge$ Minkowski Space and Lorentz Transformation \ptplus
$\blacklozenge$ Generators of Lorentz Group\ptplus
$\blacklozenge$ Irreducible Representations of Lorentz Group \\ ~~~and Weyl
Spinors \ptplus
$\blacklozenge$ SO(3,1) and SL(2,C)\ptplus
$\blacklozenge$ Chiral Transformation and Spinor Algebra\ptplus
$\blacklozenge$ Spinor space and Co-spinor space\ptplus
$\blacklozenge$ Dirac spinor and Dirac equation\ptplus
$\blacklozenge$ Invariance of the $ \gamma $ matrices in all Lorentz frames\ptplus
$\blacklozenge$ Zero Mass Limit and Helicity of Weyl spinors\ptplus
\end{frame}
\begin{frame}
\frametitle{Minkowski Space and Lorentz Transformation}
Difine the Minkowski contravariant 4-vector as:
\begin{align}
x^\mu=(x^0=ct, \vec{x}),
\end{align}
\noindent and the Minkowski covariant 4-vector as:
\begin{align}
x_\mu=(x^0=-ct, \vec{x}),
\end{align}
\noindent with the metric tensor
\begin{align}
g_{\mu\nu}=0~~~\mbox{if}~~~ \mu\neq\nu~;~~~-g_{00}=g_{11}=g_{22}=g_{33}=1.
\end{align}
\end{frame}
\begin{frame}
A linear transformation on the $ x^\mu $ given as follows
\begin{align}
x'^{\mu}=\bm{\Lambda}^\mu_\nux^\nu
\end{align}
\noindent is called Homogeneous Lorentz transformation ( HLT ), or simply LT if the following condition is met:
\begin{align}
x'^{\mu}x'_\mu=x^\nux_\mu
\end{align}
\noindent or in matrix notation as
\begin{align}\label{transform-41}
\bm{\Lambda}^T\bm{g}\bm{\Lambda}=\bm{g} ~~~\mbox{ or }~~~ \bm{\Lambda}^T\bm{g}=\bm{g}\bm{\Lambda}^{-1} .
\end{align}
\end{frame}
\begin{frame}
Since there exists an identity Lorentz transformation,$ \Lambda=\mathbf{I} $, and an inverse Lorentz transformation, $ \bm{\Lambda}^{-1} $, namely both $ \mathbf{I} $ and $ \bm{\Lambda}^{-1} $ exist.
Therefore LT forms a group SO(3,1) because:
\begin{align}\label{transform-41}
(\bm{\Lambda}_1\bm{\Lambda}_2)^Tg(\bm{\Lambda}_1\bm{\Lambda}_2)= \bm{\Lambda}_2^T\bm{\Lambda}_1^Tg\bm{\Lambda}_1\bm{\Lambda}_2=g.
\end{align}
The condition $ \mbox{det} \bm{\Lambda}=1 $ is automatically satisfied. We shall only
consider the proper LT in which $ \bm{\Lambda}^0_0\geqslant 1 $ in this lecture.
\end{frame}
\begin{frame}
Since the condition $ \bm{\Lambda}^tg\bm{\Lambda}=g $ provide 10 constraints among 16 matrix elements of $ \bm{\lambda} $, the remaining 6 independent coefficients serve as the 6 group parameters, specified as $ \bm{\Lambda}=\bm{\Lambda}(\vec{\theta},\vec{\xi}) $ and
\addtocounter{equation}{+1}
\begin{align*}
\vec{\theta}&=(\theta^1,\theta^2,\theta^3)=\mbox{ rotation},\stag{a}\\
\vec{\xi}&=(\xi^1,\xi^2,\xi^3)=\mbox{ Lorentz boost}\stag{b}.
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Generators of Lorentz Group}
The generators of the group are given:
\begin{align}
A_i=\frac{\partial}{\partial\theta^i}\bm{\Lambda}(\vec{\theta},\vec{\xi}),~~~B_i=\frac{\partial}{\partial\xi^i}\bm{\Lambda}(\vec{\theta},\vec{\xi}),\stag{a,b}
\end{align}
For the Lorentz boost along 1-axis with angle $ \xi $,
\begin{align}
\bm{\Lambda}=\begin{pmatrix}
\cosh\xi&-\sinh\xi&0&0\\
-\sinh\xi&\cosh\xi&0&0\\
0&0&1&0\\
0&0&0&1\end{pmatrix}.
\end{align}
\noindent where $ \xi=\tanh^{-1}\beta $ and
\end{frame}
\begin{frame}
\addtocounter{equation}{1}
\begin{align*}
B_1=\begin{pmatrix}
0&-1&0&0\\
-1&0&0&0\\
0&0&0&0\\
0&0&0&0\end{pmatrix}.~~~~~~\stag{a}
\end{align*}
\noindent similarly,
\begin{align*}
B_2=\begin{pmatrix}
0&0&-1&0\\
0&0&0&0\\
-1&0&0&0\\
0&0&0&0\end{pmatrix},~~~
B_3=\begin{pmatrix}
0&0&0&-1\\
0&0&0&0\\
0&0&0&0\\
-1&0&0&0\end{pmatrix}.\stag{b,c}
\end{align*}
\end{frame}
\begin{frame}
And for the generators if rotation, we have
\addtocounter{equation}{1}
\begin{align*}
A_1=\begin{pmatrix}
0&0&0&0\\
0&0&0&0\\
0&0&0&1\\
0&0&-1&0\end{pmatrix}~~~~~~\stag{a}
\end{align*}
\noindent and
\begin{align*}
\mbox{and}~~A_2=\begin{pmatrix}
0&0&0&0\\
0&0&0&-1\\
0&0&0&0\\
0&1&0&0\end{pmatrix},~~~
A_3=\begin{pmatrix}
0&0&0&0\\
0&0&1&0\\
0&-1&0&0\\
0&0&0&0\end{pmatrix}.\stag{b,c}
\end{align*}
\end{frame}
\begin{frame}
SO(3,1) Lie algebra as:
\begin{align}
[A_i,A_j]=-\epsilon_{ij}^kA_k,~~~[A_i,B_j]=-\epsilon_{ij}^kB_k,
\end{align}
Canonical formulation of algebra:
\addtocounter{equation}{+1}
\begin{align*}
M^{\mu\nu}=x^{\mu}\frac{\partial}{\partial x_\nu}-x^{\nu}\frac{\partial}{\partial x_\mu},\stag{a}
\end{align*}
\begin{align*}
[M^{\mu\nu},M^{\alpha\beta}]=-g^{\nu\beta}M^{\mu\alpha}-g^{\mu\alpha}M^{\nu\beta}
+g^{\nu\alpha}M^{\mu\beta}+g^{\mu\beta}M^{\nu\alpha}.\stag{b}
\end{align*}
\end{frame}
\begin{frame}
If we denote
\addtocounter{equation}{1}
\begin{align*}
L_i=\frac{1}{2}\left(\frac{A_i}{i}+B_i\right),~~~\mbox{and}~~~R_i=\frac{1}{2}\left(\frac{A_i}{i}-B_i\right).\stag{a,b}
\end{align*}
The algebra takes as
\addtocounter{equation}{1}
\begin{align*}
\notag[L_i,L_j]&=i\epsilon^k_{ij}L_k,\stag{a}\\
\notag[L_i,R_i]&=0,\stag{b}\\
[R_i,R_j]&=i\epsilon^k_{ij}R_k.\stag{c}
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Irreducible Representations of \\
~~~~~~~~~~~Lorentz Group and Weyl Spinors}
Consider the finite dimensional representations, denoted by $ (l, r) $ with the basis\\[-10pt]
\begin{align}
\ket{l,m}\otimes\ket{r,n}\equiv\ket{l,m;r,n}
\end{align}
\noindent where\\[-10pt]
\begin{align}
-l\leqslant m\leqslant l,~~~-r\leqslant n\leqslant r,~~~\mbox{and}~~~l,r=~\mbox{half integers.}
\end{align}
The simpliest representation of the generators, an one dimensional
$ (0,0) $-representation read as\\[-10pt]
\begin{align}
\bra{0,0;0,0}L_i \ket{0,0;0,0}=\bra{0,0;0,0}R_i \ket{0,0;0,0}=0,
\end{align}
\end{frame}
\begin{frame}
$ (\frac{1}{2}, 0) $-representation: left-handed-spinor
\begin{align}
\ket{\frac{1}{2},m;0,0}
\end{align}
$ (0, \frac{1}{2}) $-representation: right-handed-spinor
\begin{align}
\ket{0,0;\frac{1}{2},n}
\end{align}
\noindent then we have
\addtocounter{equation}{1}
\begin{align*}
L^{(\frac{1}{2},0)}_i&=\frac{1}{2}\bm{\sigma_i},&R^{(\frac{1}{2},0)}_i&=0,\tag{\arabic{chapter}.\arabic{equation}a}\\
L^{(0,\frac{1}{2})}_i&=0,&R^{(0,\frac{1}{2})}_i&=\frac{1}{2}\bm{\sigma_i},\tag{\arabic{chapter}.\arabic{equation}b}
\end{align*}
\end{frame}
\begin{frame}
\noindent which lead to
\addtocounter{equation}{1}
\begin{align*}
A^{(\frac{1}{2},0)}_i&=\frac{i}{2}\bm{\sigma_i},&B^{(\frac{1}{2},0)}_i&=\frac{1}{2}\bm{\sigma_i},\tag{\arabic{chapter}.\arabic{equation}a}\\
A^{(0,\frac{1}{2})}_i&=\frac{i}{2}\bm{\sigma_i},&B^{(0,\frac{1}{2})}_i&=-\frac{1}{2}\bm{\sigma_i},\tag{\arabic{chapter}.\arabic{equation}b}
\end{align*}
\noindent and the 2-dimensional irreducible representation of Lorentz group as
\addtocounter{equation}{1}
\begin{align*}
D^{(\frac{1}{2},0)}(\vec{\theta},\vec{\xi})=\exp\left(\frac{i}{2}\vec{\sigma}\cdot(\vec{\theta}-i\vec{\xi})\right),\stag{a}\label{dspin1}\\
D^{(0,\frac{1}{2})}(\vec{\theta},\vec{\xi})=\exp\left(\frac{i}{2}\vec{\sigma}\cdot(\vec{\theta}+i\vec{\xi})\right).\stag{b}
\end{align*}
\end{frame}
\begin{frame}
As a quick check that
\addtocounter{equation}{+1}
\begin{align*}
D^{(\frac{1}{2},0)\dag}(\vec{\theta},\vec{\xi})=\exp\left(-\frac{i}{2}\vec{\sigma}\cdot(\vec{\theta}+i\vec{\xi})\right)\neq D^{(\frac{1}{2},0)}(\vec{\theta},\vec{\xi})^{-1},
\end{align*}
\begin{align*}
D^{(0,\frac{1}{2})\dag}(\vec{\theta},\vec{\xi})=\exp\left(-\frac{i}{2}\vec{\sigma}\cdot(\vec{\theta}-i\vec{\xi})\right)\neq D^{(0,\frac{1}{2})}(\vec{\theta},\vec{\xi})^{-1},
\end{align*}
Let us perform the identifications
\addtocounter{equation}{1}
\begin{align*}
\ket{\frac{1}{2},\frac{1}{2}}_l\longmapsto\begin{pmatrix}1\\0\end{pmatrix}=e_1,~~~
\ket{\frac{1}{2},-\frac{1}{2}}_l\longmapsto\begin{pmatrix}0\\1\end{pmatrix}=e_2,\stag{a,b}
\end{align*}
\end{frame}
\begin{frame}
\noindent then\\[-10pt]
\begin{align}
\psi_l(x)=\psi^a_l(x)e_a=\begin{pmatrix}\psi^1_l(x)\\\psi^2_l(x)\end{pmatrix},
\end{align}
the Lorentz Transformation as
\begin{align}
\psi_l(x)\longmapsto\psi_l'(x')=D^{(\frac{1}{2},0)}(\vec{\theta},\vec{\xi})\psi_l(\bm{\Lambda}^{-1}x'),
\end{align}
Similarly if\\[-10pt]
\begin{align}
\psi_r(x)=\psi^a_r(x)f_a=\psi^1_r(x)\begin{pmatrix}1\\0\end{pmatrix}+ \psi^2_r(x)\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}\psi^1_r(x)\\\psi^2_r(x)\end{pmatrix},
\end{align}
\noindent the transformation reads as\\[-10pt]
\begin{align}
\psi_r(x)\longmapsto\psi_r'(x')=D^{(0,\frac{1}{2})}(\vec{\theta},\vec{\xi})\psi_r(\bm{\Lambda}^{-1}x')).
\end{align}
\end{frame}
\begin{frame}
\frametitle{SO(3,1) and SL(2,C)}
SL(2,C) transformation in $ \mathcal{C}^2 $-space:
\begin{align}
\xi'=\begin{pmatrix}\xi'^1\\\xi'^2\end{pmatrix}=\bm{L}\begin{pmatrix}\xi^1\\\xi^2\end{pmatrix}= \begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}\xi^1\\\xi^2\end{pmatrix}.
\end{align}
\noindent where
\begin{align}
\mbox{det}~\bm{L}=\left|\begin{array}{cc}a&b\\c&d\end{array}\right|=ab-bc=1,
\end{align}
If we exponentiate $ L $ by a $ 2\times 2 $ matrix $ \bm{A} $, i.e.
\begin{align}
\bm{L}=e^{\bm{A}},
\end{align}
\end{frame}
\begin{frame}
Then we have the following proposition
\proposition{
\subsection*{\textit{Proposition 1.}}
{\it If a matrix} $ \bm{L} $ {\it can be expressed as} $ \bm{L}=e^{\bm{A}} $, {\it then }
\begin{align}
\mbox{det}~\bm{L}=e^{\mbox{Tr}~\bm{A}}.
\end{align}
}
Hence we ensure that
\begin{align}
\mbox{det}~D^{(\frac{1}{2},0)}(\vec{\theta},\vec{\xi})= \mbox{det}~e^{\frac{i}{2}\vec{\sigma}\cdot(\vec{\theta}-i\vec{\xi})}=e^{\mbox{Tr}~\frac{i}{2}\vec{\sigma}\cdot(\vec{\theta}-i\vec{\xi})}=1,
\end{align}
\noindent and
\begin{align}
\mbox{det}~D^{(0,\frac{1}{2})}(\vec{\theta},\vec{\xi})= \mbox{det}~e^{\frac{i}{2}\vec{\sigma}\cdot(\vec{\theta}+i\vec{\xi})}=e^{\mbox{Tr}~\frac{i}{2}\vec{\sigma}\cdot(\vec{\theta}+i\vec{\xi})}=1.
\end{align}
\end{frame}
\begin{frame}
The isomorphism of SL(2,C) onto SO(3,1) in Lorentz transformation can be demonstrated as follows:
let
\begin{align}
\bm{X}=x^\mu\bm{\sigma}_\mu=\begin{pmatrix}-x^0+x^3&x^1-ix^2\\x^1+ix^2&-x^0-x^3\end{pmatrix},
\end{align}
\noindent and
\begin{align}
\mbox{det}~\bm{X}=(x^0)^2-\vec{x}^2.
\end{align}
\noindent which leads to the Lorentz Transformation on $ \bm{X} $ as
\begin{align}
\bm{X}'=D^{(\frac{1}{2},0)}(\vec{\theta},\vec{\xi})\bm{X}D^{(\frac{1}{2},0)\dag}(\vec{\theta},\vec{\xi}),
\end{align}
\end{frame}
\begin{frame}
\noindent because of the invariance of the length of the space-time vector,\\[-10pt]
\begin{align}
\mbox{det}~\bm{X}'=\mbox{det}~\bm{X}.
\end{align}
As an example when $ \mathcal{O}' $-frame is boost along the 3rd axis, i.e.
\begin{align}
\bm{X}'=e^{\frac{1}{2}\bm{\sigma}_3\xi}\bm{X}e^{\frac{1}{2}\bm{\sigma}_3\xi}=
\begin{pmatrix}e^{\frac{1}{2}\xi}&0\\0&e^{-\frac{1}{2}\xi}\end{pmatrix}\bm{X}
\begin{pmatrix}e^{\frac{1}{2}\xi}&0\\0&e^{-\frac{1}{2}\xi}\end{pmatrix}.
\end{align}
\noindent we regain the LT as follows
\addtocounter{equation}{1}
\begin{align*}
x'^0&=\cosh\xi x^0-\sinh\xi x^3,\stag{a}\\
x'^1&=x^1,\stag{b}\\
x'^2&=x^2,\stag{c}\\
x'^3&=-\sinh\xi x^0+\cosh\xi x^3.\stag{d}
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Chiral Transformation}
$ \Kappa =$ chiral operator, which is a discrete transformation between left handed irreducible representations and the right handed irreducible representations
It is an antilinear operator, i.e.
\begin{align}
\Kappa(a\psi+b\varphi)=a^*\Kappa\psi+b^*\Kappa\varphi,
\end{align}
\noindent as well as an antiunitary operator:
\begin{align}
(\Kappa\psi,\Kappa\varphi)=(\varphi,\psi)=(\psi,\varphi)^*.
\end{align}
\end{frame}
\begin{frame}
It is nothing to with the space-time coordinates, hence
\begin{align}
\Kappa A_i\Kappa^{-1}=A_i,~~~\Kappa B_i\Kappa^{-1}=B_i.
\end{align}
\noindnet but the operators $ L_i $ and $ R_i $ transform as follows
\addtocounter{equation}{1}
\begin{align*}
\Kappa L_i\Kappa^{-1}=\frac{1}{2}\Kappa\left(\frac{A_i}{i}+B_i\right)\Kappa^{-1}=-R_i,~~~
\Kappa R_i\Kappa^{-1}=-L_i,\stag{a,b}
\end{align*}
\noindent therefore we reach the following proposition if the basis of $ (\frac{1}{2},0) $- and $ (0,\frac{1}{2}) $-representation are abbreviated by
\addtocounter{equation}{1}
\begin{align*}
\ket{j,m;0,0}=L_{jm},\stag{a}\\
\ket{0,0,k,n}=R_{kn},\stag{b}
\end{align*}
\end{frame}
\begin{frame}
\noindent then
\proposition{
\subsection*{\textit{Proposition 2.}}
\it The vector $ \Kappa L_{jm} $ is the eigenvector of $ R^2 $ and $ R_3 $ with the eigenvalues $ j(j+1) $ and $ -m $ respectively. While the vector $ \Kappa R_{kn} $ is the eigenvector of $ L^2 $ and $ L_3 $ with the eigenvalues $ k(k+1) $ and $ -n $ respectively.
}
Since $ \KappaL^2=R^2\Kappa $, $ \Kappa\vec{L}=-\vec{R}\Kappa $
\end{frame}
\begin{frame}
\noindent then we have \\[-12pt]
\begin{align*}
R^2\Kappa L_{jm}=\Kappa L^2L_{jm}=j(j+1)\Kappa L_{jm},\\
R_3\Kappa L_{jm}=-\Kappa L_3L_{jm}=-m\Kappa L_{jm},
\end{align*}
Therefore \\[-12pt]
\begin{align*}
\Kappa L_{jm}=\gamma(m)R_{j -m}.
\end{align*}
Similarly \\[-18pt]
\begin{align*}
\Kappa R_{kn}=\delta{n}L_{k -n}.
\end{align*}
Hence we have \\[-18pt]
\begin{align}
(\Kappa L_{jm},\Kappa L_{jm'})=\gamma^*(m)\gamma(m')(R_{j,-m}, R_{j,-m'})=(L_{jm'},L_{jm}),
\end{align}
\noindent or $ \gamma^*(m)\gamma(m')\delta_{-m,-m'}=\delta_{m'm} $
\end{frame}
\begin{frame}
Take spinor for instance, we obtain that
\addtocounter{equation}{1}
\begin{align*}
\Kappa L_{\frac{1}{2} m}=\gamma(n)\Kappa R_{\frac{1}{2} -m}, ~~~~~
\Kappa R_{\frac{1}{2} n}=\delta(n)\Kappa L_{\frac{1}{2} -n}, \stag{ab}
\end{align*}
\noindent or
\addtocounter{equation}{1}
\begin{align*}
\Kappa \begin{pmatrix}1\\0\end{pmatrix}_l=\gamma(\frac{1}{2})\begin{pmatrix}0\\-1\end{pmatrix}_{r}~~~~~~~~
\Kappa \begin{pmatrix}0\\1\end{pmatrix}_l=\gamma(-\frac{1}{2})\begin{pmatrix}-1\\0\end{pmatrix}_{r}, \stag{a,b}
\end{align*}
\begin{align*}
\Kappa \begin{pmatrix}1\\0\end{pmatrix}_r=\delta(\frac{1}{2})\begin{pmatrix}0\\-1\end{pmatrix}_{l}~~~~~~~~
\Kappa \begin{pmatrix}0\\1\end{pmatrix}_r=\delta(-\frac{1}{2})\begin{pmatrix}-1\\0\end{pmatrix}_{l}, \stag{c,d}
\end{align*}
\end{frame}
\begin{frame}
Let us evaluate the following matrix elements
\begin{align}
\notag D^{(\frac{1}{2},0)}_{mm'}&=(L_{\frac{1}{2},m},e^{\vec{\theta}\cdot\vec{A}+\vec{\xi}\cdot\vec{B}}L_{\frac{1}{2},m'})
=(\Kappa e^{\vec{\theta}\cdot\vec{A}+\vec{\xi}\cdot\vec{B}}L_{\frac{1}{2},m'},\Kappa L_{\frac{1}{2},m})\\
\notag &=\gamma^*(m')\gamma(m)(e^{\vec{\theta}\cdot\vec{A}+\vec{\xi}\cdot\vec{B}}R_{\frac{1}{2},-m'},R_{\frac{1}{2},-m})\\&=\gamma(m)D^{(0,\frac{1}{2})*}_{-m,-m'}\gamma^*(m'),
\end{align}
\noindent or
\begin{align}
D^{(\frac{1}{2},0)}=\begin{pmatrix}0&\gamma(\frac{1}{2})\\\gamma(-\frac{1}{2})&0\end{pmatrix}D^{(0,\frac{1}{2})*}\begin{pmatrix}0&\gamma^*(-\frac{1}{2})\\\gamma^*(\frac{1}{2})&0\end{pmatrix}.
\end{align}
\end{frame}
\begin{frame}
Similarly,
\begin{align}
D^{(0,\frac{1}{2})}=\begin{pmatrix}0&\delta(\frac{1}{2})\\\delta(-\frac{1}{2})&0\end{pmatrix}D^{(\frac{1}{2},0)*}\begin{pmatrix}0&\delta^*(-\frac{1}{2})\\\delta^*(\frac{1}{2})&0\end{pmatrix}.
\end{align}
In order to be consistent with the LT for spinors, one chooses
\begin{align}
\gamma(\frac{1}{2})=\delta(\frac{1}{2})=-\gamma(-\frac{1}{2})=-\delta(-\frac{1}{2})=1
\end{align}
\noindent hence we have\\[-10pt]
\begin{align}
\epsilon=\begin{pmatrix}0&1\\-1&0\end{pmatrix}
\end{align}
\noindent and\\[-10pt]
\begin{align}
\epsilon\bm{\sigma}_i^*\epsilon^{-1}=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \bm{\sigma}_i^* \begin{pmatrix}0&-1\\1&0\end{pmatrix}=-\bm{\sigma}_i.
\end{align}
\end{frame}
\begin{frame}
\frametitle{Spinor space and Co-spinor space}
\begin{align}
\mbox{Let }\left\{\begin{array}{ll}\mathcal{V}_{(\frac{1}{2},0)}&:\mbox{ left-hand spinor space }\\
e_{a}&: (a=1, 2) \mbox{ two left-handed spinor }\end{array}\right.
\end{align}
A spinor in $ \mathcal{V}_{(\dot{\frac{1}{2}},0)} $
\begin{align}
\left\{\begin{array}{ll}\mathcal{V}_{(\dot{\frac{1}{2}},0)}&:\mbox{ co-left-hand spinor space }\\
e_{\dot{a}}&: (a=1, 2) \mbox{ two co-left-handed spinor }\end{array}\right.
\end{align}
\noindent which is related to the left-handed spinor by
\begin{align}
\dot{\psi}=\psi^T\epsilon^T=(\dot{\psi}_1,\dot{\psi}_1)=(\psi^1,\psi^2)\epsilon^T=(\psi^1,\psi^2)\epsilon^{-1}.
\end{align}
\end{frame}
\begin{frame}
\noindent and the corresponding LT is given as
\begin{align}
\dot{\psi}\stackrel{L.T.}{\longmapsto}\dot{\psi}'&=\psi'^T\epsilon^{-1}=
\psi^TD^{T(\frac{1}{2},0)}\epsilon^{-1}\\
&=\dot{\psi}(\epsilon D^{(\frac{1}{2},0)*}\epsilon^{-1})^{\dag}= \dot{\psi}D^{(0,\frac{1}{2})\dag}.
\end{align}
Similarly the LT for the co-right-handed spinor is given as
\begin{align}
\dot{\varphi}\stackrel{L.T.}{\longmapsto}\dot{\varphi}'=\varphi'^T\epsilon^T=
\dot{\varphi}D^{(\frac{1}{2},0)\dag}.
\end{align}
Let us construct the 4-dimensional product space as
\addtocounter{equation}{+1}
\begin{align*}
\mathcal{V}_{(\frac{1}{2},\dot{\frac{1}{2}})}:~e_a\dot{f}^b~~\mbox{as basis}\stag{a}\\
\mathcal{V}_{(\dot{\frac{1}{2}},\frac{1}{2})}:~\dot{e}_af^b~~\mbox{as basis}\stag{b}
\end{align*}
\end{frame}
\begin{frame}
then any element in $ \mathcal{V}_{(\frac{1}{2},\dot{\frac{1}{2}})} $ and in $ \mathcal{V}_{(\dot{\frac{1}{2}},\frac{1}{2})} $ can be written respectively as\\[-15pt]
\begin{align}
U^{(\frac{1}{2},\dot{\frac{1}{2}})} =\begin{pmatrix}u^1_{\dot{1}}&u^1_{\dot{2}}\\u^2_{\dot{1}}&u^2_{\dot{2}}\end{pmatrix}.
\end{align}
\noindent and\\[-15pt]
\begin{align}
U^{(\dot{\frac{1}{2}},\frac{1}{2})} =\begin{pmatrix}u^{\dot{1}}_1&u^{\dot{2}}_1\\u^{\dot{1}}_2&u^{\dot{2}}_2\end{pmatrix}.
\end{align}
It is obvious that the space-time matrix
\begin{align}
\bm{X}=x^\mu\bm{\sigma}_\mu=\begin{pmatrix}-x^0+x^3&x^1-ix^2\\x^1+ix^2&-x^0-x^3\end{pmatrix},
\end{align}
\noindent transforms as an element of $ \mathcal{V}_{(\frac{1}{2},\dot{\frac{1}{2}})} $-representation, while
\begin{align}
\bm{X}_c=\epsilon\bm{X}^*\epsilon^{-1}
\end{align}
\noindent transforms as an element of $ \mathcal{V}_{(\dot{\frac{1}{2}},\frac{1}{2})} $-representation.
\end{frame}
\begin{frame}
We are now in the position to emphasize the next proposition
\proposition{
\subsection*{\textit{Proposition 3.}}
{\it An operator of the} $ (\frac{1}{2},\dot{\frac{1}{2}}) $-{\it representation acts upon a vector of the} $ (0,\frac{1}{2}) $-{\it representation yields a vector of} $ (\frac{1}{2},0) $-{\it representation. Conversely an operator of the} $ (\dot{\frac{1}{2}},\frac{1}{2}) $-{\it representation acts upon a vector of the} $ (\frac{1}{2},0) $-{\it representation will yield a vector of the} $ (0,\frac{1}{2}) $-{\it representation.}
}
The proof goes as;
If $ \bm{A}\in (\frac{1}{2},\dot{\frac{1}{2}}) $-representation,
$ \xi\in(0,\frac{1}{2}) $-representation, then \\[-15pt]
\begin{align}
\eta\stackrel{L.T.}{\longmapsto}\eta'=\bm{A}'\xi'= D^{(\frac{1}{2},0)}\bm{A}D^{(\frac{1}{2},0)\dag}D^{(0,\frac{1}{2})}\xi=D^{(\frac{1}{2},0)}\eta,
\end{align}
\end{frame}
\begin{frame}
Hence we have a left-handed spinor, i.e.
\begin{align}
\eta=\bm{A}\xi,~~~~~~~~~~~~~
\end{align}
Similarly that
\begin{align*}
\mbox{if }\left\{\begin{array}{ll}\bm{A}&\in (\dot{\frac{1}{2}},\frac{1}{2})-\mbox{representation},\\[+8pt]
\eta&\in(\frac{1}{2},0)-\mbox{representation}.\end{array}\right.
\end{align*}
Therefore we reach as follows,
\begin{align}
\xi\stackrel{L.T.}{\longmapsto}\xi'=\bm{A}_c'\eta'= D^{(0,\frac{1}{2})}\bm{A}_cD^{(0,\frac{1}{2})\dag}D^{(\frac{1}{2},0)}\xi=D^{(0,\frac{1}{2})}\xi,
\end{align}
\end{frame}
\begin{frame}
\frametitle{Dirac spinor and Dirac equation}
Since any Lorentz 4-vector with the construction
\begin{align}
U=\sigma_\mu u^\mu,~~~~ U_c=\sigma^c_\mu u^\mu,
\end{align}
transforms as $ (\frac{1}{2},\dot{\frac{1}{2}}) $-representation and $ (\dot{\frac{1}{2}},\frac{1}{2}) $-representation respectively. Therefore,
\addtocounter{equation}{1}
\begin{align*}
\bm{P}_c\psi_l&=m_0c\psi_r,\tag{\arabic{chapter}.\arabic{equation}a}~~~~~~~~\\
\bm{P}\psi_r&=m_0c\psi_l,\tag{\arabic{chapter}.\arabic{equation}b}~~~~~~~~
\end{align*}
\noindent hence we have
\begin{align}
\begin{pmatrix}0&\bm{P}_c\\\bm{P}&0\end{pmatrix}
\begin{pmatrix}\psi_r\\\psi_l\end{pmatrix}=
m_0c\begin{pmatrix}\psi_r\\\psi_l\end{pmatrix},
\end{align}
\end{frame}
\begin{frame}
If we define the Dirac spinor $ \psi_d $ as $ \psi_r\oplus\psi_l $, then
\begin{align}\label{dirac-eq2}
\begin{pmatrix}0&\bm{P}_c\\\bm{P}&0\end{pmatrix}\psi_d(x)=m_0c\psi_d(x),~~~~~
\end{align}
\noindent or
\begin{align}
\left[\begin{pmatrix}0&i\bm{\sigma}_c^\mu\\i\bm{\sigma}^\mu&0\end{pmatrix}\partial_\mu+\frac{m_0c}{\hbar}\right]\psi_d(x)=0,
\end{align}
\noindent which can be cast into
\begin{align}\label{dirac-eq1}
\left(i\gamma^\mu\partial_\mu+\frac{m_0c}{\hbar}\right)\psi_d(x)=0,
\end{align}
\end{frame}
\begin{frame}
\noindent with $ \gamma^\mu $ defined as
\begin{align}\label{gamma-mu}
\gamma^\mu=\begin{pmatrix}0&\bm{\sigma}_c^\mu\\\bm{\sigma}^\mu&0\end{pmatrix}~~~
\mbox{or}~~~\gamma^0=\begin{pmatrix}0&\mathbf{I}\\\mathbf{I}&0\end{pmatrix},~~~
\gamma^i=\begin{pmatrix}0&-\bm{\sigma}^i\\\bm{\sigma}^i&0\end{pmatrix}.
\end{align}
The covariant formulation of Dirac equation does not imply that
\begin{align*}
\gamma_\mu\partial^\mu=\gamma'_\mu\partial'^\mu
\end{align*}
In fact that we have the following proposition.
\end{frame}
\begin{frame}
\proposition{
\subsection*{\textit{Proposition 4.}}
\it The gamma matrices $ \gamma^\mu $ are universal in all Lorentz frame, namely the Dirac equation in another Lorentz frame, i.e. the $ \mathcal{O}' $-system always takes the same gamma matrices $ \gamma^\mu $ used in $ \mathcal{O} $-system. The equation in $ \mathcal{O}' $-system is expressed as
\begin{align*}
\left(i\gamma^\mu\partial'_\mu+\frac{m_0c}{\hbar}\right)\psi_d'(x')=0.
\end{align*}
}
\end{frame}
\begin{frame}
It can be proved by defining
$$
D(\vec{\theta},\vec{\xi})= D^{(0,\frac{1}{2})}(\vec{\theta},\vec{\xi}) \oplus D^{(\frac{1}{2},0)}(\vec{\theta},\vec{\xi})
$$
\noindent and multiplying it upon the Dirac equation as follows
\begin{align}
\left(iD\gamma^\mu D^{-1}\partial_\mu+\frac{m_0c}{\hbar}\right)D\psi_d(x)=0.
\end{align}
One can show that\\[-13pt]
\begin{align}
D\gamma^\mu D^{-1}=\bm{\Lambda}^\mu_\nu\gamma^\nu,
\end{align}
\noindent and the Dirac equation in the new Lorentz frame, i.e. $ \mathcal{O}' $–frame reads as\\[-13pt]
\begin{align}
\left(i\gamma^\mu\partial'_\mu+\frac{m_0c}{\hbar}\right)\psi_d'(
x')=0,
\end{align}
\noindent by identifying $ \psi'_d(x')\equiv D(\vec{\theta},\vec{\xi})\psi_d(x)=D(\vec{\theta},\vec{\xi})\psi_d(\bm{\Lambda}^{-1}
x') $.
\end{frame}
\begin{frame}
\frametitle{Zero Mass Limit and Helicity of Weyl spinors}
Last demonstration for zero mass limit in Dirac equation
\begin{align}
\begin{pmatrix}0&\bm{P}_c\\\bm{P}&0\end{pmatrix}\psi_d(x)-m_0c\psi_d(x)=0.
\end{align}
For the limit that $ m=0 $, then
\begin{align}
\bm{P}_c\psi_l=0,~~~\bm{P}\psi_r=0.
\end{align}
\noindent and $ p^0=|\vec{p}| $, which implies that
\begin{align}
(\bm{\sigma}\cdot\hat{\bm{p}})\psi_r=\psi_r,~~~
(\bm{\sigma}\cdot\hat{\bm{p}})\psi_l=-\psi_l,
\end{align}
\end{frame}
\end{document}
英文版 kerlock 原始碼, professionalfonts 字型:
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%%%%%%%%%%%
\title{Simulation of Kerr-lens modelocking Ti:Sapphire laser}
\author{Chih-Han Lin}
\institute{Applied Physics Department, National Taiwan University}
%%%%%%
\begin{document}
\begin{frame}
\titlepage
\end{frame}
%%%%%page1
\begin{frame}{Theoretical Model}
Refer to \textit{Chin. Phys. B} \textbf{19} 014215 (2010) and \textit{Opt. Express} \textbf{12} 2731(2004), pulse propagation around cavity of KLM laser can be describe by the following equation
\begin{block}{Extended Nonlinear Scr\"{o}dinger (ENLS) equation}
\begin{align*}
\frac{\partial a(z,t)}{\partial z}&=(\hat{L}+\hat{N})a(z,t),\\
\hat{L}a(z,t)&=\frac{1}{2}(g(E_p)-2l)a(z,t)-\frac{i}{2}\left[\beta_2+i\frac{g(E_p)}{(\delta\omega)^2}\right]\frac{\partial^2}{\partial t^2}a(z,t),\\
\hat{N}a(z,t)&=\underbrace{i\gamma|a(z,t)|^2a(z,t)}_{\mbox{SPM}}-\underbrace{q(a(z,t))a(z,t)}_{\mbox{Fast SA}},\\
\mbox{where}~~&g(E_p)=\frac{g_0}{1+E_p/E_{sat}},~q(a)=\frac{q_0}{1+|a(z,t)|^2/q_{sat}}~~\mbox{and}~ E_p\simeq \int |a(z,t)|^2 dt.
\end{align*}
\end{block}
\\[+5pt]
$\hat{L}a(z,t)$ contains gain, group velocity delay (GVD) and spectrum narrowing effect. On the other hand, $\hat{N}a(z,t)$ collects nonlinear parts such as Self Phase Modulation (SPM) and fast Saturable Absorption induced by Kerr-lens effect.
\end{frame}
\begin{frame}{Numerical solution of ENLS equation}
ENLS equation is often solved by means of split-step Fourier method.
\begin{corollary}
If $\hat{N}$ and $\hat{L}$ commute, we have
\begin{align*}
a(z+\delta z,t)\simeq \exp(\hat{N}\delta z)\exp(\hat{L}\delta z)a(z,t).
\end{align*}
\end{corollary}
\\[+5pt]
It means that we can split the whole partial differential equation into linear part and nonlinear part. Firstly, we calculate $a_N(z+\delta z)=\exp(\hat{L}\delta z)a(z,t)$, and then get the next step $a(z+\delta z, t)$ by $a(z+\delta z,t)=\exp(\hat{N}\delta z)a_N(z+\delta z)$. Thus we can use different algorithm to solve the equation for optimization.
\\[+5pt]
Since $\hat{N}$ in our model includes only constant and $f(D)$ (where $D=\partal/\partial t$ is differential operator on $t$), we may efficiently calculate the $a(z+\delta z,t)=\exp(\hat{N}\delta z)$ in frequency domain, i.e. $D$ becomes $i\omega$ for simplification.
\begin{align*}
\tilde{a}(z,\omega)&=\mathcal{F}\{a(z,t)\},\\
\frac{\partial \tilde{a}}{\partial z}=\hat{N}\tilde{a}&=\left[\frac{g(E_p)}{2}-l+\frac{i\beta_2}{2}\omega^2-\frac{g(E_p)}{2(\delta\omega^2)}\omega^2\right]\tilde{a}=f(\omega)\tilde{a},\\
\tilde{a}(z+\delta z,\omega)&=\exp(f(\omega)\delta z)\tilde{a}(z,\omega).
\end{align*}
So that we have $\displaystyle a_N(z+\delta z,t)=\mathcal{F}^{-1}\{\exp[f(\omega)\delta z]\mathcal{F}\{a(z,t)\}$
\end{frame}
\begin{frame}{Experimental Setup of P-GDD ker-lens modelocking Ti:S Laser}
I will compare the simulation results with experiment data in \textit{Chin. Phys. B} \textbf{19} 014215 (2010). The output average power is about 500 mW (1.68 nJ per pulse). The output power spectrum is shown as follows:
\begin{figure}
\centering
\includegraphics[width=\textwidth]{expsetup.pdf}
\end{figure}
The simulation will focus on positive group-delay dispersion (P-GDD) regime.
\end{frame}
\begin{frame}
The relation between GDD, spectrum width and pulse duration is shown as follows, \\[+5 pt]
\begin{figure}
\centering
\includegraphics[width=\textwidth]{expsetup2.pdf}
\end{figure}
\end{frame}
\begin{frame}{RK4 method}
Instead of using $\tilde{a}(z+\delta z,\omega)&=\exp(f(\omega)\delta z)\tilde{a}(z,\omega)$ to calculate the next step field quantity, we can improve the precision of the numerical solution by 4 order Runge-Kutta (RK4) methods as follows:
\begin{corollary}
\begin{align*}
\frac{d\hat{a}}{dz}&=f(\hat{a})=f(\omega)\tilde{a},\\
k_1&=\delta z f(\hat{a}(z)),~~~~k_2=\delta z f(\hat{a}(z)+\frac{1}{2}k_1),\\
k_3&=\delta z f(\hat{a}(z)+\frac{1}{2}k_2),~~~k_4=\delta z f(\hat{a}(z)+k_3),\\
\hat{a}(z&+\delta z,\omega)=\hat{a}(z,\omega)+\frac{1}{6}(k_1+2k_2+2k_3+k_4).
\end{align*}
\end{corollary}
\\[+5pt]
When executing nonlinear step calculation, we can promote precision by using RK4 method.
\end{frame}
\begin{frame}{Structure of simulation code}
The simulation result of each round trip is stored in a 2D array \texttt{a1(cycle,size(t\_index))}. The pulse width, total energy and peak intensity are also recorded and plotted after \texttt{klm.m} is finished.\\[+5 pt]
\begin{figure}
\centering
\includegraphics[width=\textwidth]{flowchart1.pdf}
\end{figure}
\end{frame}
\begin{frame}{Simulation Parameters}
\begin{center}
\begin{tabular}{ccccccc}
\hline\hline quantity & value & unit & ~~~~ & quantity & value & unit \\
\hline $g_0$ & 100 & $m^{-1}$ & ~ & \gamma & 30$\times 10^{-5}$ & (W$\cdot$m)$^{-1}$ \\
$E_{sat}$ & 10 & nJ & ~ & $q_0$ & 5 & m$^{-1}$ \\
$\beta_2$ & 58 & fs$^2$/mm & ~ & $q_{sat}$ & 0.3 & MW \\
$\delta\omega$ & 270 & THz & ~ & l_{crystal} & 7 & mm \\
$l_{cavity}$ & 1820 &mm&&\eta &0.22&\\
\hline \hline
\end{tabular}
\end{center}\\[+10pt]
And the total GDD is between -50 fs$^2$ and 80 fs$^2$. $\beta_{2,prism}$ is calculated as,
\begin{align*}
\beta_{2,prism}=\frac{\mbox{total GDD}-2\times\beta_2 l_{crystal}}{2\times(l_{cavity}-l_{crystal})}.
\end{align*}
Both Simulation steps in P-GDD and N-GDD regime are tunable to get steady state solutions. \texttt{frame} and \texttt{cycle} in \textsc{Matlab} code control data dump rate and total round trips.\\[+5pt]
$\eta$ denotes transmission of output coupler. We can convert $\eta$ into $l$:
\begin{align*}
\frac{\partial a}{\partial z}&=-la\Longrightarrow a(z+\delta z)=\exp(-l\delta z)a(z)=\eta a(z),\\
\delta z=2\times l_{crystal}&=14~\mbox{mm}\Longrightarrow
l=\frac{-\ln\sqrt{1-\eta}}{14~\mbox{mm}}\simeq 0.0089~\mbox{mm}^{-1}.
\end{align*}
\end{frame}
\begin{frame}{Validity of Split-Step Fourier Method}
We can simulate only GVD effect and compare the pulse width broaden with analytical solution to check the validity of Split-Step Fourier method in our algorithm.
Assume that the dispersion-free pulse has Gaussian shape,
\begin{align*}
a(t)=\exp\left(-\frac{t^2}{\tau^2}\right)&\Longrightarrow
I(t)=a^2(t)=\exp\left(-\frac{2t^2}{\tau^2}\right),\\
\frac{\partial a}{\partial z}&=-\frac{iD}{2}a=\frac{i\beta_2\delta z}{2}a.
\end{align*}
FWHM of I(t) is $\tau\sqrt{2\ln 2}$. If the propagation induce D=GDD, pulse width $\tau$ will be extended to $\tau'$
\begin{align*}
\tau'=\sqrt{\tau^2+\frac{(2D)^2}{\tau^2}}=\tau\sqrt{1+\frac{(2D)^2}{\tau^4}}~~~~\mbox{where~} D=\beta_2\delta z .
\end{align*}
\end{frame}
\begin{frame}
We use \texttt{klm\_gvd\_check.m} to modified the algorithm with theoretical solution. Parameters \texttt{gain\_narrowing} and \texttt{gs} are temporal chosen as zero for pure GDD simulation.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{gvd1.pdf}
\end{figure}
The absolute error on FWHM of Gaussian pulse is approximately -3.2 fs with $\beta_2=10$ fs$^2$/mm and round trip length $l=7$ mm.
\end{frame}
\begin{frame}{Boundary condition - Limit of pulse width}
We need boundary conditions $\lim_{t\rightarrow\pm\infty}a(z,t)=0$ to avoid divergence and aliasing on spectrum. In the following case, the extended pulse does not fulfill the boundary conditions and generates ripples bouncing on both end. Generally, we have to create an temporal array that is as twice larger as the FWHM of simulated pulse.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{gvd2.pdf}
\end{figure}
\end{frame}
\begin{frame}
We also did another testing in GDD simulation. We first added positive GDD to broaden the pulse, and then we applied the same amount of negative GDD to compress the pulse. The calculation error $59.1-19.0=0.1$ fs dose not exceed the resolution of temporal array (0.1 fs).\\[+5 pt]
\begin{figure}
\centering
\includegraphics[width=\textwidth]{gvd3.pdf}
\end{figure}
\end{frame}
\end{frame}
\begin{frame}{simulation result}
Generally the simulation will stop after 2000 round trips (10 steps in P-GDD regime and 2 steps in N-GDD regime per round trip). Here is an example for GDD=13 fs$^2$ simulation:\\[+10 pt]
\begin{figure}
\centering
\includegraphics[width=\textwidth]{exc_sample.pdf}
\end{figure}
\end{frame}
\begin{frame}{Simulated Pulse duration and Power Spectrum}
The simulation in \textit{Chin. Phys. B} \textbf{19} 014215 (2010) chose GDD=18, 27, 47 fs$^2$. I did the simulation with the same parameters. Red line indicates the theoretical sech$^2$ shape function with the simulated FWHM of pulse intensity.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{compare1.pdf}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{compare12.pdf}
\end{figure}
\end{frame}
\begin{frame}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{compare2.pdf}
\end{figure}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{compare22.pdf}
\end{figure}
\end{frame}
\begin{frame}
total GDD=18 fs$^2$.\\[+10pt]
\begin{figure}
\centering
\includegraphics[width=\textwidth]{gdd18fullshape.png}
\end{figure}
\end{frame}
\begin{frame}
total GDD=27 fs$^2$.\\[+10pt]
\begin{figure}
\centering
\includegraphics[width=\textwidth]{gdd27fullshape.png}
\end{figure}
\end{frame}
\begin{frame}
total GDD=47 fs$^2$.\\[+10pt]
\begin{figure}
\centering
\includegraphics[width=\textwidth]{gdd47fullshape.png}
\end{figure}
\end{frame}
\begin{frame}
\begin{figure}
\centering
\includegraphics[width=\textwidth]{total1.pdf}
\end{figure}
\end{frame}
\begin{frame}
Pulse energy is not sensitive with GDD.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{total2.pdf}
\end{figure}
\end{frame}
\end{document}