由于我的内容表中有很多部分，因此并非所有部分都覆盖。 这是我的代码。
\documentclass[11pt,t]{beamer} %<----- top align all frame contents
\usetheme[progressbar=frametitle]{metropolis}
\setbeamertemplate{frame numbering}[fraction]
\setbeamercolor{progress bar}{fg=green,bg=blue}
\makeatletter
\setlength{\metropolis@progressinheadfoot@linewidth}{3pt}
\makeatother
\metroset{sectionpage=none} %<----- remove all section frames
\usepackage{tikz}
\usepackage{braket}
\usepackage{mdframed}
\usetikzlibrary{calc}
\usepackage[utf8]{inputenc}
\usepackage{multicol}
\usepackage{bbm}
\usepackage{colortbl}
\usepackage{textcmds}
\usepackage{slashed}
\newcolumntype{C}{>{\centering\arraybackslash}X}
\usepackage{float}
\usepackage{caption}
\usepackage{mathtools}
\usepackage[symbol]{footmisc}
\usepackage{textcomp}
\usepackage[T1]{fontenc}
\usepackage{physics}
\usepackage{textcmds}
\usepackage{lmodern}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{wrapfig}
\usepackage{float}
\usepackage[style=numeric]{biblatex}
\addbibresource{references.bib}
\setbeamertemplate{bibliography item}{\insertbiblabel} % <--- number references
\defbeamertemplate{subsection in toc}{bullets}{%
\leavevmode
\parbox[t]{1em}{\textbullet\hfill}%
\parbox[t]{\dimexpr\textwidth-1em\relax}{\inserttocsubsection}\par}
\defbeamertemplate{section in toc}{sections numbered numeric}{%
\leavevmode%
\MakeUppercase{\number\inserttocsectionnumber}.\ %
\inserttocsection\par}
\setbeamertemplate{section in toc}[sections numbered numeric]
\setbeamertemplate{subsection in toc}[bullets]
\setbeamercovered{transparent} % <----- added
\makeatletter
\setbeamertemplate{title page}{
\begin{minipage}[t][\paperheight]{\textwidth} % <---- changed from [b] to [t] alignment
% \ifx\inserttitlegraphic\@empty\else\usebeamertemplate*{title graphic}\fi % <--- commented out and moved to the end
% \vfill% % <---- commented out as not needed with top alignment
\vspace{0.45cm} % <--- move the title down by 0.45cm
\ifx\inserttitle\@empty\else\usebeamertemplate*{title}\fi
\ifx\insertsubtitle\@empty\else\usebeamertemplate*{subtitle}\fi
\vspace{-0.45cm} % <---- remove 0.45 cm of white space between title and green line
\usebeamertemplate*{title separator}
\ifx\beamer@shortauthor\@empty\else\usebeamertemplate*{author}\fi
\ifx\insertdate\@empty\else\usebeamertemplate*{date}\fi
\ifx\insertinstitute\@empty\else\usebeamertemplate*{institute}\fi
% \bigskip
\ifx\inserttitlegraphic\@empty\else\usebeamertemplate*{title graphic}\fi % <---- added the titlegraphic here
% \vfill % <---- commented out as not needed with top alignment
\vspace*{1mm}
\end{minipage}
}
\makeatother
\setbeamerfont{title}{size=\LARGE}
\setbeamerfont{author}{size=\large}
\setbeamerfont{institute}{size=\large}
\author{Anshul Sharma}
\title{Symmetry in Quantum Mechanics}
\institute {CENTRAL UNIVERSITY OF HIMACHAL PRADESH}
\titlegraphic{\includegraphics[width=0.3\linewidth]{CUHP LOGO}}
\begin{document}
\begin{frame}[plain]
\maketitle
\end{frame}
\begin{frame}
\frametitle{Overview}
\tableofcontents
\end{frame}
\section{Symmetries in Classical Physics} % <---- add sections in order to get them listed in the table of contents
\begin{frame}{\secname} % <----- \secname here used the section's name as a frametitle
One can talk symmetries in Classical Mechanics from Noether's theorem. Noether's theorem derives conservation laws from symmetries under the assumption that the principle of least action is the basic law that governs the motion of particle in classical mechanics.\\[1.5ex]
It can be phrased as-\\[3.5ex]\pause
\begin{mdframed}
{\large"The action is minimum for the path taken by the particle."}
\end{mdframed}
\end{frame}
\begin{frame}
The conservation laws that comes out when one solves the Action principle for different cases as- \\[2.5ex]
\begin{itemize}
\item \Large{Conservation of Linear Momentum gives "Homogenity of Spcae."}\pause \\[1ex]
\item \Large{Conservation of Angular Momentum gives "Isotropy of Space."}\pause\\[1ex]
\item \Large{Conservation of Energy gives "Homogenity of Time."}\\[1ex]
\end{itemize} \pause
Question arises, does these symmetry transformation holds in Quantum Mechanics too.
\end{frame}
\section{Types of Symmetry Transformation In Quantum Mechanics}
\begin{frame}{\secname}
Symmetry transformation in Quantum Mechanics are-\pause
\begin{itemize}
\item \large{Translation Symmetry}.\pause \\[2.5ex]
\item \large{Rotational Symmetry}.\pause \\[2.5ex]
\item \large{Partiy Symmetry}.\pause\\[2.5ex]
\item \large{Time Reversal Symmetry}.\pause
\end{itemize}
\end{frame}
\section{Translation Symmetry}
\begin{frame}{\secname}
Consider a $\ket x$ be a state which is well localized.\pause\\[0.5ex]
Transformation that changes $\ket x$ to $\ket {x+dx}$ such that no other factor changes as such i.e its spin value.\pause \\[0.5ex]
The transformation which is responsible for this transformation is-\pause
\begin{equation}
T(x)\ket {x}=\ket{x+dx}.
\end{equation}
Applying these transformation on the wave function i.e.\pause
\begin{equation}
T(a)\ket{\psi}=\ket{\phi},
\end{equation}
\begin{equation}
\text{where}\hspace{0.5cm} \ket{\psi}=\int {dx' \ket{x'}\braket{x'}{\psi}}.
\end{equation}
\end{frame}
\begin{frame}
Inserting Equation (3) in Equation (2) and using Equation 1, we get-
\begin{equation}
\therefore\hspace{0.5mm} \left(T(a)\psi\right)(x)=\psi(x-a).
\end{equation}
\begin{figure}[H]
\centering
\includegraphics[width=0.6\linewidth]{"translated wave"}
\caption{Representation of Translated wave function\cite{article}.}
\label{fig:translated-wave}
\end{figure}
Changing x to x+a we get,
\begin{center}
\fbox{$T(a)\psi(x+a)=\psi(x).$}
\end{center}
\end{frame}
\subsection{Properties of Translation Operator}
\begin{frame}{\subsecname}
\begin{enumerate}
\item {\bf T(0)= $\mathbbm{1}$.}\\[3.5ex]\pause
\item {\bf T$^\dagger$(dx)T$^\dagger$(dx)=$\mathbbm{1}$ or T$^\dagger$(dx)=T$^{-1}$(dx) $\implies$ Translation operator is unitary.}\\[3.5ex]\pause
\item {\bf T(dx)T(dx$'$)=T(dx+dx$'$)=T(dx$'$)T(dx) i.e. Translation operator are commutable}.\\[3.5ex]\pause
\item {\bf T(-dx)=T$^{-1}$(dx)}.
\end{enumerate}
\end{frame}
\subsection{Infinitesimal form Translation Operator}
\begin{frame}{\subsecname}
Form of infinitesimal translation operator can be written as-\pause
\begin{mdframed}
\begin{equation}
T(dx')=1-\iota k.dx',
\end{equation}
\end{mdframed}
where k is Hermitian operator.\\[0.5ex]
By using this as our translation operator form, one can derive all the properties of translation operator.\pause
For N-infinitesimal translations, one can re-write the above equation as-\pause
\begin{mdframed}
\begin{equation}
T(\Delta x,x')=\lim_{N \to \infty} (1- \frac{\iota p_x\Delta x'}{\hbar})=\exp(\frac{-\iota p_x\Delta x'}{\hbar}).
\end{equation}
\end{mdframed}
\end{frame}
\section{Applications of Translation Symmetry}
\begin{frame}{\secname}
\begin{itemize}
\item It helps us to evaluate the commutation relation between $[x_j,p_i]=\iota \hbar\delta_{ij}$ very easily.\pause\\[4.5ex]
\item Also as translation operator commutes in different direction, \\[0.5ex]
$\implies$ $[p_i,p_j]=0$. \\[2.5ex]
So whenever the generators of transformation commutes, the corresponding group are called Abelian. The trabslation group in 3D is Abelian.
\end{itemize}
Let us see how translation symmetry in Quantum Mechanics has an relation with Solid state Physics.
\end{frame}
\subsection{Lattice Translation as a Discrete Symmetry}
\begin{frame}{\subsecname}
\begin{equation}
\hspace{-2cm}\text{Consider a potential} \hspace{0.2cm} V(x\pm a)=V(x).
\end{equation}
\begin{figure}[H]
\centering
\includegraphics[width=0.7\linewidth]{"periodic potentia"}
\caption{Periodic Potential\cite{PeriodicPotential}.}
\end{figure}
Since Hamiltonian is invaruant under translation operator $T(a)$, one can digaonalize $H$ and $T$ simultaneously. So one can find eigenket and eigenvalue of translation operator $T(a)$.
\end{frame}
\begin{frame}{\subsecname}
We consider first the height of the potential barrie to be $\infty$.\\[0.2ex]
Let the particle be found at $n^{th}$ position represented by $\ket{n}$
\begin{equation}
H\ket{n}=E_{n}\ket{n}.
\end{equation}
and the wave function $\braket{x'}{n}$ is finite only in the n$^{th}$ site.\\[0.5ex]
But as potential is periodic, we see that, same similar state, properties exist too at some other well with same energy eigen value $E_{n}$.\\[0.2ex]
$\therefore$ One can say there are n-infinite well with $\infty$ ground states; n $\to$ $\infty$.\\[0.2ex]
\end{frame}
\begin{frame}{\subsecname}
But when it is applied to translation operator, we have,
\begin{equation}
T(a)\ket{n}=\ket{n+1}.
\end{equation}
which is not the eigenket for Hamiltonian, where both still commutes.\\[1ex]
So one can define simultaneous eigenket and a better representation of labelling each and every position as,
\begin{equation}
\ket{\theta}=\sum_{n=-\infty}^{\infty}e^{in\theta}\ket{n}\hspace{0.5cm} \text{; $\theta$ runs from -$\pi$ to $\pi$}.
\end{equation}
Now let us change the potential height from $\infty$ to some finite amount.
\end{frame}
\begin{frame}{\subsecname}
The wavefunction $\braket{x'}{n}$ can now be found in the other lattice sites too and thus
\begin{equation}
\bra{n'}H\ket{n}\ne 0\hspace{0.2cm} \text{and}\hspace{0.2cm} \bra{n+1}H\ket{n}=-\triangle.
\end{equation}
This approximation is called Tight binding approximation.\\[0.2ex]
$\therefore$ $\ket{n}$ is not energy eigenket, i.e.
\begin{equation}
H\ket{n}=E_{n}\ket{n}-\triangle \ket{n+1}-\ket{n-1}.
\end{equation}
\begin{equation}
\text{The quantity}\hspace{0.2cm}H\ket{\theta}=H\sum e^{\iota n \theta}\ket{n} \hspace{0.1cm}\text{equals,}
\end{equation}
\begin{equation}
H\sum e^{\iota n \theta}\ket{n}=E_{n}\sum e^{\iota n \theta}\ket{n}-2\triangle\cos(\theta)\sum e^{\iota n \theta}\ket{n}.
\end{equation}
where we see degeneracy is lifted up by $\Delta$ amount. So one has continous band of energy eigenstates.
\end{frame}
\begin{frame}{\subsecname}
\begin{equation*}
E-2\triangle < E < E+2\triangle.
\end{equation*}\vspace{-0.8cm}
\begin{enumerate}
\item When $\triangle$=0, all of the energy eigen states are zero.
\item As $\triangle$ increases, states in band gets wider.
\end{enumerate}
\begin{figure}[H]
\centering
\includegraphics[width=0.4\linewidth]{"Lifiting degenracy"}
\caption{Energy degeneracy lifted up\cite{Energy}.}
\end{figure}
\end{frame}
\begin{frame}{\subsecname}
Let us see how wavefucntion changes.
\begin{equation*}
\braket{x'}{\theta} \hspace{0.1cm} \text{or}\hspace{0.2cm} \bra{x'}T(a)\ket{\theta} \to \text{wave function of lattice translated state}
\end{equation*}
\begin{mdframed}
\begin{equation}
\therefore \hspace{0.2cm} e^{\iota k(x'-a)}u_{k}(x'-a)=e^{-\iota k a}u_{k}(x')e^{-\iota k a}.
\end{equation}
\end{mdframed}
The above equation is called {\bf Bloch theorem}.\\[0.5ex]
In 3D we can write it as,
\begin{mdframed}
\begin{equation}
\psi({\bf r'})=e^{\bf \iota k.r}u_{k}({\bf r}).
\end{equation}
\end{mdframed}
\end{frame}
\section{Rotational Symmetry}
\begin{frame}{\secname}
Rotation in different direction do not commute and as a result the corresponding group is known as Non-Abeliangroup. Rotation in quantum mechanics is an direct evidence of obtaining Angular Momentum.
\begin{figure}[H]
\centering
\includegraphics[width=0.6\linewidth]{"Screenshot (122)"}
\caption{Showing why rotations about different axis do not commute\cite{chaichian1997symmetries}.}
\label{fig:screenshot-122}
\end{figure}
\end{frame}
\begin{frame}{\secname}
Rotation affects physical system, the state ket corresponding to the rotated system is expected to look different from the state ket corresponding to original unrotated system i.e.
\begin{equation*}
\ket{\alpha}_R=D(R)\ket{\alpha},
\end{equation*}
where $\ket{\alpha}_R$ and $\ket{\alpha}$ stands for kets of the rotated and original system. \\[0.5ex]
$D(R)$ is the rotation operator.
\begin{equation}
\therefore \hspace{0.3cm}D(\hat{n},d\phi)=1-\frac{\bf J.\hat{n}}{\hbar}d\phi.
\end{equation}
$\therefore$ For a finite rotation we can write,
\begin{equation}
D_z(\phi)=\lim_{N\to\infty}\left[1-\iota\frac{J_z}{\hbar}\frac{\phi}{N}\right]^N,
\end{equation}
\end{frame}
\subsection{Properties of Rotation Operator}
\begin{frame}{\subsecname}
\begin{enumerate}
\item {\bf Identity}: As $R\mathbbm{1}=R$, $\implies D(R)\mathbbm{1}=D(R)$.\\[3.5ex]
\item {\bf Closure}: As $R_1R_2=R_3$, $\implies D(R_1)D(R_2)=D(R_3)$.\\[3.5ex]
\item {\bf Inverse}: As $RR^{-1}=1$, $\implies D(R)D(R)^{-1}=1$.\\[3.5ex]
\item {\bf Associativity}: As $R_1(R_2R_3)=(R_1R_2)R_3$, $\implies$ $D(R_1)(D(R_2)D(R_3)=(D(R_1)D(R_2))D(R_3)$.
\end{enumerate}
\end{frame}
\subsection{Commutation Result of Angular Momentum}
\begin{frame}{\subsecname}
For an infinitesimal amount, one can show that
\begin{equation*}
R_x(\phi)R_y(\phi)-R_y(\phi)R_x(\phi)=R_z(\epsilon^2)-1.
\end{equation*}
$\therefore$ We can write,
\begin{equation}
D(J_x)D(J_y)-D(J_y)D(J_x)=D(J_z^2)-1,\hspace{0.3cm}\text{which is equal to}
\end{equation}
And one applying the rotation form, one gets
\begin{equation}
\text{or}\hspace{0.3cm} [J_x,J_y]=\iota\hbar J_z.
\end{equation}
$\therefore$ Rotation about any axis gives,
\begin{equation}
[J_i,J_j]=\epsilon_{ijk}\iota\hbar J_k.
\end{equation}
\end{frame}
\section{Degeneracy}
\begin{frame}{\secname}
Consider, H being our Hamiltonian of the system, and let X be an operator such that,
\begin{equation}
[H,X]=0.
\end{equation}
where $X$, corresponds to some symmetry operator.\\[0.5ex]
Let $\ket{m}$ be the energy eigenket, having eigenvalue $E_m$. Therefore, one can say that $X\ket{m}$ is also an energy eigenket with the same energy eigenvalue, because
\begin{equation}
H(X\ket{m})=XH\ket{m}=E_{m}(X\ket{m})
\end{equation}
Say $\ket{m}$ and $X\ket{m}$ correspond to two different states. If it is that so, then these two states have same energy, i.e. they are degenerate.\\
\end{frame}
\begin{frame}{\subsecname}
To show what actually degeneracy is, a code in scilab has been made, where I have used the matrix mechaincs approach\cite{article} to find the wave function and energy eigenvalues of it.\\[2.5ex]
The results and the graph shown are for same width of the square well i.e. a=1; and the depth of the potential V0=-20. By changing the value of 'n' i.e. how many square well you need you can make a single and double square well (in our case) and even higher number of wells too.
\end{frame}
\subsection{Graph}
\begin{frame}{\subsecname}
{\bf A) For Square Well}
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{"sqaure well"}
\caption{ Wave function plot for a square well.}
\label{fig:sqaure-well}
\end{figure}
\end{frame}
\begin{frame}{\subsecname}
{\bf B) For Double Square well}
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{nwell(1,1,-20,2,10)}
\caption{(i) Wave function plot for Double Square well when width, b=1.}
\label{fig:nwell11-20210}
\end{figure}
\end{frame}
\begin{frame}{\subsecname}
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{3}
\caption{(ii)Wave function plot for Double Square well when width, b=0.5.}
\label{fig:3}
\end{figure}
\end{frame}
\begin{frame}{\subsecname}
\begin{figure}[H]
\centering
\includegraphics[width=1.1\linewidth]{3}
\caption{(ii)Wave function plot for Double Square well when width, b=0.5.}
\label{fig:3}
\end{figure}
\end{frame}
\begin{frame}{\subsecname}
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{2}
\caption{(iii) Wave function plot for Double Square well when width, b=0.05.}
\label{fig:2}
\end{figure}
\end{frame}
\begin{frame}{\subsecname}
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{4}
\caption{(iv) Wave function plot for Double Square well when width, b=0.005.}
\label{fig:4}
\end{figure}
\end{frame}
\section{Parity}
\begin{frame}{\secname}
Till now we have encoutered with continuous symmetry operations by applying successive infinitesimal transformations. But Discrete symmetry are opposite to continuous symmetry. The first one we see is, Parity as a Discrete symmetry.\\[0.5ex]
In classical mechanics the equation of motion remains invariant under {\bf r$\to$-r}.\\[0.5ex]
In quantum mechanics under Parity Transformation,
If we consider an state $\ket{\alpha}$, applying parity operator $\pi$, we get,
\begin{equation}
\ket{\alpha} \implies \pi \ket{\alpha},
\end{equation}
\begin{equation}
\therefore \hspace{0.5cm} \bra{\alpha} \pi^\dagger x \pi, \ket{\alpha}=-\bra{\alpha}x\ket{\alpha} ,
\end{equation}
\begin{equation}
\pi^\dagger x \pi=-x,
\end{equation}
Parity operator has eigenvalue of $\pm$1
\end{frame}
\begin{frame}{\secname}
$\implies$ "Translation followed by parity is equivalent to parity followed by translation in opposite direction."
\begin{center}
i.e. $\pi$T(dx$'$)=T(-dx$'$)$\pi$,\vspace{0.1mm}
\end{center}
or one can write {\Large $\pi^\dagger$p$\pi$=-p.} Hence momentum commutes with parity operator. \vspace{0.1mm}
Similarly one can obtain,
\begin{itemize}
\item $\pi^{-1}${\bf L}$\pi$={\bf L}, {\bf L} is Orbital Angular Momentum
\item $\pi^{-1}${\bf S}$\pi$={\bf S}, {\bf S} is Spin Angular Momentum
\item $\pi^{-1}${\bf J}$\pi$={\bf J}, {\bf J} is Total Angular Momentum
\end{itemize}
For spherical harmonics, $Y_{l}^{m}(\theta,\phi)$ being eigenstate of parity operator has eigenvalue $(-1)^{l}$
\begin{equation*}
\therefore \hspace{0.3cm} \pi\ket{\alpha,lm}=(-1)^l\ket{\alpha,lm}.
\end{equation*}
\end{frame}
\section{Selection Rules}
\begin{frame}{\secname}
Suppose $\ket{\alpha}$ and $\ket{\beta}$ are parity eigenstates,
if $A$ is an observable with definite parity and let $\pi A \pi=\epsilon_{A}A$,
\begin{align}
\bra{\alpha}A\ket{\beta}=\epsilon_{\alpha}\epsilon_{\beta}\bra{\alpha}\pi A\pi \ket{\beta},\\
\bra{\alpha}A\ket{\beta}=\epsilon_{\alpha}\epsilon_{\beta}\bra{\alpha}A\ket{\beta}.
\end{align}
So the above equations are equal if $\epsilon_{\alpha}\epsilon_{\beta}\epsilon_{\alpha}=1$, otherwise it is zero.
\begin{align*}\text{For eg:-}
\hspace{0.4cm} \bra{even}odd\ket{even}=0,\\
\bra{odd}odd\ket{odd}=0,\\
\bra{even}even\ket{odd}=0.
\end{align*}
\begin{equation*}
\text{So} \int \psi_{\beta}^{*}\psi_{\alpha}d\tau=0.
\end{equation*}
\hspace{2cm}iff $\psi_{\beta}^{*}$ and $\psi_{\alpha}$ have same parity.\\[0.5ex]
This rule is called {\bf Laporate rule}.
\end{frame}
\section{Invariance of Hamilonian}
\begin{frame}{\secname}
Hamiltonian is invariant under Parity Transformation, i.e.
\begin{equation}
H=\frac{p^2}{2m}+V({\bf r}).
\end{equation}
Clearly we see from Eqn.(5.19) that it commutes with the parity as,
\begin{equation*}
{\bf p\to -p}\hspace{0.3cm} \text{the kinetic energy {\bf p.p} is still invariant}
\end{equation*}
Let us see computationally how Parity can be understood by using the approach of matrix mechanics method\cite{article} again for Harmonic oscillator of the form,
\begin{equation*}
V(x)=\frac{1}{2}k(x-\frac{a}{2})^2 \hspace{0.2cm} \text{k=Spring constant and a=Distance}
\end{equation*}
\end{frame}
\subsection{Graph}
\begin{frame}{\subsecname}
The graph has been ploted for width a=2, and for spring constant k=75;
\begin{figure}[H]
\centering
\includegraphics[width=1.1\linewidth]{"Graphic window number 0"}
\caption{Plot of Potential and Wave function of Harmonic Oscillator.}
\label{fig:graphic-window-number-0}
\end{figure}
\end{frame}
\section{Time Reversal Symmetry}
\begin{frame}{\secname}
Time Reversal operator is different unlike the rest of the symmetry discussed. Time reversal operator transform Antiunitary.
If $\ket{\psi(t)}$ is a time dependent state of a system that satifies S.W.E, then
\begin{equation*}
\iota \hbar \frac{d}{dt}\ket {\psi(t)}=H\ket{\psi(t)},\end{equation*}
\begin{equation*}\text{As} \hspace{0.5mm} \hspace{0.3cm} \psi_{r}(x,t)=\psi^*(x,-t),\end{equation*}
\begin{equation*}\implies \ket{\psi_{r}(t)}=\theta\ket{\psi(-t)}.
\end{equation*}
The operator $\theta$ is the time reversal operator and does not consider time; it effects the wave function by complex conjugating it, rather than it takes kets to another kets.
\end{frame}
\subsection{Postulates}
\begin{frame}{\subsecname}
\begin{itemize}
\item Probabilities must be conserved under time reversal.
\item In classical mechanics, inital condition of motion of x(t) transform under time reversal as
\begin{center}
$(x_{0},p_{0}) \to (x_{0},-p_{0}),$
\end{center}
$\therefore$ In quantum mechanics we have,
\begin{align}
\theta {\bf x} \theta^\dagger={\bf x},\hspace{0.2cm} \text{and,}\\
\theta {\bf p} \theta^\dagger=-{\bf p}.
\end{align}
Also in such system if it occurs, then
\begin{equation}
\theta {\bf L} \theta^\dagger=-{\bf L}, \hspace{0.5cm} (\because {\bf L=r\cross p})
\end{equation}
\begin{equation}
\text{and so,} \hspace{0.3cm}\theta {\bf S}\theta^\dagger=-{\bf S},
\end{equation}
\begin{equation}
\theta {\bf J} \theta^\dagger=-{\bf J}.
\end{equation}
\end{itemize}
\end{frame}
\begin{frame}{\subsecname}
\begin{itemize}
\item $\theta$ cannot be unitary.\\[5ex]
\item Lk Decomposition rule
Given an antilinear operator A, we can write it as,\\[4ex]
\begin{center}
A=LK, where L is the Linear operator and K is antilinear
\end{center}
\end{itemize}
\end{frame}
\section{Symmetries in Dirac Equation}
\begin{frame}{\secname}
The Symmetries we discussed can be applied to Dirac equation of the form
\begin{equation}
\left(\gamma^{\mu}\left(\iota\partial_{\mu}-eA_{\mu}(x)-m\right)\right)\psi(x)=0.
\end{equation}
\begin{itemize}
\item For Translation- For translation, we have $S=\mathbbm{1}$
\begin{equation}
\therefore\hspace{0.3cm} \psi''(x)=\psi(x+a)=e^{\alpha^{\mu}\partial_\mu}\psi(x),
\end{equation}
and thus the translation operator is,
\begin{equation*}
T\equiv e^{-\iota\alpha^{\mu}\partial_\mu}=e^{\-iota \alpha^{\mu}p_{\mu}},
\end{equation*}
The translation invariance of a problem implies,
\begin{equation}
[D(A),p_{\mu}]=0,
\end{equation}
\begin{center}
Or $\implies [p_{\mu},H]=0$
\end{center}
\end{itemize}
\end{frame}
\begin{frame}{\secname}
\begin{itemize}
\item For Rotation-For rotations, as $R=e^{-\iota \phi^{k}J^{k}}$
\begin{equation*}
\text{with}\hspace{0.3cm} J=\frac{\hbar}{2}\Sigma+x\cross\frac{\hbar}{\iota}\grad.
\end{equation*}
\begin{equation}
[D(A),J]=0.
\end{equation}
\begin{equation*}
\implies [J,H]=0,\hspace{0.2cm} \text{ where $H$ is the Dirac Hamiltonian}
\end{equation*}
\item Parity- Dirac equation of the form,
\begin{equation}
\hat{H}={\pmb \alpha.\bf p}+\beta m+V({\bf r}).
\end{equation}
But as {\bf r$\to$-r} and {\bf p$\to$-p}, $\hat{H}$ doesn't remain invariant as such.
So the Parity opertor $\hat{\mathcal{P}}$ must have be of form
\begin{equation}
\hat{\mathcal{P}}=u_{p}\hat{P},
\end{equation}
\end{itemize}
\end{frame}
\begin{frame}{\secname}
So we need an extra operator (unitary operator) say $u_{p}$ such that,
\begin{equation}
\text{where}\hspace{0.3cm} \hat{\mathcal{P}}\hat{H}\hat{\mathcal P^{-1}}=\hat{H}.
\end{equation}
\begin{equation}
\therefore\hspace{0.3cm} u_p{\pmb \alpha}u_P^{-1}=-{\pmb \alpha},
\end{equation}
\begin{equation}
\text{and}\hspace{0.3cm} u_p\beta u_p=\beta,
\end{equation}
\begin{equation}
\text{and}\hspace{0.3cm}u_p^{2}=1.
\end{equation}
So as $u_{p}$ need to be a $4\cross4$ matrix and the matrix is $\gamma^{0}$
\begin{equation}
\text{i.e.}\hspace{0.3cm} u_{p}=\gamma^{0}=\begin{pmatrix}
\mathbb{1}&0\\
0&-\mathbb{1}
\end{pmatrix}=(\gamma^{0})^\dagger.
\end{equation}
\begin{equation*}
\hat{\mathcal{P}}=\beta\hat{P},\hspace{0.3cm}(\because\hspace{0.3cm}\hat{\mathcal{P}}\psi({\bf r})=\beta\psi({\bf -r}))
\end{equation*}
\end{frame}
\begin{frame}
\begin{itemize}
\item Time Reversal Symmetry- Defining Time reversal operator for Dirac equation as,
\begin{equation}
\mathcal{\hat{T}}=u_T\hat{K},
\end{equation}
where $u_T$ is unitary and $K$ is the same operator discussed in chapter 6, which effects the constant by complex conjugating them and doesn't effects on kets.
Finally one can write,
\begin{equation}
\mathcal{\hat{T}}\psi({\bf r},t)=\psi_T({\bf r},-t)=\gamma^{1}\gamma^{3}\hat{K}\psi({\bf r},t)=\gamma^{1}\gamma^{3}\psi^{*}({\bf r},t).
\end{equation}
\end{itemize}
\end{frame}
\section{Violation Of Symmetry}
Although Symmetry leads to conservation laws, nature indeed donot follow the same.\\[0.5ex]
For eg. \item Parity get violated for weak interaction.
\item In Neutral Kaon system CP violation fails, where C stands for Charge conjugation.
\item Violation of CPT theorem will lead to violation of Lorentz symmetry but till now its not been prove till yet.
\section{References}
\begin{frame}{\secname}
\printbibliography[heading=none] % <----- heading= none is added in order to prevent a duplicate heading
\end{frame}
\end{document}
答案1
您可以通过添加目录框架将目录扩展到两个框架[allowframebreaks]，如下所示：
\begin{frame}[allowframebreaks]
\frametitle{Overview}
\tableofcontents
\end{frame}