非常抱歉，我的演讲明天就要开始了，但是我Two \documentclass or \documentstyle commands. \documentclass{在 Beamer 中却遇到了这个错误。我请求你们帮帮我或者帮我生成 pdf 这是我的代码'
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\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}[allowframebreaks]
\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
Symmetry in Classical Mechanics can be described by Noether's theorem. Noether's theorem states that\\[1.5ex]\pause
\begin{mdframed}
{\large"The action is minimum for the path taken by the particle."}
\end{mdframed}\pause
The conservation laws that comes out, when one solves the Action principle for different cases are- \pause
\begin{itemize}
\item \large{Conservation of Linear Momentum gives "Homogenity of Spcae."}\pause
\item \large{Conservation of Angular Momentum gives "Isotropy of Space."}\pause
\item \large{Conservation of Energy gives "Homogenity of Time."}
\end{itemize} \pause
\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 $\ket {x+a}$ 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+a}.
\end{equation*}
Applying these transformation on the wave function i.e.\pause
\begin{equation*}
T(a)\ket{\psi}=\ket{\phi},
\end{equation*}
\begin{equation*}
\therefore\hspace{0.5mm} \left(T(a)\psi\right)(x)=\psi(x-a).
\end{equation*}
Changing x to x+a we get,
\begin{center}
\fbox{$T(a)\psi(x+a)=\psi(x).$}
\end{center}
\end{frame}
\begin{frame}{\secname}
\begin{figure}[H]
\centering
\includegraphics[width=1.0\linewidth]{"translated wave"}
\caption{Representation of Translated wave function\cite{article}.}
\label{fig:translated-wave}
\end{figure}
\end{frame}
\subsection{Properties}
\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).}\\[3.5ex]\pause
\item {\bf T(dx)T(dx$'$)=T(dx+dx$'$)=T(dx$'$)T(dx).}\\[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]
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.
\end{itemize}
\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). \pause
\end{equation*}
\begin{figure}[H]
\centering
\includegraphics[width=0.7\linewidth]{"periodic potentia"}
\caption{Periodic Potential\cite{PeriodicPotential}.}
\end{figure}
Let us find eigenket and eigenvalue of translation operator $T(a)$.
\end{frame}
\begin{frame}{\subsecname}
We consider first the height of the potential barrier to be $\infty$.\\[4.2ex]\pause
Let the particle be found at $n^{th}$ position represented by $\ket{n}$\\[3ex]
\begin{equation*}
H\ket{n}=E_{n}\ket{n}.
\end{equation*}\\[3ex]
and the wave function $\braket{x'}{n}$ is finite only in the n$^{th}$ site.\\[0.5ex]
\end{frame}
\begin{frame}{\subsecname}
But when it is applied to translation operator, we have,\pause
\begin{equation*}
T(a)\ket{n}=\ket{n+1}.
\end{equation*}
Defining simultaneous eigenket and a better representation of labelling each and every position as, \pause
\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;\\[0.5ex]\pause
\begin{equation*}
\bra{n'}H\ket{n}\ne 0\hspace{0.2cm} \text{;}\hspace{0.2cm} \bra{n+1}H\ket{n}=-\triangle.
\end{equation*}\\[3ex]\pause
This approximation is called {\bf Tight binding approximation.}\\[0.2ex]
\end{frame}
\begin{frame}{\subsecname}
\begin{equation*}
H\ket{n}=E_{n}\ket{n}-\triangle \ket{n+1}-\ket{n-1}.
\end{equation*}\\[2ex]\pause
\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*}\\[2ex]\pause
\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*}\\[3ex]
So one has continous band of energy eigenstates.
\end{frame}
\begin{frame}{\subsecname}
\begin{equation*}
E-2\triangle < E < E+2\triangle.
\end{equation*}\\[2ex]
\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}
\subsection{Wavefunction}
\begin{frame}{\subsecname}
\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}\pause
\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*}\pause
\end{mdframed}
The above equation is called {\bf Bloch theorem}.\\[0.5ex]\pause
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}
\begin{itemize}
\item Rotation in different direction do not commute and as a result the corresponding group is known as Non-Abelian group.\\[4ex] \pause
\item Rotation in quantum mechanics is an direct evidence of obtaining Angular Momentum. \\[4ex]\pause
\item Rotation affects physical system i.e. the state ket corresponding to the rotated system is expected to look different from the state ket corresponding to original unrotated system.
\end{itemize}
\end{frame}
\begin{frame}
\begin{figure}[H]
\centering
\includegraphics[width=1.0\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 of kets takes place as,
\begin{equation*}\pause
\ket{\alpha}_R=D(R)\ket{\alpha},
\end{equation*}
For infinitesimal rotation we have,\pause
\begin{equation*}
D(\hat{n},d\phi)=1-\frac{\bf J.\hat{n}}{\hbar}d\phi.
\end{equation*}
For a finite rotation we can write,\pause
\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]\pause
\item {\bf Closure}: As $R_1R_2=R_3$, $\implies D(R_1)D(R_2)=D(R_3)$.\\[3.5ex]\pause
\item {\bf Inverse}: As $RR^{-1}=1$, $\implies D(R)D(R)^{-1}=1$.\\[3.5ex]\pause
\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\pause
\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*}
\begin{equation*}
\text{or}\hspace{0.3cm} [J_x,J_y]=\iota\hbar J_z.
\end{equation*}
$\therefore$ Rotation about any axis gives,\pause
\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,\pause
\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 $E_m$.\pause
\begin{equation*}
H(X\ket{m})=XH\ket{m}=E_{m}(X\ket{m})
\end{equation*}
\end{frame}
\subsection{Pseudo code implementation of N-well}
\begin{frame}{\subsecname}
{\bf Pseudo code}
\begin{itemize}
\item Defining initial parameters and the Potential of the system.\\[0.5ex]
In my case, Depth of Potential=-20 and width of the infinite square well is 2.\pause
\item Use the algorithm\cite{article1} for the system to obtain energy eigen values and wavefunction.\pause
\item Run for different Parameters.
\end{itemize}
\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=1.0\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=1.0\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}
\subsection{Calculations}
\begin{frame}\subsecname
\begin{table}[htbp]
\centering
\caption*{{\bf Calculation of Energy diffrence of ground state and excited state when barrier width decreases}}
\begin{tabular}{rrrrrr}
\rowcolor[rgb]{ 1, 1, 0} \multicolumn{1}{l}{\textbf{Barrier width}} & \multicolumn{1}{l}{\textbf{E0}} & \multicolumn{1}{l}{\textbf{E1}} & \multicolumn{1}{l}{\textbf{E2}} & \multicolumn{1}{l}{\textbf{E3}} & \multicolumn{1}{l}{\textbf{(E0-E1) }} \\
\textbf{1.000} & \textbf{-17.108} & \textbf{-17.075} & \textbf{-8.726} & \textbf{-7.916} & \textbf{-0.033} \\
\textbf{0.500} & \textbf{-17.196} & \textbf{-17.049} & \textbf{-9.405} & \textbf{-8.646} & \textbf{-0.147} \\
\textbf{0.050} & \textbf{-18.117} & \textbf{-15.714} & \textbf{-9.388} & \textbf{-2.857} & \textbf{-2.402} \\
\textbf{0.005} & \textbf{-18.683} & \textbf{-15.138} & \textbf{-8.958} & \textbf{-0.553} & \textbf{-3.544} \\
\end{tabular}%
\label{tab:addlabel}%
\end{table}%
\end{frame}
\section{Parity}
\begin{frame}{\secname}
\begin{itemize}
\item Parity is a Discrete symmetry.\\[0.5ex]\pause
\item In classical mechanics the equation of motion remains invariant under transformation {\bf r$\to$-r}.\\[0.5ex]\pause
\item In quantum mechanics under Parity Transformation, we have\pause
\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*}\pause
\item Parity operator has eigenvalue of $\pm$
\end{itemize}
\end{frame}
\begin{frame}{\secname}
\begin{itemize}
\item "Translation followed by parity is equivalent to parity followed by translation in opposite direction."\pause
\begin{center}
i.e. $\pi$T(dx$'$)=T(-dx$'$)$\pi$,\vspace{0.1mm}
\end{center}
\begin{itemize}
\item $\pi^\dagger$p$\pi$=-p.
\item $\pi^{-1}${\bf L}$\pi$={\bf L}, {\bf L} is Orbital Angular Momentum\pause
\item $\pi^{-1}${\bf S}$\pi$={\bf S}, {\bf S} is Spin Angular Momentum\pause
\item $\pi^{-1}${\bf J}$\pi$={\bf J}, {\bf J} is Total Angular Momentum\pause
\end{itemize}
\item In spherical harmonics, $Y_{l}^{m}(\theta,\phi)$ being eigenstate of parity operator has eigenvalue $(-1)^{l}$
\begin{equation*}
\text{i.e.} \hspace{0.3cm} \pi\ket{\alpha,lm}=(-1)^l\ket{\alpha,lm}.
\end{equation*}
\end{itemize}
\end{frame}
\subsection{Selection Rules}
\begin{frame}{\subsecname}
Suppose $\ket{\alpha}$ and $\ket{\beta}$ are parity eigenstates,
if $A$ is an observable with definite parity and let $\pi A \pi=\epsilon_{A}A$;\pause
\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\pause $\epsilon_{a}\epsilon_{\beta}\epsilon_{\alpha}=1$, 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*}\pause
\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]\pause
This rule is called {\bf Laporate rule}.
\end{frame}
\section{Invariance of Hamilonian}
\begin{frame}{\secname}
Hamiltonian is invariant under Parity Transformation, i.e.\pause
\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*}
\end{frame}
\subsection{Pseudo code implementation on Simple Harmonic Oscillator.}
\begin{frame}{\subsecname}
Pseudo code Implementation
\begin{itemize}
\item Defining initial parameters and the Potential of the system.\\[0.5ex]\pause
In my case, spring constant k=75 and width of the infinite square well.\\[0.5ex]
Potential form is,\pause
\begin{equation*}
V(x)=\frac{1}{2}k(x-\frac{a}{2})^2
\end{equation*}
\item Use the algorithm\cite{article} for the system to obtain energy eigen values and wavefunction.
\end{itemize}
\end{frame}
\subsection{Graph}
\begin{frame}{\subsecname}
\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 transform by Antiunitary transformation.\\[2.5ex]\pause
If $\ket{\psi(t)}$ is a time dependent state of a system that satifies S.W.E, then,\pause
\begin{equation*}
\iota \hbar \frac{d}{dt}\ket {\psi(t)}=H\ket{\psi(t)},\end{equation*}\\[2ex]\pause
\begin{equation*}\text{As} \hspace{0.5mm} \hspace{0.3cm} \psi_{r}(x,t)=\psi^*(x,-t),\end{equation*}\\[2ex]\pause
\begin{equation*}\implies \ket{\psi_{r}(t)}=\theta\ket{\psi(-t)}.
\end{equation*}
\end{frame}
\subsection{Postulates}
\begin{frame}{\subsecname}
\begin{itemize}
\item Probabilities must be conserved under time reversal\pause.
\item In classical mechanics, inital condition of motion of x(t) transform under time reversal as\pause
\begin{center}
$(x_{0},p_{0}) \to (x_{0},-p_{0}),$
\end{center}\pause
$\therefore$ 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*}\pause
Also in such system if it occurs, then\pause
\begin{equation*}
\theta {\bf L} \theta^\dagger=-{\bf L}, \hspace{0.5cm} (\because {\bf L=r\cross p})
\end{equation*}\pause
\begin{equation*}
\text{and so,} \hspace{0.3cm}\theta {\bf S}\theta^\dagger=-{\bf S},
\end{equation*}\pause
\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]\pause
\item LK Decomposition rule\\[4ex]\pause
Given an antilinear operator A, we can write it as,\\[4ex]\pause
\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\pause
\begin{equation*}
\left(\gamma^{\mu}\left(\iota\partial_{\mu}-eA_{\mu}(x)-m\right)\right)\psi(x)=0.
\end{equation*}\pause
\begin{itemize}
\item For Translation- For translation,\pause
\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,\pause
\begin{equation*}
T\equiv e^{-\iota\alpha^{\mu}\partial_\mu}=e^{-\iota \alpha^{\mu}p_{\mu}},
\end{equation*}\pause
The translation invariance of a problem implies,\pause
\begin{equation*}
[D(A),p_{\mu}]=0,\hspace{0.2cm} \text{where}\hspace{0.1cm}D(A)\equiv\gamma^{\mu}(\iota\partial_{\mu}-eA_{\mu}),
\end{equation*}
\begin{center}
$\implies [p_{\mu},H]=0$
\end{center}
\end{itemize}
\end{frame}
\begin{frame}{\secname}
\begin{itemize}
\item For Rotation- As $R=e^{-\iota \phi^{k}J^{k}}$\pause
\begin{equation*}
\text{with}\hspace{0.3cm} J=\frac{\hbar}{2}\Sigma+x\cross\frac{\hbar}{\iota}\grad.
\end{equation*}\pause
\begin{equation*}
[D(A),J]=0.
\end{equation*}\pause
\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,\pause
\begin{equation*}
\hat{H}={\pmb \alpha.\bf p}+\beta m+V({\bf r}).
\end{equation*}\pause
So the Parity opertor $\hat{\mathcal{P}}$ must have be of form
\begin{equation*}
\hat{\mathcal{P}}=u_{p}\hat{P},
\end{equation*}\pause
\end{itemize}
\end{frame}
\begin{frame}{\secname}
So we need an extra operator (unitary operator) say $u_{p}$ such that,\pause
\begin{equation*}
\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*}
u_p\beta u_p=\beta,
\end{equation*}
\begin{equation*}
u_p^{2}=1.
\end{equation*}\pause
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}
\mathbbm{1}&0\\
0&-\mathbbm{1}
\end{pmatrix}=(\gamma^{0})^\dagger.
\end{equation*}
\begin{equation*}
\hat{\mathcal{P}}=\beta\hat{P},\hspace{0.3cm}
\end{equation*}
\end{frame}
\begin{frame}{\secname}
\begin{itemize}
\item Time Reversal Symmetry- Defining Time reversal operator for Dirac equation as,\pause
\begin{equation*}
\mathcal{\hat{T}}=u_T\hat{K},
\end{equation*}\pause
One gets,\pause
\begin{mdframed}
\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{mdframed}
\end{itemize}
\end{frame}
\section{Conclusion}
\begin{frame}{\secname}
Although symmetry transfromation in general leads to conservation laws and gives us equation of motion invaraint. But nature indeed donot follow the same.\\[0.5ex]\pause \begin{itemize}
\item Parity get violated for weak interaction.\\[1ex]\pause
\item In Neutral Kaon system CP violation fails, where C stands for Charge conjugation.\\[1ex]\pause
\item Violation of CPT theorem will lead to violation of Lorentz symmetry but till now its not been prove till yet.
\end{itemize}
\end{frame}
\section{References}
\begin{frame}[allowframebreaks]{\secname}
\printbibliography[heading=none] % <----- heading= none is added in order to prevent a duplicate heading
\end{frame}
\end{document}