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author{Anshul Sharma}
title{Symmetry in Quantum Mechanics}
institute {CENTRAL UNIVERSITY OF HIMACHAL PRADESH}
titlegraphic{includegraphics[width=0.3linewidth]{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}
    thereforehspace{0.5mm} left(T(a)psiright)(x)=psi(x-a).
    end{equation}
    begin{figure}[H]
        centering
        includegraphics[width=0.6linewidth]{"translated wave"}
        caption{Representation of Translated wave  functioncite{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_xDelta x'}{hbar})=exp(frac{-iota p_xDelta 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 $[x_j,p_i]=iota hbardelta_{ij}$ 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(xpm a)=V(x). 
        end{equation}
        begin{figure}[H]
            centering
            includegraphics[width=0.7linewidth]{"periodic potentia"}
            caption{Periodic Potentialcite{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}
            Hket{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^{intheta}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'}Hket{n}ne 0hspace{0.2cm} text{and}hspace{0.2cm} bra{n+1}Hket{n}=-triangle.
        end{equation}
        This approximation is called Tight binding approximation.\[0.2ex]
        $therefore$ $ket{n}$ is not energy eigenket, i.e.
         begin{equation}
        Hket{n}=E_{n}ket{n}-triangle ket{n+1}-ket{n-1}.
        end{equation}
         begin{equation}
        text{The quantity}hspace{0.2cm}Hket{theta}=Hsum e^{iota n theta}ket{n} hspace{0.1cm}text{equals,}
        end{equation}
    begin{equation}
    Hsum e^{iota n theta}ket{n}=E_{n}sum e^{iota n theta}ket{n}-2trianglecos(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-2triangle < E < E+2triangle.
            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=0.4linewidth]{"Lifiting degenracy"}
                caption{Energy degeneracy lifted upcite{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=0.6linewidth]{"Screenshot (122)"}
        caption{Showing why rotations about different axis do not commutecite{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},dphi)=1-frac{bf J.hat{n}}{hbar}dphi.
        end{equation}
        $therefore$ For a finite rotation we can write,
        begin{equation}
        D_z(phi)=lim_{Ntoinfty}left[1-iotafrac{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 $Rmathbbm{1}=R$, $implies D(R)mathbbm{1}=D(R)$.\[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]=iotahbar J_z.
end{equation}
$therefore$ Rotation about any axis gives,
begin{equation}
[J_i,J_j]=epsilon_{ijk}iotahbar 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 $Xket{m}$ is also an energy eigenket with the same energy eigenvalue, because
        begin{equation}
        H(Xket{m})=XHket{m}=E_{m}(Xket{m})
        end{equation}
        Say $ket{m}$ and $Xket{m}$ 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 approachcite{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=1.0linewidth]{"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=1.0linewidth]{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=1.0linewidth]{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=1.1linewidth]{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=1.0linewidth]{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=1.0linewidth]{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}xket{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} piket{alpha,lm}=(-1)^lket{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}Aket{beta}=epsilon_{alpha}epsilon_{beta}bra{alpha}pi Api ket{beta},\
    bra{alpha}Aket{beta}=epsilon_{alpha}epsilon_{beta}bra{alpha}Aket{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}oddket{even}=0,\
    bra{odd}oddket{odd}=0,\
    bra{even}evenket{odd}=0.
    end{align*}
    begin{equation*}
    text{So} int psi_{beta}^{*}psi_{alpha}dtau=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 pto -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 methodcite{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=1.1linewidth]{"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)}=Hket{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)}=thetaket{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=rcross 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(iotapartial_{mu}-eA_{mu}(x)-mright)right)psi(x)=0.
        end{equation}
        begin{itemize}
            item For Translation- For translation, we have $S=mathbbm{1}$
            begin{equation}
            thereforehspace{0.3cm} psi''(x)=psi(x+a)=e^{alpha^{mu}partial_mu}psi(x),
            end{equation}
            and thus the translation operator is,
            begin{equation*}
            Tequiv e^{-iotaalpha^{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+xcrossfrac{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}
    thereforehspace{0.3cm} u_p{pmb alpha}u_P^{-1}=-{pmb alpha},
    end{equation}
    begin{equation}
    text{and}hspace{0.3cm} u_pbeta 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 $4cross4$ 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}}=betahat{P},hspace{0.3cm}(becausehspace{0.3cm}hat{mathcal{P}}psi({bf r})=betapsi({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_That{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}

TeX - LaTeX Asked on November 21, 2021

1 Answers

One Answer

You can spread your TOC over two frames by adding [allowframebreaks] to you TOC frame, as follows:

begin{frame}[allowframebreaks]
    frametitle{Overview}
    tableofcontents
end{frame}

Answered by G. de Man on November 21, 2021

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