TeX - LaTeX Asked on November 26, 2021
Its a huge repsect please my presentation is tommorrow and i have got this error no where Two documentclass or documentstyle commands. documentclass{ in beamer. I beg of you guys please help me out or generate my pdf please Here is my code’
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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=0.3linewidth]{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*}
thereforehspace{0.5mm} left(T(a)psiright)(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=1.0linewidth]{"translated wave"}
caption{Representation of Translated wave functioncite{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_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.
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(xpm a)=V(x). pause
end{equation*}
begin{figure}[H]
centering
includegraphics[width=0.7linewidth]{"periodic potentia"}
caption{Periodic Potentialcite{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*}
Hket{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^{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;\[0.5ex]pause
begin{equation*}
bra{n'}Hket{n}ne 0hspace{0.2cm} text{;}hspace{0.2cm} bra{n+1}Hket{n}=-triangle.
end{equation*}\[3ex]pause
This approximation is called {bf Tight binding approximation.}\[0.2ex]
end{frame}
begin{frame}{subsecname}
begin{equation*}
Hket{n}=E_{n}ket{n}-triangle ket{n+1}-ket{n-1}.
end{equation*}\[2ex]pause
begin{equation*}
text{The quantity}hspace{0.2cm}Hket{theta}=Hsum e^{iota n theta}ket{n} hspace{0.1cm}text{equals,}
end{equation*}\[2ex]pause
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*}\[3ex]
So one has continous band of energy eigenstates.
end{frame}
begin{frame}{subsecname}
begin{equation*}
E-2triangle < E < E+2triangle.
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=0.4linewidth]{"Lifiting degenracy"}
caption{Energy degeneracy lifted upcite{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=1.0linewidth]{"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 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},dphi)=1-frac{bf J.hat{n}}{hbar}dphi.
end{equation*}
For a finite rotation we can write,pause
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]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 thatpause
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]=iotahbar J_z.
end{equation*}
$therefore$ Rotation about any axis gives,pause
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,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(Xket{m})=XHket{m}=E_{m}(Xket{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 algorithmcite{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=1.0linewidth]{"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=1.0linewidth]{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=1.0linewidth]{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}
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 havepause
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*}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 Momentumpause
item $pi^{-1}${bf S}$pi$={bf S}, {bf S} is Spin Angular Momentumpause
item $pi^{-1}${bf J}$pi$={bf J}, {bf J} is Total Angular Momentumpause
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} piket{alpha,lm}=(-1)^lket{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}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 ifpause $epsilon_{a}epsilon_{beta}epsilon_{alpha}=1$, 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*}pause
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]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 pto -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 algorithmcite{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=1.1linewidth]{"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 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)}=Hket{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)}=thetaket{psi(-t)}.
end{equation*}
end{frame}
subsection{Postulates}
begin{frame}{subsecname}
begin{itemize}
item Probabilities must be conserved under time reversalpause.
item In classical mechanics, inital condition of motion of x(t) transform under time reversal aspause
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, thenpause
begin{equation*}
theta {bf L} theta^dagger=-{bf L}, hspace{0.5cm} (because {bf L=rcross 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 formpause
begin{equation*}
left(gamma^{mu}left(iotapartial_{mu}-eA_{mu}(x)-mright)right)psi(x)=0.
end{equation*}pause
begin{itemize}
item For Translation- For translation,pause
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,pause
begin{equation*}
Tequiv e^{-iotaalpha^{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)equivgamma^{mu}(iotapartial_{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+xcrossfrac{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*}
thereforehspace{0.3cm} u_p{pmb alpha}u_P^{-1}=-{pmb alpha},
end{equation*}
begin{equation*}
u_pbeta u_p=beta,
end{equation*}
begin{equation*}
u_p^{2}=1.
end{equation*}pause
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}
mathbbm{1}&0\
0&-mathbbm{1}
end{pmatrix}=(gamma^{0})^dagger.
end{equation*}
begin{equation*}
hat{mathcal{P}}=betahat{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_That{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}
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