\documentclass[11pt,fleqn]{article}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site 
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% Page setup
\topmargin -1.5cm      
\oddsidemargin -0.04cm   
\evensidemargin -0.04cm  
\textwidth 16.59cm
\textheight 24cm 
\setlength{\parindent}{0cm} 
\setlength{\parskip}{0cm} 


% Fonts
\usepackage{latexsym}
\usepackage{amsmath} 
\usepackage{amssymb} 
\usepackage{bm}


% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}

% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}

% Mass symbol
\newcommand{\mass}{\mathsf{m}} 
\newcommand{\Mass}{\mathsf{M}} 

% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}} 
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}} 
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}

% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}

% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}

% heading
\newcommand{\heading}[1]{\begin{center} \Large {#1} \end{center}}
\newcommand{\auname}[1]{\begin{center} \bf Submitted by: #1 \end{center}}


\begin{document} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



\heading{E6050: Pertrubed particle in a 2D symmetric box} 
\auname{Tamir Tapuhy \& Omer Koren ; Edited by: Yaniv Oiknine}


{\bf The problem:}

\Dn

Given a particle with no spin and mass m in a box $x,y\in[-a,a]$. 

\Dn

(1) Write the wave function $\psi(x,y)$ of the ground state.

(2) A potential $V=u\delta(x)\delta(y)$ is added (relates to all questions 3-5 as well). 
Write three "lowest" (i.e correlating to lowest energies) eigenstates coupled with the ground state by the perturbation.

(3)  Write the Hamiltonian $H=H_{0}+V$  as a sum of two 4x4 matrices.

(4) Write eigenstates (as column vectors) and eigenenergies up to first order of u.

(5) Calculate ground state energy to second order.


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) 
In the case of a square box, we can write the hamilotnian of the particle as: \begin{equation}
\hat{H}=-\frac{\hbar^{2}}{2m}\left(p_{x}^{2}+p_{y}^{2}\right)\end{equation}
Boundary conditions are: \begin{equation}
\psi\left(-a,y\right)=\psi\left(a,y\right)=\psi\left(x,a\right)=\psi(x,-a)=0\end{equation}
Concluding eigenstates of a particle in a two-dimentional box:
\begin{equation}
\psi(x,y)=\begin{cases}
\begin{array}{c}
\frac{1}{a}\sin\left(k_{n,x}x\right)\sin(k_{m,y}y)\\
\frac{1}{a}\cos\left(k_{n,x}x\right)\cos(k_{m,y}y)\\
\frac{1}{a}\sin\left(k_{n,x}x\right)\cos(k_{m,y}y)\\
\frac{1}{a}\cos\left(k_{n,x}x\right)\sin(k_{m,y}y)\end{array} & \begin{array}{c}
n,m=even\\
n,m=odd\\
n=even,m=odd\\
n=odd,m=even\end{array}\end{cases}\end{equation}\\
where\\ 
\begin{equation}k_{n,x} = \frac{\pi}{2a}n \ \ ;\ \ k_{m,y} = \frac{\pi}{2a}m\end{equation} 
and the energies are :\begin{equation}
E_{n,m}=\frac{\hbar^{2}}{2m}\left(k_{n,x}^{2}+k_{m,y}^{2}\right)=\frac{\hbar^{2}\pi^{2}}{8ma^{2}}\left(n^{2}+m^{2}\right)\end{equation}
The unperturbed ground state is achieved by choosing $n,m=1$ ,
so that the wave function of the ground state is: $\frac{1}{a}\cos\left(\frac{\pi}{2a}x\right)\cos(\frac{\pi}{2a}y)$
. The energy is: $E_{0}=\frac{\hbar^{2}\pi^2}{4ma^{2}}$.
%%%%%%%%%%%%%%%
\Dn

(2)
The wave functions which are affected by the aforementioned perturbation are those for which $\psi(0,0) \neq 0$. We conclude that the 'lowest' three eigenstates are $|n=1,m=3\rangle$ $|n=3,m=1\rangle$ and $|n=3,m=3\rangle$ :\begin{equation}
\begin{array}{l}
|3,1\rangle =\left.|1\right\rangle =\frac{1}{a}\cos\left(\frac{\pi}{2a}3x\right)\cos(\frac{\pi}{2a}y)\\ \\
|1,3\rangle =\left.|2\right\rangle =\frac{1}{a}\cos\left(\frac{\pi}{2a}x\right)\cos(\frac{\pi}{2a}3y)\\ \\
|3,3\rangle =\left.|3\right\rangle =\frac{1}{a}\cos\left(\frac{\pi}{2a}3x\right)\cos(\frac{\pi}{2a}3y)\end{array}\end{equation}
and the energies are:\begin{equation}
E_{1,2}=\frac{10\hbar^{2}\pi^{2}}{8ma^{2}} ;\, E_{3}=\frac{18\hbar^{2}\pi^{2}}{8ma^{2}}\end{equation}



%%%%%%%%%%%%%%
\Dn

(3)
In order to find $V$ we will compute: $\langle n|V|m\rangle=\langle n|u\delta\left(x\right)\delta\left(y\right)|m\rangle$
.\begin{equation}
\langle0|V|0\rangle=\int_{-a}^{a}\int_{-a}^{a}\frac{1}{a}\cos\left(\frac{\pi}{2a}x\right)\cos\left(\frac{\pi}{2a}y\right)u\delta\left(x\right)\delta\left(y\right)\frac{1}{a}\cos\left(\frac{\pi}{2a}x\right)\cos\left(\frac{\pi}{2a}y\right)dxdy=\frac{u}{a^{2}}\end{equation}
It is obvious that all combinations give the same result, since the $\delta$ cooses $\cos(0) = 1$.\\
The hamiltonian is:\begin{equation}
H=H_{0}+V=E_{0}\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 5 & 0 & 0\\
0 & 0 & 5 & 0\\
0 & 0 & 0 & 9\end{array}\right)+\frac{u}{a^{2}}\left(\begin{array}{cccc}
1 & 1 & 1 & 1\\
1 & 1 & 1 & 1\\
1 & 1 & 1 & 1\\
1 & 1 & 1 & 1\end{array}\right)\end{equation}


%%%%%%%%%%%%%
\Dn

(4)
First of all, in order to use perturbation theory we must tackle degeneracies (i.e states $|1\rangle$and $|2\rangle$). Changing  basis by using Symmetric and Anti-Symmetric configurations:
\begin{equation}
\begin{array}{cc}
|A\rangle=\frac{1}{\sqrt{2}}\left(|1\rangle+|2\rangle\right) & ,\end{array}|S\rangle=\frac{1}{\sqrt{2}}\left(|1\rangle-|2\rangle\right)\end{equation}
Using the transformation matrix, we derive the new Hamiltonian:
\[
P=\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\
0 & 0 & 0 & 1\end{array}\right)\]
\begin{equation}
P^{T}HP=E_{0}\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 5 & 0 & 0\\
0 & 0 & 5 & 0\\
0 & 0 & 0 & 9\end{array}\right)+\frac{u}{a^{2}}\left(\begin{array}{cccc}
1 & \sqrt{2} & 0 & 1\\
\sqrt{2} & 2 & 0 & \sqrt{2}\\
0 & 0 & 0 & 0\\
1 & \sqrt{2} & 0 & 1\end{array}\right)=\end{equation}
\[
H_{new}=E_{0}\left(\begin{array}{cccc}
1+\frac{u}{E_{0}a^{2}} & 0 & 0 & 0\\
0 & 5+\frac{2u}{E_{0}a^{2}} & 0 & 0\\
0 & 0 & 5 & 0\\
0 & 0 & 0 & 9+\frac{u}{E_{0}a^{2}}\end{array}\right)+\frac{u}{a^{2}}\left(\begin{array}{cccc}
0 & \sqrt{2} & 0 & 1\\
\sqrt{2} & 0 & 0 & \sqrt{2}\\
0 & 0 & 0 & 0\\
1 & \sqrt{2} & 0 & 0\end{array}\right)\]

In equation (11), one can see the $1^{st}$ order correction: \\
\[
\epsilon^{[0]}=\epsilon_{n_{0}}\,;\,\epsilon^{[1]}=V_{n_{0},n_{0}}\,\rightarrow\epsilon_{0}=E_{0}+\frac{u}{a^{2}}\,;\,\epsilon_{S}=E_{1}+\frac{2u}{a^{2}}\,;\,\epsilon_{A}=E_{1}\,;\,\epsilon_{3}=E_{3}+\frac{u}{a^{2}}\]
We will compute the eigenstates to $1^{st}$ order:

\[
\Psi^{\left[1\right]}=\frac{V_{n,n_{0}}}{\epsilon_{n_{0}}-\epsilon_{n}}\]

Marking $\frac{u}{a^{2}}=\lambda$:

 \[
\Psi_{0}^{\left[1\right]}=\left(\begin{array}{c}
0\\
\frac{\sqrt{2}\frac{u}{a^{2}}}{-4E_{0}-\frac{u}{a^{2}}}\\
0\\
\frac{u}{-8E_{0}a^{2}}\end{array}\right)=\left(\begin{array}{c}
0\\
\frac{\sqrt{2}\lambda}{-4E_{0}-\lambda}\\
0\\
\frac{\lambda}{-8E_{0}a^{2}}\end{array}\right)\rightarrow\Psi_{0}^{\left[0\right]}+\Psi_{0}^{\left[1\right]}=\left(\begin{array}{c}
1\\
\frac{\sqrt{2}\lambda}{-4E_{0}-\lambda}\\
0\\
\frac{\lambda}{-8E_{0}a^{2}}\end{array}\right)\]
\begin{equation}
\Psi_{s}^{\left[1\right]}=\left(\begin{array}{c}
\frac{\sqrt{2}\lambda}{4E_{0}+\lambda}\\
0\\
0\\
\frac{\sqrt{2}\lambda}{-4E_{0}+\lambda}\end{array}\right)\rightarrow\Psi_{S}^{\left[0\right]}+\Psi_{S}^{\left[1\right]}=\left(\begin{array}{c}
\frac{\sqrt{2}\lambda}{4E_{0}+\lambda}\\
1\\
0\\
\frac{\sqrt{2}\lambda}{-4E_{0}+\lambda}\end{array}\right)\end{equation}
\[
\Psi_{A}^{\left[1\right]}=\left(\begin{array}{c}
0\\
0\\
0\\
0\end{array}\right)\rightarrow\Psi_{A}^{\left[0\right]}+\Psi_{A}^{\left[1\right]}=\left(\begin{array}{c}
0\\
0\\
1\\
0\end{array}\right)\]
\[
\Psi_{3}^{\left[1\right]}=\left(\begin{array}{c}
\frac{\lambda}{8E_{0}}\\
\frac{\sqrt{2}\lambda}{4E_{0}-2\lambda}\\
0\\
0\end{array}\right)\rightarrow\Psi_{3}^{\left[0\right]}+\Psi_{3}^{\left[1\right]}=\left(\begin{array}{c}
\frac{\lambda}{8E_{0}}\\
\frac{\sqrt{2}\lambda}{4E_{0}-2\lambda}\\
0\\
1\end{array}\right)\]

%%%%%%%%%%%%%
\Dn

(5)
In order to calculate $2^{nd}$ order correction of the \underbar{ground
state energy }we will use:\begin{equation}
\epsilon^{\left[2\right]}=\sum_{m}\frac{V_{n_{0},m}\cdot V_{m,n_{0}}}{\epsilon_{n_{0}}-\epsilon_{m}}=\left\{ n_{0}=0\right\} =\frac{V_{0,S}\cdot V_{S,0}}{-4E_{0}-\lambda}+\frac{V_{0,A}\cdot V_{A,0}}{-4E_{0}+\lambda}+\frac{V_{0,3}\cdot V_{3,0}}{-8E_{0}}=\end{equation}
\[
=\frac{2\lambda^{2}}{-4E_{0}-\lambda}+\frac{\lambda^{2}}{-8E_{0}}\]


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document} 
