\documentclass{beamer}
\usepackage[english,italian]{babel}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{listings}
\usepackage{calligra}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{cancel}
\graphicspath{ {./figure/} }
\usetheme{default}
\usecolortheme{beaver}
\usepackage{tikz}
\usetikzlibrary{shapes,arrows}

\title{interazione di configurazione}
\author{\calligra Fulvio Ciriaco}
\subject{}
\begin{document}
  \pgfdeclareimage[width=1.5cm]{logo}{uniba.png}
  \logo{\pgfuseimage{logo}}
  \begin{frame}{\titlepage}
  \tableofcontents
  \end{frame}

\begin{frame} 
  \frametitle{correlazione e interazione di configurazione}
  \begin{equation*}
    \Psi=\sum_k a_k \Psi_k
  \end{equation*}
  dove $\Psi_k$ sono determinanti di slater.

\end{frame}

\begin{frame} 
  \frametitle{determinanti}
\begin{tikzpicture}[scale=1]
\tikzstyle{orbital} = 
  [rectangle, rounded corners, fill=white, draw=black, minimum width=3.5ex, 
   minimum height=3.5ex]
\tikzstyle{el} = [thick ];
\draw[el] (0,0) -- (0.5,0) node[midway] {\textuparrow\textdownarrow}
  (0,1.5) -- ++(0.5,0) node[midway] {\textuparrow}
  ++(0.5,0.2) -- ++(0.5,0) node[midway] {\textuparrow}
  ++(0.2,0) -- ++(0.5,0)
  ++(0.2,0) -- ++(0.5,0)
  (0,2.5) -- +(0.5,0) 
  ++(0.5,0.2) -- ++(0.5,0) 
  ++(0.2,0) -- ++(0.5,0)
  ++(0.2,0) -- ++(0.5,0)
;
\end{tikzpicture}

\begin{align*}
  |1,1,1,0,1,0,0,0,0,0,0,0,0,0,0,0\rangle \\
  |1,1,1,0,1\rangle \\
  T_{4,5} |1,1,1,1\rangle
\end{align*}
\end{frame}

\begin{frame} 
  \frametitle{quanti!}
  ALLDET
  \begin{align*}
    N_{\textrm {det}} &= \binom n N \\
    \binom {20} {10} &= 184756
  \end{align*}
\end{frame}

\begin{frame} 
  \frametitle{ancora algebra lineare}
  \begin{align*}
  \left[ \begin{matrix} H_{00} & H_{01} \\
               H_{10}=H_{01} & H_{11} \end{matrix}\right] 
\left[ \begin{matrix} a_{00} \\ a_{10} \end{matrix}\right] &= 
\epsilon_0 
\left[ \begin{matrix} a_{00} \\ a_{10} \end{matrix}\right]
 \\
  \left[ \begin{matrix} H_{00} & H_{01} \\
               H_{10}=H_{01} & H_{11} \end{matrix}\right] 
\left[
  \begin{matrix} a_{01} \\ a_{11} \end{matrix}\right] &= 
\epsilon_1 
\left[ \begin{matrix} a_{01} \\ a_{11} \end{matrix}\right]
\end{align*}

possiamo riscrivere tutto riallineando le colonne:
  \begin{align*}
  \left[ \begin{matrix} E_1 & \delta \\
               \delta & E_2 \end{matrix}\right] 
      \left[ \begin{matrix} a_{00} & a_{01} \\ a_{10} & a_{11}\end{matrix}\right] &= 
      \left[ \begin{matrix}\epsilon_0 & 0 \\ 0 & \epsilon_1 \end{matrix}\right] 
      \left[ \begin{matrix} a_{00} & a_{01} \\ a_{10} & a_{11}\end{matrix}\right] \\
      A^{-1} 
  \left[ \begin{matrix} E_1 & \delta \\
               \delta & E_2 \end{matrix}\right] 
      A &= 
      \left[ \begin{matrix}\epsilon_0 & 0 \\ 0 & \epsilon_1 \end{matrix}\right] 
\end{align*}


\end{frame}

\begin{frame}
  \frametitle{ancora algebra lineare}
  \begin{align*}
      A^{-1} 
  \left[ \begin{matrix} E_1 & \delta \\
               \delta & E_2 \end{matrix}\right] 
      A &= 
      \left[ \begin{matrix}\epsilon_0 & 0 \\ 0 & \epsilon_1 \end{matrix}\right] 
  \\
\det \left[
\begin{matrix} E_1-\epsilon & \delta \\
\delta & E_2-\delta\end{matrix}\right] &=
\epsilon^2+E_1 E_2 -\epsilon(E_1+E_2)-\delta^2 = 0
    \end{align*}
che ha soluzioni
\begin{equation*}
\epsilon=\frac{E_1+E_2 \pm \sqrt{(E_2-E_1)^2+4 \delta^2}} 2
\end{equation*}
\end{frame}

\begin{frame}
  \frametitle{ancora algebra lineare, soluzione}
\begin{equation*}
\epsilon=\frac{E_1+E_2 \pm \sqrt{(E_2-E_1)^2+4 \delta^2}} 2
\end{equation*}
\begin{tikzpicture}[scale=0.8]
\tikzstyle{el} = [very thick ];
\draw[el] 
 (0,0) -- +(0.5,0)
 ++(1.5,0)
 +(0,2) -- +(0.5,2)
 +(0,-2) -- +(0.5,-2)
 +(0.25,-3) node {degenerazione $\epsilon=\frac {E_1+E_2} 2 \pm \delta$}
 ++(1.5,0)
 +(0,0) -- +(0.5,0);
\draw[densely dotted] 
 ++(0.75,0) 
 +(0,0) -- +(0.5,2)
 +(0,0) -- +(0.5,-2)
 ++(1.5,0)
 +(0,2) -- +(0.5,0)
 +(0,-2) -- +(0.5,0);
\draw[<->] (3.75,0) -- (3.75,2) node[midway,right] {$\delta$};
\begin{scope}[xshift=8cm]
\draw[el]
  +(0,-1.5) -- +(0.5,-1.5)
  ++(1.5,0)
  +(0,2) -- +(0.5,2)
  +(0,-2) -- +(0.5,-2)
  +(0.25,-3) node {perturbazione 
    $\left[\begin{matrix} \epsilon_1=E_1-k\\
    \epsilon_2=E_2+k \end{matrix} \right. $}
  ++(1.5,0)
  +(0,1.5) -- +(0.5,1.5);
\draw[densely dotted]
  ++(0.75,0)
  +(0,-1.5) -- +(0.5,2)
  +(0,-1.5) -- +(0.5,-2)
  ++(1.5,0)
  +(0,2) -- +(0.5,1.5)
  +(0,-2) -- +(0.5,1.5);
\draw[<->] (3.75,1.5) -- (3.75,2) node[midway,right] {$k$};
\end{scope}
\end{tikzpicture}

\end{frame}
\begin{frame}
  \frametitle{perchè...}
  \begin{align*}
    \epsilon &=\frac{E_1+E_2} 2 \pm \frac{E_2-E_1} 2
\sqrt{1+4\frac{\delta^2}{(E_2-E_1)^2}} 
\approx \left\{
\begin{matrix}
E_1-k\\
E_1+k\end{matrix}\right.
\\
    k &=\frac{\delta^2}{(E_2-E_1)}
  \end{align*}
\end{frame}
\end{document}

