\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}

\newcommand{\nz}[1]{{n^{(#1)}}}
\newcommand{\ez}[1]{{E^{(#1)}_n}}

\title{teoria delle perturbazioni statiche}
\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{problema}
  \begin{align*}
    H\Psi_n &= E_n \Psi_n \\
    (H+h) (\Psi_n+\psi_n) &=(E_n+e_n)(\Psi_n+\psi_n)
  \end{align*}

\end{frame}

\begin{frame} 
  \frametitle{approccio}
  \begin{align*}
    (H+\lambda h) \psi_n(\lambda) &= E_n(\lambda)\psi_n(\lambda) \\
    \psi_n(\lambda) &= |\nz 0 \rangle + \lambda |\nz 1 \rangle + \lambda^2 |\nz 2 \rangle... \\
    E(\lambda) &= \ez 0 + \lambda \ez 1 + \lambda^2 \ez 2 ...
  \end{align*}
\pause
ricostruendo
\begin{equation*}
  (H-\ez 0) | \nz 0 + \lambda \nz 1  ... \rangle =
  (\lambda \ez 1 + \lambda^2 \ez 2 ...)|\nz 0 + \lambda \nz 1 + ... \rangle
\end{equation*}
\end{frame}

\begin{frame} 
  \frametitle{soluzione}
\begin{equation*}
  (H-\ez 0) | \nz 0 + \lambda \nz 1  ... \rangle =
  (\lambda \ez 1 + \lambda^2 \ez 2 ...)|\nz 0 + \lambda \nz 1 + ... \rangle
\end{equation*}

come esprimere $|\nz {k>0}\rangle$?
\pause
\begin{equation*}
  |\nz k\rangle = \sum_{m \neq n} |m\rangle\langle m|\nz k\rangle
\end{equation*}
\pause 
chiudo con $\langle \nz 0|$ per ottenere:
\begin{align*}
  0=\langle \nz 0| H - \ez 0| \nz 0 + \lambda \nz 1 ...\rangle &=
  \langle \nz 0| \lambda \ez 1 + \lambda^2 \ez 2 ...| \nz 0 + \lambda \nz 1 ...\rangle \\
  \ez k &= \langle \nz 0 | h | \nz {k-1}
  \rangle
\end{align*}
\end{frame}

\begin{frame} 
  \frametitle{soluzione}
  \begin{equation*}
  \ez k = \langle \nz 0 | h | \nz {k-1}
  \end{equation*}
  al primo ordine:
  \begin{equation*}
    \ez 1 = \langle \nz 0 | h | \nz 0 \rangle
  \end{equation*}
  per ottenere gli ordini superiori bisogna scalare le relazioni $\nz k \rightarrow \ez {k+1}
  \rightarrow \nz {k+1} ... $. Per ottenere le relazioni per $\nz k$ dobbiamo chiudere con
  $\langle k=k^{(0)}|$.
  \begin{align*}
    \langle k | H - \ez 0| \nz 1\rangle &=
    - \langle k | h | \nz 0\rangle \\
    \langle k | \nz 1\rangle &= \frac {\langle k | h | \nz 0\rangle} {\ez 0 - E_k^{(0)}} \implies \\
    | \nz 1 \rangle &= \sum_k |k\rangle \frac {\langle k | h | n\rangle}{\ez 0 - E_k^{(0)}}
  \end{align*}
\end{frame}

\begin{frame} 
  \frametitle{soluzione}
  \begin{align*}
    | \nz 1 \rangle &= \sum_k \frac {\langle k | h | n\rangle}{\ez 0 - E_k^{(0)}} \\
    \ez 2 &=
    \langle \nz 0 | h |\nz 1\rangle = \sum_k \frac
    {\langle k | h | n\rangle \langle n | h | k\rangle}
    {\ez 0 - E_k^{(0)}}
  \end{align*}
\end{frame}

\begin{frame} 
  \frametitle{MPi}
  calcolo perturbativo dell'energia di correlazione
  \begin{align*}
    F \Psi^{\textrm{monodet}}_n &{=} E^{\textrm{noncorr}}_n \Psi_n \\
    (F + (H-F)) \Phi_n &\stackrel{?}{=} E^{\textrm{corr}}_n \Phi_n
  \end{align*}
\end{frame}

\end{document}


