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\begin{document}

\title{Project 3,  FYS-KJM4480, Fall 2009}
%\author{}
\maketitle
\section*{Project 3, deadline 12pm Friday November 20}
This is the last project. We present a simplified Hamiltonian (which has
many common traits with the Lipkin model of project 1) consisting of  an unperturbed Hamiltonian and a so-called 
 pairing interaction term. It is a model which to a large extent mimicks some central features of
atomic nuclei, certain atoms and systems which exhibit superfluiditity or superconductivity.  To study this system, 
we will use a mix of many-body perturbation theory, Hartree-Fock theory and the configuration interaction method. The latter will also provide us with the exact answer. 
When setting up the Hamiltonian matrix you will need to solve an eigenvalue problem (as you did for the Lipkin model). This can easily be done with either
octave or Matlab. 

We define first the Hamiltonian, with a definition of the model space and
the single-particle basis. Thereafter, we present the various exercises.
\subsection*{Introduction and Hamiltonian}


The Hamiltonian acting in the complete Hilbert space (usually infinite
dimensional) consists of an unperturbed one-body part, $\hat{H}_0$,
and a perturbation $\hat{V}$. 

We limit ourselves to at most two-body interactions, our Hamiltonian  is 
then represented by the following operators
\[
\hat{H} = \sum_{\alpha\beta}\langle \alpha |h_0|\beta\rangle a_{\alpha}^{\dagger}a_{\beta} +\frac{1}{4}\sum_{\alpha\beta\gamma\delta}\langle \alpha\beta| V|\gamma\delta\rangle a_{\alpha}^{\dagger}a_{\beta}^{\dagger}a_{\delta}a_{\gamma},
\]
where $a_{\alpha}^{\dagger}$ and $a_{\alpha}$ etc.~are standard fermion creation and annihilation operators, respectively,
and $\alpha\beta\gamma\delta$ represent all possible single-particle quantum numbers. 
The full single-particle space is defined by the completeness relation
$\hat{{\bf 1}} = \sum_{\alpha =1}^{\infty}|\alpha \rangle \langle \alpha|$.
In our calculations  we will let  the single-particle states $|\alpha\rangle$
be eigenfunctions of  the one-particle operator $\hat{h}_0$. 


The above Hamiltonian 
acts in turn on various many-body Slater determinants constructed from the single-basis defined by the one-body
operator $\hat{h}_0$.    
As an example, 
the two-particle model space $\mathcal{P}$ is defined by an operator 
\[
\hat{P} =   \sum_{\alpha\beta =1}^{m}|\alpha\beta \rangle \langle \alpha\beta|,
\]
where we assume that $m=\dim(\mathcal{P})$ and the full space is defined by
\[
\hat{P}+\hat{Q}=\hat{{\bf 1}},
\]
with  the projection operator 
\[
\hat{Q} =   \sum_{\alpha\beta =m+1}^{\infty}|\alpha\beta \rangle \langle \alpha\beta|,
\]
being the complement of $\hat{P}$. 


Our specific model consists of $N$ doubly-degenerate and equally spaced
single-particle levels labelled by $p=1,2,\dots$ and spin $\sigma=\pm
1$.  These states are schematically portrayed in
Fig.~\ref{fig:schematic}.  The first two single-particle levels
define a possible model space, indicated by the label $\mathcal{P}$.  
The
remaining states span the excluded space $\mathcal{Q}$.

We write
the Hamiltonian as 
\[ \hat{H} = \hat{H}_0 + \hat{V} , \]
where
\[
\hat{H}_0=\xi\sum_{p\sigma}(p-1)a_{p\sigma}^{\dagger}a_{p\sigma}
\]
and 
\[
\hat{V}=-\frac{1}{2}g\sum_{pq}a^{\dagger}_{p+}
a^{\dagger}_{p-}a_{q-}a_{q+}.
\]
Here, $H_0$ is the unperturbed Hamiltonian with a spacing between
successive single-particle states given by $\xi$, which we will set to
a constant value $\xi=1$ without loss of generality. The two-body
operator $\hat{V}$ has one term only. It represents the
pairing contribution and carries a constant strength $g$. The indices
$\sigma=\pm$ represent the two possible spin values. The 
interaction can only couple pairs and excites therefore only two
particles at the time, as indicated by the rightmost four-particle
state in Fig.~\ref{fig:schematic}. There one of the  pairs is excited to the
state with $p=9$ and the other to the state $p=7$. The two middle possibilities are not possible with the 
present model.  
We label single-particle states within the model space as
hole-states. The single-particle states outside the model space are
then particle states. The model is not so different from the Lipkin model studied in project 1.

In our model we have kept both the interaction strength and the single-particle level as constants.
In a realistic system like an atom or the atomic  nucleus this is not the case. 





\begin{figure*}[htbp]
\vspace{1.0cm}
 \setlength{\unitlength}{1cm}
 \begin{picture}(15,14)
 \thicklines
\put(-0.6,1){\makebox(0,0){$p=1$}}
\put(-0.6,2){\makebox(0,0){$p=2$}}
\put(-0.6,3){\makebox(0,0){$p=3$}}
\put(-0.6,4){\makebox(0,0){$p=4$}}
\put(-0.6,5){\makebox(0,0){$p=5$}}
\put(-0.6,6){\makebox(0,0){$p=6$}}
\put(-0.6,7){\makebox(0,0){$p=7$}}
\put(-0.6,8){\makebox(0,0){$p=8$}}
\put(-0.6,9){\makebox(0,0){$p=9$}}
\put(-0.6,10){\makebox(0,0){$p=10$}}
\put(-0.6,11){\makebox(0,0){$p=\dots$}}
\put(16,8){\makebox(0,0){$\mathcal{Q}$}}
\put(16,2){\makebox(0,0){$\mathcal{P}$}}
% first 4-particle state
\put(0.8,1){\circle*{0.3}}
\put(0.8,2){\circle*{0.3}}
\put(1.7,1){\circle*{0.3}}
\put(1.7,2){\circle*{0.3}}
% second 4-particle state
\put(5.0,1){\circle*{0.3}}
\put(5.9,1){\circle*{0.3}}
\put(5.0,4){\circle*{0.3}}
\put(5.0,2){\circle{0.3}}
\put(5.9,3){\circle*{0.3}}
\put(5.9,2){\circle{0.3}}
% third 4-particle state
\put(9.2,1){\circle*{0.3}}
\put(10.1,3){\circle*{0.3}}
\put(9.2,4){\circle*{0.3}}
\put(10.1,8){\circle*{0.3}}
\put(10.1,1){\circle{0.3}}
\put(9.2,2){\circle{0.3}}
\put(10.1,2){\circle{0.3}}
% third 4-particle state
\put(13.4,7){\circle*{0.3}}
\put(14.3,7){\circle*{0.3}}
\put(13.4,9){\circle*{0.3}}
\put(14.3,9){\circle*{0.3}}
\put(13.4,1){\circle{0.3}}
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\put(14.3,2){\circle{0.3}}
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\dashline[+1]{2.5}(0,2)(15,2)
\dashline[+1]{2.5}(0,3)(15,3)
\dashline[+1]{2.5}(0,4)(15,4)
\dashline[+1]{2.5}(0,5)(15,5)
\dashline[+1]{2.5}(0,6)(15,6)
\dashline[+1]{2.5}(0,7)(15,7)
\dashline[+1]{2.5}(0,8)(15,8)
\dashline[+1]{2.5}(0,9)(15,9)
\dashline[+1]{2.5}(0,10)(15,10)
\thinlines
\dashline{0.1}(0,2.5)(15,2.5)
\dashline{0.1}(0,11)(15,11)
 \end{picture}
\caption{Schematic plot of the possible single-particle levels with double degeneracy.
The filled circles indicate occupied particle states while the empty circles 
represent vacant particle(hole) states.
The spacing between each level $p$ is constant in this picture. 
The first two single-particle levels define our possible model space, indicated by the label $\mathcal{P}$.
The remaining states span the excluded space $\mathcal{Q}$. 
The first state to the left represents
a possible ground state representation for a four-fermion system. In the second state to the left,
one pair is broken. This possibility is however not included in our interaction. \label{fig:schematic}}
\end{figure*}

\subsection*{Exercises}
\begin{enumerate}
\item Show that the  
unperturbed Hamiltonian  $\hat{H}_0$ and $\hat{V}$ commute
with both the spin projection $\hat{S}_z$ and the total spin
$\hat{S}^2$, given by
\[
  \hat{S}_z := \frac{1}{2}\sum_{p\sigma} \sigma a^\dag_{p\sigma}a_{p\sigma}
\]
and
\[
  \hat{S}^2 := \hat{S}_z^2 + \frac{1}{2}(\hat{S}_+\hat{S}_- +
  \hat{S}_-\hat{S}_+),
\]
where
\[
  \hat{S}_\pm := \sum_{p} a^\dag_{p\pm} a_{p\mp}.
\]

This is an important feature of our system that allows us to block-diagonalize
the full Hamiltonian. We will focus on total spin $S=0$.
In this case, it is convenient to define the so-called pair creation and pair
annihilation operators
\[
\hat{P}^{+}_p = a^\dag_{p+}a^\dag_{p-},
\]
and
\[
\hat{P}^{-}_p = a_{p-}a_{p+},
\] 
respectively.

Show that you can rewrite the Hamiltonian (with $\xi=1$) as  
\[
\hat{H}=\sum_{p\sigma}(p-1)a_{p\sigma}^{\dagger}a_{p\sigma}
-\frac{1}{2}g\sum_{pq}\hat{P}^{+}_p\hat{P}^{-}_q.
\]
Show also that Hamiltonian commutes with the product of the pair creation and annihilation operators.
This model corresponds to a system with no broken pairs. This means that the Hamiltonian can only link two-particle states in so-called spin-reversed states. 


\item Construct thereafter the Hamiltonian matrix for a system with no broken pairs and spin $S=0$ for the case of the four lowest single-particle levels  
indicated in the Fig.~\ref{fig:schematic}. Our system consists of four particles only.
Our single-particle space consists of only the four lowest levels 
$p=1,2,3,4$.  You need to set up all possible Slater determinants. 
Find all eigenvalues by diagonalizing the Hamiltonian matrix.
Vary your results for values of $g\in [-1,1]$. 
We  refer to this as the exact calculation. Comment the behavior of the ground state as function of $g$. 
\item
Instead of setting up all possible Slater determinants, construct only an approximation to the ground state
(where we assume that the four particles are in the two lowest single-particle orbits only)
which includes at most two-particle-two-hole excitations. Diagonalize this matrix and compare with the 
exact calculation and comment your results. Can you set up which diagrams this approximation corresponds to?
\item Hereafter we will define our model space to consist of the single-particle levels $p=1,2$. 
The remaining levels $p=3,4$ define our excluded space. 
This means that our ground state Slater determinant consists of four particles which can be placed in the doubly
degenerate orbits $p=1$ and $p=2$. 

We will now study the system using non-degenerate Rayleigh-Schr\"odinger perturbation theory to third order in the interaction.
If we exclude the first order contribution, all 
possible diagrams (Hugenholz diagrams where the vertices have been opened)  are shown in Fig.~\ref{fig:diagrams}.
\begin{figure}[hbtp]
\includegraphics[width=.6\textwidth]{diagrams.eps}
\caption{Diagrams to third order in the interaction. The first order term is excluded.\label{fig:diagrams}}
\end{figure}

Based on the form of the interaction, which diagrams contribute to the binding energy of the ground state?
Write down the expressions for the diagrams that contribute and find the contribution to the ground state
energy as function $g\in [-1,1]$. Comment your results.  Compare these results with those you obtained
in 2) and 3).
\item The diagrams with only two single particle states as intermediate states (for example diagrams 1 and 4
in Fig.~\ref{fig:diagrams})
can be summed to infinite order since they can be expressed as a geometric series.
Find this contribution and compare the final energy with the results from 2) and 3). Comment your results. 
You can also perform a resummation of diagrams like diagram 5 with hole lines as intermediate states only between
various vertices. Can you find this result as well? Compare now the final results with the resummed
two-particle and two-hole diagrams with the results from 2) and 3).
 
\item 
We will now set up the Hartree-Fock equations by varying the coefficients of the 
single-particle functions. The single-particle basis functions
are defined as 
\[
\psi_p  = \sum_{\lambda} C_{p\lambda}\psi_{\lambda}.
\]
where in our case $p=1,2,3,4$ and $\lambda=1,2,3,4$, that is the first four lowest single-particle orbits
of Fig.~\ref{fig:schematic}.
Set up the Hartree-Fock equations for this system by varying the coefficients $C_{p\lambda}$ 
and solve them for values of $g\in [-1,1]$. 
Comment your results and compare with the exact solution. Discuss also which diagrams in Fig.~\ref{fig:diagrams}
that can be affected by a Hartree-Fock basis. Compute the total binding energy using a Hartree-Fock basis and
comment your results. 
\item To fourth order in perturbation theory we can produce diagrams with so-called four-particle-four-hole
excitations. An example is given in Fig.~\ref{fig:fourthorder}.
\begin{figure}[hbtp]
\includegraphics[width=.4\textwidth]{fig8.ps}
\caption{An example of a fourth-order diagram with an intermediate state involving four-particle-four-hole 
excitations.
\label{fig:fourthorder}}
\end{figure}
Find the contribution to the binding energy of the ground state from this type of contributions and compare with your previous results with and without a Hartree-Fock basis. Discuss in particular the connection with the results in exercise 2) with the full diagonalization where Slater determinants involving 
four-particle-four-hole excitations are involved.
\item When summing over all intermediate states in diagram 1 or 4 of Fig.~\ref{fig:diagrams}, we have limited
the sum over intermedate particle states to include the states $p=3$ and $p=4$ only. Compute this sum by taking the limit $p=\infty$. Comment your results.
\end{enumerate}



 \end{document}


