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\begin{document}

\title{Midterm exam FYS-KJM4480, Fall 2011}
%\author{}
\maketitle
\section*{Deadline 12pm Friday October 14. Use only your candidate number. It counts 30\% of the final grade.}
We present a simplified Hamiltonian 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 Hartree-Fock theory and exact diagonalization. 
The latter will also provide us with the exact answer. 
When setting up the Hamiltonian matrix you will need to solve an eigenvalue problem. 
This can easily be done with either
octave or Matlab. 

We define first the Hamiltonian, with a definition of 
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 two-body interaction  $\hat{V}$. 

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-particle basis defined by the one-body
operator $\hat{h}_0$.    

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
will define the states up to the Fermi level (our hole states), 
indicated by the label $\mathcal{F}$.  
The
remaining states are particle (virtual) states.

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.  

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){Particles}}
\put(16,2){\makebox(0,0){$\mathcal{F}$}}
% 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 hole states, indicated by the label $\mathcal{F}$.
The remaining states are particle states. 
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 
Our system will now consist of four particles only and the 
single-particle space consists of only the four lowest levels 
$p=1,2,3,4$ in the Fig.~\ref{fig:schematic}. Define the Slater determinant 
ansaztz for the ground state.  Find the reference energy $E_0$ and write down the diagrams which contribute.
Make sure that the two-body matrix elements are properly anti-symmetrized.

\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}.  Write down the diagrams which contribute.
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 for $g\in [-1,1]$ and comment your results. Can you set up which diagrams this approximation corresponds to?  Do you get contributions of the type one-particle-two-hole?
 
\item Set up (the general expressions) the Hartree-Fock equations in second quantization. 
Write down the corresponding diagrams and find the expression for the Hartree-Fock operator.
Discuss also the stability of the Hartree-Fock ground state energy.  
Insert then our model and find the expressions for the Hartree-Fock operator. Comment your results.

\item
We will now set up the Hartree-Fock equations by varying the coefficients of the 
single-particle functions (one of the standard ways of solving these equations). 
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, in particular in the light of the discussion in the previous exercise. 
Compare with the exact solution. Compute the total binding energy using a Hartree-Fock basis and
comment your results. 
\end{enumerate}



 \end{document}


