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\section*{Second midterm project fall 2013 FYS-KJM4480/9480, deadline November 25 at 12pm (noon). It counts 30\% of the final mark.  Use your student number only. }


{\bf Hand in an individual answer only with your student number. The
  box for handing in the answer is at the reception of the Department
  of Physics (see letter box marked fys-kjm4480/9480.)}\newline\newline
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
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.

We define first the Hamiltonian, with a definition of the model space
and the single-particle basis. Thereafter, we present the various
exercises.


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}} \put(14.3,1){\circle{0.3}}
\put(13.4,2){\circle{0.3}} \put(14.3,2){\circle{0.3}}
\dashline[+1]{2.5}(0,1)(15,1) \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. All interaction vertices represent anti-symmetrized matrix elements.\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 
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 full diagonalization where Slater determinants involving
four-particle-four-hole excitations are involved.
\end{enumerate}



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