\documentclass[prc]{revtex4}
\usepackage[dvips]{graphicx}
\usepackage{mathrsfs}
\usepackage{amsfonts}
\usepackage{lscape}

\usepackage{epic,eepic}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[dvips]{epsfig}
\usepackage[T1]{fontenc}
\usepackage{hyperref}
\usepackage{bezier}
\usepackage{pstricks}
\usepackage{dcolumn}% Align table columns on decimal point
\usepackage{bm}% bold math
%\usepackage{braket}
\usepackage[dvips]{graphicx}
\usepackage{pst-plot}

\newcommand{\One}{\hat{\mathbf{1}}}
\newcommand{\eff}{\text{eff}}
\newcommand{\Heff}{\hat{H}_\text{eff}}
\newcommand{\Veff}{\hat{V}_\text{eff}}
\newcommand{\braket}[1]{\langle#1\rangle}
\newcommand{\Span}{\operatorname{sp}}
\newcommand{\tr}{\operatorname{trace}}
\newcommand{\diag}{\operatorname{diag}}
\newcommand{\bra}[1]{\left\langle #1 \right|}
\newcommand{\ket}[1]{\left| #1 \right\rangle}
\newcommand{\element}[3]
    {\bra{#1}#2\ket{#3}}

\newcommand{\normord}[1]{
    \left\{#1\right\}
}

\usepackage{amsmath}
%\newcommand{\be}{\begin{equation}}
%\newcommand{\ee}{\end{equation}}
%\newcommand{\OP}[1]{{\bf\widehat{#1}}}

\begin{document}
\section*{Midterm project fall 2010 FYS-KJM4480, deadline Friday 15 at 12pm (noon). It counts 30\% of the final mark.  Use your student number only. }

In this project we will develop two simple models for studying the 
helium atom (with two electrons) and the beryllium atom with four electrons.

After having introduced the  Born-Oppenheimer approximation which effectively freezes out the nucleonic degrees
of freedom, the Hamiltonian for $N$ electrons takes the following form 
\[
  \hat{H} = \sum_{i=1}^{N} t(x_i) 
  - \sum_{i=1}^{N} k\frac{Ze^2}{r_i} + \sum_{i<j}^{N} \frac{ke^2}{r_{ij}},
\]
with $k=1.44$ eVnm. Througout this work we will use atomic units, this means
that $\hbar=c=e=m_e=1$. The constant $k$ becomes also equal 1. 
The resulting energies have to be multiplied by $2\times 13.6$ eV
in order to obtain energies in eletronvolts.

 We can rewrite our Hamiltonians as
\begin{equation}
    \hat{H} = \hat{H_0} + \hat{H_I} 
    = \sum_{i=1}^{N}\hat{h}_0(x_i) + \sum_{i<j}^{N}\frac{1}{r_{ij}},
\label{H1H2}
\end{equation}
where  we have defined $r_{ij}=| {\bf r}_i-{\bf r}_j|$ and
$\hat{h}_0(x_i) =  \hat{t}(x_i) - \frac{Z}{r_i}$
The variable $x$ contains both the spatial coordinates and the spin values.
The first term of Eq.~(\ref{H1H2}), $H_0$, is the sum of the $N$
\emph{one-body} Hamiltonians $\hat{h}_0$. Each individual
Hamiltonian $\hat{h}_0$ contains the kinetic energy operator of an
electron and its potential energy due to the attraction of the
nucleus. The second term, $H_I$, is the sum of the $N(N-1)/2$
two-body interactions between each pair of electrons. Note that the double sum carries a restriction $i<j$.

As basis functions for our calculations we will use hydrogen-like single-particle functions.  This means the onebody operator is diagonal in this basis for states $i,j$ with quantum numbers $nlm_lsm_s$ with  
energies 
\[\langle i|\hat{h}_0| j\rangle =  -Z^2/2n^2\delta_{i,j}.\]  
The quantum number $n$ refers to the number of nodes 
of the wave function.  Observe that this expectation value is independent of spin.

We will in all calculations here restrict ourselves to only so-called $s$ -waves,
that is the orbital momentum $l$ is zero. We will also limit the quantum number $n$ to $n\le 4$.  It means that every $ns$ state can accomodate two electrons due to the spin degeneracy. This is illustrated in Fig.~\ref{fig:fig1} here.
\begin{figure}[hbtp]
\vspace{1.0cm}
 \setlength{\unitlength}{1cm}
 \begin{picture}(15,5)
 \thicklines
\put(-0.6,1){\makebox(0,0){$1s$}}
\put(-0.6,2){\makebox(0,0){$2s$}}
\put(-0.6,3){\makebox(0,0){$3s$}}
\put(-0.6,4){\makebox(0,0){$4s$}}
% first 2-particle state
\put(0.8,1){\circle*{0.3}}
\put(1.7,1){\circle*{0.3}}
% second 2-particle state
\put(5.0,2){\circle*{0.3}}
\put(5.9,2){\circle*{0.3}}
% third 2-particle state
\put(9.2,1){\circle*{0.3}}
\put(10.1,3){\circle*{0.3}}
% third 2-particle state
\put(13.4,2){\circle*{0.3}}
\put(14.3,4){\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)
 \end{picture}
\caption{Schematic plot of the possible single-particle levels with double degeneracy.
The filled circles indicate occupied particle states.
We show some possible two-particle states which can describe states in the helim atom. \label{fig:fig1}}
\end{figure}


In the calculations you 
will need the Coulomb interaction with matrix elements
involving single-particle wave functions with $l=0$ only, the so-called $s$-waves.
We need only the radial part since the 
spherical harmonics for the $s$-waves are rather simple. We omit single-particle states with $l> 0$.
Our radial wave functions are
\[
R_{n0}(r)=\left(\frac{2Z}{n}\right)^{3/2}\sqrt{\frac{(n-1)!}{2n\times n!}}L_{n-1}^1(\frac{2Zr}{n})\exp{(-\frac{Zr}{n})},
\]
where $L_{n-1}^1(r)$ are the so-called Laguerre polynomials.
These wave functions can then be used to compute the direct part of the
Coulomb interaction
\[
\langle \alpha\beta| V| \gamma\delta\rangle = \int r_1^2dr_1 \int r_2^2dr_2R_{n_{\alpha}0}^*(r_1) R_{n_{\beta}0}^*(r_2) 
  \frac{1}{| {\bf r}_1-{\bf r}_2|}R_{n_{\gamma}0}(r_1)R_{n_{\delta}0}(r_2)
\]
Observe that this is only the radial integral and that the labels $\alpha\beta\gamma\delta$ refer only to the quantum numbers $nlm_l$, with $m_l$ the projection of the orbital momentum $l$. 
A similar expression can be found for the exchange part. Since we have restricted ourselves to only $s$-waves, these integrals are straightforward but tedious to calculate. As an addendum to this project we list all closed-form expressions for the relevant matrix elements. Note well that these matrix elements do not include spin. When setting up the final antisymmetrized matrix elements you need to consider the spin degrees of freedom as well. Please pay in particular special attention to the exchange part and the pertinent spin values of the single-particle states.


We will also, for both helium and beryllium assume that the many-particle states we construct have always the same total spin projection $M_S=0$. This means that if we excite one or two particles from the ground state, the spins of the various single-particle states should always sum up to zero. 

\begin{enumerate}
\item[1a)] We start with the helium atom and define our single-particle Hilbert space to consist of the single-particle orbits $1s$, $2s$, $3s$ and $4s$, with their corresponding spin degeneracies, see Fig.~\ref{fig:fig1}. 

Set up the ansatz for the ground state $|c\rangle = |\Phi_0\rangle$ in second 
quantization and define a table of single-particle states. Construct thereafter
all possible one-particle-one-hole excitations  $|\Phi_i^a\rangle$ where $i$ refer to levels below the Fermi level (define this level) and $a$ refers to particle states. Define particles and holes. The Slater determinants have to be written in terms of the respective creation and annihilation operators.
The states you construct should all have total spin projection $M_S=0$. 
Construct also all possible two-particle-two-hole states $|\Phi_{ij}^{ab}\rangle$  in a second quantization representation. 


\item[1b)]  Define the Hamiltonian in a second-quantized form and use this to
compute the expectation value of the ground state (defining the so-called reference energy and later our Hartree-Fock functional) of the helium atom. 
Show that it is given by
\[
  E[\Phi_0] = \langle c | \hat{H}| c \rangle 
  = \sum_{i} \langle i | \hat{h}_0 | i\rangle+ \frac{1}{2}\sum_{ij}\left[\langle ij |\frac{1}{r}|ij\rangle-\langle ij |\frac{1}{r}|ji\rangle\right].
\]
Define properly the sums keeping in mind that the states $ij$ refer to all
quantum numbers $nlm_lsm_s$.
Use the values for the various matrix elements listed at the end of the project to find the value of $E$
as function of $Z$ and compute the minimum of $E$ as function of $Z$.
Can you give a physical interpretation to the value of $Z$ where the energy is a minimum? 

\item[1c)]
Hereafter we will limit ourselves to a system which now contains only one-particle-one-hole
excitations beyond the chosen state $|c\rangle$.
Using the possible Slater determinants from exercise a) for the helium atom,   
compute also the expectation values (without inserting the explicit values for the matrix elements first) of 
\[
\langle c | \hat{H}| \Phi_i^a \rangle,
\] 
and 
\[
\langle \Phi_i^a | \hat{H}| \Phi_j^b \rangle.
\]
Represent these expectation values in a diagrammatic form, both for the onebody part and the two-body part of the Hamiltonian. 
 
Insert then the explicit values for the various matrix elements and 
set up the final Hamiltonian matrix and diagonalize it using for example
Octave, Matlab, Python, C++ or Fortran as programming tools.

Compare your results from those of exercise b) and comment your results. 
The exact energy with our Hamiltonian is $-2.9037$ atomic units for helium. This value is also close to the experimental energy.
\item[1d)] We repeat exercises b) and c) but now for the beryllium atom.
Define the ansatz for $|c\rangle$ and limit yourself again to one-particle-one-hole excitations.   Compute the reference energy 
$\langle c | \hat{H}| c \rangle $ and vary it as function of $Z$. Find the minimum.  Thereafter you will need to set up the appropriate Hamiltonian matrix
which involves also one-particle-one-hole excitations. Diagonalize this matrix
and compare your eigenvalues with the minimum of $\langle c | \hat{H}| c \rangle$ as function of $Z$ and comment your results. 
The exact energy with our Hamiltonian is $-14.6674$ atomic units for beryllium. This value is again close to the experimental energy.\newline\newline\newline
With a given energy functional, we can perform at least two types of variational strategies.
\begin{itemize}
\item Vary the Slater determinant by changing the spatial part of the single-particle
wave functions themselves. 
\item   Expand the single-particle functions in a known basis  and vary the coefficients, 
that is, the new function single-particle wave function $|p\rangle$ is written as a linear expansion
in terms of a fixed basis $\phi$ (harmonic oscillator, Laguerre polynomials etc)
\[
\psi_p  = \sum_{\lambda} C_{p\lambda}\phi_{\lambda},
\]
 \end{itemize}
Both cases lead to a new Slater determinant which is related to the previous via  a unitary transformation.
Below we will set up the Hartree-Fock equations using the second option.  
We assume that our basis is still formed by the hydrogen-like wave functions. 


\item[1e)] Consider a Slater determinant built up of single-particle orbitals $\phi_{\lambda}$ where the indices $\lambda$ refer to specific single-particle states.  As an example, you could think of the ground state ansatz for the Beryllium atom. 

The unitary transformation
\[
\psi_p  = \sum_{\lambda} C_{p\lambda}\phi_{\lambda},
\]
brings us into the new basis $\psi$.  Show that the new basis is orthonormal.
Show that the new Slater determinant constructed from the new single-particle wave functions can be
written as the determinant based on the previous basis and the determinant of the matrix $C$.
Show that the old and the new Slater determinants are equal up to a complex constant with absolute value unity.
(Hint, $C$ is a unitary matrix).   Which assumptions are you making in order to obtain this result? 


\item[1f)]
Minimizing with respect to $C^*_{p\alpha}$, remembering that $C^*_{p\alpha}$ and $C_{p\alpha}$ (and that the indices contain all single-particle quantum numbers including spin)
are independent and defining
\[
h_{\alpha\gamma}^{HF}=\langle \alpha | h | \gamma \rangle+
\sum_{p}\sum_{\beta\delta} C^*_{p\beta}C_{p\delta}\langle \alpha\beta|V|\gamma\delta\rangle_{AS},
\]
show that you can write the Hartree-Fock  equations as 
\[
\sum_{\gamma}h_{\alpha\gamma}^{HF}C_{p\gamma}=\epsilon_p^{\mathrm{HF}}C_{p\alpha}.
\label{eq:newhf}
\]
Explain the meaning of the different terms and define the Hartree-Fock 
operator in second quantization. Write down its diagrammatic representation as well.  The greek letters refer to the wave functions in the original basis (in our case the hydrogen-like wave functions) while roman letters refer to the new basis. 
\item[1g)]
The Hartree-Fock equations with a variation of the coefficients $C_{p\alpha}$
lead to an eigenvalue problem whose eigenvectors are the coefficients
$C_{p\alpha}$ and eigenvalues are the new single-particle energies. 
Use the single-particle orbits $1s-4s$ and set up the Hartree-Fock matrix for both the helium atom and the beryllium atom. Find after the first diagonalization the new single-particle energies and the new ground state energy.
Compare these results with  those you obtained under the minimization of the ground states as functions of $Z$ and the full diagonalization. When setting up the Hartree-Fock matrix 
in the first iteration, our guess for the coefficients $C_{p\beta}$ etc.~is $C_{p\beta}=1$ for $p=\beta$ and zero else. 
\item[1h)]
The final stage is to 
set up an iterative scheme where you use the new wave functions
determined via the coefficients $C_{p\alpha}$ to solve iteratively the Hartree-Fock equations till a given self-consistency is reached. A typical way of doing this is to compare the single-particle energies from the previous iteration with those obtained from the new diagonalization. If the total difference is smaller than a prefixed value, the iterative process is stopped. 

Compare these results with the those you obtained under the minimization of the ground states as functions of $Z$ and the full diagonalization. Discuss your results. 


\end{enumerate}

We conclude by listing the matrix elements for the radial integrals to be used for the direct part and the exchange part. Note again that these integrals do not include spin. The nomenclature is $1=1s$, $2=2s$, $3=3s$ and $4=4s$, with no
spin degrees of freedom. 
\begin{verbatim}
<11|V|11> = (5*Z)/8
<11|V|12> = (4096*Sqrt[2]*Z)/64827
<11|V|13> = (1269*Sqrt[3]*Z)/50000
<11|V|14> = (416415744*Z)/15083778125
<11|V|21> = (4096*Sqrt[2]*Z)/64827
<11|V|22> = (16*Z)/729
<11|V|23> = (110592*Sqrt[6]*Z)/24137569
<11|V|24> = (98304*Sqrt[2]*Z)/19487171
<11|V|31> = (1269*Sqrt[3]*Z)/50000
<11|V|32> = (110592*Sqrt[6]*Z)/24137569
<11|V|33> = (189*Z)/32768
<11|V|34> = (1808400384*Sqrt[3]*Z)/852891037441
<11|V|41> = (416415744*Z)/15083778125
<11|V|42> = (98304*Sqrt[2]*Z)/19487171
<11|V|43> = (1808400384*Sqrt[3]*Z)/852891037441
<11|V|44> = (22848*Z)/9765625
<12|V|11> = (4096*Sqrt[2]*Z)/64827
<12|V|12> = (17*Z)/81
<12|V|13> = (1555918848*Sqrt[6]*Z)/75429903125
<12|V|14> = (288456704*Sqrt[2]*Z)/14206147659
<12|V|21> = (16*Z)/729
<12|V|22> = (512*Sqrt[2]*Z)/84375
<12|V|23> = (2160*Sqrt[3]*Z)/823543
<12|V|24> = (376832*Z)/129140163
<12|V|31> = (110592*Sqrt[6]*Z)/24137569
<12|V|32> = (29943*Sqrt[3]*Z)/13176688
<12|V|33> = (1216512*Sqrt[2]*Z)/815730721
<12|V|34> = (423788544*Sqrt[6]*Z)/762939453125
<12|V|41> = (98304*Sqrt[2]*Z)/19487171
<12|V|42> = (108838912*Z)/44840334375
<12|V|43> = (10148806656*Sqrt[6]*Z)/19073486328125
<12|V|44> = (39*Z)/(32768*Sqrt[2])
<13|V|11> = (1269*Sqrt[3]*Z)/50000
<13|V|12> = (1555918848*Sqrt[6]*Z)/75429903125
<13|V|13> = (815*Z)/8192
<13|V|14> = (11694850770862080*Sqrt[3]*Z)/702392443647273463
<13|V|21> = (110592*Sqrt[6]*Z)/24137569
<13|V|22> = (2160*Sqrt[3]*Z)/823543
<13|V|23> = (37826560*Sqrt[2]*Z)/22024729467
<13|V|24> = (487489536*Sqrt[6]*Z)/762939453125
<13|V|31> = (189*Z)/32768
<13|V|32> = (1216512*Sqrt[2]*Z)/815730721
<13|V|33> = (617*Z)/(314928*Sqrt[3])
<13|V|34> = (30254432256*Z)/41426511213649
<13|V|41> = (1808400384*Sqrt[3]*Z)/852891037441
<13|V|42> = (10148806656*Sqrt[6]*Z)/19073486328125
<13|V|43> = (90581886173184*Z)/129457847542653125
<13|V|44> = (74450880*Sqrt[3]*Z)/285311670611
<14|V|11> = (416415744*Z)/15083778125
<14|V|12> = (288456704*Sqrt[2]*Z)/14206147659
<14|V|13> = (11694850770862080*Sqrt[3]*Z)/702392443647273463
<14|V|14> = (22513*Z)/390625
<14|V|21> = (98304*Sqrt[2]*Z)/19487171
<14|V|22> = (376832*Z)/129140163
<14|V|23> = (487489536*Sqrt[6]*Z)/762939453125
<14|V|24> = (5053*Z)/(3538944*Sqrt[2])
<14|V|31> = (1808400384*Sqrt[3]*Z)/852891037441
<14|V|32> = (423788544*Sqrt[6]*Z)/762939453125
<14|V|33> = (30254432256*Z)/41426511213649
<14|V|34> = (1243165779*Sqrt[3]*Z)/4564986729776
<14|V|41> = (22848*Z)/9765625
<14|V|42> = (39*Z)/(32768*Sqrt[2])
<14|V|43> = (74450880*Sqrt[3]*Z)/285311670611
<14|V|44> = (1804351488*Z)/6179146071875
<21|V|11> = (4096*Sqrt[2]*Z)/64827
<21|V|12> = (16*Z)/729
<21|V|13> = (110592*Sqrt[6]*Z)/24137569
<21|V|14> = (98304*Sqrt[2]*Z)/19487171
<21|V|21> = (17*Z)/81
<21|V|22> = (512*Sqrt[2]*Z)/84375
<21|V|23> = (29943*Sqrt[3]*Z)/13176688
<21|V|24> = (108838912*Z)/44840334375
<21|V|31> = (1555918848*Sqrt[6]*Z)/75429903125
<21|V|32> = (2160*Sqrt[3]*Z)/823543
<21|V|33> = (1216512*Sqrt[2]*Z)/815730721
<21|V|34> = (10148806656*Sqrt[6]*Z)/19073486328125
<21|V|41> = (288456704*Sqrt[2]*Z)/14206147659
<21|V|42> = (376832*Z)/129140163
<21|V|43> = (423788544*Sqrt[6]*Z)/762939453125
<21|V|44> = (39*Z)/(32768*Sqrt[2])
<22|V|11> = (16*Z)/729
<22|V|12> = (512*Sqrt[2]*Z)/84375
<22|V|13> = (2160*Sqrt[3]*Z)/823543
<22|V|14> = (376832*Z)/129140163
<22|V|21> = (512*Sqrt[2]*Z)/84375
<22|V|22> = (77*Z)/512
<22|V|23> = (5870679552*Sqrt[6]*Z)/669871503125
<22|V|24> = (31363072*Sqrt[2]*Z)/4202539929
<22|V|31> = (2160*Sqrt[3]*Z)/823543
<22|V|32> = (5870679552*Sqrt[6]*Z)/669871503125
<22|V|33> = (73008*Z)/9765625
<22|V|34> = (14739259392*Sqrt[3]*Z)/6131066257801
<22|V|41> = (376832*Z)/129140163
<22|V|42> = (31363072*Sqrt[2]*Z)/4202539929
<22|V|43> = (14739259392*Sqrt[3]*Z)/6131066257801
<22|V|44> = (424*Z)/177147
<23|V|11> = (110592*Sqrt[6]*Z)/24137569
<23|V|12> = (2160*Sqrt[3]*Z)/823543
<23|V|13> = (37826560*Sqrt[2]*Z)/22024729467
<23|V|14> = (487489536*Sqrt[6]*Z)/762939453125
<23|V|21> = (29943*Sqrt[3]*Z)/13176688
<23|V|22> = (5870679552*Sqrt[6]*Z)/669871503125
<23|V|23> = (32857*Z)/390625
<23|V|24> = (55508689880137728*Sqrt[3]*Z)/5049196699148208943
<23|V|31> = (1216512*Sqrt[2]*Z)/815730721
<23|V|32> = (73008*Z)/9765625
<23|V|33> = (6890942464*Sqrt[2/3]*Z)/1210689028125
<23|V|34> = (69158928384*Sqrt[2]*Z)/34271896307633
<23|V|41> = (423788544*Sqrt[6]*Z)/762939453125
<23|V|42> = (14739259392*Sqrt[3]*Z)/6131066257801
<23|V|43> = (36645380390912*Sqrt[2]*Z)/24984212408264457
<23|V|44> = (145503*Sqrt[3/2]*Z)/134217728
<24|V|11> = (98304*Sqrt[2]*Z)/19487171
<24|V|12> = (376832*Z)/129140163
<24|V|13> = (487489536*Sqrt[6]*Z)/762939453125
<24|V|14> = (5053*Z)/(3538944*Sqrt[2])
<24|V|21> = (108838912*Z)/44840334375
<24|V|22> = (31363072*Sqrt[2]*Z)/4202539929
<24|V|23> = (55508689880137728*Sqrt[3]*Z)/5049196699148208943
<24|V|24> = (4043*Z)/78732
<24|V|31> = (10148806656*Sqrt[6]*Z)/19073486328125
<24|V|32> = (14739259392*Sqrt[3]*Z)/6131066257801
<24|V|33> = (69158928384*Sqrt[2]*Z)/34271896307633
<24|V|34> = (2496169683*Sqrt[3/2]*Z)/1677721600000
<24|V|41> = (39*Z)/(32768*Sqrt[2])
<24|V|42> = (424*Z)/177147
<24|V|43> = (145503*Sqrt[3/2]*Z)/134217728
<24|V|44> = (21252608*Sqrt[2]*Z)/35595703125
<31|V|11> = (1269*Sqrt[3]*Z)/50000
<31|V|12> = (110592*Sqrt[6]*Z)/24137569
<31|V|13> = (189*Z)/32768
<31|V|14> = (1808400384*Sqrt[3]*Z)/852891037441
<31|V|21> = (1555918848*Sqrt[6]*Z)/75429903125
<31|V|22> = (2160*Sqrt[3]*Z)/823543
<31|V|23> = (1216512*Sqrt[2]*Z)/815730721
<31|V|24> = (10148806656*Sqrt[6]*Z)/19073486328125
<31|V|31> = (815*Z)/8192
<31|V|32> = (37826560*Sqrt[2]*Z)/22024729467
<31|V|33> = (617*Z)/(314928*Sqrt[3])
<31|V|34> = (90581886173184*Z)/129457847542653125
<31|V|41> = (11694850770862080*Sqrt[3]*Z)/702392443647273463
<31|V|42> = (487489536*Sqrt[6]*Z)/762939453125
<31|V|43> = (30254432256*Z)/41426511213649
<31|V|44> = (74450880*Sqrt[3]*Z)/285311670611
<32|V|11> = (110592*Sqrt[6]*Z)/24137569
<32|V|12> = (29943*Sqrt[3]*Z)/13176688
<32|V|13> = (1216512*Sqrt[2]*Z)/815730721
<32|V|14> = (423788544*Sqrt[6]*Z)/762939453125
<32|V|21> = (2160*Sqrt[3]*Z)/823543
<32|V|22> = (5870679552*Sqrt[6]*Z)/669871503125
<32|V|23> = (73008*Z)/9765625
<32|V|24> = (14739259392*Sqrt[3]*Z)/6131066257801
<32|V|31> = (37826560*Sqrt[2]*Z)/22024729467
<32|V|32> = (32857*Z)/390625
<32|V|33> = (6890942464*Sqrt[2/3]*Z)/1210689028125
<32|V|34> = (36645380390912*Sqrt[2]*Z)/24984212408264457
<32|V|41> = (487489536*Sqrt[6]*Z)/762939453125
<32|V|42> = (55508689880137728*Sqrt[3]*Z)/5049196699148208943
<32|V|43> = (69158928384*Sqrt[2]*Z)/34271896307633
<32|V|44> = (145503*Sqrt[3/2]*Z)/134217728
<33|V|11> = (189*Z)/32768
<33|V|12> = (1216512*Sqrt[2]*Z)/815730721
<33|V|13> = (617*Z)/(314928*Sqrt[3])
<33|V|14> = (30254432256*Z)/41426511213649
<33|V|21> = (1216512*Sqrt[2]*Z)/815730721
<33|V|22> = (73008*Z)/9765625
<33|V|23> = (6890942464*Sqrt[2/3]*Z)/1210689028125
<33|V|24> = (69158928384*Sqrt[2]*Z)/34271896307633
<33|V|31> = (617*Z)/(314928*Sqrt[3])
<33|V|32> = (6890942464*Sqrt[2/3]*Z)/1210689028125
<33|V|33> = (17*Z)/256
<33|V|34> = (2486755845603328*Z)/(158298797548828125*Sqrt[3])
<33|V|41> = (30254432256*Z)/41426511213649
<33|V|42> = (69158928384*Sqrt[2]*Z)/34271896307633
<33|V|43> = (2486755845603328*Z)/(158298797548828125*Sqrt[3])
<33|V|44> = (2560158144*Z)/678223072849
<34|V|11> = (1808400384*Sqrt[3]*Z)/852891037441
<34|V|12> = (423788544*Sqrt[6]*Z)/762939453125
<34|V|13> = (30254432256*Z)/41426511213649
<34|V|14> = (1243165779*Sqrt[3]*Z)/4564986729776
<34|V|21> = (10148806656*Sqrt[6]*Z)/19073486328125
<34|V|22> = (14739259392*Sqrt[3]*Z)/6131066257801
<34|V|23> = (69158928384*Sqrt[2]*Z)/34271896307633
<34|V|24> = (2496169683*Sqrt[3/2]*Z)/1677721600000
<34|V|31> = (90581886173184*Z)/129457847542653125
<34|V|32> = (36645380390912*Sqrt[2]*Z)/24984212408264457
<34|V|33> = (2486755845603328*Z)/(158298797548828125*Sqrt[3])
<34|V|34> = (621550729*Z)/13841287201
<34|V|41> = (74450880*Sqrt[3]*Z)/285311670611
<34|V|42> = (145503*Sqrt[3/2]*Z)/134217728
<34|V|43> = (2560158144*Z)/678223072849
<34|V|44> = (413631006610176000*Sqrt[3]*Z)/249430673908303812379
<41|V|11> = (416415744*Z)/15083778125
<41|V|12> = (98304*Sqrt[2]*Z)/19487171
<41|V|13> = (1808400384*Sqrt[3]*Z)/852891037441
<41|V|14> = (22848*Z)/9765625
<41|V|21> = (288456704*Sqrt[2]*Z)/14206147659
<41|V|22> = (376832*Z)/129140163
<41|V|23> = (423788544*Sqrt[6]*Z)/762939453125
<41|V|24> = (39*Z)/(32768*Sqrt[2])
<41|V|31> = (11694850770862080*Sqrt[3]*Z)/702392443647273463
<41|V|32> = (487489536*Sqrt[6]*Z)/762939453125
<41|V|33> = (30254432256*Z)/41426511213649
<41|V|34> = (74450880*Sqrt[3]*Z)/285311670611
<41|V|41> = (22513*Z)/390625
<41|V|42> = (5053*Z)/(3538944*Sqrt[2])
<41|V|43> = (1243165779*Sqrt[3]*Z)/4564986729776
<41|V|44> = (1804351488*Z)/6179146071875
<42|V|11> = (98304*Sqrt[2]*Z)/19487171
<42|V|12> = (108838912*Z)/44840334375
<42|V|13> = (10148806656*Sqrt[6]*Z)/19073486328125
<42|V|14> = (39*Z)/(32768*Sqrt[2])
<42|V|21> = (376832*Z)/129140163
<42|V|22> = (31363072*Sqrt[2]*Z)/4202539929
<42|V|23> = (14739259392*Sqrt[3]*Z)/6131066257801
<42|V|24> = (424*Z)/177147
<42|V|31> = (487489536*Sqrt[6]*Z)/762939453125
<42|V|32> = (55508689880137728*Sqrt[3]*Z)/5049196699148208943
<42|V|33> = (69158928384*Sqrt[2]*Z)/34271896307633
<42|V|34> = (145503*Sqrt[3/2]*Z)/134217728
<42|V|41> = (5053*Z)/(3538944*Sqrt[2])
<42|V|42> = (4043*Z)/78732
<42|V|43> = (2496169683*Sqrt[3/2]*Z)/1677721600000
<42|V|44> = (21252608*Sqrt[2]*Z)/35595703125
<43|V|11> = (1808400384*Sqrt[3]*Z)/852891037441
<43|V|12> = (10148806656*Sqrt[6]*Z)/19073486328125
<43|V|13> = (90581886173184*Z)/129457847542653125
<43|V|14> = (74450880*Sqrt[3]*Z)/285311670611
<43|V|21> = (423788544*Sqrt[6]*Z)/762939453125
<43|V|22> = (14739259392*Sqrt[3]*Z)/6131066257801
<43|V|23> = (36645380390912*Sqrt[2]*Z)/24984212408264457
<43|V|24> = (145503*Sqrt[3/2]*Z)/134217728
<43|V|31> = (30254432256*Z)/41426511213649
<43|V|32> = (69158928384*Sqrt[2]*Z)/34271896307633
<43|V|33> = (2486755845603328*Z)/(158298797548828125*Sqrt[3])
<43|V|34> = (2560158144*Z)/678223072849
<43|V|41> = (1243165779*Sqrt[3]*Z)/4564986729776
<43|V|42> = (2496169683*Sqrt[3/2]*Z)/1677721600000
<43|V|43> = (621550729*Z)/13841287201
<43|V|44> = (413631006610176000*Sqrt[3]*Z)/249430673908303812379
<44|V|11> = (22848*Z)/9765625
<44|V|12> = (39*Z)/(32768*Sqrt[2])
<44|V|13> = (74450880*Sqrt[3]*Z)/285311670611
<44|V|14> = (1804351488*Z)/6179146071875
<44|V|21> = (39*Z)/(32768*Sqrt[2])
<44|V|22> = (424*Z)/177147
<44|V|23> = (145503*Sqrt[3/2]*Z)/134217728
<44|V|24> = (21252608*Sqrt[2]*Z)/35595703125
<44|V|31> = (74450880*Sqrt[3]*Z)/285311670611
<44|V|32> = (145503*Sqrt[3/2]*Z)/134217728
<44|V|33> = (2560158144*Z)/678223072849
<44|V|34> = (413631006610176000*Sqrt[3]*Z)/249430673908303812379
<44|V|41> = (1804351488*Z)/6179146071875
<44|V|42> = (21252608*Sqrt[2]*Z)/35595703125
<44|V|43> = (413631006610176000*Sqrt[3]*Z)/249430673908303812379
<44|V|44> = (19541*Z)/524288
\end{verbatim}
\end{document}

