# Fock expansion for two-electron atoms. High order angular coefficients.

Evgeny Z. Liverts111Corresponding author. E-mail address:
<EMAIL_ADDRESS>Racah Institute of Physics, The Hebrew
University, Jerusalem 91904, Israel  Rajmund Krivec  Department of Theoretical
Physics, J. Stefan Institute, 1000 Ljubljana, Slovenia

###### Abstract

The Fock expansion, which describes the properties of two-electron atoms near
the nucleus, is studied. The angular Fock coefficients
$\psi_{k,p}(\alpha,\theta)$ with the maximum possible value of subscript $p$
are calculated on examples of the coefficients with $5\leq k\leq 10$. The
presented technique makes it possible to calculate such angular coefficients
for any arbitrarily large $k$. The mentioned coefficients being leading in the
logarithmic power series representing the Fock expansion may be indispensable
for the development of simple methods for calculating the helium-like
electronic structure. The theoretical results obtained are verified by other
suitable methods. The Wolfram Mathematica is used extensively.

##  I Introduction

The properties of a two-electron atomic (helium-like) system with an
infinitely massive nucleus of the charge $Z$ and non-relativistic energy $E$
are defined by the wave function (WF) $\Psi(r_{1},r_{2},r_{12})$, where
$r_{1}$ and $r_{2}$ are the electron-nucleus distances, and $r_{12}$ is the
distance between the electrons. The behavior of the ground state WF in the
vicinity of the nucleus located at the origin is determined by the Fock
expansion FOCK

$\bar{\Psi}(r_{1},r_{2},r_{12})\equiv\Psi(R,\alpha,\theta)=\sum_{k=0}^{\infty}R^{k}\sum_{p=0}^{[k/2]}\psi_{k,p}(\alpha,\theta)\ln^{p}R,$
(1)

where the hyperspherical coordinates $R,~{}\alpha$ and $\theta$ are defined by
the relations:

$R=\sqrt{r_{1}^{2}+r_{2}^{2}},~{}~{}~{}~{}\alpha=2\arctan\left(\frac{r_{2}}{r_{1}}\right),~{}~{}~{}~{}\theta=\arccos\left(\frac{r_{1}^{2}+r_{2}^{2}-r_{12}^{2}}{2r_{1}r_{2}}\right).$
(2)

The convergence of expansion (1) had been proven in Ref.MORG . The angular
Fock coefficients (AFC) $\psi_{k,p}$ satisfy the Fock recurrence relation
(FRR)

$\left[\Lambda^{2}-k(k+4)\right]\psi_{k,p}(\alpha,\theta)=h_{k,p}(\alpha,\theta)$
(3)

with the RHS of the form AB1 , LEZ4 :

$h_{k,p}=2(k+2)(p+1)\psi_{k,p+1}+(p+1)(p+2)\psi_{k,p+2}-2V\psi_{k-1,p}+2E\psi_{k-2,p}.$
(4)

The dimensionless Coulomb potential representing the electron-electron and
electron-nucleus interactions is

$V\equiv\frac{R}{r_{12}}-Z\left(\frac{R}{r_{1}}+\frac{R}{r_{2}}\right)=\frac{1}{\xi}-\frac{2Z\eta}{\sin\alpha},$
(5)

where we have introduced the important (in what follows) angular quantities:

$\xi=\sqrt{1-\sin\alpha\cos\theta},~{}~{}~{}\eta=\sqrt{1+\sin\alpha}.$ (6)

The hyperspherical angular momentum operator, projected on $S$ states, is
defined as

$\Lambda^{2}=-4\left[\frac{\partial^{2}}{\partial\alpha^{2}}+2\cot\alpha\frac{\partial}{\partial\alpha}+\frac{1}{\sin^{2}\alpha}\left(\frac{\partial^{2}}{\partial\theta^{2}}+\cot\theta\frac{\partial}{\partial\theta}\right)\right].$
(7)

It is clear that all circumnuclear features of the two-electron atoms (ions)
are defined by the Fock expansion (1). There are a large number of methods for
calculating the electronic structure of the two-electron atomic systems. An
excellent review on this topic can be found in Refs. NAK ; ROD ; FOR ; DRK ;
AB1 . However, we are known only one technique that correctly represents the
WF, $\Psi(r_{1},r_{2},r_{12})$ near the nucleus. It is the so called
correlation function hyperspherical harmonic method (CFHHM) HM0 ; HM1 ; HM2 .
Unfortunately, the CFHHM is extremely difficult to implement. The expansion in
hyperspherical harmonics (HHs) provides the correct representation of the
AFCs. However, the HH-expansion is known to converge very slowly. Although
this method makes it possible to increase the convergence of the HH-expansion,
a sufficiently good accuracy requires a large HHs basis size, which, in turn,
creates great computational difficulties.

Thus, the conclusion suggests itself that it would be extremely useful to
develop a much simpler method for calculating the WF with correct behavior
near the nucleus. In this regard, we would like to emphasize the following
important peculiarities of the Fock expansion (FE). It follows from definition
(1) that the FE can be splitted into individual power series (lines)
associated with definite power of $\textrm{ln}R$. In other words, the FE can
be represented in the form:

$\displaystyle\Psi=(\ln
R)^{0}\left(\psi_{0,0}+R\psi_{1,0}+R^{2}\psi_{2,0}+...\right)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle+(\ln
R)^{1}R^{2}\left(\psi_{2,1}+R\psi_{3,1}+R^{2}\psi_{4,1}+...\right)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle+(\ln
R)^{2}R^{4}\left(\psi_{4,2}+R\psi_{5,2}+R^{2}\psi_{6,2}+...\right)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle+(\ln
R)^{3}R^{6}\left(\psi_{6,3}+R\psi_{7,3}+R^{2}\psi_{8,3}+...\right)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle+(\ln
R)^{4}R^{8}\left(\psi_{8,4}+R\psi_{9,4}+R^{2}\psi_{10,4}+...\right)+...~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(8)

It is seen that the leading term of each line represents the product $(\ln
R)^{k/2}R^{k}\psi_{k,k/2}(\alpha,\theta)$ with even $k$. The first AFCs
($\psi_{0,0}=1$) corresponding to $k=0,2,4$ are well-known (see, e.g., AB1 ;
LEZ4 ):

$\psi_{1,0}=\frac{1}{2}\xi-Z\eta,$ (9)
$\psi_{2,1}=-\frac{Z(\pi-2)}{3\pi}(1-\xi^{2}),$ (10)
$\psi_{3,1}=\frac{Z(\pi-2)}{36\pi}\left[6Z\eta(1-\xi^{2})+\xi(5\xi^{2}-6)\right],$
(11)
$\psi_{4,2}=\frac{Z^{2}(\pi-2)(5\pi-14)}{540\sqrt{\pi}}\left[Y_{40}(\alpha,\theta)+\sqrt{2}~{}Y_{42}(\alpha,\theta)\right].$
(12)

The normalized HHs are

$Y_{40}(\alpha,\theta)=\pi^{-3/2}(4\cos^{2}\alpha-1),~{}~{}~{}~{}~{}Y_{42}(\alpha,\theta)=2\sqrt{2}\pi^{-3/2}\sin^{2}\alpha
P_{2}(\cos\theta),~{}~{}~{}$ (13)

where $P_{n}(x)$ denotes the Legendre polynomials.

In this paper, we present the theoretical calculations of the AFCs
$\psi_{5,2}(\alpha,\theta)$, $\psi_{6,3}(\alpha,\theta)$,
$\psi_{7,3}(\alpha,\theta)$, $\psi_{8,4}(\alpha,\theta)$ and
$\psi_{9,4}(\alpha,\theta)$ included into the $k=4$, $k=6$ and $k=8$ ”lines”
of the expansion (I), and also the AFC $\psi_{10,5}(\alpha,\theta)$
representing the leading term of $k=10$ ”line”. It is important to note that
all mentioned angular coefficients represent the AFCs $\psi_{k,p}$ with the
maximum possible $p$ for a given $k$.

##  II Derivation of the angular Fock coefficient $\psi_{5,2}(\alpha,\theta)$

The FRR (3)-(4) for $k=5$ and $p=2$ reduces to the form

$\left(\Lambda^{2}-45\right)\psi_{5,2}(\alpha,\theta)=h_{5,2}(\alpha,\theta),$
(14)

where

$h_{5,2}(\alpha,\theta)=-2V\psi_{4,2}(\alpha,\theta).$ (15)

Using Eqs.(5)-(6) and (12)-(13), it is convenient to represent the RHS of
Eq.(14) in the form

$h_{5,2}(\alpha,\theta)=-\frac{Z^{2}(\pi-2)(5\pi-14)}{270\sqrt{\pi}}\left[3\pi^{-3/2}(2h_{1}+h_{2})-2Z(h_{3}+\sqrt{2}h_{4})\right],$
(16)

where

$h_{1}=\frac{(1-\xi^{2})^{2}}{\xi},~{}~{}~{}h_{2}=\frac{\cos(2\alpha)}{\xi},~{}~{}~{}h_{3}=\frac{\eta
Y_{40}(\alpha,\theta)}{\sin\alpha}~{}~{}~{}h_{4}=\frac{\eta
Y_{42}(\alpha,\theta)}{\sin\alpha}.$ (17)

Accordingly, we obtain the solution of Eq.(14) in the identical form

$\psi_{5,2}(\alpha,\theta)=-\frac{Z^{2}(\pi-2)(5\pi-14)}{270\sqrt{\pi}}\left[3\pi^{-3/2}(2f_{1}+f_{2})-2Z(f_{3}+\sqrt{2}f_{4})\right],$
(18)

where the AFC-components $f_{i}$ satisfy the individual Fock recurrence
relations (IFRR)

$\left(\Lambda^{2}-45\right)f_{i}=h_{i}.~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(i=1,2,3,4)$
(19)

We will sequentially find solutions to each of the IFRRs (19) using various
methods presented in Ref.LEZ4 .

###  II.1 Solution of the IFRR $(\Lambda^{2}-45)f_{3}=\eta
Y_{40}/\sin\emph{}\alpha$

Moving from simpler to more complex solutions, let’s start with IFRR

$\left(\Lambda^{2}-45\right)f_{3}=h_{3}.$ (20)

The RHS $h_{3}\equiv h_{3}(\alpha)$ represents the function of only one angle
variable $\alpha$. It was shown LEZ4  that the solution of the corresponding
IFRR (20) reduces to solution $g(\rho)=f_{3}(\alpha)$ of the inhomogeneous
differential equation

$(\rho^{2}+1)^{2}g^{\prime\prime}(\rho)+2\rho^{-1}(\rho^{2}+1)g^{\prime}(\rho)+45g(\rho)=-\textrm{h}(\rho),$
(21)

where $\rho=\tan(\alpha/2)$, and

$\textrm{h}(\rho)\equiv
h_{3}(\alpha)=\frac{(4\cos^{2}\alpha-1)\sqrt{1+\sin\alpha}}{\pi^{3/2}\sin\alpha}=\frac{(\rho+1)(3\rho^{4}-10\rho^{2}+3)}{2\pi^{3/2}\rho(\rho^{2}+1)^{3/2}}.$
(22)

For convenience, we will solve the Eq.(21) with the RHS $\textrm{h}(\rho)$ not
containing the multiplier $\pi^{-3/2}$. The final solution $f_{3}$ will be
multiplied by this factor.

Using the method of variation of parameters, one obtains LEZ4  the particular
solution of Eq.(21) in the form

$g^{(p)}(\rho)=v_{50}(\rho)\int\frac{u_{50}(\rho)\textrm{h}(\rho)d\rho}{(\rho^{2}+1)^{2}W_{0}(\rho)}-u_{50}(\rho)\int\frac{v_{50}(\rho)\textrm{h}(\rho)d\rho}{(\rho^{2}+1)^{2}W_{0}(\rho)},$
(23)

where

$W_{0}(\rho)=-(\rho^{2}+1)/\rho^{2}.$ (24)

The independent solutions of the homogeneous equation associated with Eq.(21)
are LEZ4 :

$u_{50}(\rho)=\frac{(\rho^{2}+1)^{9/2}}{\rho}~{}_{2}F_{1}\left(4,\frac{7}{2};\frac{1}{2};-\rho^{2}\right)=\frac{1-7\rho^{2}(3-5\rho^{2}+\rho^{4})}{\rho(\rho^{2}+1)^{5/2}},$
(25)
$v_{50}(\rho)=(\rho^{2}+1)^{9/2}~{}_{2}F_{1}\left(4,\frac{9}{2};\frac{3}{2};-\rho^{2}\right)=\frac{1-35\rho^{2}+21\rho^{4}-\rho^{6}}{7(\rho^{2}+1)^{5/2}},$
(26)

where $~{}_{2}F_{1}(...)$ is the Gauss hypergeometric function. Substitution
of Eqs.(24)-(26) into the general representation (23) yields

$g^{(p)}(\rho)=\frac{-7\rho\left\\{\rho\left[5\rho(3\rho-4)(3\rho+5)-24\right]+23\right\\}-23}{420\rho(\rho^{2}+1)^{5/2}}.$
(27)

The general solution of the inhomogeneous equation can be expressed as the sum
of the general solution of the associated homogeneous (complementary) equation
and the particular solution of the inhomogeneous equation, whence

$g(\rho)=g^{(p)}(\rho)+c_{u}u_{50}(\rho)+c_{v}v_{50}(\rho),$ (28)

where the coefficients $c_{u}$ and $c_{v}$ are currently undetermined. To
choose these coefficients, it is necessary to determine the behavior of all
independent solutions on the boundaries of the domain $[0,\infty]$. We easily
obtain:

$g^{(p)}(\rho)\underset{\rho\rightarrow
0}{=}-\frac{23}{420\rho}-\frac{23}{60}+\frac{451\rho}{840}+O(\rho^{2}),$ (29)
$g^{(p)}(\rho)\underset{\rho\rightarrow\infty}{=}-\frac{3}{4\rho}-\frac{1}{4\rho^{2}}+\frac{85}{24\rho^{3}}+O(\rho^{-4}),$
(30) $u_{50}(\rho)\underset{\rho\rightarrow
0}{=}\frac{1}{\rho}-\frac{47\rho}{2}+O(\rho^{3}),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(31)
$u_{50}(\rho)\underset{\rho\rightarrow\infty}{=}-7+\frac{105}{2\rho^{2}}+O(\rho^{-4}),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(32) $v_{50}(\rho)\underset{\rho\rightarrow
0}{=}1-\frac{15\rho^{2}}{2}+O(\rho^{4}),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(33)
$v_{50}(\rho)\underset{\rho\rightarrow\infty}{=}-\frac{\rho}{7}+\frac{47}{14\rho}+O(\rho^{-3}).~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(34)

It is seen that the particular solution $g^{(p)}(\rho)$ is divergent at
$\rho=0$, whereas the solutions of the homogeneous equation associated with
Eq. (21) are divergent, at $\rho=0$ and $\rho=\infty$ for $u_{50}(\rho)$ and
$v_{50}(\rho)$, respectively. Thus, to avoid the divergence on the whole range
of definition, one should set $c_{u}=23/420$ and $c_{v}=0$ in the general
solution (28). Then the final physical solution becomes

$\displaystyle
f_{3}(\alpha)=-\frac{(\rho+1)(23\rho^{4}+22\rho^{3}-122\rho^{2}+22\rho+23)}{60\pi^{3/2}(\rho^{2}+1)^{5/2}}=~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle=-\frac{1}{60\pi^{3/2}}\left[11\sin\alpha+21\cos(2\alpha)+2\right]\sqrt{1+\sin\alpha}.~{}~{}~{}~{}~{}~{}$
(35)

###  II.2 Solution of the IFRR $(\Lambda^{2}-45)f_{4}=\eta Y_{42}/\sin\alpha$

It was shown in Ref.LEZ4  that the solution $f_{4}\equiv f_{4}(\alpha,\theta)$
of the IFRR

$\left(\Lambda^{2}-45\right)f_{4}=2\pi^{-3/2}\eta\sqrt{2}\sin\alpha
P_{2}(\cos\theta)$ (36)

can be found in the form

$f_{4}=2\sqrt{2}~{}\pi^{-3/2}\sin^{2}\alpha P_{2}(\cos\theta)g_{4}(\rho),$
(37)

where the function $g_{4}(\rho)$ satisfies the equation

$(\rho^{2}+1)^{2}g_{4}^{\prime\prime}(\rho)+2\rho^{-1}[1+\rho^{2}+2(1-\rho^{4})]g_{4}^{\prime}(\rho)+13g_{4}(\rho)=-\textrm{h}_{4}(\rho),$
(38)

with

$\textrm{h}_{4}(\rho)=\frac{\eta}{\sin\alpha}=\frac{(\rho+1)\sqrt{\rho^{2}+1}}{2\rho}.$
(39)

Using the method of variation of parameters, one obtains LEZ4  the particular
solution of Eq. (38) in the form

$g_{4}^{(p)}(\rho)=v_{52}(\rho)\int\frac{u_{52}(\rho)\textrm{h}_{4}(\rho)d\rho}{(\rho^{2}+1)^{2}W_{2}(\rho)}-u_{52}(\rho)\int\frac{v_{52}(\rho)\textrm{h}_{4}(\rho)d\rho}{(\rho^{2}+1)^{2}W_{2}(\rho)},$
(40)

where

$W_{2}(\rho)=-\frac{5}{\rho}\left(\frac{\rho^{2}+1}{\rho}\right)^{5}.$ (41)

The independent solutions of the homogeneous equation associated with Eq.(38)
are:

$u_{52}(\rho)=\frac{(\rho^{2}+1)^{13/2}}{\rho^{5}}~{}_{2}F_{1}\left(4,\frac{3}{2};-\frac{3}{2};-\rho^{2}\right)=\frac{1+11\rho^{2}+99\rho^{4}-231\rho^{6}}{\rho^{5}\sqrt{\rho^{2}+1}},$
(42)
$v_{52}(\rho)=(\rho^{2}+1)^{13/2}~{}_{2}F_{1}\left(4,\frac{13}{2};\frac{7}{2};-\rho^{2}\right)=\frac{231-99\rho^{2}-11\rho^{4}-\rho^{6}}{231\sqrt{\rho^{2}+1}}.$
(43)

Thus, the particular solution (40) reduces to the form:

$g_{4}^{(p)}(\rho)=-\frac{11(21\rho^{5}+9\rho^{4}+\rho^{2})+1}{2772\rho^{5}\sqrt{\rho^{2}+1}}.$
(44)

Considering the series expansions for $g_{4}^{(p)}(\rho)$, $u_{52}(\rho)$ and
$v_{52}(\rho)$ on the boundaries of the range of definition
($\rho\in[0,\infty]$), it can be shown that function

$g_{4}^{(p)}(\rho)+\frac{1}{2772}u_{52}(\rho)=-\frac{\rho+1}{12\sqrt{\rho^{2}+1}}=-\frac{1}{12}\sqrt{1+\sin\alpha}$
(45)

represents the physical (finite) solution of Eq.(38), and hence we finally
obtain:

$f_{4}(\alpha,\theta)=-\frac{\sqrt{2}}{6\pi^{3/2}}(\sin\alpha)^{2}\sqrt{1+\sin\alpha}~{}P_{2}(\cos\theta).$
(46)

###  II.3 Solution of the IFRR $(\Lambda^{2}-45)f_{1}=(1-\xi^{2})^{2}/\xi$

It is important to note that the RHS $h_{1}$ (see Eq.(17)) of the IFRR

$\left(\Lambda^{2}-45\right)f_{1}=h_{1}(\xi)$ (47)

is a function of $\xi$ (only) defined by Eq.(6). For this case LEZ4 ; LEZ2
the solution of Eq.(47) coincides with solution of the inhomogeneous
differential equation

$(\xi^{2}-2)f_{1}^{\prime\prime}(\xi)+\xi^{-1}(5\xi^{2}-4)f_{1}^{\prime}(\xi)-45f_{1}(\xi)=h_{1}(\xi).$
(48)

A particular solution of Eq. (48) can be found by the method of variation of
parameters in the form LEZ2

$f_{1}^{(p)}(\xi)=\frac{1}{7\sqrt{2}}\left[u_{5}(\xi)\int
v_{5}(\xi)w(\xi)d\xi-v_{5}(\xi)\int u_{5}(\xi)w(\xi)d\xi\right],$ (49)

where

$w(\xi)=h_{1}(\xi)\xi^{2}\sqrt{2-\xi^{2}}.$ (50)

The linearly independent solutions of the homogeneous equation associated with
Eq.(48) are defined by the relations

$u_{5}(\xi)=\frac{P_{13/2}^{1/2}(\xi/\sqrt{2})}{\xi\sqrt[4]{2-\xi^{2}}}=\frac{2^{1/4}(8\xi^{6}-28\xi^{4}+28\xi^{2}-7)}{\sqrt{\pi(2-\xi^{2})}},$
(51)
$v_{5}(\xi)=\frac{Q_{13/2}^{1/2}(\xi/\sqrt{2})}{\xi\sqrt[4]{2-\xi^{2}}}=\frac{\sqrt{\pi}(-8\xi^{6}+20\xi^{4}-12\xi^{2}+1)}{2^{3/4}\xi}.$
(52)

where $P_{\nu}^{\mu}(x)$ and $Q_{\nu}^{\mu}(x)$ are the associated Legendre
functions of the first and second kind, respectively. Substitution of the
representations (50)-(52) into (49) yields for the particular solution:

$f_{1}^{(p)}(\xi)=-\frac{1}{60}\xi(13\xi^{4}-30\xi^{2}+15).$ (53)

It can be verified that the particular solution $f_{1}^{(p)}(\xi)$ is finite
on the whole range of definition ($\xi\in[0,\sqrt{2}]$), whereas the solutions
of the homogeneous equation associated with Eq.(48) are divergent at
$\xi=\sqrt{2}$ ($\alpha=\pi/2,\theta=\pi$) and $\xi=0$
($\alpha=\pi/2,\theta=0$) for $u_{5}(\xi)$ and $v_{5}(\xi)$, respectively.
Thus, we can conclude that the final physical solution of the IFFR (47)
coincides with the particular solution (53), whence

$f_{1}=-\frac{1}{60}\xi(13\xi^{4}-30\xi^{2}+15)=-\frac{1}{60}\sqrt{1-\sin\alpha\cos\theta}\left[\sin\alpha\cos\theta(4+13\sin\alpha\cos\theta)-2\right].~{}~{}~{}~{}$
(54)

###  II.4 Solution of the IFRR $(\Lambda^{2}-45)f_{2}=\cos(2\alpha)/\xi$

To solve the IFRR

$\left(\Lambda^{2}-45\right)f_{2}=h_{2},$ (55)

with the RHS $h_{2}$ defined by Eq.(17), first of all, it is necessary to
recall the Sack’s representation SACK  (see also AB1  and LEZ4 ) for
$\xi^{\nu}$ with $\nu=-1$:

$\xi^{-1}=\sum_{l=0}^{\infty}P_{l}(\cos\theta)\left(\frac{\sin\alpha}{2}\right)^{l}F_{l}(\rho),$
(56)

where

$F_{l}(\rho)=~{}_{2}F_{1}\left(\frac{l}{2}+\frac{1}{4},\frac{l}{2}+\frac{3}{4};l+\frac{3}{2};\frac{4\rho^{2}}{(\rho^{2}+1)^{2}}\right)=\left\\{\begin{array}[]{c}~{}~{}{\mathfrak{F}_{l}(\rho)}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}0\leq\rho\leq
1\\\ {\mathfrak{F}_{l}(1/\rho)}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\rho\geq
1\end{array}\right.$ (57)

with

$\mathfrak{F}_{l}(\rho)=(\rho^{2}+1)^{l+\frac{1}{2}}.$ (58)

This enables us to present the RHS of Eq.(55) in the form

$h_{2}\equiv\frac{\cos(2\alpha)}{\xi}=\sum_{l=0}^{\infty}P_{l}(\cos\theta)(\sin\alpha)^{l}\textrm{h}_{l}(\rho),$
(59)

where

$\textrm{h}_{l}(\rho)=2^{-l}F_{l}(\rho)\cos(2\alpha)=2^{-l}F_{l}(\rho)\left[1-\frac{8\rho^{2}}{(\rho^{2}+1)^{2}}\right].$
(60)

In turn, it was shown in Ref.LEZ4  that in case of the RHS is determined by
Eq. (59), the solution of the corresponding IFRR (55) can be found in the form

$f_{2}(\alpha,\theta)=\sum_{l=0}^{\infty}P_{l}(\cos\theta)(\sin\alpha)^{l}\sigma_{l}(\rho),$
(61)

where the function $\sigma_{l}(\rho)$ satisfies the inhomogeneous differential
equation

$(\rho^{2}+1)^{2}\sigma_{l}^{\prime\prime}(\rho)+2\rho^{-1}[1+\rho^{2}+l(1-\rho^{4})]\sigma_{l}^{\prime}(\rho)+(5-2l)(9+2l)\sigma_{l}(\rho)=-\textrm{h}_{l}(\rho).$
(62)

The linearly independent solutions of the homogeneous equation associated with
Eq. (62) are:

$\displaystyle
u_{5l}(\rho)=\frac{(\rho^{2}+1)^{l+9/2}}{\rho^{2l+1}}~{}_{2}F_{1}\left(4,\frac{7}{2}-l;\frac{1}{2}-l;-\rho^{2}\right)=\frac{(\rho^{2}+1)^{l-5/2}}{\rho^{2l+1}}\times~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle\left[\rho^{6}\left(1+\frac{120}{2l-5}-\frac{120}{2l-3}+\frac{24}{2l-1}\right)+3\rho^{4}\left(1+\frac{40}{2l-3}-\frac{24}{2l-1}\right)+3\rho^{2}\left(1+\frac{8}{2l-1}\right)+1\right],~{}~{}~{}$
(63) $\displaystyle
v_{5l}(\rho)=(\rho^{2}+1)^{l+9/2}~{}_{2}F_{1}\left(4,\frac{9}{2}+l;\frac{3}{2}+l;-\rho^{2}\right)=(\rho^{2}+1)^{l-5/2}\times~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle\left[\rho^{6}\left(1-\frac{24}{2l+3}+\frac{120}{2l+5}-\frac{120}{2l+7}\right)+3\rho^{4}\left(1+\frac{24}{2l+3}-\frac{40}{2l+5}\right)+3\rho^{2}\left(1-\frac{8}{2l+3}\right)+1\right].~{}~{}~{}~{}$
(64)

The method of variation of parameters enables us to obtain the particular
solution of the inhomogeneous differential equation (62) in the form

$\sigma_{l}^{(p)}(\rho)=v_{5l}(\rho)\int\frac{u_{5l}(\rho)\textrm{h}_{l}(\rho)d\rho}{(\rho^{2}+1)^{2}W_{l}(\rho)}-u_{5l}(\rho)\int\frac{v_{5l}(\rho)\textrm{h}_{l}(\rho)d\rho}{(\rho^{2}+1)^{2}W_{l}(\rho)},$
(65)

where

$W_{l}(\rho)=-\frac{2l+1}{\rho}\left(\frac{\rho^{2}+1}{\rho}\right)^{2l+1}.$
(66)

Note that due to different representations for the function $F_{l}(\rho)$ (see
Eq.(57)) at values of $\rho$ less and greater than $1$, we obtain special
representations for a particular solution in these two regions:

$\sigma_{l}^{(0)}(\rho)=\frac{(\rho^{2}+1)^{l-5/2}}{2^{l+1}(2l-1)(2l-3)}\left[(2l-3)\rho^{4}-4(l-2)\rho^{2}-\frac{4l^{2}+4l-27}{3(2l-5)}\right],~{}~{}~{}~{}~{}0\leq\rho\leq
1~{}$ (67)
$\sigma_{l}^{(1)}(\rho)=\frac{\rho^{-2l-1}(\rho^{2}+1)^{l-5/2}}{2^{l+1}(2l+3)(2l+5)}\left[\frac{4l^{2}+4l-27}{3(2l+7)}+4(l+3)\rho^{2}-(2l+5)\rho^{4}\right],~{}~{}~{}~{}~{}~{}~{}\rho\geq
1~{}$ (68)

It can be verified that both functions (67) and (68) have no singularities on
their domains of definition. On the other hand, function $u_{5l}(\rho)$ is
singular at $\rho=0$, whereas $v_{5l}(\rho)$ is singular at $\rho=\infty$.
This means that one should search the general solution of Eq.(62) in the form:

$\sigma_{l}(\rho)=\sigma_{l}^{(0)}(\rho)+c_{5l}^{(v)}v_{5l}(\rho),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}0\leq\rho\leq
1~{}$ (69)
$\sigma_{l}(\rho)=\sigma_{l}^{(1)}(\rho)+c_{5l}^{(u)}u_{5l}(\rho),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\rho\geq
1~{}$ (70)

Note that two coefficients $c_{5l}^{(v)}$ and $c_{5l}^{(u)}$ are presently
undetermined. To calculate them, we need to find two equations relating these
coefficients. The first equation is quite obvious. It follows from the
condition that the representations (69) and (70) are coincident at the common
point $\rho=1$, that is

$\sigma_{l}^{(0)}(1)+c_{5l}^{(v)}v_{5l}(1)=\sigma_{l}^{(1)}(1)+c_{5l}^{(u)}u_{5l}(1).$
(71)

This relationship reduces to the first desired equation:

$c_{5l}^{(u)}(2l+3)(2l+5)(2l+7)=c_{5l}^{(v)}(2l-5)(2l-3)(2l-1)+\frac{27-4l-4l^{2}}{3\times
2^{l+1}}.$ (72)

It can be verified that $\textrm{h}_{l}(0)=\textrm{h}_{l}(\infty)=2^{-l}$. It
follows from this relations that
$\sigma_{l}^{(0)}(0)+c_{5l}^{(v)}v_{5l}(0)=\sigma_{l}^{(1)}(\infty)+c_{5l}^{(u)}u_{5l}(\infty)$.
It can be assumed that the last equation represents the second desired
equation. However, this assumption turns out to be false, because it again
leads to Eq.(72).

We propose the following method to find the second desired equation. Remind
that any suitable function of the angles $\alpha$ and $\theta$ may be expanded
into HHs since they form a complete set:

$f(\alpha,\theta)=\sum_{n=0(2)}^{\infty}\sum_{l=0}^{n/2}f_{n,l}Y_{n,l}(\alpha,\theta),$
(73)

where (see, e.g., AB1 )

$f_{n,l}=\int f(\alpha,\theta)Y_{n,l}(\alpha,\theta)d\Omega$ (74)

with

$d\Omega=\pi^{2}\sin^{2}\alpha\sin\theta d\alpha
d\theta.~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\alpha\in[0,\pi],~{}~{}\theta\in[0,\pi]$
(75)

For the function $f(\alpha,\theta)=f_{2}(\alpha,\theta)$ represented by
Eq.(61), the expansion coefficient with $n=2l$ becomes

$f_{2l,l}=\pi^{2}\int_{0}^{\pi}\int_{0}^{\pi}f_{2}(\alpha,\theta)Y_{2l,l}(\alpha,\theta)\sin^{2}\alpha\sin\theta
d\alpha
d\theta=\frac{2\pi^{2}N_{2l,l}}{2l+1}\left[K_{0}(l)+c_{l}^{(v)}K_{v}(l)+c_{l}^{(u)}K_{u}(l)\right],$
(76)

where

$\displaystyle
K_{0}(l)=\int_{0}^{\pi/2}(\sin\alpha)^{2l+2}\sigma_{l}^{(0)}(\rho)d\alpha+\int_{\pi/2}^{\pi}(\sin\alpha)^{2l+2}\sigma_{l}^{(\infty)}(\rho)d\alpha=~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle=\frac{\sqrt{2}\left(-24l^{3}-100l^{2}+198l+249\right)}{(2l-5)(2l-3)(2l+3)(2l+5)(2l+7)(2l+9)},~{}~{}~{}~{}~{}~{}~{}~{}$
(77)
$K_{v}(l)=\int_{0}^{\pi/2}(\sin\alpha)^{2l+2}v_{5l}(\rho)d\alpha=\frac{2^{l+3/2}(2l-1)}{(2l+5)(2l+9)},~{}~{}~{}~{}~{}~{}~{}$
(78)
$K_{u}(l)=\int_{\pi/2}^{\pi}(\sin\alpha)^{2l+2}u_{5l}(\rho)d\alpha=\frac{2^{l+3/2}(2l+3)(2l+7)}{(2l-5)(2l-3)(2l+9)}.$
(79)

To derive the results (76)-(79), we used the representation
$Y_{2l,l}(\alpha,\theta)=N_{2l,l}\sin^{l}\alpha P_{l}(\cos\theta)$ for the
particular case of the HHs, and the orthogonality property for the Legendre
polynomials. It should be noted that the explicit form of the normalization
constant $N_{2l,l}$ is not required.

On the other hand, expanding $f_{2}(\alpha,\theta)$ in HHs, and inserting this
expansion into the LHS of the IFRR (55), we obtain

$\left(\Lambda^{2}-45\right)f_{2}(\alpha,\theta)=\sum_{n=0(2)}^{\infty}\sum_{l=0}^{n/2}f_{n,l}\left[n(n+4)-45\right]Y_{n,l}(\alpha,\theta).$
(80)

To derive the last equation we used the fact that $Y_{n,l}(\alpha,\theta)$ is
an eigenfunction of the operator $\Lambda^{2}$ with an eigenvalue equal to
$n(n+4)$, that is

$\Lambda^{2}Y_{n,l}(\alpha,\theta)=n(n+4)Y_{n,l}(\alpha,\theta).$ (81)

The HH-expansion of the RHS of Eq. (55) is

$h_{2}(\alpha,\theta)=\sum_{n=0(2)}^{\infty}\sum_{l=0}^{n/2}h_{n,l}Y_{n,l}(\alpha,\theta).$
(82)

Whence,

$f_{2l,l}=\frac{h_{2l,l}}{4l(l+2)-45}.$ (83)

Using again the Sack’s representation (56)-(57) and Eq.(74), we obtain the
expansion coefficient $h_{2l,l}$ in explicit form:

$h_{2l,l}=\pi^{2}\int_{0}^{\pi}\int_{0}^{\pi}h_{2}(\alpha,\theta)Y_{2l,l}(\alpha,\theta)\sin^{2}\alpha\sin\theta
d\alpha
d\theta=-\frac{2^{7/2}\pi^{2}N_{2l,l}(4l^{2}+24l+19)}{(2l+1)(2l+3)(2l+5)(2l+7)}.~{}~{}~{}~{}$
(84)

Thus, inserting (84) into the RHS of Eq. (83) and equating the result to the
RHS of Eq.(76) we obtain the desired _second equation_ in the form:

$c_{5l}^{(u)}(2l+3)(2l+5)(2l+7)=-c_{5l}^{(v)}(2l-5)(2l-3)(2l-1)-2^{-l-1}(2l+1).$
(85)

Solving the system of two linear equations (72) and (85) gives the desired
coefficients:

$c_{5l}^{(u)}=-\frac{2^{-l-1}(l+4)(2l-3)}{3(2l+3)(2l+5)(2l+7)},~{}~{}~{}~{}~{}c_{5l}^{(v)}=\frac{2^{-l-1}(l-3)(2l+5)}{3(2l-5)(2l-3)(2l-1)}.$
(86)

It should be noted that the method described above for calculating the
coefficients $c_{5l}^{(u)}$ and $c_{5l}^{(v)}$ is very reliable, but quite
complex. A much simpler method is based on the statement that the point
$\rho=1$ represents the match point for the functions defined by Eqs. (69) and
(70). This means that not only these functions, but also their first (at
least) derivatives must coincide at this point. Thus, the second required
equation relating the coefficients $c_{5l}^{(u)}$ and $c_{5l}^{(v)}$ is:

$\frac{d\sigma_{l}^{(0)}(\rho)}{d\rho}\Big{|}_{\rho=1}+c_{5l}^{(v)}\frac{dv_{5l}(\rho)}{d\rho}\Big{|}_{\rho=1}=\frac{d\sigma_{l}^{(1)}(\rho)}{d\rho}\Big{|}_{\rho=1}+c_{5l}^{(u)}\frac{du_{5l}(\rho)}{d\rho}\Big{|}_{\rho=1}.$
(87)

The solution of the system of two equations (72) and (87) again gives the
coefficients defined by Eq.(86). Substituting these coefficients into the
representations (69) and (70), we finally obtain:

$f_{2}(\alpha,\theta)=\frac{1}{6}\sum_{l=0}^{\infty}\frac{\zeta_{l}(\rho)P_{l}(\cos\theta)}{(2l-1)(2l+3)},$
(88)

where

$\zeta_{l}(\rho)=\left\\{\begin{array}[]{c}{\chi_{l}(\rho)},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}0\leq\rho\leq
1\\\ {\chi_{l}(1/\rho)},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\rho\geq
1\end{array}\right.$ (89)

with

$\chi_{l}(\rho)=\frac{\rho^{l}}{(\rho^{2}+1)^{5/2}}\left[\frac{(l-3)(2l-1)\rho^{6}}{2l+7}+9l\rho^{4}-9(l+1)\rho^{2}-\frac{(l+4)(2l+3)}{2l-5}\right].$
(90)

It is clear that only the function (69) is required for calculating the
function $\chi_{l}(\rho)$. Thus, in fact, we need to calculate only one
coefficient $c_{5l}^{(v)}$ to define this function. In this regard, it is
important to emphasize that the representation (89) reflects the fact that the
WF of a two-electron atomic system must preserve its parity with interchanging
the electrons. For the singlet S-states (which include the ground state) this
means that the AFC and/or its component preserves its form (including the
sign) under the transformation $\alpha\rightleftarrows\pi-\alpha$. For the
AFC-component $f_{2}(\alpha,\theta)$, represented by the series expansion
(61), this property corresponds (in terms of variable $\rho$) to the
relationship:

$\sigma_{l}^{(0)}(\rho^{-1})+c_{5l}^{(v)}v_{5l}(\rho^{-1})=\sigma_{l}^{(1)}(\rho)+c_{5l}^{(u)}u_{5l}(\rho).$
(91)

Elimination of the RHSs between equations (71) and (91) for $\rho=1$ yields
the identity, whereas the use of Eq.(87) instead of Eq.(71) yields the
required equation:

$\frac{d\sigma_{l}^{(0)}(\rho)}{d\rho}\Big{|}_{\rho=1}+c_{5l}^{(v)}\frac{dv_{5l}(\rho)}{d\rho}\Big{|}_{\rho=1}=\frac{d\sigma_{l}^{(0)}(\rho^{-1})}{d\rho}\Big{|}_{\rho=1}+c_{5l}^{(v)}\frac{dv_{5l}(\rho^{-1})}{d\rho}\Big{|}_{\rho=1}.$
(92)

Solution of the last equation gives the coefficients $c_{5l}^{(v)}$ presented
by Eq.(86). Note that the coefficient $c_{5l}^{(u)}$ can then be calculated by
the use of Eq.(71) if needed.

In the general case, we cannot sum the infinite series (88), to obtain the
function $f_{2}(\alpha,\theta)$ in an explicit closed form. However, this can
be done for some special angles $\alpha$ and/or $\theta$. For example, it is
worth noting that the angles $\theta=0,\pi$ correspond to the collinear
configuration LEZ6  of the two-electron atomic system in question. For these
cases we obtain

$f_{2}(\alpha,0)=\pm\frac{(\rho-1)(12\rho^{4}-13\rho^{3}-88\rho^{2}-13\rho+12)}{90(\rho^{2}+1)^{5/2}},$
(93)
$f_{2}(\alpha,\pi)=-\frac{(\rho+1)(12\rho^{4}+13\rho^{3}-88\rho^{2}+13\rho+12)}{90(\rho^{2}+1)^{5/2}}.$
(94)

Sign ”$+$” in Eq.(93) corresponds to $0\leq\alpha\leq\pi/2$ ($0\leq\rho\leq
1$), whereas ”$-$” to $\pi/2\leq\alpha\leq\pi$ ($\rho\geq 1$). The list of
special $\theta$-angles can be supplemented with an intermediate angle
$\theta=\pi/2$:

$f_{2}\left(\alpha,\frac{\pi}{2}\right)=-\frac{2(\rho^{4}-3\rho^{2}+1)}{15(\rho^{2}+1)^{2}}.$
(95)

It is worth noting that for the important cases of the nucleus-electron and
electron-electron coalescence, Eq.(88) respectively reduce to:

$f_{2}\left(0,\theta\right)=-\frac{2}{15},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}f_{2}\left(\frac{\pi}{2},0\right)=0.$
(96)

To derive the results (93)-(96) we used the relationships:

$P_{n}(0)=\sqrt{\pi}\Gamma^{-1}\left(\frac{1-n}{2}\right)\Gamma^{-1}\left(\frac{n}{2}+1\right),~{}~{}~{}~{}~{}P_{n}(1)=1,~{}~{}~{}~{}~{}P_{n}(-1)=(-1)^{n},$
(97)

where $\Gamma(x)$ is the gamma function.

##  III Derivation of the angular Fock coefficient $\psi_{6,3}(\alpha,\theta)$

We start this section by considering the FRR (3)-(4) for $k=6$ and $p=2$:

$\left(\Lambda^{2}-60\right)\psi_{6,2}=48\psi_{6,3}-2V\psi_{5,2}+2E\psi_{4,2}.$
(98)

Next, let’s expand each function in Eq.(98) into HHs, using Eq.(73). This
gives

$\psi_{k,p}=\sum_{n=0(2)}^{\infty}\sum_{l=0}^{n/2}c_{nl}^{(kp)}Y_{n,l}(\alpha,\theta),$
(99)

with $\\{k,p\\}=\\{6,3\\},\\{6,2\\},\\{4,2\\}$, and

$V\psi_{5,2}=\sum_{n=0(2)}^{\infty}\sum_{l=0}^{n/2}\textrm{f}_{nl}Y_{n,l}(\alpha,\theta),$
(100)

where the dimensionless potential $V$ is defined by Eq.(5), whereas the
expansion coefficient $\textrm{f}_{nl}$ can be calculated by the formula

$\textrm{f}_{nl}=\int V\psi_{5,2}Y_{n,l}(\alpha,\theta)d\Omega,$ (101)

according to Eq.(74).

It follows from Eq.(4) that $h_{k,k/2}=0$ for even $k$. Using additionally
Eq.(81), we can conclude that the AFC $\psi_{k,k/2}$ (with even $k$)
represents the linear combination of the HHs, $Y_{k,l}(\alpha,\theta)$.
Whence,

$c_{nl}^{(63)}=0~{}~{}~{}~{}for~{}~{}~{}~{}n\neq 6,$ (102)
$c_{nl}^{(42)}=0~{}~{}~{}~{}for~{}~{}~{}~{}n\neq 4.$ (103)

Equating the coefficients for the HHs, $Y_{6,l}(\alpha,\theta)$ in both sides
of Eq.(98), we obtain:

$0=48c_{6l}^{(63)}-2\textrm{f}_{6l}.$ (104)

Whence (using additionally Eq.(101)),

$c_{6l}^{(63)}=\frac{1}{24}\int V\psi_{5,2}Y_{6,l}(\alpha,\theta)d\Omega.$
(105)

Note that the LHS of Eq.(104) equals zero, because $(\Delta^{2}-60)Y_{6,l}=0$
as follows from Eq.(81).

Thus, according to Eqs.(99) and (102) the AFC $\psi_{6,3}(\alpha,\theta)$
represents a linear combination of four HHs, $Y_{6,l}(\alpha,\theta)$ with
$l=0,1,2,3$. The contribution of each HH is determined by the coefficient
$c_{6l}^{(63)}$ given by Eq.(105). However, it is easy to prove that only the
coefficients with odd values of $l$ are nonzero for
$\psi_{6,3}(\alpha,\theta)$. Indeed, this has already been mentioned in Sec.
II.4 that the WF of a two-electron atom/ion must preserve its parity with
interchanging the electrons. For the singlet S-states this means that only the
HHs, which preserve the sign under transformation
$\alpha\leftrightarrows\pi-\alpha$, differ from zero in the expansion of the
WF, and hence in the expansion of any AFC. In turn, it is easy to show that
only $Y_{n,l}(\alpha,\theta)$ with even values of $(n/2-l)$ satisfy the above
property. Whence, the AFC in question, becomes

$\psi_{6,3}(\alpha,\theta)=a_{61}Y_{6,1}(\alpha,\theta)+a_{63}Y_{6,3}(\alpha,\theta),$
(106)

where we denoted $a_{6l}\equiv c_{6l}^{(63)}$ ($l=1,3$) for convenience and
simplicity, and where the normalized HHs are

$Y_{6,1}(\alpha,\theta)=\frac{2\left[\sin\alpha+3\sin(3\alpha)\right]\cos\theta}{\pi^{3/2}\sqrt{5}},~{}~{}~{}~{}~{}Y_{6,3}(\alpha,\theta)=\frac{8\sin^{3}\alpha
P_{3}(\cos\theta)}{\pi^{3/2}\sqrt{5}}.~{}~{}~{}~{}~{}$ (107)

Using formula (105) and taking into account the representations (5) and (18)
for the dimensionless potential $V$ and the AFC $\psi_{5,2}$, respectively, we
can represent the desired coefficients in the form

$a_{6l}=-\frac{(\pi-2)(5\pi-14)}{6480}\left(I_{l,4}Z^{4}+I_{l,3}Z^{3}+I_{l,2}Z^{2}\right),~{}~{}~{}(l=1,3)$
(108)

where

$I_{l,4}=4\pi^{3/2}\int_{0}^{\pi}\int_{0}^{\pi}\left[f_{3}(\alpha,\theta)+\sqrt{2}f_{4}(\alpha,\theta)\right]\eta
Y_{6,l}(\alpha,\theta)\sin\alpha\sin\theta d\alpha d\theta,$ (109)
$\displaystyle
I_{l,3}=-2\int_{0}^{\pi}\int_{0}^{\pi}\left\\{\frac{3\eta\left[2f_{1}(\alpha,\theta)+f_{2}(\alpha,\theta)\right]}{\sin\alpha}+\frac{\pi^{3/2}\left[f_{3}(\alpha,\theta)+\sqrt{2}f_{4}(\alpha,\theta)\right]}{\xi}\right\\}\times~{}~{}~{}~{}~{}$
$\displaystyle\times Y_{6,l}(\alpha,\theta)\sin^{2}\alpha\sin\theta d\alpha
d\theta,$ (110)
$I_{l,2}=3\int_{0}^{\pi}\int_{0}^{\pi}\left[2f_{1}(\alpha,\theta)+f_{2}(\alpha,\theta)\right]\xi^{-1}Y_{6,l}(\alpha,\theta)\sin^{2}\alpha\sin\theta
d\alpha d\theta.$ (111)

To calculate the integral (III), it is useful to separate the contributions
which include the functions $f_{1},f_{3},f_{4}$ represented by the explicit
closed expressions, and the function $f_{2}$ represented by the infinite
series (88). We obtain:

$I_{l,3}=I_{l,3}^{(134)}-6I_{l,3}^{(2)},$ (112)

where

$\displaystyle
I_{l,3}^{(134)}=-2\int_{0}^{\pi}\int_{0}^{\pi}\left\\{\frac{6\eta
f_{1}(\alpha,\theta)}{\sin\alpha}+\frac{\pi^{3/2}\left[f_{3}(\alpha,\theta)+\sqrt{2}f_{4}(\alpha,\theta)\right]}{\xi}\right\\}Y_{6,l}(\alpha,\theta)\sin^{2}\alpha\sin\theta
d\alpha d\theta,~{}~{}$
$I_{l,3}^{(2)}=\int_{0}^{\pi}\int_{0}^{\pi}f_{2}(\alpha,\theta)\eta
Y_{6,l}(\alpha,\theta)\sin\alpha\sin\theta d\alpha d\theta.$ (114)

The integrals (III) can be taken in explicit (closed) form that gives:

$I_{1,3}^{(134)}=\frac{3(45\pi-122)}{35\pi^{3/2}\sqrt{5}},~{}~{}~{}~{}~{}~{}~{}~{}~{}I_{3,3}^{(134)}=\frac{245\pi-816}{70\pi^{3/2}\sqrt{5}}.$
(115)

The problem of calculating the integrals (114) is that the corresponding
integrands contain the function $f_{2}(\alpha,\theta)$ represented by the
infinite series (88). Fortunately, using the orthogonality relationship for
the Legendre polynomials, we can get these integrals also in explicit form.
Changing the order of summation and integration, we easily obtain:

$\displaystyle
I_{1,3}^{(2)}=\frac{\pi^{-3/2}}{3\sqrt{5}}\sum_{l=0}^{\infty}\int_{0}^{\pi}[\sin\alpha+3\sin(3\alpha)]~{}\eta\left[\frac{\zeta_{l}(\rho)}{(2l-1)(2l+3)}\right]\sin\alpha
d\alpha\int_{0}^{\pi}P_{l}(\cos\theta)\cos\theta\sin\theta d\theta~{}~{}$
$\displaystyle=\frac{2\pi^{-3/2}}{45\sqrt{5}}\int_{0}^{\pi}[\sin\alpha+3\sin(3\alpha)]~{}\eta~{}\zeta_{1}(\rho)\sin\alpha
d\alpha=\frac{7\pi+22}{210\pi^{3/2}\sqrt{5}},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(116) $\displaystyle
I_{3,3}^{(2)}=\frac{4\pi^{-3/2}}{3\sqrt{5}}\sum_{l=0}^{\infty}\int_{0}^{\pi}\eta\left[\frac{\zeta_{l}(\rho)}{(2l-1)(2l+3)}\right]\sin^{4}\alpha~{}d\alpha\int_{0}^{\pi}P_{l}(\cos\theta)P_{3}(\cos\theta)\sin\theta
d\theta~{}~{}~{}~{}$
$\displaystyle=\frac{8\pi^{-3/2}}{945\sqrt{5}}\int_{0}^{\pi}\eta~{}\zeta_{3}(\rho)\sin^{4}\alpha~{}d\alpha=\frac{3\pi-32}{180\pi^{3/2}\sqrt{5}}.~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(117)

Recall that $\eta\equiv\eta(\alpha)$ is defined by Eq.(6) and
$\rho=\tan(\alpha/2)$.

It can be shown (using fairly long non-trivial derivations) that the integrals
$I_{l,2}$ and $I_{l,4}$ vanish both for $l=1$ and $l=3$. This means that
(according to the representations (106) and (108)) the AFC,
$\psi_{6,3}(\alpha,\theta)$ is proportional to the third power of the nucleus
charge $Z$ (only), which is in full agreement with formula (13) from Ref.LEZ4
.

Thus, combining the results of this Section, we obtain the nonzero
coefficients $a_{6,l}$ in the simple final form:

$a_{61}=-\frac{(\pi-2)(5\pi-14)(32\pi-97)}{56700\pi^{3/2}\sqrt{5}}Z^{3},$
(118)
$a_{63}=-\frac{(\pi-2)(5\pi-14)(357\pi-1112)}{680400\pi^{3/2}\sqrt{5}}Z^{3}.$
(119)

##  IV Derivation of the angular Fock coefficients $\psi_{7,3}(\alpha,\theta)$
and $\psi_{8,4}(\alpha,\theta)$

In Sections (II) and (III) we have detailed the derivation of the AFCs
$\psi_{5,2}(\alpha,\theta)$ and $\psi_{6,3}(\alpha,\theta)$, respectively.
Therefore, for the AFCs $\psi_{7,3}(\alpha,\theta)$ and
$\psi_{8,4}(\alpha,\theta)$ we give only abbreviated derivations, and include
extended explanations only in case of significant differences.

###  IV.1 The AFC $\psi_{7,3}(\alpha,\theta)$

The FRR (3)-(4) for $k=7$ and $p=3$ reduces to the form

$\left(\Lambda^{2}-77\right)\psi_{7,3}(\alpha,\theta)=h_{7,3}(\alpha,\theta),$
(120)

where

$h_{7,3}(\alpha,\theta)=-2V\psi_{6,3}(\alpha,\theta).$ (121)

Using Eqs.(106), (107) and (5) the RHS of Eq.(120) can be represented in the
form:

$h_{7,3}(\alpha,\theta)=\frac{(\pi-2)(5\pi-14)Z^{3}}{340200\sqrt{5}\pi^{3/2}}\left\\{\frac{\breve{h}_{1}+\breve{h}_{2}}{\sqrt{5}\pi^{3/2}}-2Z\left[12(32\pi-97)\breve{h}_{3}+(357\pi-1112)\breve{h}_{4}\right]\right\\},$
(122)

where

$\breve{h}_{1}=20\xi^{-1}\left[12(32\pi-97)(1-\xi^{2})+(357\pi-1112)(1-\xi^{2})^{3}\right],$
(123)
$\breve{h}_{2}=60(688-255\pi)\xi^{-1}\sin^{3}\alpha\cos\theta,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(124) $\breve{h}_{3}=\eta\sin^{-1}\alpha
Y_{6,1}(\alpha,\theta),~{}~{}~{}~{}~{}\breve{h}_{4}=\eta\sin^{-1}\alpha
Y_{6,3}(\alpha,\theta).~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$ (125)

Accordingly, the solution of the FRR (120) can be found in the form:

$\psi_{7,3}(\alpha,\theta)=\frac{(\pi-2)(5\pi-14)Z^{3}}{340200\sqrt{5}\pi^{3/2}}\left\\{\frac{\breve{f}_{1}+\breve{f}_{2}}{\sqrt{5}\pi^{3/2}}-2Z\left[12(32\pi-97)\breve{f}_{3}+(357\pi-1112)\breve{f}_{4}\right]\right\\},$
(126)

where the AFC-components $\breve{f}_{i}$ satisfy the IFRRs

$\left(\Lambda^{2}-77\right)\breve{f}_{i}=\breve{h}_{i}.~{}~{}~{}~{}~{}~{}~{}~{}~{}(i=1,2,3,4)$
(127)

Note that the components $\breve{h}_{i}$ of the RHS $h_{7,3}$ of the FRR (120)
for the AFC $\psi_{7,3}$ are reasonably close to the components $h_{i}$ of the
RHS $h_{5,2}$ of the FRR (14). Therefore, we will only briefly dwell on the
conclusions of the corresponding results, as we noted earlier.

It is seen from Eq.(123) that the RHS $\breve{h}_{1}$ is a function of a
single variable $\xi$ defined by Eq.(6). The solution of the corresponding
IFRR have been described in Sec. IV of Ref. LEZ4  (see also Sec. II of Ref.
LEZ2 ) and illustrated (among others) in Sec. II.3 of the current article.
Thus, following the technique mentioned above, we obtain:

$\displaystyle\breve{f}_{1}=\left(\frac{41437\pi}{12}-\frac{74342}{7}\right)\xi^{7}+\left(36476-\frac{35588\pi}{3}\right)\xi^{5}+~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle+\frac{5}{2}(4931\pi-15156)\xi^{3}+5(2276-741\pi)\xi.$ (128)

It can be verified that the RHSs $\breve{h}_{3}$ and $\breve{h}_{4}$ represent
functions of the form $f(\alpha)P_{l}(\cos\theta)$ with $l$ equals $1$ and
$3$, respectively. The solution of the corresponding IFRR have been described
in Sec. V of Ref. LEZ4  and illustrated in Sections II.1 and II.2 of the
current article. This enables us to obtain:

$\breve{f}_{3}=-\frac{\rho(1+\rho)\left(29+\rho\left\\{16+\rho[\rho(16+29\rho)-114]\right\\}\right)\cos\theta}{9\sqrt{5}~{}\pi^{3/2}(\rho^{2}+1)^{7/2}},$
(129)
$\breve{f}_{4}=-\frac{\sin^{3}\alpha\sqrt{1+\sin\alpha}}{2\sqrt{5}~{}\pi^{3/2}}P_{3}(\cos\theta).$
(130)

Remind that the $\rho$ variable was defined previously in Sec. II.1.

The RHS $\breve{h}_{2}$ represented by Eq.(124) is slightly more complicated
than $h_{2}$ discussed in Sec. II.4. In this regard, it would be useful to
clarify two points.

First, using representation (56) for $\xi^{-1}$ we can rewrite Eq.(124) in the
form:

$\breve{h}_{2}=60(688-225\pi)\bar{h}_{2}$ (131)

where

$\bar{h}_{2}=\sum_{l=0}^{\infty}2^{-l}(\sin\alpha)^{l+3}F_{l}(\rho)\cos\theta
P_{l}(\cos\theta),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(132)

and where $F_{l}(\rho)$ is defined by Eqs.(57)-(58). In order to apply to the
solution of the corresponding IFRR by the method described in Sec. II.4 (see
also LEZ4 ), the $\theta$-dependent $l$-component in the series expansion of
$\bar{h}_{2}$ must be pure $P_{l}(\cos\theta)$. To solve the problem one could
use the general formula representing the Clebsch-Gordan series for product of
two _spherical_ harmonics. However, in our simple case, it is easier to use
the recurrence relation for the Legendre polynomials

$(l+1)P_{l+1}(x)-(2l+1)xP_{l}(x)+lP_{l-1}(x)=0,$ (133)

which enables us to represent $\bar{h}_{2}$ in the desired form:

$\bar{h}_{2}=\sum_{l=0}^{\infty}\bar{\textrm{h}}_{l}(\rho)(\sin\alpha)^{l}P_{l}(\cos\theta),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(134)

where

$\bar{\textrm{h}}_{l}(\rho)=\frac{l}{2^{l-1}(2l-1)}\sin^{2}\alpha
F_{l-1}(\rho)+\frac{l+1}{2^{l+1}(2l+3)}\sin^{4}\alpha F_{l+1}(\rho).$ (135)

The second point is related to the calculation of the coefficient

$\bar{h}_{2l,l}=\pi^{2}\int_{0}^{\pi}\int_{0}^{\pi}\bar{h}_{2}(\alpha,\theta)Y_{2l,l}(\alpha,\theta)\sin^{2}\alpha\sin\theta
d\alpha d\theta$ (136)

in HH-expansion of $\bar{h}_{2}$ (see the corresponding Eq.(84) for
calculation of $\psi_{5,2}(\alpha,\theta)$). Of course, we can use
representation (134)-(135) and then apply the orthogonality condition for the
Legendre polynomials. However, the simpler way is to use the original
representation (132) taking into account that $\cos\theta\equiv
P_{1}(\cos\theta)$. In this case, we can apply the well-known formula for the
integral of three Legendre polynomials

$\int_{-1}^{1}P_{l}(x)P_{L}(x)P_{l^{\prime}}(x)dx=2\begin{pmatrix}l&L&l^{\prime}\\\
0&0&0\\\ \end{pmatrix}^{2},$ (137)

where the RHS represents twice the square of the Wigner 3-j symbol.

Thus, applying the methodology outlined in Sec. II.4, and given the above
features, one obtains

$\bar{f}_{2}=\frac{1}{48}\sum_{l=0}^{\infty}\frac{\bar{\zeta}_{l}(\rho)P_{l}(\cos\theta)}{(2l-1)(2l+3)},$
(138)

where

$\bar{\zeta}_{l}(\rho)=\left\\{\begin{array}[]{c}{\bar{\chi}_{l}(\rho)},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}0\leq\rho\leq
1\\\ {\bar{\chi}_{l}(1/\rho)},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\rho\geq
1\end{array}\right.$ (139)

with

$\displaystyle\bar{\chi}_{l}(\rho)=-\frac{\rho^{l}}{(\rho^{2}+1)^{7/2}}\Big{\\{}\frac{(32l^{2}+26l-25)\rho^{6}}{2l+5}\left[\frac{(2l-1)\rho^{2}}{2l+9}+4\right]+~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle+\frac{1}{2l-3}\left[\frac{6(84l^{2}+84l-95)\rho^{4}}{2l+5}-(32l^{2}+38l-19)\left(\frac{2l+3}{2l-7}+4\rho^{2}\right)\right]\Big{\\}}.~{}~{}~{}~{}~{}~{}~{}~{}$
(140)

Recall that the component $\breve{f}_{2}$ in the RHS of Eq.(126) is equal to
$60(688-225\pi)\bar{f}_{2}$ according to representation (131).

As in the case of the AFC $\psi_{5,2}(\alpha,\theta)$, there are combinations
of special hyperspherical angles $\alpha$ and $\theta$ for which the component
$\bar{f}_{2}\equiv\bar{f}_{2}(\alpha,\theta)$ of the AFC
$\psi_{7,3}(\alpha,\theta)$ can be obtained in closed form. In particular, one
obtains:

$\bar{f}_{2}(\alpha,0)=\mp\frac{(\rho-1)(95\rho^{6}+1166\rho^{5}-1879\rho^{4}-8844\rho^{3}-1879\rho^{2}+1166\rho+95)}{5040(\rho^{2}+1)^{7/2}},$
(141)
$\bar{f}_{2}(\alpha,\pi)=\frac{(\rho+1)(95\rho^{6}-1166\rho^{5}-1879\rho^{4}+8844\rho^{3}-1879\rho^{2}-1166\rho+95)}{5040(\rho^{2}+1)^{7/2}},$
(142)
$\bar{f}_{2}(\alpha,\frac{\pi}{2})=\frac{19\rho^{4}+10\rho^{2}+19}{1008(\rho^{2}+1)^{2}}.$
(143)

Sign ”$-$” in Eq.(141) corresponds to $0\leq\alpha\leq\pi/2$ ($0\leq\rho\leq
1$), whereas ”$+$” to $\pi/2\leq\alpha\leq\pi$ ($\rho\geq 1$).

For the important cases of the nucleus-electron and electron-electron
coalescence, representation (138)-(IV.1) is simplified to:

$\bar{f}_{2}\left(0,\theta\right)=\frac{19}{1008},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\bar{f}_{2}\left(\frac{\pi}{2},0\right)=0.$
(144)

###  IV.2 The AFC $\psi_{8,4}(\alpha,\theta)$

Having at our disposal the AFC $\psi_{7,3}\equiv\psi_{7,3}(\alpha,\theta)$, we
can calculate the AFC $\psi_{8,4}\equiv\psi_{8,4}(\alpha,\theta)$ using the
FRR (3)-(4) for $k=8$ and $p=3$:

$\left(\Lambda^{2}-96\right)\psi_{8,3}=80\psi_{8,4}-2V\psi_{7,3}+2E\psi_{6,3}.$
(145)

It follows from Eq.(81) and the FRR (3)-(4) for $k=8$ and $p=4$ that the AFC,
$\psi_{8,4}$ is a linear combination of the HHs, $Y_{8,l}\equiv
Y_{8,l}(\alpha,\theta)$. Moreover, given that only $Y_{n,l}(\alpha,\theta)$
with even values of $n/2-l$ are suitable for singlet S-states, we obtain:

$\psi_{8,4}=a_{80}Y_{8,0}+a_{82}Y_{8,2}+a_{84}Y_{8,4}.$ (146)

For further derivations, it is advisable to represent the HHs in the form

$Y_{8,l}(\alpha,\theta)=y_{8l}(\alpha)P_{l}(\cos\theta),$ (147)

where

$y_{80}(\alpha)=\pi^{-3/2}\left[2\cos(4\alpha)+2\cos(2\alpha)+1\right],$ (148)
$y_{82}(\alpha)=\frac{2}{\pi^{3/2}}\sqrt{\frac{10}{7}}\sin^{2}\alpha\left[4\cos(2\alpha)+3\right],$
(149) $y_{84}(\alpha)=\frac{8}{\pi^{3/2}}\sqrt{\frac{2}{7}}\sin^{4}\alpha.$
(150)

It was found in Sec. III that $\psi_{6,3}\equiv\psi_{6,3}(\alpha,\theta)$ is
the linear combination of the HHs $Y_{6,l}(\alpha,\theta)$. Thus, expanding
each function of Eq.(145) in HHs, and equating the coefficients for $Y_{8,l}$,
we obtain (see the corresponding result (105) for $a_{6l}$)

$a_{8l}=\frac{1}{40}\int V\psi_{7,3}Y_{8,l}d\Omega,$ (151)

where the potential $V$ is defined by Eq.(5). When deriving the last equation,
it was taken into account that $(\Delta^{2}-96)Y_{8,l}=0$, as follows from
Eq.(81).

Direct substitution of the representations (5), (126) and (147) into the RHS
of Eq. (151), yields:

$\displaystyle
a_{8l}=\frac{(\pi-2)(5\pi-14)Z^{3}}{13608000\sqrt{5}~{}\pi^{3/2}}\times~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle\times\int\left(\frac{1}{\xi}-\frac{2Z\eta}{\sin\alpha}\right)\left\\{\frac{\breve{f}_{1}+\breve{f}_{2}}{\sqrt{5}\pi^{3/2}}-2Z\left[12(32\pi-97)\breve{f}_{3}+(357\pi-1112)\breve{f}_{4}\right]\right\\}Y_{8,l}(\alpha,\theta)d\Omega.~{}~{}$

It follows from Eq.(13) of Ref.LEZ4  that only the coefficients at $Z^{4}$ are
nonzero on the RHS of the last equation. Whence, Eq.(IV.2) reduces to the
form:

$a_{8l}=-\frac{(\pi-2)(5\pi-14)Z^{4}}{6804000\sqrt{5}~{}\pi^{3/2}}\left[\frac{S_{1l}+S_{2l}}{\sqrt{5}\pi^{3/2}}+12(32\pi-97)S_{3l}+(357\pi-1112)S_{4l}\right],$
(153)

where

$S_{1l}=\pi^{2}\int_{0}^{\pi}\int_{0}^{\pi}\breve{f}_{1}(\xi)Y_{8,l}(\alpha,\theta)\eta\sin\alpha\sin\theta
d\alpha d\theta,$ (154) $\displaystyle
S_{2l}=\pi^{2}\int_{0}^{\pi}\int_{0}^{\pi}\breve{f}_{2}(\alpha,\theta)Y_{8,l}(\alpha,\theta)\eta\sin\alpha\sin\theta
d\alpha
d\theta=~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle=\frac{5\pi^{2}(688-225\pi)}{(2l-1)(2l+1)(2l+3)}\int_{0}^{\pi/2}\bar{\chi}_{l}(\rho)y_{8l}(\alpha)\eta\sin\alpha
d\alpha,~{}~{}~{}~{}~{}~{}$ (155)
$S_{nl}=\pi^{2}\int_{0}^{\pi}\int_{0}^{\pi}\breve{f}_{n}(\alpha,\theta)Y_{8,l}(\alpha,\theta)\xi^{-1}\sin^{2}\alpha\sin\theta
d\alpha d\theta.~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(n=3,4)$ (156)

The identifiers $\xi$ and $\eta$ are defined by Eq.(6), whereas functions
$\bar{\chi}_{l}(\rho)$ can be calculated by the formula (IV.1). When deriving
Eq.(IV.2) we applied the orthogonality condition for the Legendre polynomials.
Fortunately, all integrals (154)-(156) can be taken in closed form. Thus, by
collecting these results and substituting them into the RHS of Eq.(153), we
finally obtain the desired coefficients in the form:

$a_{8l}=\frac{Z^{4}(\pi-2)(5\pi-14)}{\pi^{5/2}}b_{8l},$ (157)

with

$\displaystyle
b_{80}=\frac{\pi(150339\pi-927292)+1430792}{19289340000},~{}~{}~{}b_{82}=\frac{\pi(751965\pi-4654046)+7200976}{1928934000\sqrt{70}},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle
b_{84}=\frac{\pi(3190317\pi-19828996)+30802176}{25719120000\sqrt{14}}.~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(158)

##  V Results and Discussions

The angular Fock coefficients $\psi_{k,p}\equiv\psi_{k,p}(\alpha,\theta)$ with
the _maximum possible_ value of subscript $p$ were calculated on examples of
the coefficients with $5\leq k\leq 8$. The AFCs $\psi_{9,4}$ and $\psi_{10,5}$
are additionally presented in the Appendix without derivations. The presented
technique makes it possible to calculate such AFCs for any arbitrarily large
$k$. These coefficients are leading in the logarithmic power series
representing the Fock expansion (see Eq.(I)). As such, they may be
indispensable for the development of simple methods for calculating the
helium-like electronic structure.

The proposed technique, as well as the final results, are quite complex.
Therefore, both require verification. We are aware of two ways of the above-
mentioned verification. The first one is to use the Green’s function (GF)
approach (see, Ref. FOCK  and also Ref. LEZ7 , Sec. 4) which enables us to
calculate (at least, numerically) the AFC (or its component) by the following
integral representation:

$\psi_{k,p}(\alpha,\theta)=\frac{1}{8\pi}\int_{0}^{\pi}d\alpha^{\prime}\sin^{2}\alpha^{\prime}\int_{0}^{\pi}d\theta^{\prime}\sin\theta^{\prime}~{}h_{k,p}(\alpha^{\prime},\theta^{\prime})\int_{0}^{\pi}\frac{\cos\left[\left(\frac{k}{2}+1\right)\omega\right]}{\sin\omega}(1-\lambda)d\varphi,$
(159)

where $\omega$ is an angle defined by the relation

$\cos\omega=\cos\alpha\cos\alpha^{\prime}+\sin\alpha\sin\alpha^{\prime}\left(\cos\theta\cos\theta^{\prime}+\sin\theta\sin\theta^{\prime}\cos\varphi\right),$
(160)

whereas

$\lambda=\left\\{\begin{array}[]{c}0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}k~{}~{}~{}odd\\\
\omega/\pi~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}k~{}~{}~{}even\\\
\end{array}\right..$ (161)

For even $k$ and maximum value of $p=k/2$ the RHS $h_{k,k/2}$ of the FRR (3)
equals zero. This implies that the GF formula (159) cannot be applied in this
case. Hence, only the AFCs $\psi_{k,p}$ with odd values of $k$ (and maximum
$p$) can be verified with the GF method. Thus, numerically calculating (for
various combinations of angles $\alpha$ and $\theta$) the triple integrals
(159) representing the AFCs $\psi_{5,2}(\alpha,\theta)$,
$\psi_{7,3}(\alpha,\theta)$ and $\psi_{9,4}(\alpha,\theta)$, we have verified
that the representations obtained for them in Sections II, IV.1 and in the
Appendix are correct.

The second method of the verification under consideration, covering all
possible combinations of angles, is much more complicated. This is the CFHH
method mentioned in Introduction. It is based on decomposing the full WF into
a form

$\Psi^{\textrm{CFHH}}(r_{1},r_{2},r_{12})=\exp\left[f(r_{1},r_{2},r_{12})\right]\Phi^{\textrm{CFHH}}(R,\alpha,\theta),$
(162)

where the so called correlation function $f$ can be taken in a simple linear
form

$f(r_{1},r_{2},r_{12})=c_{1}r_{1}+c_{2}r_{2}+c_{12}r_{12}.$ (163)

The so called ”cusp parametrization”

$c_{1}=c_{2}=-Z,~{}~{}~{}~{}~{}~{}~{}c_{12}=1/2$ (164)

is used as a rule. For small enough hyperspherical radius $R$, the function
$\Phi$ is represented as

$\Phi^{\textrm{CFHH}}(R,\alpha,\theta)=\frac{1}{d_{0,0}(\alpha,\theta)}\sum_{k=0}^{K}(2\kappa
R)^{k}\sum_{p=0}^{[k/2]}d_{k,p}(\alpha,\theta)\ln^{p}(2\kappa R),$ (165)

where $\kappa=\sqrt{-2E}$, and functions $d_{k,p}(\alpha,\theta)$ are expanded
in $N$ (basis size) HHs. It follows from representation (165) that the AFCs
$\psi_{k,p}(\alpha,\theta)$ can be expressed in terms of the functions
$d_{k^{\prime},p^{\prime}}(\alpha,\theta)$ calculated by the CFHHM. For
example, for the AFCs in questions one obtains:

$\psi_{k,k/2}^{\textrm{CFHH}}(\alpha,\theta)=\frac{(2\kappa)^{k}d_{k,k/2}(\alpha,\theta)}{d_{0,0}(\alpha,\theta)}.$
(166)

We have calculated all AFCs discussed in this article using CFHHM with $K=18$
and $N=1600$. The angles $0\leq\alpha\leq\pi$ and $0\leq\theta\leq\pi$ with
step $\pi/6$ were considered. The relative difference
$|1-\psi_{k,p}(\alpha,\theta)/\psi_{k,p}^{\textrm{CFHH}}(\alpha,\theta)|$ was
less than $10^{-7}$ for all examined cases, including $1\leq Z\leq 5$. This
indicates that all our theoretical calculations were correct.

## Appendix A

In Sections II-III the AFCs $\psi_{5,2}(\alpha,\theta)$ and
$\psi_{6,3}(\alpha,\theta)$ were calculated with detailed derivations. In Sec.
IV the AFCs $\psi_{7,3}(\alpha,\theta)$ and $\psi_{8,4}(\alpha,\theta)$ were
presented with a very brief derivations. In this Appendix, we present the AFCs
$\psi_{9,4}\equiv\psi_{9,4}(\alpha,\theta)$ and
$\psi_{10,5}\equiv\psi_{10,5}(\alpha,\theta)$ without derivation. To find the
latter AFCs, the methods described in the main sections were used.

So, the first AFC under consideration can be represented as:

$\psi_{9,4}=2Z^{4}\left[2ZX_{1}(\alpha,\theta)-X_{2}(\alpha,\theta)\right],$
(167)

where

$X_{1}(\alpha,\theta)=\check{a}_{80}\check{f}_{1}+\check{a}_{82}\check{f}_{2}+\check{a}_{84}\check{f}_{3},$
(168)
$X_{2}(\alpha,\theta)=\frac{35}{\pi^{3/2}}\sqrt{\frac{2}{7}}~{}\check{a}_{84}\check{f}_{4}+\frac{(\pi-2)(5\pi-14)}{123451776000\pi^{4}}\left[c_{5}\check{f}_{5}+c_{6}\check{f}_{6}+c_{7}\check{f}_{7}+16(c_{8}\check{f}_{8}+c_{9}\check{f}_{9})\right].$
(169)

Here $\check{a}_{8l}=Z^{-4}a_{8l}$, where the coefficients $a_{8l}$ are
defined by Eqs.(157), (IV.2), whereas the other coefficients are:

$\displaystyle
c_{5}=\pi(29757524-4780401\pi)-46286848,~{}~{}~{}c_{6}=\pi(9581100\pi-59458928)+92239360,~{}~{}~{}~{}~{}$
$\displaystyle
c_{7}=\pi(28060+10149\pi)-167168,~{}~{}~{}~{}c_{8}=9\pi(134543\pi-828732)+11488128,~{}~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle
c_{9}=\pi(4804833\pi-29773780)+46119680.~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(170)

The functions $\check{f}_{i}\equiv\check{f}_{i}(\alpha,\theta)$ are:

$\check{f}_{1}=-\frac{(\rho+1)(563\rho^{8}+1012\rho^{7}-8932\rho^{6}-3668\rho^{5}+23954\rho^{4}-3668\rho^{3}-8932\rho^{2}+1012\rho+563)}{1260\pi^{3/2}(\rho^{2}+1)^{9/2}},$
(171)
$\check{f}_{2}=-\left(\frac{8}{\pi^{3/2}}\sqrt{\frac{10}{7}}\right)\frac{\rho^{2}(1+\rho)(126+49\rho-424\rho^{2}+49\rho^{3}+126\rho^{4})}{300(\rho^{2}+1)^{9/2}}P_{2}(\cos\theta),$
(172)
$\check{f}_{3}=-\frac{2}{5\pi^{3/2}}\sqrt{\frac{2(1+\sin\alpha)}{7}}\sin^{4}\alpha~{}P_{2}(\cos\theta),$
(173)
$\check{f}_{4}=-\frac{\xi(315-1680\xi^{2}+2814\xi^{4}-1854\xi^{6}+419\xi^{8})}{1260},$
(174)
$\check{f}_{5}=-\frac{\xi}{60}(2\xi^{2}-3)(2\xi^{2}-1)(4\xi^{4}-10\xi^{2}+5).$
(175)

The remaining $\check{f}$-functions are represented by series:

$\check{f}_{j}=\frac{(\rho^{2}+1)^{-9/2}}{k_{j}}\sum_{l=0}^{\infty}\frac{\rho^{l}\zeta_{jl}(\rho)}{(2l-1)(2l+3)}P_{l}(\cos\theta),~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(j=6,7,8,9)$
(176)

where

$k_{6}=6,~{}~{}~{}~{}~{}~{}k_{7}=60,~{}~{}~{}~{}~{}~{}k_{8}=24,~{}~{}~{}~{}~{}~{}k_{9}=40,$
(177)

and the corresponding $\zeta$-functions are:

$\displaystyle\zeta_{6l}(\rho)=\frac{(2l-15)(2l-1)(l+1)\rho^{10}}{(2l+7)(2l+11)}+\frac{(22l^{2}-5l-12)\rho^{8}}{(2l+7)}+\frac{10(2l^{2}+11l+3)\rho^{6}}{2l+7}-~{}$
$\displaystyle-\frac{10(2l^{2}-7l-6)\rho^{4}}{2l-5}-\frac{(22l^{2}+49l+15)\rho^{2}}{2l-5}-\frac{l(2l+3)(2l+17)}{(2l-9)(2l-5)},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(178)
$\displaystyle\zeta_{7l}(\rho)=\frac{(2l-1)(4l^{2}+160l-189)\rho^{10}}{(2l+7)(2l+11)}+\frac{35(4l^{2}+40l-9)\rho^{8}}{(2l+7)}-\frac{350(4l^{2}+16l+3)\rho^{6}}{2l+7}+~{}$
$\displaystyle+\frac{350(4l^{2}-8l-9)\rho^{4}}{2l-5}-\frac{35(4l^{2}-32l-45)\rho^{2}}{2l-5}-\frac{(2l+3)(4l^{2}-152l-345)}{(2l-9)(2l-5)},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(179)
$\displaystyle\zeta_{8l}(\rho)=-\frac{(2l-1)(56l^{3}+250l^{2}+338l+171)\rho^{10}}{(2l+5)(2l+7)(2l+11)}-\frac{(136l^{3}+314l^{2}-110l-153)\rho^{8}}{(2l+5)(2l+7)}-~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle-\frac{2(80l^{4}+652l^{3}+566l^{2}-1824l-873)\rho^{6}}{(2l-3)(2l+5)(2l+7)}+\frac{2(80l^{4}-332l^{3}-910l^{2}+1320l+945)\rho^{4}}{(2l-5)(2l-3)(2l+5)}+~{}~{}~{}$
$\displaystyle+\frac{(136l^{3}+94l^{2}-330l-135)\rho^{2}}{(2l-5)(2l-3)}+\frac{(2l+3)(56l^{3}-82l^{2}+6l-27)}{(2l-9)(2l-5)(2l-3)},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(180)
$\displaystyle\zeta_{9l}(\rho)=\frac{(2l-1)(24l^{3}-150l^{2}-670l-439)\rho^{10}}{(2l+5)(2l+7)(2l+11)}+\frac{5(72l^{3}+162l^{2}-70l-103)\rho^{8}}{(2l+5)(2l+7)}+~{}~{}~{}~{}~{}~{}~{}$
$\displaystyle+\frac{10(16l^{4}+220l^{3}+222l^{2}-804l-423)\rho^{6}}{(2l-3)(2l+5)(2l+7)}-\frac{10(16l^{4}-156l^{3}-342l^{2}+652l+399)\rho^{4}}{(2l-5)(2l-3)(2l+5)}-~{}~{}~{}$
$\displaystyle-\frac{5(72l^{3}+54l^{2}-178l-57)\rho^{2}}{(2l-5)(2l-3)}-\frac{(2l+3)(24l^{3}+222l^{2}-298l-57)}{(2l-9)(2l-5)(2l-3)},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}$
(181)

It is important to emphasize that the representations (176)-(A) are valid only
for $0\leq\rho\leq 1$. For values $\rho>1$, one should replace $\rho$ with
$1/\rho$, which is equivalent to simply redefining $\rho$ as $\cot(\alpha/2)$.

The second AFC under consideration is of the form:

$\psi_{10,5}=-\frac{Z^{5}(\pi-2)(5\pi-14)}{\pi^{7/2}}\left[b_{10,1}Y_{10,1}(\alpha,\theta)+b_{10,3}Y_{10,3}(\alpha,\theta)+b_{10,5}Y_{10,5}(\alpha,\theta)\right],$
(182)

where

$b_{10,1}=\frac{\pi[3\pi(6840010557\pi-63828704998)+595609133656]-617517605744}{401025378600000\sqrt{105}},$
(183)
$b_{10,3}=\frac{\pi[\pi(9194460432\pi-85833963053)+267084629592]-277009842768}{100256344650000\sqrt{30}},$
(184)
$b_{10,5}=\frac{\pi[\pi(622341848670\pi-5812646794643)+18095537797140]-18776793358080}{10025634465000000\sqrt{42}},$
(185)

and $Y_{10,l}(\alpha,\theta)$ with $l=1,3,5$ are the normalized HHs.

## References

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  * (2) J. D. Morgan III, Convergence properties of Fock’s expansion for S-state eigenfunctions of the helium atom, Theor. Chim. Acta 69, 181-223 (1986).
  * (3) P. C. Abbott and E. N. Maslen, Coordinate systems and analytic expansions for three-body atomic wavefunctions: I. Partial summation for the Fock expansion in hyperspherical coordinates, J. Phys. A: Math. Gen. 20, 2043-2075 (1987).
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