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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 12 "Atomic Model" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 290 "A simple model fo
r the structure of few-electron atoms is presented. A model potential \+
for the helium atom is introduced (it represents a good approximation \+
to the so-called Hartee-Fock potential for the helium atom). The eigen
energies and eigenfunctions of this potential are determined by" }}
{PARA 0 "" 0 "" {TEXT -1 96 "a) the variational method for the lowest-
lying eigenstate for a given angular momentum symmetry;" }}{PARA 0 "" 
0 "" {TEXT -1 62 "b) the numerical solution of the radial Schroedinger
 equation;" }}{PARA 0 "" 0 "" {TEXT -1 62 "c) a matrix diagonalization
 using a Slater-type orbital basis." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 255 "A simple model potential for the ground
 state of helium can be obtained as follows: assume that in the ground
 state the wavefunction corresponds to an 1s^2 configuration (one spin
-up, one spin-down electron). What potential should the electron exper
ience?" }}{PARA 0 "" 0 "" {TEXT -1 96 "The nuclear attraction -2/r and
 a repulsive potential due to the presence of the other electron." }}
{PARA 0 "" 0 "" {TEXT -1 133 "What is the repulsive potential due to t
he other electron (which has the same 1s-wavefunction, it has just opp
osite spin projection)?" }}{PARA 0 "" 0 "" {TEXT -1 426 "We have an ch
icken-and-egg problem here: given a potential, we can determine the |1
s> state, and calculate the repulsive part of the potential correctly.
 This is the objective of a self-consistent field calculation. Then we
 would need to repeat the calculation of the new |1s> state using the \+
new potential. Repeating this procedure one can come up with a |1s> st
ate and corresponding potential that provide the lowest energy." }}
{PARA 0 "" 0 "" {TEXT -1 89 "We break the closed circuit by stating an
 approximate answer for the potential in helium:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 44 "V:=r->-2/r+1/r*(1-exp(-3.36*r)*(1+1.665*r));
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VGR6#%\"rG6\"6$%)operatorG%&ar
rowGF(,&*&\"\"\"F.9$!\"\"!\"#*&,&F.F.*&-%$expG6#,$F/$!$O$F1F.,&F.F.*&$
\"%l;!\"$F.F/F.F.F.F0F.F/F0F.F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 200 "This is the potential experienced by an 1s-electron in the hel
ium atom: at short distances the electron just feels the full nuclear \+
attraction (-2/r) as the second expression approaches a constant as " 
}{TEXT 272 1 "r" }{TEXT -1 76 " goes to zero. At large distances the e
lectronic repulsion term goes like 1/" }{TEXT 273 1 "r" }{TEXT -1 80 "
 and screens the nucleus by one unit so that the overall potential goe
s like -1/" }{TEXT 274 1 "r" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "We can check for the consistenc
y by a simple variational calculation:" }}{PARA 0 "" 0 "" {TEXT -1 
112 "We start with an unnormalized 1s state that depends on a 'charge'
 parameter (cf. the hydrogen-like wavefunction)" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 20 "chi:=r*exp(-beta*r);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$chiG*&%\"rG\"\"\"-%$expG6#,$*&%%betaGF'F&F'!\"\"F'" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "We are using the radial wavefu
nction (i.e. r*R_nl(r)) so that the radial kinetic energy is just prop
ortional to the second derivative." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "assume(beta>0);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 84 "E1sT:=Int(chi*(-1/2*Diff(chi,r$2)+V(r)*chi),r=0..infi
nity)/Int(chi^2,r=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%
E1sTG*&-%$IntG6$*(%\"rG\"\"\"-%$expG6#,$*&%&beta|irGF+F*F+!\"\"F+,&-%%
DiffG6$*&F*F+F,F+-%\"$G6$F*\"\"##F2F;*(,&*&F+F+F*F2!\"#*&,&F+F+*&-F-6#
,$F*$!$O$F@F+,&F+F+*&$\"%l;!\"$F+F*F+F+F+F2F+F*F2F+F+F*F+F,F+F+F+/F*;
\"\"!%)infinityGF+-F'6$*&)F*F;F+)F,F;F+FNF2" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 18 "E1sT:=value(E1sT);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%E1sTG,$*&*&,,$!'/Ff\"\"!\"\"\"*&$\"&+D'F+F,)%&beta|irG\"\"%F,
F,*&$\"&+]'F+F,)F1\"\"$F,F,*&$\"'D*=&F+F,)F1\"\"#F,!\"\"*&$\"'[?wF+F,F
1F,F=F,F1F,F,,**$F6F,$\"&Dc\"F+*&$\"&](yF+F,F;F,F,*&$\"'+B8F+F,F1F,F,$
\"&)3uF+F,F=$\"++++]7!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 
"plot(E1sT,beta=1.55..1.8);" }}{PARA 13 "" 1 "" {GLPLOT2D 693 299 299 
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)o*)F-7$$\"3i;z%*f%)Q!o\"F*$!3gJ<:X?Fp*)F-7$$\"3+](oza'=&o\"F*$!3I*)f
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BELSG6$Q&beta|ir6\"Q!6\"-%%VIEWG6$;$\"$b\"!\"#$\"#=F^[l%(DEFAULTG" 1 
2 0 1 10 0 2 9 1 4 2 1.000000 44.000000 45.000000 0 0 "Curve 1" }}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "beta0:=fsolve(diff(E1sT,beta
),beta=1..2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&beta0G$\"+PG\\(o\"
!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "E0:=subs(beta=beta0
,E1sT);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E0G$!+K$o%p*)!#5" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 606 "Note that the beta-value is less \+
than 2. The value of beta=2 would be obtained if there was only one 1s
-electron (the hydrogen-like solution for the ground state of the He+ \+
ion). The fact that the electron wants to have a slightly more diffuse
 wavefunction reflects the so-called inner screening: the electrostati
c repulsion when combined with the simple model of two identical elect
rons (apart from the spin projection) leads to two electrons which rep
el each other on average (independent electron model). Both electrons \+
experience the same common central potential and are bound by the same
 eigenenergy." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Now we would lik
e to check two things:" }}{PARA 0 "" 0 "" {TEXT -1 45 "1) how accurate
 is this variational solution?" }}{PARA 0 "" 0 "" {TEXT -1 85 "2) how \+
close is the used potential to the potential produced by this 1s-wavef
unction?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
98 "Part 1 can be answered by using dsolve[numeric]. For part 2 we nee
d to solve the Poisson equation." }}{PARA 0 "" 0 "" {TEXT -1 119 "Let \+
us start with the first question: To solve the SE we start the integra
tion not at zero, but at some small value of " }{TEXT 257 5 "r=eta" }
{TEXT -1 26 " to avoid the singularity." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "eta:=10^(-8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$e
taG#\"\"\"\"*++++\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "IC:=
phi(eta)=eta,D(phi)(eta)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6
$/-%$phiG6##\"\"\"\"*++++\"F*/--%\"DG6#F(F)F+" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 12 "Et:=-0.9042;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%#EtG$!%U!*!\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "SE:=-1/
2*diff(phi(r),r$2)+(V(r)-Et)*phi(r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%#SEG,&-%%diffG6$-%$phiG6#%\"rG-%\"$G6$F,\"\"##!\"\"F0*&,(*&\"\"\"F
6F,F2!\"#*&,&F6F6*&-%$expG6#,$F,$!$O$F7F6,&F6F6*&$\"%l;!\"$F6F,F6F6F6F
2F6F,F2F6$\"%U!*!\"%F6F6F)F6F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 57 "sol:=dsolve(\{SE,IC\},phi(r),numeric,output=listprocedure):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phir:=subs(sol,phi(r)):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "plot(phir,0..10);" }}
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45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E
1s:=Et;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E1sG$!%U!*!\"%" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 272 "The numerically exact solution fo
r the radial SE yields a slightly lower eigenvalue of E_1s = -0.9042 a
.u. (1 a.u. = 27.12 eV) compared to the variational result of E_1s_v =
-0.897 a.u.. This means that the true eigenfunction is somewhat differ
ent from the hydrogenic form." }}{PARA 0 "" 0 "" {TEXT -1 243 "The num
erical eigenfunction is not normalized properly. To compare the graphs
 we simply change the normalization of the variational answer. In fact
, the variational state was normalized such that the derivative of the
 function equals unity at " }{TEXT 258 1 "r" }{TEXT -1 3 "=0." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "eval(subs(r=0,diff(subs(beta
=beta0,chi),r)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"\"\"\"\"!" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}{PARA 7 "" 
1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined\n
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "P1:=plot(phir,0..6,colo
r=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "P2:=plot(subs(be
ta=beta0,chi),r=0..6,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "display([P1,P2]);" }}{PARA 13 "" 1 "" {GLPLOT2D 695 
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LABELSG6%Q!6\"F]hm%(DEFAULTG-%%VIEWG6$;FdblFegmF_hm" 1 2 0 1 10 0 2 9 
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{PARA 0 "" 0 "" {TEXT -1 152 "Note that the agreement for the energy w
as at the level of  7/900 , i.e. in the 1 % range. The deviation betwe
en the wavefunctions is in the 10 % range." }}{PARA 0 "" 0 "" {TEXT 
-1 154 "Nevertheless, we can state that the simple hydrogenic wavefunc
tion catches the main feature of the numerical solution, namely the mo
st likely location in " }{TEXT 259 1 "r" }{TEXT -1 21 " for the 1s ele
ctron." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "The result for the bin
ding energy for one of the two helium electrons is the first quantity \+
that can be compared with experiment:" }}{PARA 0 "" 0 "" {TEXT -1 4 "T
he " }{TEXT 260 20 "ionization potential" }{TEXT -1 82 " of helium is \+
measured to be 24.481 eV (cf. R. Liboff, table 12.2). Our answer is:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "-Et*27.12*_eV;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#,$%$_eVG$\")/>_C!\"'" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 149 "For any atom the eigenenergy of the highest occupie
d orbital should equal the negative of the ionization potential. Our r
esult is indeed quite close." }}{PARA 0 "" 0 "" {TEXT -1 328 "There is
 another quantity that can be measured, namely the total energy of the
 atom (equal to the sum of ionization energies for both electrons). Th
is is not simply twice the eigenenergy, since after ionizing on of the
 two He-electrons, the other is left in a hydrogen-like state (energy \+
of -2 a.u. due to Z=2, and E_1s=-Z^2/2)." }}{PARA 0 "" 0 "" {TEXT -1 
97 "Combining this answer with the calculated 1s binding energy we hav
e for the total binding energy:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 17 "(Et-2)*27.12*_eV;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%$_eVG$
!)/>wy!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 261 28 "The electrostatic \+
repulsion:" }}{PARA 0 "" 0 "" {TEXT -1 256 "Now we look at the questio
n as to what potential is associated with the approximate wavefunction
. For this we need a solution to the Poisson equation for a sphericall
y symmetric charge distribution based on the properly normalized varia
tional wavefunction." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "A1s
:=1/sqrt(int(chi^2,r=0..infinity));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%$A1sG,$*$)%&beta|irG#\"\"$\"\"#\"\"\"F+" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 33 "rho:=subs(beta=beta,(A1s*chi)^2);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%$rhoG,$*()%&beta|irG\"\"$\"\"\")%\"rG\"\"#F*)-%$ex
pG6#,$*&F(F*F,F*!\"\"F-F*\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 23 "int(rho,r=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"
\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "Note that the 4Pi from t
he integration over theta and phi are cancelled by the square of the s
pherical harmonic Y00!" }}{PARA 0 "" 0 "" {TEXT -1 91 "The solution to
 Poisson's equation in multipole expansion leads to the monopole expre
ssion:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "V0:=unapply(simpl
ify(int(rho,r=0..R)/R+int(rho/r,r=R..infinity)),R);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#V0GR6#%\"RG6\"6$%)operatorG%&arrowGF(,$*&,(*(%&bet
a|irG\"\"\"9$F1-%$expG6#,$*&F0F1F2F1!\"#F1F1F3F1F1!\"\"F1F2F9F9F(F(F(
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "plot([V(r)+2/r,subs(bet
a=beta0,V0(r))],r=0..5,color=[red,blue],view=[0..5,0..2]);" }}{PARA 
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l6&F[^lF_^lF_^lF\\^l-%+AXESLABELSG6$Q\"r6\"Q!6\"-%%VIEWG6$;F_^lFc]l;F_
^l$\"\"#Fe]l" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 
"Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 300 "We recogn
ize that the electronic repulsion in our helium atom of independent el
ectrons is modelled after the simple hydrogen-like solution to the pro
blem. The above potential shows how asymptotically the charge distribu
tion of electron 1 screens one of the protons for electron 2 (located \+
at a large " }{TEXT 262 1 "r" }{TEXT -1 8 "-value)." }}{PARA 0 "" 0 "
" {TEXT -1 216 "A sophisticated central-field or Hartree-Fock calculat
ion take the electrostatic repulsion due to the numerically obtained c
harge density and re-calculates the eigenenergy/eigenfunction until co
nvergence is achieved." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 263 22 "Electronic excitations" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 443 "We proceed to calculat
e the energy levels for the 2s and 2p states. We simply assume that we
 can use the potential obtained for the ground state, and calculate th
e energy spectrum for this potential. For the 2s-state we will not car
ry out a variational calculation as the energy will not be guranteed t
o be above the exact eigenenergy for the given potential. We repeat ou
r trial-and-error procedure to find a radial function with one node at
 " }{TEXT 264 1 "r" }{TEXT -1 3 ">0:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "IC:=phi(eta)=eta,D(phi)(eta)=1;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#ICG6$/-%$phiG6##\"\"\"\"*++++\"F*/--%\"DG6#F(F)F+" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Et:=-0.15768;" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#EtG$!&od\"!\"&" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "SE:=-1/2*diff(phi(r),r$2)+(V(r)-Et)*phi(r);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#SEG,&-%%diffG6$-%$phiG6#%\"rG-%\"$G6$F,\"
\"##!\"\"F0*&,(*&\"\"\"F6F,F2!\"#*&,&F6F6*&-%$expG6#,$F,$!$O$F7F6,&F6F
6*&$\"%l;!\"$F6F,F6F6F6F2F6F,F2F6$\"&od\"!\"&F6F6F)F6F6" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "sol:=dsolve(\{SE,IC\},phi(r),numeri
c,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 
"phir:=subs(sol,phi(r)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 
"plot(phir,0..20,-0.5..0.5);" }}{PARA 13 "" 1 "" {GLPLOT2D 694 299 
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@[7F\\[l$!+9NfTLF*7$$\"+3'HKH\"F\\[l$!+f?brFF*7$$\"+xanL8F\\[l$!+sv#oL
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6$*!#77$$\"+U9C#e\"F\\[l$!+c9zbwF[_l7$$\"+1*3`i\"F\\[l$!+X%[W='F[_l7$$
\"+$*zym;F\\[l$!+Oz)o(\\F[_l7$$\"+^j?4<F\\[l$!+#33^\"RF[_l7$$\"+jMF^<F
\\[l$!+D2_+IF[_l7$$\"+q(G**y\"F\\[l$!+M%p^D#F[_l7$$\"+9@BM=F\\[l$!+v*R
U[\"F[_l7$$\"+`v&Q(=F\\[l$!+Uz1m%)!#87$$\"+Ol5;>F\\[l$!+vwho>Fdal7$$\"
+/Uac>F\\[l$\"+7fyxTFdal7$\"#?$\"+=AA&4\"F[_l-%'COLOURG6&%$RGBG$\"#5!
\"\"$\"\"!F[clFjbl-%+AXESLABELSG6$Q!6\"F_cl-%%VIEWG6$;Fjbl$F`blF[cl;$!
\"&Fibl$\"\"&Fibl" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E
2s:=Et;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E2sG$!&od\"!\"&" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "Now let us carry out the calculat
ion for the 2p state. First we carry out the hydrogen-like variational
 calculation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "l:=1;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"lG\"\"\"" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 146 "The trial function is hydrogen-like: the radial functi
on picks up a factor of r^l, and we add the centrifugal potential to t
he radial Hamiltonian:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "c
hi:=r^(1+l)*exp(-beta*r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$chiG*&
)%\"rG\"\"#\"\"\"-%$expG6#,$*&%&beta|irGF)F'F)!\"\"F)" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "E2pT:=Int(chi*(-1/2*Diff(chi,r$2)+
(V(r)+1/2*l*(l+1)/r^2)*chi),r=0..infinity)/Int(chi^2,r=0..infinity);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E2pTG*&-%$IntG6$*()%\"rG\"\"#\"\"
\"-%$expG6#,$*&%&beta|irGF-F+F-!\"\"F-,&-%%DiffG6$*&F*F-F.F--%\"$G6$F+
F,#F4F,*(,(*&F-F-F+F4!\"#*&,&F-F-*&-F/6#,$F+$!$O$FAF-,&F-F-*&$\"%l;!\"
$F-F+F-F-F-F4F-F+F4F-*&F-F-*$F*F-F4F-F-F*F-F.F-F-F-/F+;\"\"!%)infinity
GF--F'6$*&)F+\"\"%F-)F.F,F-FQF4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
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#DF0$\"#UF0F-F:F-FC$\"++++]7!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 26 "plot(E2pT,beta=0.45..0.6);" }}{PARA 13 "" 1 "" {GLPLOT2D 693 
299 299 {PLOTDATA 2 "6%-%'CURVESG6$7S7$$\"35+++++++X!#=$!3+kqp*=!**[7F
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3!)**\\7jZ!>y&F*$!31na<ev<]7F*7$$\"3()*\\(=(4bM\"eF*$!3)[J6V&oI[7F*7$$
\"32++]xlWUeF*$!3U?si\"=3lC\"F*7$$\"3/+]i&3uc(eF*$!32&)R&p*GNW7F*7$$\"
36*****\\;$R0fF*$!32_$z#f.MU7F*7$$\"3\")*\\(=-*zq$fF*$!3qqIA/k5S7F*7$$
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-%'COLOURG6&%$RGBG$\"#5!\"\"$\"\"!F_[lF^[l-%+AXESLABELSG6$Q&beta|ir6\"
Q!6\"-%%VIEWG6$;$\"#X!\"#$\"\"'F][l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 41 "beta0:=fsolve(diff(E2pT,beta),beta=0..2);" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "E2pV:=subs(beta=beta0,E2pT);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%&beta0G$\"+hBY\\^!#5" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%E2pVG$!+9h^o7!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 66 "Now we verify the variational calculation by a numerical soluti
on:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "IC:=phi(eta)=eta^(l+
1),D(phi)(eta)=(l+1)*eta^l;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6
$/-%$phiG6##\"\"\"\"*++++\"#F+\"2++++++++\"/--%\"DG6#F(F)#F+\")+++]" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Et:=-0.12699;" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#EtG$!&*p7!\"&" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 59 "SE:=-1/2*diff(phi(r),r$2)+(V(r)+1/2*l*(l+1)/r^2-Et)*p
hi(r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG,&-%%diffG6$-%$phiG6#%
\"rG-%\"$G6$F,\"\"##!\"\"F0*&,**&\"\"\"F6F,F2!\"#*&,&F6F6*&-%$expG6#,$
F,$!$O$F7F6,&F6F6*&$\"%l;!\"$F6F,F6F6F6F2F6F,F2F6*&F6F6*$)F,F0F6F2F6$
\"&*p7!\"&F6F6F)F6F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "sol
:=dsolve(\{SE,IC\},phi(r),numeric,output=listprocedure):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phir:=subs(sol,phi(r)):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "plot(phir,0..25);" }}{PARA 13 "" 1 
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\"+qA!GN)Fio$\"+&4[%zlFA7$$\"+$e'3I))Fio$\"+n'*etdFA7$$\"+.<G&Q*Fio$\"
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\"+')[$H4\"Fiz$\"+hhYcIFA7$$\"+Ol]Y6Fiz$\"+U#**\\c#FA7$$\"+N?q&>\"Fiz$
\"+oOFv@FA7$$\"+Egw[7Fiz$\"+s9<9=FA7$$\"+*f%)QI\"Fiz$\"+$41n\\\"FA7$$
\"+![l=N\"Fiz$\"+$pSAE\"FA7$$\"+Xho.9Fiz$\"+Kt4Z5FA7$$\"+i>Ad9Fiz$\"+*
zgug)F*7$$\"+;jf4:Fiz$\"+iVQ'3(F*7$$\"+&>r-c\"Fiz$\"+]JbceF*7$$\"+4q`;
;Fiz$\"+FUAEZF*7$$\"+YV4n;Fiz$\"+2!3\"))QF*7$$\"+%4v5s\"Fiz$\"+x&**y9$
F*7$$\"+u'*)*p<Fiz$\"+-\"3Ef#F*7$$\"+JiYB=Fiz$\"+HD\")*3#F*7$$\"+/Nyt=
Fiz$\"+A%)G*p\"F*7$$\"+^&zj#>Fiz$\"+#***Rh8F*7$$\"+-=!y(>Fiz$\"+Le$z3
\"F*7$$\"+LhjJ?Fiz$\"+cW=.&)F37$$\"+#*\\[$3#Fiz$\"+@<a!f'F37$$\"+Qz]O@
Fiz$\"+me'G$\\F37$$\"+H=4*=#Fiz$\"+N6Y:NF37$$\"+i4TPAFiz$\"+(GsxN#F37$
$\"+V,z#H#Fiz$\"+\")=$G9\"F37$$\"+U>KUBFiz$\"+0jZ(3\"F-7$$\"+qJ8&R#Fiz
$!+<'*e)*)*F-7$$\"+b-oXCFiz$!+Mr'e3#F37$\"#D$!+#H=*pLF3-%'COLOURG6&%$R
GBG$\"#5!\"\"$\"\"!F_dlF^dl-%+AXESLABELSG6$Q!6\"Fcdl-%%VIEWG6$;F^dl$Fd
clF_dl%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 
0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E2p:=Et;" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E2pG$!&*p7!\"&" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 115 "The answer is just slightly below the value obta
ined by the variational calculation with a hydrogenic wavefunction." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "(E2p-E1s)*27.12*_eV;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%$_eVG$\"*_$z2@!\"(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 122 "We learn that the model predicts an exci
tation energy of slightly more than 21 eV. What does this have to do w
ith reality?" }}{PARA 0 "" 0 "" {TEXT -1 318 "Unfortunately the connec
tion is not very direct: inspection of a spectroscopic table for the H
e atom reveals that it is complicated, and that many levels correspond
 to the primitive (1s 2p) configuration: this is the result of complic
ations arising from coupling orbital and spin angular momenta for the \+
two electrons." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 418 "Nevertheless, if we are generous, we can compare the cal
culated excitation energy with the average result which in some sense \+
ignores the spin-orbit interactions (these are also called magnetic or
 fine structure interactions). The model does provide a realistic asse
ssment for the average transition energy (which is in the UV regime). \+
For an understanding of the visible spectrum one cannot ignore the fin
e structure." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 98 " One important piece of information that does emerge from this \+
calculation is the lifting of the (" }{TEXT 265 2 "nl" }{TEXT -1 196 "
)-degeneracy observed in the hydrogen atom spectrum: the 2s and 2p sta
tes are no longer degenerate, and thus, there is the possibility of ob
serving photons that correspond to 2p-2s de-excitations." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 266 21 "Matrix re
presentation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 210 "We proceed with the calculation of approximate spectra in an a
ngular momentum symmetry sector. For this purpose we first define ours
elves a suitable basis. We defined so-called Slater-type orbitals for \+
a given " }{TEXT 267 1 "l" }{TEXT -1 66 "-symmetry, and use a Gram-Sch
midt procedure to orthogonalize them." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 103 "Our radial functions are real-valued
, and therefore we have the simple inner product and normalization:" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "IP:=(f,g)->int(f*g,r=0..in
finity):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "NO:=xi->xi/sqrt
(IP(xi,xi));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NOGR6#%#xiG6\"6$%)o
peratorG%&arrowGF(*&9$\"\"\"-%%sqrtG6#-%#IPG6$F-F-!\"\"F(F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "STO:=(n,l,beta)->r^(n+l)*exp
(-beta*r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$STOGR6%%\"nG%\"lG%%be
taG6\"6$%)operatorG%&arrowGF**&)%\"rG,&9$\"\"\"9%F3F3-%$expG6#,$*&9&F3
F0F3!\"\"F3F*F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "L:=0;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG\"\"!" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 5 "N:=8;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"NG\"
\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "B1:=[seq(STO(n,L,17/
10),n=1..N)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#B1G7**&%\"rG\"\"\"
-%$expG6#,$F'#!#<\"#5F(*&)F'\"\"#F(F)F(*&)F'\"\"$F(F)F(*&)F'\"\"%F(F)F
(*&)F'\"\"&F(F)F(*&)F'\"\"'F(F)F(*&)F'\"\"(F(F)F(*&)F'\"\")F(F)F(" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "These states are linearly independ
ent, but not orthogonal!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 
"IP(B1[1],B1[2]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"%]P\"&@N)" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 119 "The Gram-Schmidt procedure takes \+
a list of functions (state vectors) and orthonormalizes them using the
 two procedures." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "GS:=pro
c(vecs) local i,n,j,res,xi;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "n:=n
ops(vecs); res:=[NO(vecs[1])];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "f
or i from 2 to n do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "xi:=vecs[i]
-add(IP(vecs[i],res[j])*res[j],j=1..i-1);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "res:=[op(res),NO(xi)]; od: end:" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 13 "B1ON:=GS(B1):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "IP(B1ON[5],B1ON[5]);" }}{PARA 11 "" 1 "" {XPPMATH 20 
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 162 "Now we recycle the code from Matr
ixRep.mws to generate the Hamiltonian matrix: we define procedures to \+
calculate the kinetic and potential energy matrix elements:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "Tkin:=(phi,psi)->-1/2*int(ex
pand(phi*diff(psi,r$2)),r=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%TkinGR6$%$phiG%$psiG6\"6$%)operatorG%&arrowGF),$-%$intG6$-%'e
xpandG6#*&9$\"\"\"-%%diffG6$9%-%\"$G6$%\"rG\"\"#F6/F>;\"\"!%)infinityG
#!\"\"F?F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Vpot:=(ph
i,psi)->int(expand(phi*psi*(V(r)+L*(L+1)/(2*r^2))),r=0..infinity);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%VpotGR6$%$phiG%$psiG6\"6$%)operator
G%&arrowGF)-%$intG6$-%'expandG6#*(9$\"\"\"9%F5,&-%\"VG6#%\"rGF5*&*(#F5
\"\"#F5%\"LGF5,&F@F5F5F5F5F5*$)F;F?F5!\"\"F5F5/F;;\"\"!%)infinityGF)F)
F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "with(LinearAlgebra): \+
Digits:=15:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "HM:=Matrix(N
):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "We make use of the symmetr
y of the hamiltonian matrix: it allows to save almost a factor of 2 in
 computation time." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "for \+
i from 1 to N do: for j from 1 to i do: HM[i,j]:=Tkin(B1ON[i],B1ON[j])
+Vpot(B1ON[i],B1ON[j]);  od: od:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 71 "for i from 1 to N do: for j from i+1 to N do: HM[i,j]
:=HM[j,i]: od: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "HMf:=
map(evalf,HM):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "evals:=Eigenvalue
s(HMf, output='list'):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ev_s:=sor
t(map(Re,evals));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%ev_sG7*$!3NOe`
Vu*>/*!#=$!33&\\3&3Ch^:F($\"3%emF;NE\"pG!#>$\"3E0m`%yZ/!QF($\"3mn;#y1I
h4\"!#<$\"3zU-AM/B3FF2$\"3.12M.UyUtF2$\"3#)*=&)3zU>K$!#;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 134 "We see that the chosen basis set can rep
roduce the 1s and 2s states, but that it does not have a prediction fo
r a bound 3s eigenstate." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 268 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 121 "Incr
ease the matrix size and observe the stability of the 1s and 2s eigenv
alues. Do you find an acceptable 3s eigenvalue?" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 269 11 "Exercise 2:" }}{PARA 0 "
" 0 "" {TEXT -1 158 "Change the value of the parameter that controls t
he Slater type orbital (STO) basis from the chosen value of 17/10. Fin
d the best basis for the 3s eigenvalue." }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "Now the eigenfunctions. We \+
need a sorting procedure to arrange the result from the eigenvector ca
lculation in proper order." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
36 "VE:=Eigenvectors(HMf,output='list'):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 66 "Vp:=[seq([Re(VE[i][1]),VE[i][2],map(Re,VE[i][3])],i=1
..nops(VE))]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 344 "Min:=proc
(x,y); if type(x,numeric) and type(y,numeric) then if x<=y then RETURN
(true): else RETURN(false): fi; elif type(x,list) and type(y,list) and
 type(x[1],numeric) and type(y[1],numeric) then if x[1]<=y[1] then RET
URN(true): else RETURN(false): fi; elif convert(x,string)<=convert(y,s
tring) then RETURN(true): else RETURN(false): fi: end:\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "VEs:=sort(Vp,Min):\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 252 "Suppose that we would like to see the ei
genfunctions corresponding to the four lowest-lying eigenvalues. The e
igenvector for a given eigenvalue contains the expansion coefficients \+
for the expansion of the eigenstate in terms of the chosen basis state
s." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i from 1 to 4 do:
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "psi0:=add(VEs[i][3][j]*B1ON[j],
j=1..N):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "No:=1/sqrt(int(expand(p
si0^2),r=0..infinity));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "phi_a[i]
:=add(No*VEs[i][3][j]*B1ON[j],j=1..N): od:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 66 "plot([seq(phi_a[i],i=1..4)],r=0..15,color=[red,blue
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$Q\"r6\"Q!6\"-%%VIEWG6$;F(F_bl%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve \+
4" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "The basis functions do not \+
have a good span at large " }{TEXT 270 1 "r" }{TEXT -1 88 " where the \+
higher states wish to reside. A smaller value of beta should improve m
atters." }}{PARA 0 "" 0 "" {TEXT -1 133 "For this reason one mixes usu
ally STO's with different values of beta. The Gram-Schmidt process tak
es care of the orthonormalization." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 271 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 
300 "Calculate the eigenfunctions for a few low-lying eigenstates in t
he L=2 and L=3 symmetry sectors. Observe how the eigenenergies approac
h the hydrogenic result in this case of E_n=-1/(2n^2). Can you explain
 this behaviour? Hint: graph the effective potential, and compare with
 that of a hydrogen atom." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}}{MARK "0 0 0" 12 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }