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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));
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 200 "This is the potential experi
enced by an 1s-electron in the helium atom: at short distances the ele
ctron just feels the full nuclear attraction (-2/r) as the second expr
ession approaches a constant as " }{TEXT 272 1 "r" }{TEXT -1 76 " goes
 to zero. At large distances the electronic repulsion term goes like 1
/" }{TEXT 273 1 "r" }{TEXT -1 80 " and screens the nucleus by one unit
 so that the overall potential goes 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 consistency by a simple variational calculati
on:" }}{PARA 0 "" 0 "" {TEXT -1 112 "We start with an unnormalized 1s \+
state that depends on a 'charge' parameter (cf. the hydrogen-like wave
function)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "chi:=r*exp(-be
ta*r);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "We are using the radia
l wavefunction (i.e. r*R_nl(r)) so that the radial kinetic energy is j
ust proportional 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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 18 "E1sT:=value(E1sT);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "plot(E1sT,beta=1.55..1.8);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 41 "beta0:=fsolve(diff(E1sT,beta),beta=1..2);" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 26 "E0:=subs(beta=beta0,E1sT);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 606 "Note that the beta-value is less than 2. The v
alue of beta=2 would be obtained if there was only one 1s-electron (th
e hydrogen-like solution for the ground state of the He+ ion). The fac
t that the electron wants to have a slightly more diffuse wavefunction
 reflects the so-called inner screening: the electrostatic repulsion w
hen combined with the simple model of two identical electrons (apart f
rom the spin projection) leads to two electrons which repel each other
 on average (independent electron model). Both electrons experience th
e same common central potential and are bound by the same eigenenergy.
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Now we would like to check tw
o things:" }}{PARA 0 "" 0 "" {TEXT -1 45 "1) how accurate is this vari
ational solution?" }}{PARA 0 "" 0 "" {TEXT -1 85 "2) how close is the \+
used potential to the potential produced by this 1s-wavefunction?" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 98 "Part 1 ca
n be answered by using dsolve[numeric]. For part 2 we need to solve th
e Poisson equation." }}{PARA 0 "" 0 "" {TEXT -1 119 "Let us start with
 the first question: To solve the SE we start the integration not at z
ero, 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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "IC:=ph
i(eta)=eta,D(phi)(eta)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "Et:=-0.9042;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "SE:=-1/
2*diff(phi(r),r$2)+(V(r)-Et)*phi(r);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 57 "sol:=dsolve(\{SE,IC\},phi(r),numeric,output=listproce
dure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phir:=subs(sol,ph
i(r)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "plot(phir,0..10);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E1s:=Et;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 272 "The numerically exact solution for the r
adial 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 different fro
m the hydrogenic form." }}{PARA 0 "" 0 "" {TEXT -1 243 "The numerical \+
eigenfunction is not normalized properly. To compare the graphs we sim
ply change the normalization of the variational answer. In fact, the v
ariational state was normalized such that the derivative of the functi
on 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)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):
" }}}{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]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 152 "Note that the agreement for the energy was at the level of  7/
900 , i.e. in the 1 % range. The deviation between the wavefunctions i
s in the 10 % range." }}{PARA 0 "" 0 "" {TEXT -1 154 "Nevertheless, we
 can state that the simple hydrogenic wavefunction catches the main fe
ature of the numerical solution, namely the most likely location in " 
}{TEXT 259 1 "r" }{TEXT -1 21 " for the 1s electron." }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 129 "The result for the binding energy for one of t
he two helium electrons is the first quantity that can be compared wit
h experiment:" }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 260 20 "ioni
zation 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;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 149 "For any atom the eigenenergy of the highest occupied orbital s
hould equal the negative of the ionization potential. Our result is in
deed quite close." }}{PARA 0 "" 0 "" {TEXT -1 328 "There is another qu
antity that can be measured, namely the total energy of the atom (equa
l to the sum of ionization energies for both electrons). This is not s
imply twice the eigenenergy, since after ionizing on of the two He-ele
ctrons, 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 "Combinin
g this answer with the calculated 1s binding energy we have for the to
tal binding energy:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "(Et-
2)*27.12*_eV;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 261 28 "The electrostat
ic repulsion:" }}{PARA 0 "" 0 "" {TEXT -1 256 "Now we look at the ques
tion as to what potential is associated with the approximate wavefunct
ion. For this we need a solution to the Poisson equation for a spheric
ally symmetric charge distribution based on the properly normalized va
riational wavefunction." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "
A1s:=1/sqrt(int(chi^2,r=0..infinity));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "rho:=subs(beta=beta,(A1s*chi)^2);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 23 "int(rho,r=0..infinity);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 116 "Note that the 4Pi from the integration over thet
a and phi are cancelled by the square of the spherical harmonic Y00!" 
}}{PARA 0 "" 0 "" {TEXT -1 91 "The solution to Poisson's equation in m
ultipole expansion leads to the monopole expression:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 68 "V0:=unapply(simplify(int(rho,r=0..R)/R+in
t(rho/r,r=R..infinity)),R);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
81 "plot([V(r)+2/r,subs(beta=beta0,V0(r))],r=0..5,color=[red,blue],vie
w=[0..5,0..2]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 300 "We recognize \+
that the electronic repulsion in our helium atom of independent electr
ons is modelled after the simple hydrogen-like solution to the problem
. The above potential shows how asymptotically the charge distribution
 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 calculatio
n take the electrostatic repulsion due to the numerically obtained cha
rge density and re-calculates the eigenenergy/eigenfunction until conv
ergence 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 calculate \+
the energy levels for the 2s and 2p states. We simply assume that we c
an use the potential obtained for the ground state, and calculate the \+
energy spectrum for this potential. For the 2s-state we will not carry
 out a variational calculation as the energy will not be guranteed to \+
be above the exact eigenenergy for the given potential. We repeat our \+
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;" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 13 "Et:=-0.15768;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 43 "SE:=-1/2*diff(phi(r),r$2)+(V(r)-Et)*phi(r);" }}}
{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 27 "plot(phir,0..20,-0.5..0.5);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "E2s:=Et;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "Now \+
let us carry out the calculation for the 2p state. First we carry out \+
the hydrogen-like variational calculation." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 5 "l:=1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "The t
rial function is hydrogen-like: the radial function picks up a factor \+
of r^l, and we add the centrifugal potential to the radial Hamiltonian
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "chi:=r^(1+l)*exp(-beta
*r);" }}}{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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "E2pT:=val
ue(E2pT);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot(E2pT,beta
=0.45..0.6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "beta0:=fsol
ve(diff(E2pT,beta),beta=0..2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "E
2pV:=subs(beta=beta0,E2pT);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "No
w we verify the variational calculation by a numerical solution:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "IC:=phi(eta)=eta^(l+1),D(phi
)(eta)=(l+1)*eta^l;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Et:=
-0.12699;" }}}{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)*phi(r);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 57 "sol:=dsolve(\{SE,IC\},phi(r),numeric,output=list
procedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phir:=subs(s
ol,phi(r)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "plot(phir,0.
.25);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E2p:=Et;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "The answer is just slightly below
 the value obtained by the variational calculation with a hydrogenic w
avefunction." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "(E2p-E1s)*2
7.12*_eV;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "We learn that the m
odel predicts an excitation energy of slightly more than 21 eV. What d
oes this have to do with reality?" }}{PARA 0 "" 0 "" {TEXT -1 318 "Unf
ortunately the connection is not very direct: inspection of a spectros
copic table for the He atom reveals that it is complicated, and that m
any levels correspond to the primitive (1s 2p) configuration: this is \+
the result of complications arising from coupling orbital and spin ang
ular momenta for the two electrons." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 418 "Nevertheless, if we are generous, we ca
n compare the calculated excitation energy with the average result whi
ch in some sense ignores the spin-orbit interactions (these are also c
alled magnetic or fine structure interactions). The model does provide
 a realistic assessment for the average transition energy (which is in
 the UV regime). For an understanding of the visible spectrum one cann
ot ignore the fine 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 spect
rum: the 2s and 2p states are no longer degenerate, and thus, there is
 the possibility of observing photons that correspond to 2p-2s de-exci
tations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 266 21 "Matrix representation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 210 "We proceed with the calculation of appr
oximate spectra in an angular momentum symmetry sector. For this purpo
se we first define ourselves a suitable basis. We defined so-called Sl
ater-type orbitals for a given " }{TEXT 267 1 "l" }{TEXT -1 66 "-symme
try, and use a Gram-Schmidt procedure to orthogonalize them." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "Our radial fun
ctions 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..infinity):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 27 "NO:=xi->xi/sqrt(IP(xi,xi));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "STO:=(n,l,beta)->r^(n+l)*exp(-beta*r);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "L:=0;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "N:=8;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "B1
:=[seq(STO(n,L,17/10),n=1..N)];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
58 "These states are linearly independent, but not orthogonal!" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "IP(B1[1],B1[2]);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 119 "The Gram-Schmidt procedure takes a list \+
of functions (state vectors) and orthonormalizes them using the two pr
ocedures." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "GS:=proc(vecs)
 local i,n,j,res,xi;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "n:=nops(vec
s); res:=[NO(vecs[1])];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i fr
om 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]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 20 "IP(B1ON[4],B1ON[5]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
8 "B1ON[2];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot([seq(B1
ON[i],i=1..4)],r=0..10,color=[red,blue,green,black]);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 162 "Now we recycle the code from MatrixRep.m
ws to generate the Hamiltonian matrix: we define procedures to calcula
te the kinetic and potential energy matrix elements:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 67 "Tkin:=(phi,psi)->-1/2*int(expand(phi*diff
(psi,r$2)),r=0..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
75 "Vpot:=(phi,psi)->int(expand(phi*psi*(V(r)+L*(L+1)/(2*r^2))),r=0..i
nfinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "with(LinearAlg
ebra): Digits:=15:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "HM:=M
atrix(N):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "We make use of the \+
symmetry 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:=Eigenval
ues(HMf, output='list'):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ev_s:=s
ort(map(Re,evals));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "We see th
at the chosen basis set can reproduce the 1s and 2s states, but that i
t does not have a prediction for a bound 3s eigenstate." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 11 "Exercise 1:" }}
{PARA 0 "" 0 "" {TEXT -1 121 "Increase the matrix size and observe the
 stability of the 1s and 2s eigenvalues. 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 the Slater type orbital (STO) basis fro
m the chosen value of 17/10. Find 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 th
e result from the eigenvector calculation 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,n
umeric) then if x<=y then RETURN(true): else RETURN(false): fi; elif t
ype(x,list) and type(y,list) and type(x[1],numeric) and type(y[1],nume
ric) then if x[1]<=y[1] then RETURN(true): else RETURN(false): fi; eli
f convert(x,string)<=convert(y,string) 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 eigenfunctions corresponding to the four lowe
st-lying eigenvalues. The eigenvector for a given eigenvalue contains \+
the expansion coefficients for the expansion of the eigenstate in term
s of the chosen basis states." }}}{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(psi0^2),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,green,black]);" }}}{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 matters." }}{PARA 0 "" 0 "" 
{TEXT -1 133 "For this reason one mixes usually STO's with different v
alues of beta. The Gram-Schmidt process takes care of the orthonormali
zation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 271 
11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 300 "Calculate the eigenfu
nctions for a few low-lying eigenstates in the L=2 and L=3 symmetry se
ctors. Observe how the eigenenergies approach the hydrogenic result in
 this case of E_n=-1/(2n^2). Can you explain this behaviour? Hint: gra
ph the effective potential, and compare with that of a hydrogen atom.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "103" 0 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }