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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Numerical solution of the \+
Schroedinger equation for partial waves:" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 125 "scattering from a spherical poten
tial [we choose the attractive exponential potential below, but any po
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FAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve
 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "In the
 r>R region we see a mixture of sin and cos with the same wavenumber. \+
Why a mixture?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 35 "How do we extract the phase shift? " }}{PARA 0 "" 0 "" 
{TEXT -1 65 "[we could fit to A*F(k*r) + B*G(k*r) at r=R, then tan(del
ta)=B/A]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
21 "L=0 should be simple." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 19 "a simpler strategy:" }}{PARA 0 "" 0 "" {TEXT -1 
76 "u'/u should go like k*cos(k*r-0.5*L*pi+delta0)/sin(cos(k*r-0.5*L*p
i+delta0))" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "pi:=evalf(Pi)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#piG$\"/)*e`EfTJ!#8" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "d0:=arctan(1/(u1R/uR/k))-k*R-0.5*L*
pi;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#d0G$!/)yx)R`\\>!#8" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Does this make sense?" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "How accurate is th
e answer?" }}{PARA 0 "" 0 "" {TEXT -1 13 "R-dependence?" }}{PARA 0 "" 
0 "" {TEXT -1 23 "possibly eta-dependence" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "d0+1*pi;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/5\"e
me?>\"!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "uR:=uS(2*R);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#uRG$!3SWK!\\'R:x#)!#>" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "u1R:=u1S(2*R);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$u1RG$\"31V,,Jb9Kb!#=" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "d0:=arctan(1/(u1R/uR/k))-2*k*R-0.5*L*pi;" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#d0G$!/brG0nu]!#8" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "d0+2*pi;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/TYyZ
^37!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 12 "uR:=uS(3*R);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#uRG$\"3'R$\\Ppe=%G(!#=" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "u1R:=u1S(3*R);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$
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=arctan(1/(u1R/uR/k))-3*k*R-0.5*L*pi;" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#d0G$!/eDkUA;#)!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "d
0+3*pi;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/O^'p`&37!#8" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 87 "Try the integral expression (2.32): the denominator constant (w
hen u goes like r^(L+1))" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 
"Dcon:=doublefactorial(2*L+1)/k^(L+1);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%DconG$\"/++++++?!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "
Now try to get the phase shift from the integral expression. We need F
 and G and numerical integration. Eqs 1.12 and 1.13 define them" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "pi;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"/)*e`EfTJ!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
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" 1 "" {XPPMATH 20 "6#>%\"FGf*6%%\"kG%\"rG%\"LG6\"6$%)operatorG%&arrow
GF**&-%%sqrtG6#**$\"\"&!\"\"\"\"\"%#piGF69$F69%F6F6-%(BesselJG6$,&9&F6
#F6\"\"#F6*&F8F6F9F6F6F*F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 56 "G:=(k,r,L)->sqrt(0.5*pi*k*r)*(-1)^L*BesselJ(-L-1/2,k*r);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"GGf*6%%\"kG%\"rG%\"LG6\"6$%)operat
orG%&arrowGF**(-%%sqrtG6#**$\"\"&!\"\"\"\"\"%#piGF69$F69%F6F6)F59&F6-%
(BesselJG6$,&F;F5#F6\"\"#F5*&F8F6F9F6F6F*F*F*" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 2 "k;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"\"&!\"\"
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Show the regular and irregula
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t([F(k,r,2),G(k,r,2)],r=0..20,color=[red,blue],view=[0..20,-2..5],thic
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**HLLBKlX\"Fho$\"3SZ:uLk(p@'Fho7$F\\p$\"3+9HMei$Q([Fho7$$\"3-++vV^\"\\
)=Fho$\"3SD9$*esqsRFho7$Fap$\"37y%e%\\fBQLFho7$Ffp$\"3O-@<j!>_d#Fho7$F
[q$\"3G<MPT()p+@Fho7$F`q$\"3/S*Hkgiky\"Fho7$Feq$\"3=s9LvQXp:Fho7$Fjq$
\"33(*yvGG/*R\"Fho7$F_r$\"31Xc&3PV0F\"Fho7$Fdr$\"3#>E>JsiT8\"Fho7$Fir$
\"3+dHkdA<a**FG7$F^s$\"3MVNY0,IP&)FG7$Fcs$\"37ii\"oKzh:(FG7$Fhs$\"3CTZ
@Opy(Q&FG7$F]t$\"3+yB`Zx&4z$FG7$Fgt$\"3spXRGB$f%=FG7$Fau$\"3uU?S<D'f(z
F_o7$F[v$!37h(\\()ys!e=FG7$Fev$!3?\\qh,_P^OFG7$Fjv$!3&zV,u;`*3aFG7$F_w
$!3G_gCmSDooFG7$Fdw$!38t!*elLX@#)FG7$Fiw$!39/=\")3ZEH$*FG7$F_x$!31qDl8
f\"3+\"Fho7$Fdx$!3!\\DC?L^7/\"Fho7$Fix$!3URMRXX3X5Fho7$F^y$!3!*fnyZCx5
5Fho7$Fcy$!3Tctx?s$RU*FG7$Fhy$!3glyQ]hT$G)FG7$F]z$!3qiq+!3g;&pFG7$Fbz$
!3!H&)[Z0W/E&FG7$Fgz$!3p*o:xD)pTNFG7$Fa[l$!3zMA;IDrG:FG7$Fe\\l$\"30T<h
rsKKUF*7$F_]l$\"3!yX3M#p.YCFG7$Fi]l$\"3+\\eE%y4/L%FG7$F^^l$\"30OIYhY/D
hFG7$Fc^l$\"3fg2k8r*4h(FG7$Fh^l$\"3Fx(>bt]=#))FG7$F]_l$\"3ut=kVGGk'*FG
7$Fg_l$\"31(f*>.\"p#45Fho7$Fa`l$\"3J[/Ug%fb,\"Fho7$F[al$\"3l-]6/-(4#)*
FG7$F`al$\"3UlSR)*eIq!*FG7$Feal$\"3#)pZ'G'*)>)*zFG7$Fjal$\"3*)yMP*\\Ip
]'FG-F`bl6&FbblFfblFfblFcbl-%*THICKNESSG6#\"\"#-%+AXESLABELSG6$%\"rGQ!
6\"-%%VIEWG6$;F\\blF[bl;!\"#\"\"&" 1 2 0 1 10 2 2 9 1 4 2 1.000000 
45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 57 "What does it mean to do one of the integrals to infinit
y>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Rcut:=10;" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%%RcutG\"#5" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 64 "NumI:=-2/k*evalf(Int(F(k,r,L)*V*uS(r),r=0..Rcut,metho
d=_Gquad));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%NumIG$\"/Ul8A]P5!#8
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "DenI:=2/k*evalf(Int(G(k
,r,L)*V*uS(r),r=0..Rcut,method=_Gquad));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%DenIG$!/C%yLzng\"!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 
"With the integrals and the denominator constant defined (using the no
rmalization of u_L) we can implement (2.32)" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 22 "K_L:=NumI/(Dcon+DenI);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$K_LG$\"/bm)Gt%QE!#8" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "arctan(K_L);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/7
Ibs^37!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "2*R;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
124 "We get more than 5 digit agreement with the result obtained from \+
fitting the numerical solution at R to the asymptotic form." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "Why do the resu
lts depend on Rcut?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Rcut:
=20;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%RcutG\"#?" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 64 "NumI:=-2/k*evalf(Int(F(k,r,L)*V*uS(r),r=0
..Rcut,method=_Gquad));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%NumIG,$*
&$\"/++++++S!#8\"\"\"-%$IntG6$,$**$\"/++++++5F)F*-%$sinG6#,$*&$\"\"&!
\"\"F*%\"rGF*F*F*-%$expG6#,$*&$F*\"\"!F*F:F*F9F*-%#uSG6#F:F*F9/F:;$FAF
A$\"#?FAF*F9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "NumI:=-2/k*
evalf(Int(F(k,r,L)*V*uS(r),r=0..Rcut,method=_Gquad,digits=8));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%NumIG$\"/+++!yu.\"!#8" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "DenI:=2/k*evalf(Int(G(k,r,L)*V*uS(r
),r=0..Rcut,method=_Gquad,digits=8));" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%%DenIG$!/++!)[$og\"!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "K_L_20:=NumI/(Dcon+DenI);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'K_
L_20G$\"/%RuV%yQE!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "[ar
ctan(K_L),arctan(K_L_20)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$\"/7I
bs^37!#8$\"/juKjb37F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "Some ju
dgement will have to be applied as to where to stop the analysis where
 R is increased until convergence is achieved." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Altern
ative to Gaussian quadrature in Maple?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "Np:=1000;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NpG\"%
+5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Dr:=(Rcut/Np);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#DrG#\"\"\"\"#]" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 64 "NumI:=evalf(-2/k*Dr*add(eval(F(k,r,L)*V*uS(r
),r=i*Dr),i=1..Np));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%NumIG$\"/e$
4.yu.\"!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 534 "We seem to have ch
osen a sufficiently small Dr for the simple Riemann sum to work (it wo
rks, since the integrand satisifies some properties, such as equal fun
ction value, and perhaps even derivative values at both endpoints). In
 an automated procedure one has to make a choice as to what is more ef
ficient: a powerful algorithm with automated accuracy check (which wil
l sometimes fail to our annoyance), or something simple that always 'w
orks', but which sometimes will give a number that is either inaccurat
e, or even outright wrong!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "Now we would like to find \+
a consistent phaseshift  as a function of wavenumber k." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "It is logical to sta
rt at large k (we know the phaseshift should approach zero)." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 256 "We could eithe
r wrap our phaseshift calculation by a do loop, or we define a procedu
re that calculates the phaseshift for given k, and then work with a ta
ble. Note that for each wavenumber k the SE needs to be re-defined. We
 use local and global variables:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "PhSh:=p
roc(k) local SE,sol,uS,Rcut,NumI,DenI,Dcon; global V,L,IC;" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 48 "SE:=diff(u(r),r$2)+(k^2-2*V-L*(L+1)/r^2)*u(
r)=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "sol:=dsolve(\{SE,IC\},nume
ric,output=listprocedure):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "uS:=e
val(u(r),sol):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Rcut:=50; # this \+
is somewhat arbitrary!!!" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "NumI:=-
2/k*evalf(Int(F(k,r,L)*V*uS(r),r=0..Rcut,method=_Gquad,digits=8));" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "DenI:=2/k*evalf(Int(G(k,r,L)*V*uS(r
),r=0..Rcut,method=_Gquad,digits=8));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 37 "Dcon:=doublefactorial(2*L+1)/k^(L+1);" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "arctan(NumI/(Dcon+DenI)); end:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "
PhSh(0.5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/eBzjb37!#8" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "This agrees to many digits with th
e Rmax=20 result, so we are happy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 60 "We can generalize the procedure to work o
n a list of values." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "PhSh
([0.5,1.0]);" }}{PARA 8 "" 1 "" {TEXT -1 110 "Error, (in f) unable to \+
store '-.100000000000000002e-2*[.5, 1.0]^2-.19980009996667e-2' when da
tatype=float[8]\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "We need to d
o something to allow repeated calculations in PhSh." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 68 "PhSh:=proc() local k,SE,sol,uS,Rcut,NumI,
DenI,Dcon,N; global V,L,IC;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "if t
ype(args[1],numeric) then k:=args[1];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 48 "SE:=diff(u(r),r$2)+(k^2-2*V-L*(L+1)/r^2)*u(r)=0;" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 50 "sol:=dsolve(\{SE,IC\},numeric,output=listproce
dure):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "uS:=eval(u(r),sol):" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Rcut:=50; # this is somewhat arbitr
ary!!!" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "NumI:=-2/k*evalf(Int(F(k,
r,L)*V*uS(r),r=0..Rcut,method=_Gquad,epsilon=1E-5));" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 76 "DenI:=2/k*evalf(Int(G(k,r,L)*V*uS(r),r=0..Rcut,m
ethod=_Gquad,epsilon=1E-5));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Dco
n:=doublefactorial(2*L+1)/k^(L+1);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "arctan(NumI/(Dcon+DenI)); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "e
lif type(args[1],list) then N:=nops(args[1]); [seq(PhSh(args[1][j]),j=
1..N)];" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end:" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 4 "end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "PhSh(0.5),PhSh(1.0);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"/'y]Kc&37!#8$\"/$\\z\\F8!y!#9" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "This still works! We did change t
he accuracy control check to use the relative error tolerance instead \+
of digits to allow calculation at larger k." }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 16 "PhSh([0.5,1.0]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#7$$\"/'y]Kc&37!#8$\"/$\\z\\F8!y!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
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" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "kL:=[0.05,0.1,0.15,0.2,
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{XPPMATH 20 "6#>%#kLG70$\"\"&!\"#$\"\"\"!\"\"$\"#:F($\"\"#F+$\"#DF($\"
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-1 94 "as k decreases, the phaseshift reaches pi/2; the lowest datapoi
nts should be shifted up by pi." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 67 "It looks like the 'true curve' wants to r
each pi as k goes to zero." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 112 "According to Levinson's theorem this should impl
y that the s-wave sector (L=0) supports exactly one bound state." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "How do we
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{MPLTEXT 1 0 20 "with(LinearAlgebra):" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 52 "What are acceptable basis function on [0,infinity) ?" }}
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F-7$$\"3ILe9\"*Hk'Q\"Fbs$!3D-r49W-X**F-7$$\"33](=#\\W`\"R\"Fbs$!3#fR*Q
%o/z(**F-7$F`fm$!3%Qb?KRgg***F-7$$\"3F$ek`O<8S\"Fbs$!31Pi!zuk%****F-7$
$\"3/+vVB)3iS\"Fbs$!3_#3$eCF6))**F-7$$\"3k;/^\"G+6T\"Fbs$!3%p\\Ac2@?'*
*F-7$F^v$!3=,*3\\IG7#**F-7$F_cl$!3Gc/^$G\"3s&*F-7$Fcv$!3]#*\\Zx;S`*)F-
7$Fgcl$!3$GL11B`19)F-7$Fhv$!3TIuDS#3\\7(F-7$F_dl$!3QQrtPvqteF-7$F]w$!3
'=;q[;.DY%F-7$Fgdl$!3+\"*ec&4v\\'HF-7$Fbw$!3UGEQ#>O-R\"F-7$Fdgm$\"3')z
E)HF!ekHF07$Fgw$\"3-`M^=Fpu>F-7$Fahm$\"39QGsoxIQNF-7$F\\x$\"3yM]Pi+D3]
F-7$Fcim$\"3Iq&)zh(oSP'F-7$Fax$\"3MR795,VjvF-7$F`jm$\"3rDSUY:BO&)F-7$F
fx$\"351E%ek2mF*F-7$Fbgo$\"3[#*4Y>@hM(*F-7$F[y$\"39/$\\k%Gto**F-7$$\"3
gL3-)omaz\"Fbs$\"3K&)y+]Am$***F-7$$\"3*pmTgg/5!=Fbs$\"3E5epG()o****F-7
$$\"3Q+D1CDa1=Fbs$\"3Eo7[$)3!o)**F-7$F_ho$\"3!G$4h\"4B]&**F-7$$\"3a+]7
yi:B=Fbs$\"37)eUSrt]$)*F-7$F`y$\"3@qPk*pZ2k*F-7$Fgho$\"30UF!G;$e7\"*F-
7$Fey$\"3_@3^.2@k$)F-7$F_io$\"3_#p5`E#=WtF-7$Fjy$\"38/g2)*oXAhF-7$Fd[n
$\"3oalEcu7&z%F-7$F_z$\"3))=*GadLrM$F-7$F\\\\n$\"39,I]([M#)p\"F-7$Fdz$
\"3Mint&*RBBhFhz-Fjz6&F\\[l$\")w6%H(F_[l$\")LLLLF_[l$\"))4XF)F_[l-%+AX
ESLABELSG6$%\"rGQ!6\"-%%VIEWG6$;F(Fdz%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 
2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curv
e 4" "Curve 5" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "Perhaps not th
e smartest basis, since the amplitude of oscillation is constant over \+
[0,Rb], but maybe good enough?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 56 "What are the orthogonality and normalizat
ion properties?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "OL:=(m,n
)->int(sin(kb(n)*r)*sin(kb(m)*r),r=0..Rb);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#OLGf*6$%\"mG%\"nG6\"6$%)operatorG%&arrowGF)-%$intG6$
*&-%$sinG6#*&-%#kbG6#9%\"\"\"%\"rGF9F9-F26#*&-F66#9$F9F:F9F9/F:;\"\"!%
#RbGF)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "seq(OL(i,i),i
=1..5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6'\"#5F#F#F#F#" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "BF:=n->sqrt(2/Rb)*sin(kb(n)*r);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#BFGf*6#%\"nG6\"6$%)operatorG%&arrow
GF(*&-%%sqrtG6#,$*&\"\"#\"\"\"%#RbG!\"\"F3F3-%$sinG6#*&-%#kbG6#9$F3%\"
rGF3F3F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "OL:=(m,n)->
int(BF(n)*BF(m),r=0..Rb);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#OLGf*6
$%\"mG%\"nG6\"6$%)operatorG%&arrowGF)-%$intG6$*&-%#BFG6#9%\"\"\"-F26#9
$F5/%\"rG;\"\"!%#RbGF)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
20 "seq(OL(i,i),i=1..5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6'\"\"\"F#F#
F#F#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "seq(OL(1,i),i=1..5)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6'\"\"\"\"\"!F$F$F$" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 39 "It looks like the basis is orthonormal." 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "KE:=(m,n)->-1/2*int(BF(m)
*diff(BF(n),r$2),r=0..Rb);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#KEGf*
6$%\"mG%\"nG6\"6$%)operatorG%&arrowGF),$*&#\"\"\"\"\"#F0-%$intG6$*&-%#
BFG6#9$F0-%%diffG6$-F76#9%-%\"$G6$%\"rGF1F0/FC;\"\"!%#RbGF0!\"\"F)F)F)
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "Nb:=20:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "TM:=Matrix(Nb,Nb):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "for m from 1 to Nb do: for n from 1
 to Nb do: TM[m,n]:=KE(m,n): od: od:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "print(map(evalf,SubMatrix(TM,1..5,1..5)));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#-%'RTABLEG6%\")7X,))-%'MATRIXG6#7'7'$\"/i8]0
qL7!#:$\"\"!F0F/F/F/7'F/$\"/Za+A![$\\F.F/F/F/7'F/F/$\"/E7&\\I.6\"!#9F/
F/7'F/F/F/$\"/z@!)3#R(>F7F/7'F/F/F/F/$\"//Mv8D%3$F7%'MatrixG" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "The kinetic energy matrix is diag
onal, because the sine functions are eigenfunctions of the ree-particl
e Hamiltonian operator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 51 "The potential energy matrix is obviously symmetric.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "VM:=Matrix(Nb,Nb,shape=
symmetric):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "HM:=Matrix(N
b,Nb,shape=symmetric):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "f
or m from 1 to Nb do: for n from 1 to m do: VM[m,n]:=int(BF(m)*V*BF(n)
,r=0..Rb): od: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "#prin
t(map(evalf,VM));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "for m
 from 1 to Nb do: for n from 1 to m do: HM[m,n]:=evalf(VM[m,n]): if m=
n then HM[m,n]:=HM[m,n]+evalf(TM[m,m]): fi: od: od:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 36 "convert(sort(Eigenvalues(HM)),list);" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#76$!3'o'\\$RfoVq*!#?$\"3$Qs1pk!\\AD!#>
$\"3&e)GBqz5M\")F)$\"3I6$\\D(QiQ;!#=$\"3)3;+siZ.s#F.$\"3'R2\"Go/KbSF.$
\"3%4!)4wyX<k&F.$\"3/K^o,WYyuF.$\"3q$G)GQ7nk&*F.$\"3K%H$e\\&z**=\"!#<$
\"3C>MQKQM[9F;$\"3@!)>7nQ`J<F;$\"31Dd(o!f`R?F;$\"3kiigN!\\BP#F;$\"38=m
n%)y)*HFF;$\"3&Gp/n\\)[7JF;$\"3F1B:y&>*>NF;$\"3EbF^&)pS_RF;$\"3CyfE!y5
-T%F;$\"3SLX#>))eT*[F;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 201 "The ma
trix eigenvalues are upper bounds to the true differential equation ei
genvalues. We can have certainty over the accuracy of the bound part o
f the spectrum after performing a convergence analysis." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "We seem to have only
 one bound state with an energy of E=-0.0097 units." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 237 "Note that matrix diago
nalization results in a 'discretized continuum' of E>0 eigenenergies. \+
These are called pseudo-continuum states, because the eigensolutions v
anish at the box edge set by Rb, and are not meaningful outside of the
 box." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 51 "
Let us generate the differential equation solution:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "E0:=-.970436859393496686e-2;" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%#E0G$!3'o'\\$RfoVq*!#?" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 38 "SE:=-0.5*diff(u(r),r$2)+(V-E0)*u(r)=0;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG/,&*&$\"\"&!\"\"\"\"\"-%%diffG6$
-%\"uG6#%\"rG-%\"$G6$F2\"\"#F+F**&,&-%$expG6#,$F2F*F*$\"3'o'\\$RfoVq*!
#?F+F+F/F+F+\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "IC:=u(
0)=0,D(u)(0)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6$/-%\"uG6#\"
\"!F*/--%\"DG6#F(F)\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
50 "sol:=dsolve(\{SE,IC\},numeric,output=listprocedure):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "uS:=eval(u(r),sol):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "uS(1);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"3ScWu!R013)!#=" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "
plot(uS(r),r=0..Rb,thickness=2);" }}{PARA 13 "" 1 "" {GLPLOT2D 635 
291 291 {PLOTDATA 2 "6&-%'CURVESG6$7_o7$$\"\"!F)F(7$$\"/ML3x&)*3\"!#9$
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+\"!\")F(F(-%*THICKNESSG6#\"\"#-%+AXESLABELSG6$%\"rGQ!6\"-%%VIEWG6$;F(
F_`l%(DEFAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 47.000000 45.000000 0 
0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "One should do the \+
following checks:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 301 "1) the matrix diagonalization solution is still affected
 by the finite value of Rb (the cut-off distance at large r); why? the
 numerical solution is showing a node before r=20, which means that it
erating on E0 to generate a smooth approach towards the axis would yie
ld a slightly lower E0 *ASSIGNMENT*" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 116 "    one check that should be applied: w
hat is the classical turning point for the given E and V? this is done
 below." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
195 "2) basis size will play a role when the real axis is extended (Rb
 is increased) further. The matrix diagonalization ground-state energy
 should converge towards the 'exact' numerical ODE solution." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 230 "3) a further c
heck would be: does this one bound state remain the only one? When the
 matrix size is increased, is it possible that another E<0 solution ap
pears (in the L=0 sector)? What if the potential constant V0=1 is incr
eased?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
285 "ASSIGNMENT: Choose a deeper potential (either by increasing V0 or
 by changing the r-scaling, exp(-r) can be changed to exp(-0.5*r), etc
.; such that two bound states exist for L=0. Then re-do the phase shif
t calculation as a function of k, and observe whether it starts at 2*p
i for k=0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 28 "The classical turning point:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "R_tp:=solve(E0=V,r);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#$\"//NC\"z^j%!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "The weakly
 bound state has a long probability amplitude tail in the tunneling re
gion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "
Can we measure the scattering length a?" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 113 "The definition according to eq. (3.
3) on p.44 says that tan(delta) goes like -k*a (we can ignore the shif
t by pi)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "kL[1],PS_L[1];" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6$$\"\"&!\"#$!/bVah:1U!#9" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "a_guess:=-tan(PS_L[1])/kL[1];" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%(a_guessG$\"/Su/FAY*)!#8" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "-tan(PS_L[2])/kL[2];" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"/XM&>d')z*!#8" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "We shou
ld fit the data to make an accurate estimate:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 7 "k:='k';" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG
F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "with(Statistics):" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "X := Vector([seq(kL[j],j=1
..3)], datatype=float);\nY := Vector([seq(tan(PS_L[j]),j=1..3)], datat
ype=float);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"XG-%'RTABLEG6%\")!o
*Gr-%'MATRIXG6#7%7#$\"3G+++++++]!#>7#$\"3/+++++++5!#=7#$\"3%**********
****\\\"F4&%'VectorG6#%'columnG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"YG-%'RTABLEG6%\")KuGr-%'MATRIXG6#7%7#$!33+?P_86tW!#=7#$!33+XM&>d')z*
F07#$!3'**HSOZA\\u\"!#<&%'VectorG6#%'columnG" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 34 "LinearFit([k, k^2, k^3], X, Y, k);" }}{PARA 7 "
" 1 "" {TEXT -1 43 "Warning, there are zero degrees of freedom\n" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&$\"3!=\"Q8>I^v!*!#<\"\"\"%\"kGF(!
\"\"*&$\"3Bl*pPV0.C\"!#;F()F)\"\"#F(F(*&$\"3Y`1)*4'\\M'>!#:F()F)\"\"$F
(F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "We need more data, and we \+
should concentrate on lower wavenumber values!" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 27 "kL1:=[0.01,0.02,0.03,0.04];" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 17 "PS_L1:=PhSh(kL1);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%$kL1G7&$\"\"\"!\"#$\"\"#F($\"\"$F($\"\"%F(" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&PS_L1G7&$!/?rU+A\"o)!#:$!/8gQm6H<!#9$!/9U\\SEwDF+$!/
JvG?g.MF+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "X := Vector([
seq(kL[j],j=1..4)], datatype=float);\nY := Vector([seq(tan(PS_L[j]),j=
1..4)], datatype=float);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "LinearF
it([k, k^2, k^3], X, Y, k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"XG-
%'RTABLEG6%\")CDGr-%'MATRIXG6#7&7#$\"3G+++++++]!#>7#$\"3/+++++++5!#=7#
$\"3%**************\\\"F47#$\"35+++++++?F4&%'VectorG6#%'columnG" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"YG-%'RTABLEG6%\")7RHr-%'MATRIXG6#7
&7#$!33+?P_86tW!#=7#$!33+XM&>d')z*F07#$!3'**HSOZA\\u\"!#<7#$!35+#)*=$R
]TJF7&%'VectorG6#%'columnG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&$\"3
?%4\"3V)G02\"!#;\"\"\"%\"kGF(!\"\"*&$\"3$edo&=j5VZF'F()F)\"\"#F(F(*&$
\"3d>A!y-]vh$!#:F()F)\"\"$F(F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 92 
"We see that the fit is far from converged! Probably a good place for \+
an assignment question!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 113 "Theory (page 51) also points out that the tangent o
f the phase shift has linear plus cubic terms at lowest orders" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "LinearFit([k, k^3, k^5], X, \+
Y, k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&$\"3Mf+sOmna))!#<\"\"\"%
\"kGF(!\"\"*&$\"3/t>\")y>:@j!#;F()F)\"\"$F(F**&$\"3:U\"elV3=q#!#9F()F)
\"\"&F(F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 108 "Note that the minus
 sign convention in the scattering length is not universal, some autho
rs don't put it in." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 72 "Now let us check that the E=0 solution for the SE has a
 node at about a." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "E0:=1E-10; # going for zero \+
energy" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E0G$\"\"\"!#5" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "SE:=-0.5*diff(u(r),r$2)+(V-E0)*u(r)
=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG/,&*&$\"\"&!\"\"\"\"\"-%%
diffG6$-%\"uG6#%\"rG-%\"$G6$F2\"\"#F+F**&,&-%$expG6#,$F2F*F*$F+!#5F*F+
F/F+F+\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "IC:=u(0)=0,D
(u)(0)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6$/-%\"uG6#\"\"!F*/
--%\"DG6#F(F)\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "sol:
=dsolve(\{SE,IC\},numeric,output=listprocedure):" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 19 "uS:=eval(u(r),sol):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 6 "uS(1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"3;zV
qv\\4_!)!#=" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot(uS(r),r
=0..2*Rb,thickness=2);" }}{PARA 13 "" 1 "" {GLPLOT2D 784 292 292 
{PLOTDATA 2 "6&-%'CURVESG6$7Y7$$\"\"!F)F(7$$\"/LLL3VfV!#9$\"/!Rf,(pRTF
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Le'40j\"F8$\"/A&)=:(z-\"F87$$\"/m\"H_(zV=F8$\"/\\&4SRI1\"F87$$\"/+](Q&
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1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 169 "This verifies the claim that the scattering length corre
sponds to the first node location in the E=0 solution. Question: are t
here more nodes for the E=0 solution above?" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
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