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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 36 "Hydrogenic positive-energ
y solutions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 4 "The " }{TEXT 19 3 "E>0" }{TEXT -1 136 " solutions for the hydrog
en atom problem represent scattering of electrons from protons. These \+
states are defined for continuous energy " }{TEXT 19 1 "E" }{TEXT -1 
241 ", and are therefore called continuum states to distinguish them f
rom the bound states whose energy is discrete. States corresponding to
 continuous eigenenergies cannot be normalized in the usual way (cf. t
he plane-wave free-particle states)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 127 "We can gain some understanding of thes
e solutions by first studying the Rydberg states, i.e., the high-lying
 bound states with " }{TEXT 19 14 "E=-Z^2/(2*n^2)" }{TEXT -1 12 ", e.g
., for " }{TEXT 19 3 "n>5" }{TEXT -1 138 ". The dependence of these ei
genstates on the energy becomes very weak, because their eigenenergies
 approach zero. Following H. Friedrich: " }{TEXT 257 26 "Theoretical A
tomic Physics" }{TEXT -1 114 " (Springer-Verlag 1990) we graph them wi
th a choice of normalization such that they all begin with equal slope
 at " }{TEXT 19 3 "r=0" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 55 "The bound states are defined as (Frie
drich, Eq. 1.137):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "resta
rt: Digits:=14:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "Phi:=(a
,n,l)->simplify(sqrt((n-l-1)!/a/(n+l)!)/n*(2*r/n/a)^(l+1)*LaguerreL(n-
l-1,2*l+1,2*r/n/a)*exp(-r/n/a));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
$PhiGj+6%%\"aG%\"nG%\"lG6\"6$%)operatorG%&arrowGF*-%)simplifyG6#*,-%%s
qrtG6#*(-%*factorialG6#,(9%\"\"\"9&!\"\"F;F=F;9$F=-F76#,&F:F;F<F;F=F;F
:F=),$**\"\"#F;%\"rGF;F:F=F>F=F;,&F<F;F;F;F;-%*LaguerreLG6%F9,&*&FEF;F
<F;F;F;F;FCF;-%$expG6#,$*(FFF;F:F=F>F=F=F;F*F*F*6%\"\"!FSFS" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "Note: without the " }{TEXT 19 8 "s
implify" }{TEXT -1 99 " command some parts below will calculate incorr
ectly (the derivatives of the Laguerre polynomials)!" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "Here the scale paramete
r for the distance is given as the Bohr radius a:" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 22 "a0:=hbar^2/mu/(Z*e^2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#a0G**%%hbarG\"\"#%#muG!\"\"%\"ZGF)%\"eG!\"#" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "We introduce Bohr units in which \+
the charge of the electron and its mass is set equal to unity, as is P
lanck's constant divided by " }{TEXT 19 4 "2*Pi" }{TEXT -1 68 ". We in
troduce the reduced mass by approximating the proton mass as " }{TEXT 
19 8 "M=1836*m" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 20 "e:=1: hbar:=1: m:=1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 "M:=1836: mu:=m*M/(m+M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#mu
G#\"%O=\"%P=" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "We select hydrog
en (as opposed to a hydrogen-like ion), and check that the wavefunctio
n defined above represents a radial orbital, " }{TEXT 19 11 "Phi = r*P
si" }{TEXT -1 79 ", which means that the normalization integral is to \+
be carried out without the " }{TEXT 19 3 "r^2" }{TEXT -1 149 " part of
 the volume element in spherical polar coordinates. The orbital is to \+
be multiplied by a spherical harmonic to form the overall wavefunction
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "Z:=1;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%\"ZG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 33 "int(Phi(a0,3,2)^2,r=0..infinity);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "Let us v
erify the eigenvalue problem for this state:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 74 "Hrad:=(phi,l)->simplify(-1/2/mu*diff(phi,r$2)/ph
i+l*(l+1)/(2*mu*r^2)-Z/r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%HradG
j+6$%$phiG%\"lG6\"6$%)operatorG%&arrowGF)-%)simplifyG6#,(*&#\"\"\"\"\"
#F3*(%#muG!\"\"-%%diffG6$9$-%\"$G6$%\"rGF4F3F;F7F3F7*&#F3F4F3**9%F3,&F
CF3F3F3F3F6F7F?!\"#F3F3*&%\"ZGF3F?F7F7F)F)F)6$\"\"!FI" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "n0:=3: l0:=2: Hrad(simplify(Phi(a0,
n0,l0)),l0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##!$-\"\"%P=" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "simplify(-Z^2/2/3^2*mu);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6##!$-\"\"%P=" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 303 "Note how the radial Schroedinger equation involves two
 parts of the kinetic energy: the radial and centrifugal parts. The ce
ntrifugal part acts like a potential energy (very much like in classic
al mechanics we have an effective potential). Due to the knowledge of \+
the angular parts as eigenfunctions of " }{TEXT 19 3 "L^2" }{TEXT -1 
126 ", namely the spherical harmonics, we have radial eigenvalue equat
ions we predefined centrifugal potential that depends on the " }{TEXT 
258 1 "l" }{TEXT -1 35 " quantum number. The eigenvalue of " }{TEXT 
19 3 "L^2" }{TEXT -1 4 " is " }{TEXT 19 14 "hbar^2*l*(l+1)" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 "A
lso note how the naive (infinite-mass proton) eigenvalue " }{TEXT 19 
12 "-Z^2/(2*n^2)" }{TEXT -1 292 " is multiplied by the reduced mass, a
nd that the eigenenergy does not depend on the angular momentum quantu
m number. The latter part is a specialty of the pure Coulomb potential
 (accidental degeneracy related to the fact that there is an additiona
l conserved quantity, the Runge-Lenz vector)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 11 "Exercise 1:" }}{PARA 0 "
" 0 "" {TEXT -1 41 "Verify the eigenvalue problem for other (" }{TEXT 
260 1 "n" }{TEXT -1 2 ", " }{TEXT 259 1 "l" }{TEXT -1 252 ") eigenstat
es. Graph the radial probability densities for the chosen states. Supe
rimpose on one graph the probability drawn over the baseline of the ei
genenergy and the effective potential energy. Indicate the locations o
f the classical turning points." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 197 "Now we will normalize the radial \+
functions differently, such that they will not allow a direct probabil
istic interpretation, but will have comparable height of the first max
ima and minima at small " }{TEXT 263 1 "r" }{TEXT -1 23 " in the limit
 of large " }{TEXT 262 1 "n" }{TEXT -1 44 ". The Rydberg in our Bohr u
nits is given as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Ry:=hb
ar^2/mu/a0^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#RyG#\"%O=\"%P=" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "fn:=n->sqrt(n^3/2/Ry);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fnGj+6#%\"nG6\"6$%)operatorG%&arrow
GF(-%%sqrtG6#,$*&#\"\"\"\"\"#F2*&9$\"\"$%#RyG!\"\"F2F2F(F(F(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 6 "l0:=0;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#l0G\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 50 "P4:=plot(fn(4)*Phi(a0,4,l0),r=0..50*a0,color=red):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "P5:=plot(fn(5)*Phi(a0,5,l0),r=0..50
*a0,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "P6:=plo
t(fn(6)*Phi(a0,6,l0),r=0..50*a0,color=green):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 52 "P7:=plot(fn(7)*Phi(a0,7,l0),r=0..50*a0,color=bro
wn):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "P8:=plot(fn(8)*Phi(
a0,8,l0),r=0..50*a0,color=magenta):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 52 "P9:=plot(fn(9)*Phi(a0,9,l0),r=0..50*a0,color=black):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "P10:=plot(fn(10)*Phi(a0
,10,l0),r=0..50*a0,color=yellow):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "display
(P4,P5,P6,P7,P8,P9,P10);" }}{PARA 13 "" 1 "" {GLPLOT2D 949 403 403 
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}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "Why do the functions approach a \+
limiting behavior at small " }{TEXT 264 1 "r" }{TEXT -1 100 " ? The ra
dial eigenvalue problem becomes the solution of the differential equat
ion with eigenenergy " }{TEXT 265 1 "E" }{TEXT -1 73 "=0 approximately
. This graph follows Fig. 1.5 in Harald Friedrich's book." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 11 "Exercise 2:" }}
{PARA 0 "" 0 "" {TEXT -1 76 "Repeat the above graph for higher values \+
of angular momentum quantum number " }{TEXT 266 1 "l" }{TEXT -1 44 ". \+
Why does the radial probability avoid the " }{TEXT 267 1 "r" }{TEXT 
-1 45 "=0 region for higher angular momentum values?" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Now we let Mapl
e solve the differential equation for us in the " }{TEXT 269 1 "E" }
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%(DEFAULTG" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "C
urve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "We recognize
 that the second solution is not acceptable for small " }{TEXT 19 1 "r
" }{TEXT -1 48 ", since it approaches a constant in the case of " }
{TEXT 270 1 "l" }{TEXT -1 47 "=0. Thus the wavefunction itself is not \+
finite." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
35 "We also recognize that in the high-" }{TEXT 271 1 "n" }{TEXT -1 
53 " limit the eigenfunctions become proportional to the " }{TEXT 272 
1 "E" }{TEXT -1 24 "=0 solution found above." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 393 "We notice that the pos
itive-energy solutions to the Schroedinger equation do not represent a
n eigenvalue problem in the same sense as for the bound states: we can
 fix the energy value (given that it is continuous, we can choose any \+
value we want) and ask for the corresponding solution. Thus, we re-def
ine the Schroedinger equation to allow for the positive energy value a
s an input parameter." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "SE:=E->-1/2/mu*diff(u(r),r$2)+l*(l+
1)/(2*mu*r^2)*u(r)-Z/r*u(r)-E*u(r)=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#SEGj+6#%\"EG6\"6$%)operatorG%&arrowGF(/,**&#\"\"\"\"\"#F0*&%#muG
!\"\"-%%diffG6$-%\"uG6#%\"rG-%\"$G6$F;F1F0F0F4*&#F0F1F0*,%\"lGF0,&FBF0
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" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "sol:=dsolve(SE(0.1));" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"uG6#%\"rG,&*&%$_C1G\"\"
\"-%+WhittakerMG6%*&^##!\"'\"%P=F-\"'N%o%#F-\"\"#,&%\"lGF-F7F-*(^##\"#
7\"%&=*F-F6F7F)F-F-F-*&%$_C2GF--%+WhittakerWGF0F-F-" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 92 "We expect again that the solutions are made up \+
of regular and irregular ones at the origin (" }{TEXT 273 1 "r" }
{TEXT -1 258 "=0), and that we will choose only the one with a finite \+
probability at this point. The question of normalizability of this sol
ution should now be addressed, and then we will graph the solutions fo
r different energies to produce the equivalent of Fig. 1.6 in " }
{TEXT 274 26 "Theoretical Atomic Physics" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 158 "The first part of
 the solution represents a standing wave, which is regular at the orig
in. It can be made real by an imaginary choice of integration constant
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "sol1:=eval(op(1,rhs(so
l)),\{_C1=-I,l=l0\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol1G*&^#!
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%&=*F(F1F2%\"rGF(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 151 "The second part of the solution i
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e (it has out-of-phase real and imaginary parts:" }}}{EXCHG {PARA 0 ">
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[bn$\"/y3Pen$>\"Fen7$$\"/We1KPgXFen$\"/9RZ@G)e\"Fen7$F`bn$\"/9'z6Et%>F
en7$Fjbn$\"/I>lh)H]#Fen7$Fdcn$\"/'4OlD(4HFen7$Ficn$\"/p+j`u\\IFen7$F^d
n$\"/c9Z'eO:$Fen7$$\"/QXBaC`ZFen$\"/Enh5kpJFen7$Fcdn$\"/xA$QMt=$Fen7$$
\"/'>*>amzZFen$\"/]$=(e]zJFen7$Fhdn$\"/a6JRp(>$Fen7$Fben$\"/u]wOP7IFen
7$F\\fn$\"/b1.05)o#Fen7$Fafn$\"/d@^\"[%RCFen7$Fffn$\"/f6)oo86#Fen7$F[g
n$\"/O1m/,\\<Fen7$F`gn$\"/xhz$GqV\"Fen-Fegn6&FggnF[hnF[hnFhgn-%*THICKN
ESSG6#\"\"#-%+AXESLABELSG6$Q\"r6\"Q!Fbdq-%%VIEWG6$;F[hnF`gn%(DEFAULTG
" 1 2 0 1 10 2 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "C
urve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "The " }{TEXT 275 1 "E" 
}{TEXT -1 238 ">0 solutions are usually expressed in terms of the conf
luent hypergeometric series. They are called Coulomb functions, and Ap
pendix A4 in Friedrich's book provides a brief summary. The standard r
eference is M. Abramowitz and I.A. Stegun: " }{TEXT 276 34 "Handbook o
f Mathematical Functions" }{TEXT -1 14 " (Dover 1970)." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 360 "The positive-energy
 solutions are difficult to compute for Maple at large arguments, beca
use they are obtained from a power series. They are not normalizable i
n the sense of bound states; they are not normalizable, i.e., they are
 not members of the usual Hilbert space of square-integrable functions
 L2. They can be delta-function normalized in the sense that " }{TEXT 
19 18 "<phi(E) | phi(E')>" }{TEXT -1 20 " is proportional to " }{TEXT 
19 11 "delta(E-E')" }{TEXT -1 70 ". It is also possible to characteriz
e them in terms of the wavenumber " }{TEXT 277 1 "k" }{TEXT -1 7 " (wi
th " }{TEXT 19 12 "E=k^2/(2*mu)" }{TEXT -1 48 "), and to change the no
rmalization condition to " }{TEXT 19 18 "<phi(k) | phi(k')>" }{TEXT 
-1 30 " chosen to be proportional to " }{TEXT 19 11 "delta(k-k')" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 13 "In the large-" }{TEXT 278 1 "r" }{TEXT -1 66 " asymptotic
 regime the radial functions are chosen to behave like " }{TEXT 19 14 
"sin(k*r+gamma)" }{TEXT -1 30 ". At these largest values of  " }{TEXT 
279 1 "r" }{TEXT -1 123 " both the centrifugal potential and the Coulo
mb potential vanish and the solution is obviously given by sine and co
sine of " }{TEXT 19 3 "k*r" }{TEXT -1 18 ". The phase shift " }{TEXT 
19 5 "gamma" }{TEXT -1 78 " has some peculiarity due to the long-range
 nature of the Coulomb interaction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 101 "We define the confluent hypergeometric \+
series (it is also built into Maple in a more general form as " }
{TEXT 19 9 "hypergeom" }{TEXT -1 41 "). We truncate it at some finite \+
order N." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "F:=(a,b,z,N)->e
valf(GAMMA(b)/GAMMA(a)*add(GAMMA(a+n)/GAMMA(b+n)*z^n/n!,n=0..N));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FGj+6&%\"aG%\"bG%\"zG%\"NG6\"6$%)o
peratorG%&arrowGF+-%&evalfG6#*(-%&GAMMAG6#9%\"\"\"-F46#9$!\"\"-%$addG6
$**-F46#,&F:F7%\"nGF7F7-F46#,&F6F7FCF7F;)9&FCF7-%*factorialG6#FCF;/FC;
\"\"!9'F7F+F+F+6&FNFNFNFN" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "F(1,2,1.5,100);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/?*o/E6K#!#8
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "hypergeom([1],[2],1.5);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/?*o/E6K#!#8" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 12 "We will use " }{TEXT 19 9 "hypergeom" }{TEXT 
-1 46 ", as it has a convergence check built into it." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eta:=-1/(k*a0);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%$etaG,$*(\"%O=\"\"\"\"%P=!\"\"%\"kGF*F*" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "Fl:=(k,l,z)->2^l*exp(Pi/(2*k*a0))/
(2*l+1)!*abs(GAMMA(l+1-I/k/a0))*exp(-I*z)*z^(l+1)*hypergeom([l+1+I/k/a
0],[2*l+2],2*I*z);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FlGj+6%%\"kG%
\"lG%\"zG6\"6$%)operatorG%&arrowGF**0)\"\"#9%\"\"\"-%$expG6#,$*&#F2F0F
2*(%#PiGF29$!\"\"%#a0GF<F2F2F2-%*factorialG6#,&*&F0F2F1F2F2F2F2F<-%$ab
sG6#-%&GAMMAG6#,(F1F2F2F2*(^#F<F2F;F<F=F<F2F2-F46#*&FKF29&F2F2)FO,&F1F
2F2F2F2-%*hypergeomG6%7#,(F1F2F2F2*(F;F<F=F<^#F2F2F27#,&*&F0F2F1F2F2F0
F2*&^#F0F2FOF2F2F*F*F*6%\"\"!FinFin" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 37 "We should be ready for our graph now." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 20 "kE:=E->sqrt(2*mu*E);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#kEGj+6#%\"EG6\"6$%)operatorG%&arrowGF(-%%sqrtG6#,$*(
\"\"#\"\"\"%#muGF29$F2F2F(F(F(6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 78 "The energy-normalized Coulomb functions are given as Eq. \+
(1.150) in Friedrich:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "FE
:=(k,l,r)->evalf(sqrt(2*mu/Pi/hbar^2/k)*Fl(k,l,k*r));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#FEGj+6%%\"kG%\"lG%\"rG6\"6$%)operatorG%&arrowGF
*-%&evalfG6#*&-%%sqrtG6#,$*,\"\"#\"\"\"%#muGF8%#PiG!\"\"%%hbarG!\"#9$F
;F8F8-%#FlG6%F>9%*&F>F89&F8F8F*F*F*6%\"\"!FFFF" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 67 "To generate the E=0 solution we select a small value
 of the energy:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "s:=10;" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"sG\"#5" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 51 "P0:=plot(FE(kE(1E-5),l0,r),r=0..50*a0,thickness=2)
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "P1:=plot(0.1*s+FE(kE(0
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0 "" {MPLTEXT 1 0 68 "P2:=plot(0.2*s+FE(kE(0.2),l0,r),r=0..50*a0,color
=green,thickness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "P3:
=plot(0.3*s+FE(kE(0.3),l0,r),r=0..50*a0,color=black,thickness=2):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "display(P0,P1,P2,P3);" }}
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1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve \+
4" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "One notices the similarity \+
of the continuum states for small " }{TEXT 280 1 "r" }{TEXT -1 105 " w
here the energy balance is dominated by the Coulomb potential. The ene
rgy unit corresponds to 27.12 eV." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 282 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 
78 "Generate the regular Coulomb solutions for higher angular momentum
 symmetries " }{TEXT 281 1 "l" }{TEXT -1 75 ">0 and compare the wavefu
nctions to the corresponding bound states of high " }{TEXT 283 1 "n" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT 284 20 "Asymptotic behavior:" }}{PARA 0 "" 0 "" {TEXT -1 243 
"The regular Coulomb functions have an asymptotic behavior which is pa
rametrized in terms of a phase shift where the logarithmically diverge
nt term is removed. The divergence arises as a result of the long-rang
e nature of the Coulomb potential." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 51 "sigma:=(k,l)->evalf(argument(GAMMA(l+1-I/(k*a0))));" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sigmaGj+6$%\"kG%\"lG6\"6$%)operat
orG%&arrowGF)-%&evalfG6#-%)argumentG6#-%&GAMMAG6#,(9%\"\"\"F8F8*(^#!\"
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{MPLTEXT 1 0 11 "sigma(1,0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/=x
rs\"p,$!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "Fas:=(k,l,r)-
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1 0 59 "P1a:=plot(Fas(kE(0.1),l0,r),r=50*a0..150*a0,color=magenta):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(P1A,P1a);" }}
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j!*RbvK9FL$!3aH;7MNC$e)Fagq7$$\"3fjDh<9rP9FL$!31V[<Zu]vkFagq7$$\"3$H9f
sT`.W\"FL$!3b8q>/pl0_Fagq7$$\"3bAd!pT&*HW\"FL$!39_8A!*)>#eQFagq7$$\"3)
=I_lTPcW\"FL$!3?>e?%f5LX#Fagq7$$\"3A\")))>;%z#[9FL$!37e`wMU*=,\"Fagq7$
$\"3cga%eT@4X\"FL$\"3'Hd84Lf_W%F`iq7$$\"3!*R?\\:Mc`9FL$\"3%Ry:6YHU*=Fa
gq7$$\"3_>'Q^T0iX\"FL$\"3Svm\\8Eg:LFagq7$$\"3'))>&y9u%)e9FL$\"39ja#e]g
uo%Fagq7$$\"3=M&eb9w8Y\"FL$\"3lo'4D7Y_$fFagq7$$\"3yp=Lw[!RY\"FL$\"3]%=
!\\]4\">5(Fagq7$$\"370_52OVm9FL$\"3#\\J_ZKR:<)Fagq7$$\"3WS&yyLi*o9FL$
\"3(*GtmK[aH\"*Fagq7$$\"336_U*z>SZ\"FL$\"37M\\sIe,m5Fi[r7$$\"3u\")=(4E
x!z9FL$\"3%GKP3zk5;\"Fi[r7$$\"3KUED%*fV![\"FL$\"3e17Kv3(f<\"Fi[r7$$\"3
#HSLvs%z\"[\"FL$\"39!y%zK.C'=\"Fi[r7$$\"3]jT\"3Y`J[\"FL$\"3/oZ.'4L=>\"
Fi[r7$$\"3QC\\4%>7X[\"FL$\"3/)4f\"Qvs#>\"Fi[r7$$\"3%eWc1mHs[\"FL$\"3q'
4B&ovU!=\"Fi[r7$$\"3-nz@Fr%**[\"FL$\"3?`'fJ[P&\\6Fi[r7$$\"3m45Mg?Q&\\
\"FL$\"3A1UbI2BM5Fi[r7$$\"3I_SY$*p\"3]\"FL$\"3$4$)*o&p]1a)Fagq-Fafq6&F
cfqFffqFdfqFffq-%+AXESLABELSG6%Q\"r6\"Q!Fc^x-%%FONTG6#%(DEFAULTG-%%VIE
WG6$;F(F\\fqFh^x" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 
0 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "
eval([FE(kE(0.1),l0,r),evalf(Fas(kE(0.1),l0,r))],r=300);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#7$^$$\"/RSKr)e9\"!#8$!/$fZs.(ey!#G$\"/]%[,A1;\"F
'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "eval([FE(kE(0.1),l0,r)
,evalf(Fas(kE(0.1),l0,r))],r=1000);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#7$^$$!/Y;lG!)\\>!#9$\"/g`'*RyWq!#H$!/?1#\\\"**))=F'" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 57 "eval([FE(kE(0.2),l0,r),evalf(Fas(kE(0.2),l0,r))],r=1000);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#7$^$$\"/zrk4_2K!#9$!/r=w**Q7k!#D$\"/67
\"R!GIKF'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "The numbers are susp
iciously close." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "eval([FE
(kE(0.4),l0,r),evalf(Fas(kE(0.4),l0,r))],r=1000);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7$^$$!/?1bU'o'y!#9$\"/WySkPmy!#D$!/HZ/olpyF'" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "eval([FE(kE(0.2),l0,r),evalf
(Fas(kE(0.2),l0,r))],r=2000);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$^$$
\"/@R:!)H-5!#8$!/R3%R+tK#!#G$\"/2G<W%H+\"F'" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 74 "The asymptotic limit of the solution is reached at rath
er large values of " }{TEXT 285 1 "r" }{TEXT -1 1 "." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 288 11 "Exercise 4:" }}{PARA 
0 "" 0 "" {TEXT -1 105 "Check the validity of the asymptotic expressio
n for different choices of angular momentum quantum number " }{TEXT 
287 1 "l" }{TEXT -1 23 " at different energies." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
286 28 "Verification of the solution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 101 "Now we compare the solution as stated \+
above with the numerical solution of the Schroedinger equation." }}
{PARA 0 "" 0 "" {TEXT -1 88 "The Schroedinger equation is formulated f
or a radial orbital which starts with psi(0)=0:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 86 "SE:=(E,l)->-hbar^2/(2*mu)*diff(psi(r),r$2)+(hb
ar^2/(2*mu)*l*(l+1)/r^2-Z/r-E)*psi(r)=0;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#SEGj+6$%\"EG%\"lG6\"6$%)operatorG%&arrowGF)/,&*&#\"\"\"\"\"#F
1*(%%hbarGF2%#muG!\"\"-%%diffG6$-%$psiG6#%\"rG-%\"$G6$F=F2F1F1F6*&,(*&
#F1F2F1*,F4F2F5F69%F1,&FFF1F1F1F1F=!\"#F1F1*&%\"ZGF1F=F6F69$F6F1F:F1F1
\"\"!F)F)F)6$FLFL" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E0:=0.2
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E0G$\"\"#!\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "SE(E0,l0);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/,&*&#\"%P=\"%sO\"\"\"-%%diffG6$-%$psiG6#%\"rG-%\"$G6$F
0\"\"#F)!\"\"*&,&$F4F5F5*&F)F)F0F5F5F)F-F)F)\"\"!" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 11 "k0:=kE(E0);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%#k0G$\"/#[,lLGK'!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "We ma
ke a choice of initial condition which is compatible with the analytic
 solution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "zero:=1E-10;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%zeroG$\"\"\"!#5" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "FE(k0,l0,zero);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#^$$\"/-kV)f*)*>!#B$!/Mj?M=xb!#Z" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 12 "psi0:=Re(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%%psi0G$\"/-kV)f*)*>!#B" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
38 "FEp:=eval(diff(FE(k0,l0,r),r),r=zero);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$FEpG^$$\"//WV)f*)*>!#8$!\"\"!#B" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 13 "psi0p:=Re(%);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%&psi0pG$\"//WV)f*)*>!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 38 "IC:=psi(zero)=psi0,D(psi)(zero)=psi0p;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#ICG6$/-%$psiG6#$\"\"\"!#5$\"/-kV)f*)*>!#B/--%\"DG6#F
(F)$\"//WV)f*)*>!#8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "sol:
=dsolve(\{SE(E0,l0),IC\},numeric,output=listprocedure,maxfun=0):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Fnum:=eval(psi(r),sol):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Fnum(10),FE(k0,l0,10);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6$$!3iMoaAes0&)!#=^$$!/,v-Xr0&)!#9$\"/'o
hV\\Vr\"!#E" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 183 "For a large value
 of r the numerical solution can be faster to evaluate than the analyt
ical one. The power series for the hypergeometric function converges s
lowly for large arguments:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
11 "Fnum(1000);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"3:iy\"=MDz?$!#=
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "FE(k0,l0,1000);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#^$$\"/zrk4_2K!#9$!/sj>X!yS'!#D" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "The imaginary part is an artefact.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "This
 also raises the question of accuracy. How well do our solutions compa
re with the asymptotic anwer?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 "evalf(Fas(k0,l0,1000));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/6
7\"R!GIK!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "Fnum(5000),F
E(k0,l0,5000),evalf(Fas(k0,l0,5000));" }}{PARA 7 "" 1 "" {TEXT -1 33 "
Warning, computation interrupted\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 37 "The numerical solver works for large " }{TEXT 289 1 "r" }{TEXT 
-1 20 " values only if the " }{TEXT 19 6 "maxfun" }{TEXT -1 152 " opti
on is set to iterate as long as required to reach the pre-set toleranc
e. The accuracy of the power series could be increased by an increase \+
of the " }{TEXT 19 6 "Digits" }{TEXT -1 58 " parameter. (This would sl
ow down the numerical solver in " }{TEXT 19 6 "dsolve" }{TEXT -1 204 "
, as it would no longer be using the floating point hardware of the CP
U). For the given settings the convergence of the numerical solution t
o the asymptotic result has been demonstrated in the line above." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 218 "The nume
rical approach can be used for potentials that occur in models of many
-electron atoms, where the pure Coulomb potential is replaced by a scr
eened Coulomb potential for which no analytical solution can be found.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 290 11 "Exer
cise 5:" }}{PARA 0 "" 0 "" {TEXT -1 81 "Verify the solutions for other
 energies and also for angular momentum symmetries " }{TEXT 291 1 "l" 
}{TEXT -1 3 ">0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "103 0 0" 0 
}{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }