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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 16 "Quantum Pendulum" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 598 "The prob
lem of quantizing the nonlinear mathematical pendulum was taken up by \+
E U Condon (Phys Rev 31, 891). It seems of academic interest at first,
 and is not pursued in quantum texts. It has had some attention in Ame
rican Journal of Physics articles: a visualization of the solutions in
 terms of Mathieu functions was given in AJP 71, 233; applications in \+
molecular oscillations and hindered rotations due to electrostatic res
toring torques were discussed in AJP 70, 525; semiclassical quantizati
on was considered in AJP 48, 660; further pedagogical AJP references c
an be found in these articles." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 228 "The pendulum problem is interesting as i
t connects two distinct limits: for small-amplitude oscillations the w
ell-known harmonic oscillator is obtained; for large eigenenergies the
 states should approach those of a free rotator." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 326 "These solutions are also
 of interest in the problem of periodic potential wells with wrap-arou
nd boundary conditions, such as discussed, e.g., in the QM text of Gri
ffiths. Bloch's theorem can be seen at work in this context when repla
cing the angular variable of the quantum pendulum by a one-dimensional
 Cartesian coordinate." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 338 "The quantum pendulum Schroedinger equation for an o
bject of rotational inertia m*L^2 is cast into dimensionless form. and
 is then connected with the Mathieu differential equation. Our aim is \+
to find the allowed eigenenergies from a matrix representation in a Fo
urier basis. The eigenfunctions are then obtained from the Mathieu fun
ctions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "restart; Digits:=15: with(LinearAlg
ebra):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "a:=Pi;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"aG%#PiG" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 199 "We selected a standard interval for the dynamical variab
le, i.e., [-Pi,Pi). The re-scaled potential energy is chosen such that
 in the oscillator limit the eigenenergies are given as the odd intege
rs." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "V0:=2;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#V0G\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "V:=x->V0*(1-cos(Pi*x/a));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"VGj+6#%\"xG6\"6$%)operatorG%&arrowGF(*&%#V0G\"\"\",
&F.F.-%$cosG6#*(%#PiGF.9$F.%\"aG!\"\"F7F.F(F(F(6#\"\"!" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Vmax:=V(a);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%VmaxG\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "W
e pick a single well (our program allows to create multiple regions ad
jacent to each other):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "Nw
:=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NwG\"\"\"" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 88 "Vmw:=piecewise(seq(op([x>a*(2*j-1) and x<
a*(2*j+1),V(x-2*a*j)]),j=-(Nw-1)/2..(Nw-1)/2));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$VmwG-%*PIECEWISEG6$7$,&\"\"#\"\"\"*&F*F+-%$cosG6#%\"
xGF+!\"\"32,$%#PiGF1F02F0F57$\"\"!%*otherwiseG" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 44 "PTS:=[seq((2*j-1)*a,j=-(Nw-1)/2..(Nw+1)/2)];" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$PTSG7$,$%#PiG!\"\"F'" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 35 "We want to have periodicity with 2a" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "Pmw:=plot(Vmw,x=PTS[1]..PTS[
nops(PTS)],numpoints=500):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 200 "A \+
periodic basis would allow for left- and right-travelling states with \+
periodicity conditions at a, and -a. We would solve within a single we
ll, but the states would be of travelling plane-wave type." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "psi:=n->eval(exp(I*Pi*n*x/(Nw*a)));
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$psiGj+6#%\"nG6\"6$%)operatorG%&
arrowGF(-%%evalG6#-%$expG6#*.%#PiG\"\"\"9$F4%\"xGF4%#NwG!\"\"%\"aGF8^#
F4F4F(F(F(6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "int(co
njugate(psi(0))*psi(2),x=PTS[1]..PTS[nops(PTS)]);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "L:=
int(conjugate(psi(1))*psi(1),x=PTS[1]..PTS[nops(PTS)]);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"LG,$*&\"\"#\"\"\"%#PiGF(F(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 68 "int(-diff(psi(3),x$2)*conjugate(psi(3)),x
=PTS[1]..PTS[nops(PTS)])/L;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"*" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "4*3^2*Pi^2/(L)^2;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "We need posit
ive and negative n, including 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 24 "n_max:=20; N:=2*n_max+1;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&n_maxG\"#?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"NG
\"#T" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "VME:=unapply(int(V(
x)*exp(Pi*I*dn*x/(Nw*a)),x=x1..x2),x1,x2,dn);" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#>%$VMEGj+6%%#x1G%#x2G%#dnG6\"6$%)operatorG%&arrowGF*,$*
*\"\"#\"\"\",2*(^#!\"\"F1-%$expG6#*(9&F19$F1^#F1F1F1)F:F0F1F1*&F6F1F<F
1F1**F=F1F6F1-%$cosG6#F;F1F<F1F1*(F6F1-%$sinGFBF1F:F1F1*(-F76#*(F:F19%
F1F<F1F1F=F1F<F1F1*&F4F1FGF1F1**F4F1F=F1FGF1-FA6#FJF1F1*(FGF1-FEFNF1F:
F1F5F1F:F5,&*$F=F1F1F1F5F5F5F*F*F*6%\"\"!FTFT" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 23 "#VME(PTS[1],PTS[2],21);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 33 "#limit(VME(PTS[1],PTS[2],s),s=0);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "kh:=1; # for the cosine potential w
e put the strength into V(x)." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#kh
G\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "H:=Matrix(N,N,sh
ape=hermitian,datatype=complex[8]):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 144 "We should have a 1/sqrt(2*a) = 1/sqrt(L) normalization factor!
 We can fix this by dividing the matrix by L=n*2a, where n is the numb
er of wells." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "for i1 fro
m 1 to N do: n1:=-n_max+i1-1: for i2 from 1 to i1 do: n2:=-n_max+i2-1:
 if i2=i1 or n1-n2=Nw then" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 209 "H[i1
,i2]:=evalf(1/L*kh*add(limit(VME(PTS[i0],PTS[i0+1],s),s=n1-n2),i0=1..N
w)): else H[i1,i2]:=evalf(1/L*kh*add(VME(PTS[i0],PTS[i0+1],n1-n2),i0=1
..Nw)): fi: od: H[i1,i1]:=H[i1,i1]+evalf(4*n1^2*Pi^2/(L)^2): od: " }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "Eigenvalues(H);" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "EV:=convert(%,list);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#-%'RTABLEG6*\")Wg&Q#%'sfloatG&%'VectorG6#%'columnG%,rectangularG
%.Fortran_orderG7\"\"\"\";F/\"#T" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%
#EVG7K$\"0%eAaHq)H*!#:$\"0v&zc-s'o#!#9$\"07R'3(osq$F+$\"0\"4`A)3I6'F+$
\"0(HqEZXihF+$\"0#oi&GNd5\"!#8$\"0+(o7)[d5\"F4$\"0\"yWy*yJ!=F4$\"0<Ui)
*yJ!=F4$\"05_CH@?q#F4$\"0&RX#H@?q#F4$\"0jPo&*)R,QF4$\"0hQo&*)R,QF4$\"0
HR(zd-,^F4$\"0JR(zd-,^F4$\"0#3MvVy+mF4$\"0*3MvVy+mF4$\"0IE_A>1I)F4$\"0
\"oAD#>1I)F4$\"0:4p7]+-\"!#7$\"0;4p7]+-\"FQ$\"0*\\w39/I7FQ$\"0.l(39/I7
FQ$\"0nRJyM+Y\"FQ$\"0pRJyM+Y\"FQ$\"0Td*H'H+r\"FQ$\"0Zd*H'H+r\"FQ$\"0$R
*HaD+)>FQ$\"0-%*HaD+)>FQ$\"0YzqCA+F#FQ$\"0^zqCA+F#FQ$\"0-O/b>+e#FQ$\"0
1O/b>+e#FQ$\"0]#3;t,5HFQ$\"0`#3;t,5HFQ$\"0[sSW:+E$FQ$\"0dsSW:+E$FQ$\"0
,e@%R,IOFQ$\"00e@%R,IOFQ$\"0wX#GjD?SFQ$\"0yX#GjD?SFQ" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 223 "We could construct the eigenfunctions from the
 Eigenvectors command. The calculations with these states are tedious \+
(slow). We prefer to connect with the Mathieu differential equation. T
his requires us to rescale theta=2*x." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "q:=2*V0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG\"\"
%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "MDE:=a->diff(u(x),x$2)
+(a+2*q*cos(2*x))*u(x)=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$MDEGj+
6#%\"aG6\"6$%)operatorG%&arrowGF(/,&-%%diffG6$-%\"uG6#%\"xG-%\"$G6$F4
\"\"#\"\"\"*&,&9$F9*(F8F9%\"qGF9-%$cosG6#,$*&F8F9F4F9F9F9F9F9F1F9F9\"
\"!F(F(F(6#FD" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "Here a is the e
igenvalue scaled according to a=4*(EV-V0), and the potential strength \+
parameter q=2*V0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 16 "aM:=e->4*(e-V0);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#aMGj+6#%\"eG6\"6$%)operatorG%&arrowGF(,&*&\"\"%\"\"
\"9$F/F/*&F.F/%#V0GF/!\"\"F(F(F(6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "sol:=dsolve(MDE(aM(EV[1])));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$solG/-%\"uG6#%\"xG,&*&%$_C1G\"\"\"-%)MathieuCG6%#!0$
[:4%f-9#\"/++++++]!\"%F)F-F-*&%$_C2GF--%)MathieuSGF0F-F-" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 297 "The MathieuC and MathieuS functions are \+
symmetric and antisymmetric respectively, i.e., behave like sine and c
osine. We invoke a physics argument which states that the eigenstates \+
of the problem are either symmetric or antisymmetric, as the Hamiltoni
an is symmetric. The ground state is symmetric." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 97 "plot([eval(rhs(sol),\{_C1=1,_C2=0\}),eval(rhs(
sol),\{_C2=1,_C1=0\})],x=-Pi/2..Pi/2,color=[red,blue]);" }}{PARA 13 "
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "The antisymmetric solution is not
 acceptable, it is not periodic. The ground state is unique, it has a \+
symmetric eigenfunction." }}{PARA 0 "" 0 "" {TEXT -1 122 "Let us plot \+
a probability using the eigenenergy as a baseline, and superimpose it \+
on the potential as a function of theta." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 93 "P0:=plot([EV[1],EV[1]+(eval(rhs(sol),\{_C1=1,_C2=0,x=
s/2\}))^2],s=-Pi..Pi,color=[black,green]):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 23 "plots[display](Pmw,P0);" }}{PARA 13 "" 1 "" 
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" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=dsolve(MDE(aM(EV[
2])));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"uG6#%\"xG,&*&%$_
C1G\"\"\"-%)MathieuCG6%#\".$=F5)ou#\".++++++\"!\"%F)F-F-*&%$_C2GF--%)M
athieuSGF0F-F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "We expect the f
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Now the symmetric solution is not
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Fbbx-%%FONTG6#%(DEFAULTG-%%VIEWG6$;$!0z*e`EfTJF^bu$\"0z*e`EfTJF^buFgbx
" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "C
urve 2" "Curve 3" "Curve 4" "Curve 5" }}}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 179 "This looks quite analogous to the harmonic oscillator re
sult except that we can see the considerable lowering of the first exc
ited state (cf E_HO=3) as the potential turns around." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 212 "Based upon the harmon
ic oscillator we might expect the next level to be a rotator state as \+
E_HO=5 is above the potential barrier. Our numerically obtained eigenv
alue is below the barrier. Let us see what happens." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=dsolve(MDE(aM(EV[3])));" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#>%$solG/-%\"uG6#%\"xG,&*&%$_C1G\"\"\"-%)Mathieu
CG6%#\"/8\"p'\\\"eO\"\".++++++#!\"%F)F-F-*&%$_C2GF--%)MathieuSGF0F-F-
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "We find a bound state which i
s symmetric. The antisymmetric function is not periodic." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "plot([eval(rhs(sol),\{_C1=1,_C2=0\}
),eval(rhs(sol),\{_C2=1,_C1=0\})],x=-Pi/2..Pi/2,color=[red,blue]);" }}
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257 21 "Exercise 1 (trivial):" }}{PARA 0 "" 0 "" {TEXT -1 110 "Complet
e the graph of the probability density for all bound states superimpos
ed on the potential energy curve." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 42 "Now we get into the rotator regime (E>V0)
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=dsolve(MDE(aM(EV[
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{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fS:=eva
l(rhs(sol),\{_C1=1,_C2=0\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fSG
-%)MathieuCG6%#\"/4`A)3I6%\".+++++]#!\"%%\"xG" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 33 "fA:=eval(rhs(sol),\{_C2=1,_C1=0\});" }}{PARA 11 
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i/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"0!eMjw/\\8!#9" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "evalf(limit(fS,x=Pi/2));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"0!eMjw/\\8!#9" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 33 "evalf(limit(diff(fS,x),x=-Pi/2));" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#$!0d<)G$Qla\"!#:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "evalf(limit(diff(fS,x),x=Pi/2));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"0d<)G$Qla\"!#:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
46 "The symmetric function is not really periodic!" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 25 "evalf(limit(fA,x=-Pi/2));" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#$\"0s3*zO1vF!#F" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "evalf(limit(fA,x=Pi/2));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$!0s3*zO1vF!#F" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 33 "evalf(limit(diff(fA,x),x=-Pi/2));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"0R\"RPlj7u!#:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "e
valf(limit(diff(fA,x),x=Pi/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"
0R\"RPlj7u!#:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "The anti-symmetr
ic function is perfectly acceptable! Now look at the next eigenvalue:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=dsolve(MDE(aM(EV[5
])));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"uG6#%\"xG,&*&%$_C
1G\"\"\"-%)MathieuCG6%#\"0>\"o!*=)\\m\"\"/++++++5!\"%F)F-F-*&%$_C2GF--
%)MathieuSGF0F-F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 163 "Again, it l
ooks like both functions are periodic. However, this is probably not t
rue. The antisymmetric function is not matching at the boundaries, onl
y almost so." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "plot([eval(
rhs(sol),\{_C1=1,_C2=0\}),eval(rhs(sol),\{_C2=1,_C1=0\})],x=-Pi/2..Pi/
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m\"\"/++++++5!\"%%\"xG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "f
A:=eval(rhs(sol),\{_C2=1,_C1=0\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
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{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalf(limit(fS,x=-Pi/2));" }}{PARA 
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" {MPLTEXT 1 0 24 "evalf(limit(fS,x=Pi/2));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"0`&QtQpT8!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 25 "evalf(limit(fA,x=-Pi/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$
!0vPq&G0]5!#;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "evalf(limi
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "What is our finding?" }}{PARA 0 "
" 0 "" {TEXT -1 414 "In the regime where we expect rotator solutions w
e are getting closely spaced anti-symmetric and symmetric solutions. T
hese look like bound states. We cannot form linear combinations that w
ould correspond to travelling waves as the energies are not degenerate
. Rotation is inhibited, because in quantum scattering both reflection
 and transmission are probable for energies not too high above the pot
ential barrier." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 258 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 342 "Explore wha
t happens for the higher energies. Do true rotator states appear? If t
he symmetric and anti-symmetric solutions appeat at identical energies
, then one can form linear combinations along the lines of exp(I*s) = \+
cos(s) + I*sin(s). Both signs are possible resulting in counterclockwi
se and clockwise rotator states at the same energy." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 11 "Exercise 3:" }}{PARA 
0 "" 0 "" {TEXT -1 137 "What is the energy spectrum of the pure rotato
r. How close are the higher states to the expected m-dependence, where
 m counts the states." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
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