{VERSION 6 0 "IBM INTEL NT" "6.0" }
{USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 
1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }
{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 
258 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 
0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Tim
es" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }}
{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;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 199 "We selected a standard interval for the \+
dynamical variable, i.e., [-Pi,Pi). The re-scaled potential energy is \+
chosen such that in the oscillator limit the eigenenergies are given a
s the odd integers." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "V0:=2
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "V:=x->V0*(1-cos(Pi*x/a
));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Vmax:=V(a);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "We pick a single well (our program
 allows to create multiple regions adjacent to each other):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "Nw:=1;" }}}{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));" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 44 "PTS:=[seq((2*j-1)*a,j=-(Nw-1)/2..(Nw+1)/2)];" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "We want to have periodicity with 2
a" }}}{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 state
s with periodicity conditions at a, and -a. We would solve within a si
ngle well, 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/(N
w*a)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "int(conjugate(ps
i(0))*psi(2),x=PTS[1]..PTS[nops(PTS)]);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 58 "L:=int(conjugate(psi(1))*psi(1),x=PTS[1]..PTS[nops(PT
S)]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "int(-diff(psi(3),x
$2)*conjugate(psi(3)),x=PTS[1]..PTS[nops(PTS)])/L;" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 17 "4*3^2*Pi^2/(L)^2;" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "We need \+
positive 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;" }}}{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);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "#VME
(PTS[1],PTS[2],21);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "#lim
it(VME(PTS[1],PTS[2],s),s=0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 64 "kh:=1; # for the cosine potential we put the strength into V(x).
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "H:=Matrix(N,N,shape=her
mitian,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 number of we
lls." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "for i1 from 1 to N
 do: n1:=-n_max+i1-1: for i2 from 1 to i1 do: n2:=-n_max+i2-1: if i2=i
1 or n1-n2=Nw then" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 209 "H[i1,i2]:=ev
alf(1/L*kh*add(limit(VME(PTS[i0],PTS[i0+1],s),s=n1-n2),i0=1..Nw)): els
e 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);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 223 "We could construct the eigenfunctions from the Eigenvect
ors command. The calculations with these states are tedious (slow). We
 prefer to connect with the Mathieu differential equation. This requir
es us to rescale theta=2*x." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
8 "q:=2*V0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "MDE:=a->diff
(u(x),x$2)+(a+2*q*cos(2*x))*u(x)=0;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 102 "Here a is the eigenvalue 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);" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=dsolve(MDE(aM(EV[1])))
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 297 "The MathieuC and MathieuS f
unctions are symmetric and antisymmetric respectively, i.e., behave li
ke sine and cosine. We invoke a physics argument which states that the
 eigenstates of the problem are either symmetric or antisymmetric, as \+
the Hamiltonian 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
]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "The antisymmetric solutio
n is not acceptable, it is not periodic. The ground state is unique, i
t has a symmetric eigenfunction." }}{PARA 0 "" 0 "" {TEXT -1 122 "Let \+
us plot a probability using the eigenenergy as a baseline, and superim
pose 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);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 28 "sol:=dsolve(MDE(aM(EV[2])));" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 52 "We expect the first excited state is anti
-symmetric." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "plot([eval(r
hs(sol),\{_C1=1,_C2=0\}),eval(rhs(sol),\{_C2=1,_C1=0\})],x=-Pi/2..Pi/2
,color=[red,blue]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Now the s
ymmetric solution is not acceptable, it is not periodic. The first exc
ited state is unique, it has an anti-symmetric eigenfunction." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "P1:=plot([EV[2],EV[2]+(eval(
rhs(sol),\{_C1=0,_C2=3,x=s/2\}))^2],s=-Pi..Pi,color=[black,magenta]):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plots[display](Pmw,P0,P
1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 179 "This looks quite analogou
s to the harmonic oscillator result except that we can see the conside
rable lowering of the first excited 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 harmonic oscillator we might expect the n
ext level to be a rotator state as E_HO=5 is above the potential barri
er. Our numerically obtained eigenvalue is below the barrier. Let us s
ee what happens." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=ds
olve(MDE(aM(EV[3])));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "We find \+
a bound state which is symmetric. The antisymmetric function is not pe
riodic." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "plot([eval(rhs(s
ol),\{_C1=1,_C2=0\}),eval(rhs(sol),\{_C2=1,_C1=0\})],x=-Pi/2..Pi/2,col
or=[red,blue]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 257 21 "Exercise 1 (t
rivial):" }}{PARA 0 "" 0 "" {TEXT -1 110 "Complete the graph of the pr
obability density for all bound states superimposed on the potential e
nergy 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[4])));" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 42 "It looks like both functions are 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=[re
d,blue]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fS:=eval(rhs(sol),\{_C1=1,_C2=0\});
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fA:=eval(rhs(sol),\{_C2
=1,_C1=0\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalf(limit
(fS,x=-Pi/2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "evalf(lim
it(fS,x=Pi/2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "evalf(li
mit(diff(fS,x),x=-Pi/2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
32 "evalf(limit(diff(fS,x),x=Pi/2));" }}}{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));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "evalf(limit(fA,x=Pi/2));" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "evalf(limit(diff(fA,x),x=-P
i/2));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "evalf(limit(diff(
fA,x),x=Pi/2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "The anti-symme
tric function is perfectly acceptable! Now look at the next eigenvalue
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=dsolve(MDE(aM(EV[
5])));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 163 "Again, it looks like b
oth functions are periodic. However, this is probably not true. The an
tisymmetric function is not matching at the boundaries, only 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/2,color=[re
d,blue]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fS:=eval(rhs(s
ol),\{_C1=1,_C2=0\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fA
:=eval(rhs(sol),\{_C2=1,_C1=0\});" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "evalf(limit(fS,x=-Pi/2));" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 24 "evalf(limit(fS,x=Pi/2));" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 25 "evalf(limit(fA,x=-Pi/2));" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 24 "evalf(limit(fA,x=Pi/2));" }}}{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 we are getting cl
osely spaced anti-symmetric and symmetric solutions. These look like b
ound states. We cannot form linear combinations that would correspond \+
to travelling waves as the energies are not degenerate. Rotation is in
hibited, because in quantum scattering both reflection and transmissio
n are probable for energies not too high above the potential barrier.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 11 "Exer
cise 2:" }}{PARA 0 "" 0 "" {TEXT -1 342 "Explore what happens for the \+
higher energies. Do true rotator states appear? If the symmetric and a
nti-symmetric solutions appeat at identical energies, then one can for
m linear combinations along the lines of exp(I*s) = cos(s) + I*sin(s).
 Both signs are possible resulting in counterclockwise and clockwise r
otator 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 rotator. How close are th
e higher states to the expected m-dependence, where m counts the state
s." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 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 "2 0 0" 17 }
{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }