{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 "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 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 "Times" 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 }{PSTYLE "Maple Out put" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Plot" -1 13 1 {CSTYLE "" -1 -1 "Time s" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 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;" }}{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 " 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$\"0\"yWy*yJ!=F4$\"01I)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 " " 1 "" {GLPLOT2D 882 331 331 {PLOTDATA 2 "6&-%'CURVESG6$7en7$$!0([zEjz q:!#9$\"0/mK[X/T&!#;7$$!0r&RZ&=B]\"F*$\"01T(zk\"oc&F-7$$!0>*o(ROFW\"F* $\"0@YcJMC'fF-7$$!0rT)Q+tv8F*$\"0c$\\+B+8nF-7$$!0P_$=&z#38F*$\"0%[dI:4 DyF-7$$!0(GTy&\\6C\"F*$\"0]Fy$3fG$*F-7$$!0]e3`6*y6F*$\"0g%e)3-?6\"!#:7 $$!0O%ePwY96F*$\"0E[vEy/\"F*$\"0m$[5GFL7$$!0k'e#y'\\I\"*FL$\"0eMUNH\\T#FL7$$!0!\\5v\\fG& )FL$\"05]&\\1*3&GFL7$$!0$e$**4'*4&yFL$\"0J%H-Dd2MFL7$$!0'HoQ\\hqrFL$\" 0M&)*=wXLSFL7$$!0T_#=;%\\^'FL$\"02K)zyp#p%FL7$$!0![nGu_>fFL$\"01IQ)HAH `FL7$$!/OdnB_6_F*$\"0soKI-c6'FL7$$!/Uk6#\\F*$\"0(*pzS\"o.$*FL7$$!/#G!>s`+8F*$\"07Jd;S3p *FL7$$!0]j1'yK9**F-$\"0tJNtI!>)*FL7$$!.X5`$GBoF*$\"05s/BGQ\"**FL7$$!0+ 1N3Hi:&F-$\"0jSnO!p]**FL7$$!0+nfju\"*[$F-$\"0M^m;(Qx**FL7$$!0+G%)=?@#= F-$\"0&H0Rv#Q***FL7$$!-*)3ul]:F*$\"0yL>Gb*****FL7$$\"0+()HJNld\"F-$\"0 ]??')y`***FL7$$\"0+joOO\"3LF-$\"0'Q!zGq'z**FL7$$\"0+R2UP(R]F-$\"0opi]( )G&**FL7$$\".:YZQ8x'F*$\"0zW28H^\"**FL7$$\"0]?,4mgy*F-$\"0)*H%*GPO#)*F L7$$\"/Ecq$z+G\"F*$\"0W+c!)G.q*FL7$$\"/Tu`qFJ>F*$\"0%oxIA?K$*FL7$$\"/) 4-)z-/EF*$\"0%Q6Ja^A))FL7$$\"/eE7N=iKF*$\"0_SHVK7A)FL7$$\"/qDJG)*)*QF* $\"0W8,ml)ovFL7$$\"/]o.*Rgg%F*$\"09jWr3\\z'FL7$$\"/`*38i8C&F*$\"0[4,58 @3'FL7$$\"/kD:Uq>fF*$\"07%)3\")H!H`FL7$$\"/^9lHQMlF*$\"0%R4f$fCn%FL7$$ \"/sq'>*Q1sF*$\"0JM\\<&)*)*RFL7$$\"/p$**o%pQyF*$\"0O?>))4$=MFL7$$\"/$3 47O'*\\)F*$\"08x^:jK(GFL7$$\"/vxyr#e9*F*$\"0Ij*G$[XS#FL7$$\"/i\\I$HB#) *F*$\"0S)\\:lk\")>FL7$$\"0#*y18)QZ5F*$\"0IAY&oDP;FL7$$\"0fG3t=S6\"F*$ \"0%4D^x\\W8FL7$$\"0'z'ff(4!=\"F*$\"0`]zM-#36FL7$$\"0fX.@<3C\"F*$\"0.! yw.5P$*F-7$$\"0KF1u3/J\"F*$\"0q9a\"[7%y(F-7$$\"0&z\"\\B_EP\"F*$\"0[rP% 3obnF-7$$\"0F%4#y;!R9F*$\"0\\2c@7a*fF-7$$\"0Un;7OD]\"F*$\"0(>py1#ec&F- 7$$\"0([zEjzq:F*F+-%'COLOURG6&%$RGBG$\"*++++\"!\")$\"\"!Fg]lFf]l-F$6$7 W7$F($!0t(Q8x-\\aF*7$$!0H&4PubO:F*$!0V\\i%*RR&[F*7$F/$!0`v)4]pGVF*7$$! 0XUDZFDZ\"F*$!0*y^XFQARF*7$F4$!07!3.(>$eNF*7$F9$!0k$>%HRC(GF*7$F>$!0J6 @WpZL#F*7$FC$!0HfM#pV=>F*7$FH$!0t-t(HB:;F*7$FN$!0nR'p2)pO\"F*7$FS$!0$e (\\1P^;\"F*7$FX$!0/_FQoo+\"F*7$Fgn$!0(R5I`y\"y)FL7$F\\o$!0[az0Nf'yFL7$ Fao$!08j='fJ;qFL7$Ffo$!0JRu)yg-jFL7$F[p$!0Q5Q(=-/dFL7$F`p$!0JDf)Gj2_FL 7$Fep$!0([W^uqYYFL7$Fjp$!0Ok%y!yt<%FL7$F_q$!0#eKC)\\2i$FL7$Fdq$!0%yqB1 J3JFL7$Fiq$!0)3r9P]=DFL7$F^r$!0<'*zVk(G>FL7$Fcr$!0Z#R/'QsG\"FL7$F]s$!0 (f**>cs.oF-7$Fat$!09zu4b1b\"!#<7$Feu$\"0,]=&=A_nF-7$F_v$\"0=n^`)Qn7FL7 $Fdv$\"0xS:CQ!*)=FL7$Fiv$\"0yV$3'y^]#FL7$F^w$\"0/31Cd\"zIFL7$Fcw$\"0,% e$H)R;\"F*7$F jz$\"0Vdb:]aO\"F*7$F_[l$\"0.t8uz.i\"F*7$Fd[l$\"0'RwBSi;>F*7$Fi[l$\"0aj ,kc(\\BF*7$F^\\l$\"0uWc&=&[%GF*7$Fc\\l$\"0h#eqYh:NF*7$$\"0%3)=Xw2Z\"F* $\"0R9%z\"f)**QF*7$Fh\\l$\"0R?55H=L%F*7$$\"09JUAmm`\"F*$\"0a+du:d&[F*7 $F]]l$\"0t(Q8x-\\aF*-F`]l6&Fb]lFf]lFf]lFc]l-%+AXESLABELSG6$Q\"x6\"Q!Fi hl-%%VIEWG6$;$!0!\\zEjzq:F*$\"0!\\zEjzq:F*%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "The antisymmetric solution is not acceptable, it is not periodic. The ground state is unique, it has a \+ symmetric eigenfunction." }}{PARA 0 "" 0 "" {TEXT -1 122 "Let us plot \+ a probability using the eigenenergy as a baseline, and superimpose it \+ on the potential as a function of theta." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "P0:=plot([EV[1],EV[1]+(eval(rhs(sol),\{_C1=1,_C2=0,x= s/2\}))^2],s=-Pi..Pi,color=[black,green]):" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 23 "plots[display](Pmw,P0);" }}{PARA 13 "" 1 "" {GLPLOT2D 1020 263 263 {PLOTDATA 2 "6'-%'CURVESG6$7`jl7$$!3]I(*e`EfTJ! #<$\"\"%\"\"!7$$!34q9yr&=%GJF*$\"3oX'ygWE)**RF*7$$!3#*R:kte&p6$F*$\"3Q -VxfIR**RF*7$$!3)*>qtq[1/JF*$\"3]J-zI=f)*RF*7$$!3')zt*yT)3\"4$F*$\"3GW THk)\\u*RF*7$$!3y\\H()ROmIF*$\"37&* HvP'=V*RF*7$$!31+>)Q'>!Q0$F*$\"3%)3MhVxH#*RF*7$$!39!*\\=5*z4/$F*$\"3)) )o\\Ch&))*)RF*7$$!3****o(p(*)>GIF*$\"3%z@;A0:IF*$ \"3[da$H()3S)RF*7$$!34SU9UDZ.IF*$\"3YO4sVJ&4)RF*7$$!3')*H+=dO/*HF*$\"3 %RhC!4`>xRF*7$$!3'*p'4X2Zt(HF*$\"39qv.!)R3tRF*7$$!3Aq_W=HtkHF*$\"3s#\\ \\B3-)oRF*7$$!37]5Gc!yK&HF*$\"3U_(\\qGUY'RF*7$$!3#)4)Hs6d'RHF*$\"3g$o7 6Ug$fRF*7$$!3))*)=>&R=\"GHF*$\"3%Rik*R:gaRF*7$$!3\")f@P=hp9HF*$\"3c/gJ ]%Q([RF*7$$!3'*>%p!H^\"G!HF*$\"3eW:@pcDVRF*7$$!3-?)o2!)z(*)GF*$\"3#za9 dkCp$RF*7$$!3'*>+k&3nt(GF*$\"3M^dx%3!fIRF*7$$!3+SCe>dTkGF*$\"3$em&pbNm BRF*7$$!3!*pEcAB__GF*$\"3dWkd/'=q\"RF*7$$!3/5Z8(o$pRGF*$\"3b9Y*z;Z&4RF *7$$!3%)ptMS$oj#GF*$\"3`dT6=PX,RF*7$$!3\")Hnz'eoZ\"GF*$\"3o6$G$>M8%*QF *7$$!3$**Rw^`SA!GF*$\"3ICufb5%f)QF*7$$!3)*p=)>$yH*y#F*$\"3'H5_:qmr(QF* 7$$!3#*>j%>#fjwFF*$\"3o'=*z4%y#oQF*7$$!3)*\\O#H([QkFF*$\"3_QZv5MRfQF*7 $$!3+S1tr@y]FF*$\"3)*\\+gS7?\\QF*7$$!3y\\TqR&f&QFF*$\"3dVKr<7vRQF*7$$! 375t9r#4bs#F*$\"3Oy!*G\\zNHQF*7$$!3&)>N=)y$o8FF*$\"3z#RF#*)4n>QF*7$$!3 *)HZy6av+FF*$\"3u#)Gmj%*y3QF*7$$!3#)*RDr\"3f)o#F*$\"3#pV#)Qqu#)z$F*7$$ !3=qxj5`(en#F*$\"3CMGXD$**py$F*7$$!3))HJ')yNWjEF*$\"3cp#*HVipvPF*7$$!3 )*>!3yqG/l#F*$\"3q&>vg&*oNw$F*7$$!3zRy7SQ*yj#F*$\"3gQ!3U^1;v$F*7$$!3-! ztu7v]i#F*$\"3FdG*>d)3RPF*7$$!3?]DBgDO7EF*$\"3IGMZc?REPF*7$$!3%)4S@J5o +EF*$\"3?pdIa#zWr$F*7$$!3%)*)))p!o#H(e#F*$\"3t^h=8#Q0q$F*7$$!3!*zWR!)z JvDF*$\"3u%*ybo3\"yo$F*7$$!37+/G_/biDF*$\"3it``DV(Rn$F*7$$!3yR`x)HI.b# F*$\"3gD\\S%Qv/m$F*7$$!39q&z()Qhn`#F*$\"3wy'*>7h>XOF*7$$!3))>$R(4<,DDF *$\"38'Qw]d?5Aj#QjQ#F*$\"3Yk^cxV>cMF*7$$!3-]7sN#\\KP#F*$ \"3B\\bUyd7QMF*7$$!39I=\"R^+9O#F*$\"3;5(=Szc:U$F*7$$!3!*>d?eaI[BF*$\"3 lpq@%*G,.MF*7$$!3#)>StwEwNBF*$\"3/Z@AZa-&Q$F*7$$!3**pMuC/DBBF*$\"3!=v7 NokoO$F*7$$!3#*fO/H8=5BF*$\"3+?pFbsmZLF*7$$!33])*z9U9)H#F*$\"3^9B97?yH LF*7$$!3A!H(Hjl\"eG#F*$\"37f\"=SRl7J$F*7$$!3;q=Cr`?sAF*$\"3yb@+)[*e!H$ F*7$$!3-]os^:))fAF*$\"3C4is)zi;F$F*7$$!3yzY*y8ysC#F*$\"3oNN$4)p5_KF*7$ $!3$F*7$$!3!)>0hnql(>#F*$\"37Kvn-0@tJF*7$$!3))fz aq_K%=#F*$\"3CWH`dF^^JF*7$$!3/+?`>KJF*7$$!3;S[5xk)*e @F*$\"3\\&R:'pCr4JF*7$$!3!)H(4W$ptY@F*$\"3G41K,yC*3$F*7$$!30!4)>_mhM@F *$\"3UB%\\z?Q)oIF*7$$!3:gQ9bMe@@F*$\"3Qy)>jA;n/$F*7$$!38q47Ov\\3@F*$\" 3^F^U%)eKCIF*7$$!3ypfS(G\"f'4#F*$\"3BhPN*[,Q+$F*7$$!3+]V9!QZR3#F*$\"3) z5(\\m(\\=)HF*7$$!3#)*G=C;3;2#F*$\"3yTfkmcFgHF*7$$!35?nA?BBe?F*$\"3\\n j&G2Cn$HF*7$$!3-Ix%y\\em/#F*$\"3SUY4***4i\"HF*7$$!3e(\\#*GF*7$$!3(*f%=[Yp3-#F*$\"3%oTQ(zF1qGF*7$$!3$)R)Qg3%\\3?F*$\"3u! 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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. 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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. 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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. 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