{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 Input" 2 19 "" 0 1 255 0 0 1 0 0 0 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 "" 0 1 0 0 0 0 0 0 1 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 "" 1 16 0 0 0 0 0 1 0 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 } {PSTYLE "Maple Output" -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 Plot " -1 13 1 {CSTYLE "" -1 -1 "Times" 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 258 58 "Eigenvalues by solving th e ODE as an initial value problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 345 "We solve the Schroedinger equation numer ically for eigenvalues/eigenfunctions by imposing symmetry at x=0, and using the energy value as a trial value. It is giving results for arb itrary symmetric potential shapes. The eigenenergy is selected by dema nding wavefunctions that vanish as they approach infinity so that they are square-normalizable." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 182 "We choose for a potential a shape that shares som e features with the 3d Coulomb potential. It has a long-range tail, bo und states for E<0 and continuous scattering solutions for E>0." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "restart; Digits:=14:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "V:=-1/sqrt(1+x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,$*&\" \"\"F'*$,&F'F'*$)%\"xG\"\"#F'F'#F'F-!\"\"F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "plot(V,x=-5..5,thickness=3,axes=boxed,view=[-5..5, -1..0]);" }}{PARA 13 "" 1 "" {GLPLOT2D 458 257 257 {PLOTDATA 2 "6'-%'C URVESG6$7\\o7$$!\"&\"\"!$!3WS=Q^8;h>!#=7$$!3YLLLe%G?y%!#<$!3--i7oo)o/# F-7$$!3OmmT&esBf%F1$!3.3/Pm]mF@F-7$$!3ALL$3s%3zVF1$!31^a<6EFEAF-7$$!3_ LL$e/$QkTF1$!3(Ru**z^R\\L#F-7$$!3ommT5=q]RF1$!3mGKM]*3QX#F-7$$!3ILL3_> f_PF1$!3Kp#fX%['\\d#F-7$$!3K++vo1YZNF1$!3]W&3GbzJr#F-7$$!3;LL3-OJNLF1$ !3_$\\APa8>(GF-7$$!3p***\\P*o%Q7$F1$!3K+ZvOxx[IF-7$$!3Kmmm\"RFj!HF1$!3 wD`,w;c`KF-7$$!33LL$e4OZr#F1$!3miI*y4Y'*HcF-7$$!3#****\\(=t)eC\"F1$!3q7l *)=,]fiF-7$$!3!ommmh5$\\5F1$!3!3%=0Gl#*)*oF-7$$!3S$***\\(=[jL)F-$!3M\\ 7&**yt5o(F-7$$!3)f***\\iXg#G'F-$!3kJ)Q$G#RO&*F-7$$!3%\\mmTg=><#F-$!3G[$R x.o@x*F-7$$!3FK$3Fpy7k\"F-$!3_F/#Htrz')*F-7$$!3g***\\7yQ16\"F-$!3LOQ>g ))))Q**F-7$$!3iK$3_D)=`%)!#>$!3#pVhm9iW'**F-7$$!3Epm\"zp))**z&F^t$!3!> 3nMQAK)**F-7$$!3#f+D19*yYJF^t$!3!32!4LD0&***F-7$$!3vDMLLe*e$\\!#?$!3E9 u)o=y)****F-7$$\"3+l;a)3RBE#F^t$!3Snf%G*=W(***F-7$$\"3bsmTgxE=]F^t$!3; 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global Xmax,SE;" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "IC:=u(0)=1,D(u)(0)=0; sol:=dsolve( \{SE(E),IC\},numeric,output=listprocedure):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "U:=eval(u(x),sol): " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "U(Xmax); end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "Xmax: =15;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%XmaxG\"#:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "EVeven(-0.669);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!3y&)pRgXO\\9!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "EVeven(-0.67);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\" 38u.!p=N&4U!#:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "E0:=fsolv e(EVeven,-0.7..-0.6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E0G$!/)[Jb sxp'!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "For your benefit: gra ph the ground state energy superimposed on the potential energy plot t o realize the amount of zero-point fluctuations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "Now find the next eigenvalue for even symmetry:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E2:=fsolve(EVeven,-0.66..-0.1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E2G$!/tnX " 0 "" {MPLTEXT 1 0 54 "sol:=dsolve(\{SE(E2),IC\},numeric,output=listprocedur e):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "U:=eval(u(x),sol):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plot(U(x),x=0..Xmax,view= [0..Xmax,-2..2],thickness=3);" }}{PARA 13 "" 1 "" {GLPLOT2D 453 189 189 {PLOTDATA 2 "6&-%'CURVESG6$7]o7$$\"\"!F)$\"\"\"F)7$$\"/+D1k'p3%!#: $\"/%)4$QIe)**!#97$$\"/+]7G$R<)F/$\"/p%F2$\"/([Tq[RA)F27$$\"/+v= 7T9hF2$\"/Yf'f;A3(F27$$\"/]7.-29xF2$\"/G`pJbgbF27$$\"/+](=HPJ*F2$\"/\" 3'\\LDcQF27$$\"/++v\"*R#4\"!#8$\"/wb4zVC?F27$$\"/+DJaU`7F]o$\"/gMAEKo8 F/7$$\"/D\"yN'o89F]o$!/U]s6*fu\"F27$$\"/]P%GZRd\"F]o$!/co)oD8f$F27$$\" /]7y'HDs\"F]o$!/t81-#*R_F27$$\"/](=276(=F]o$!/*HeS%z2oF27$$\"/]PM0'\\- #F]o$!/q%>!)4,L)F27$$\"/](o**3)y@F]o$!/?#f:Zyt*F27$$\"/](oH>zL#F]o$!/! *=11^16F]o7$$\"/](ofHq\\#F]o$!/1_`*3dA\"F]o7$$\"/]Pf'HU\"GF]o$!/2Bg7rA 9F]o7$$\"/+]7*309$F]o$!/6d!3(zq:F]o7$$\"/+Dce*yU$F]o$!/S)e$>,f;F]o7$$ \"/](=ng'*e$F]o$!/4\"GyzGp\"F]o7$$\"/+]([D9v$F]o$!/Vpu>Q;,Xj \"F]o7$$\"/++]>0)H&F]o$!/jBBRFs:F]o7$$\"/](=-p6j&F]o$!/5w2.'*)[\"F]o7$ $\"/++vS.EfF]o$!/Ov Js)F27$$\"/]7yG>6\")F]o$!/DHJ1\"Fgz$!/_y!)H'pK$F2 7$$\"/+vQ(zS4\"Fgz$!/=UNByFHF27$$\"/v=-,FC6Fgz$!/U*o&3W&e#F27$$\"/v$4t Fe:\"Fgz$!/())\\8o\"fAF27$$\"/+D\"3\"o'=\"Fgz$!/7mu6ao>F27$$\"/voz;)*= 7Fgz$!/v7'e(H\"p\"F27$$\"/++&*44]7Fgz$!/R1)oNvW\"F27$$\"/]7jZ!>G\"Fgz$ !/#Hoh*)*=7F27$$\"/v=(4bMJ\"Fgz$!/I+)>z-,\"F27$$\"/+]xlWU8Fgz$!/KDACR> $)F/7$$\"/]i&3ucP\"Fgz$!/Jhxoz1kF/7$$\"/++lJR09Fgz$!/j6W\"e:z%F/7$$\"/ v=-*zqV\"Fgz$!/1A[\\!Q9$F/7$$\"/D\"G:3uY\"Fgz$!/mNXxY:;F/7$$\"#:F)$\"/ 7:b8pC9!#D-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"x6\"Q!Ff` l-%*THICKNESSG6#\"\"$-%%VIEWG6$;F(Ff_l;$!\"#F)$\"\"#F)" 1 2 0 1 10 3 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 143 "This looks a little suspicious. The tail of th e wavefunction does not approach the axis smoothly (with negligible sl ope). We ought to increase " }{TEXT 19 4 "Xmax" }{TEXT -1 39 " and see how it affects the eigenvalue." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 256 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 113 "Find the first four eigenenergies in the even-symmetry sector to \+ four significant digits. Be careful to increase " }{TEXT 19 4 "Xmax" } {TEXT -1 91 " sufficiently for the higher states by looking at the wav efunctions. Report your choice of " }{TEXT 19 4 "Xmax" }{TEXT -1 136 " , and observe which way the eigenvalue shifted. Can you explain why it changes and how it changes when a node is forced at too small an " } {TEXT 19 1 "x" }{TEXT -1 7 "-value?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT 257 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 260 "Write a modified procedure to calculate the four lowest eigens tates in the odd-symmetry sector. Ensure that you know the eigenvalues to four significant digits. Again explore what happens to the eigenva lues when the wavefunction is cut off to zero at a finite " }{TEXT 19 1 "x" }{TEXT -1 7 "-value." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "24" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }