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We start with the harmonic oscillator, for which we develop solutions to Schroedinger's equation in analytic form, and then we move on to solv e the problem numerically for an anharmonic oscillator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "Without further mot ivation we state that the allowed energies for a quantum particle boun d by a potential " }{TEXT 259 1 "V" }{TEXT -1 1 "(" }{TEXT 258 1 "x" } {TEXT -1 57 ") is given by a differential-equation eigenvalue problem: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "V:=k/2*x^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,$*&%\"kG\"\"\")%\"xG\"\"#F(#F(F+" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "SE:=-h_^2/(2*m)*diff(psi(x), x$2)+V*psi(x)=lambda*psi(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG /,&*&*&)%#h_G\"\"#\"\"\"-%%diffG6$-%$psiG6#%\"xG-%\"$G6$F3F+F,F,%\"mG! \"\"#F8F+**#F,F+F,%\"kGF,)F3F+F,F0F,F,*&%'lambdaGF,F0F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "Let us make life simple by adopting Bohr \+ units (h_=m_e=e=1), and by choosing k=1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "h_:=1: m:=1: k:=1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "SE;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&-%%diffG6$-% $psiG6#%\"xG-%\"$G6$F+\"\"##!\"\"F/*(#\"\"\"F/F4)F+F/F4F(F4F4*&%'lambd aGF4F(F4" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "The physical constra ints on this problem are that the wavefunction psi has to vanish at in finity, i.e., for large positive and negative values of " }{TEXT 257 1 "x" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 112 "We will be inter ested only in such solutions, i.e., we will discard solutions that do \+ not vanish asymptotically." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "On the other hand one can show that the solution s have to be either symmetric, or antisymmetric functions, i.e., " }} {PARA 0 "" 0 "" {TEXT -1 50 "either: psi(x) = psi(-x), or: psi(x) = \+ - psi(-x)" }}{PARA 0 "" 0 "" {TEXT -1 76 "for the acceptable solutions . These conditions translate into conditions at " }{TEXT 260 1 "x" } {TEXT -1 16 " = 0 as follows:" }}{PARA 0 "" 0 "" {TEXT -1 12 "either: \+ psi(" }{TEXT 261 1 "x" }{TEXT -1 58 ") = const, D(psi)(0)=0 ; or: psi (0)=0, D(psi)(0) = const." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 251 "The two undetermined constants fix the amplitude \+ of the solution, and we recognize that this amplitude is arbitrary fro m a mathematical point of view. This is the consequence of solving a h omogeneous problem. Let us start with the symmetric solutions:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICs:=psi(0)=1,D(psi)(0)=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% $ICsG6$/-%$psiG6#\"\"!\"\"\"/--%\"DG6#F(F)F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sol:=dsolve(\{SE,ICs\},psi(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG6\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "So, it's not as simple as that in Maple... Let us substitute a nu mber for lambda (we are using special units, so the lambda - value had dimensions of energy." }}{PARA 0 "" 0 "" {TEXT -1 209 "Let us try whe ther the energy can be zero, i.e., whether we can have a particle sitt ing at the bottom of the well, which corresponds to having the mass si mply at rest in the equilibrium position of the spring:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "sol:=dsolve(\{subs(lambda=0,SE),ICs \},psi(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%$psiG6#%\"xG, &**-%&GAMMAG6##\"\"$\"\"%\"\"\"-%%sqrtG6#\"\"#F2-F46#F)F2-%(BesselIG6$ #F2F1,$*$)F)F6F2#F2F6F2F@*&*(F,F2F7F2-%(BesselKGF;F2F2%#PiG!\"\"F2" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot(rhs(sol),x=-5..5,view= [-5..5,0..5]);" }}{PARA 13 "" 1 "" {GLPLOT2D 755 178 178 {PLOTDATA 2 " 6%-%'CURVESG6$7S7$$\"+!pCR&>!#7$\"+%*********!#57$$\"+!QYP%))F*$\"+&** *******F-7$$\"+2oNt:!#6$\"+/+++5!\"*7$$\"+v*QBE#F6$\"+@+++5F97$$\"+8LI SOF6$\"+X,++5F97$$\"+]wE=]F6$\"+G0++5F97$$\"+Dj>uxF6$\"+VI++5F97$$\"++ D,`5F-$\"+X-,+5F97$$\"+N#)>/;F-$\"+*=b++\"F97$$\"*(RQb@F9$\"+f)z,+\"F9 7$$\"+v:+:JF-$\"+QZy+5F97$$\"*=>Y2%F9$\"+h\")H-5F97$$\"*yXu9'F9$\"+r<$ >,\"F97$$\"*\\y))G)F9$\"+H)p'R5F97$$\"+i_QQ5F9$\"+sL\"*)4\"F97$$\"+!y% 3T7F9$\"+&HOi?\"F97$$\"+O![hY\"F9$\"+))z'zT\"F97$$\"+#Qx$o;F9$\"+::RS< F97$$\"+u.I%)=F9$\"+r-(3J#F97$$\"+(pe*z?F9$\"+(yL4;$F97$$\"+C\\'QH#F9$ \"+!p3pr%F97$$\"+8S8&\\#F9$\"+J*>&RsF97$$\"+0#=bq#F9$\"+E\"[=>\"!\")7$ $\"+2s?6HF9$\"+p!Gh.#Fhr7$$\"+IXaEJF9$\"+\"4\\#[PFhr7$$\"+l*RRL$F9$\"+ /^_rqFhr7$$\"+`<.YNF9$\"+'o$y=9!\"(7$$\"+8tOcPF9$\"+VcQlHF]t7$$\"+\\Qk \\RF9$\"+L;s!3'F]t7$$\"+3ASgSF9$\"+MZSQ$*F]t7$$\"+p0;rTF9$\"+3FZ_9!\"' 7$$\"+mTAqUF9$\"+8(y%z@Fbu7$$\"+lxGpVF9$\"+!3^OI$Fbu7$$\"+A-\"\\Z%F9$ \"+)QB^?&Fbu7$$\"+!oK0e%F9$\"+p8l&H)Fbu7$$\"+k(z5j%F9$\"+;X2T5!\"&7$$ \"+[oi\"o%F9$\"+k9%*48F\\w7$$\"+KRk'pIF\\w7$$\"+30O\"*[F9$\"+k5U%\\$F\\w7$$\"+\"Q?&=\\F9 $\"+`h&4)RF\\w7$$\"+a-oX\\F9$\"+ielQXF\\w7$$\"+\">g#f\\F9$\"+7\\_Z[F\\ w7$$\"+F,%G(\\F9$\"+9(*Qy^F\\w7$$\"+k+U')\\F9$\"+#e!)G`&F\\w7$$\"\"&\" \"!$\"+qHv7fF\\w-%'COLOURG6&%$RGBG$\"#5!\"\"$F\\[lF\\[lFf[l-%+AXESLABE LSG6$Q\"x6\"Q!6\"-%%VIEWG6$;$F\\wF\\[lFjz;Ff[lFjz" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 27 "evalf(subs(x=-1,rhs(sol)));" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#^#$\"+[F$[3\"!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "We don't get an acceptable answer for negative " }{TEXT 264 1 " x" }{TEXT -1 43 ", but presumably we could use the positive-" }{TEXT 263 1 "x" }{TEXT -1 66 " results, and use the symmetry to extend the s olution to negative " }{TEXT 262 1 "x" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 40 "The reason for not plotting at negative " }{TEXT 267 1 "x" }{TEXT -1 33 " is that the solution has a sqrt(" }{TEXT 268 1 "x " }{TEXT -1 42 ") factor, which is imaginary for negative " }{TEXT 269 1 "x" }{TEXT -1 8 " values." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 64 "The main point is that the solution is no t acceptable for large " }{TEXT 265 1 "x" }{TEXT -1 97 ". It blows up. The question is: Can we find energy values such that the solution is \+ normalizable?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "Strategy: try lambda=1/4, and lambda=1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol:=dsolve(\{subs(lambda=1/4,SE),ICs\},p si(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%$psiG6#%\"xG,&*&* &-%&GAMMAG6##\"\"&\"\")\"\"\"-%+WhittakerWG6%#F3F2#F3\"\"%*$)F)\"\"#F3 F3F3*&-%%sqrtG6#%#PiGF3-F?6#F)F3!\"\"F3*&**F " 0 "" {MPLTEXT 1 0 41 "plot(rhs(sol),x=-5..5,view=[-5..5,0..5]); " }}{PARA 13 "" 1 "" {GLPLOT2D 755 178 178 {PLOTDATA 2 "6%-%'CURVESG6$ 7S7$$\"+!pCR&>!#7$\"+^/******!#57$$\"+!QYP%))F*$\"+oW!)****F-7$$\"+2oN t:!#6$\"+'>\"Q****F-7$$\"+v*QBE#F6$\"+,2s)***F-7$$\"+8LISOF6$\"+#p)o'* **F-7$$\"+]wE=]F6$\"+*=5P***F-7$$\"+Dj>uxF6$\"+(oC\\)**F-7$$\"++D,`5F- $\"+[URs**F-7$$\"+N#)>/;F-$\"+'*GGO**F-7$$\"*(RQb@!\"*$\"+1:(e))*F-7$$ \"+v:+:JF-$\"++r:m(*F-7$$\"*=>Y2%FZ$\"+L*Q.h*F-7$$\"*yXu9'FZ$\"+LI7%=* 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the Schroedinger wave equation result we compute the probabi lity density for this state. The square of the wave function is provid ing the probability for finding a particle at location " }{TEXT 293 1 "x" }{TEXT -1 140 ". For this interpretation to hold we need to normal ize the solution of the homogeneous Schroedinger eigenvalue problem in the following way:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "int( rhs(sol)^2,x=-infinity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# *$-%%sqrtG6#%#PiG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 " WF:=1/sqrt(%)*rhs(sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#WFG*&-%$ expG6#,$*$)%\"xG\"\"#\"\"\"#!\"\"F-F.*$)%#PiG#F.\"\"%F.F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "int(WF^2,x=-infinity..infinity);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 259 "Now we graph the probability in such a way as to show th at there is a finite probability to find the quantum particle outside \+ the classically allowed range. For this we plot the density superimpos ed on the potential while using the energy value as a baseline:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plot([V,1/2,WF^2+1/2],x=-2.. 2,color=[red,blue,green]);" }}{PARA 13 "" 1 "" {GLPLOT2D 694 299 299 {PLOTDATA 2 "6'-%'CURVESG6$7S7$$!\"#\"\"!$\"\"#F*7$$!3MLLL$Q6G\">!#<$ \"36!z/TpB%H=F07$$!3bmm;M!\\p$=F0$\"3YCH1x3>(o\"F07$$!3MLLL))Qj^#*oO(Q\"F07$$!3wmm;C2G!e\"F0$\"3I ije$eV'[7F07$$!3OLL$3yO5]\"F0$\"3?[sq3dbE6F07$$!3&*****\\nU)*=9F0$\"3 \"eD1d'***FV$\"3K&3%\\A%*>'*\\FV7$$!3E++++0\"*H\"*FV$ \"3o70!pGjx;%FV7$$!35++++83&H)FV$\"3R%[!o)o=/W$FV7$$!3\\LLL3k(p`(FV$\" 3uh'))*o1ISGFV7$$!3Anmmmj^NmFV$\"3Ug]hsQ],AFV7$$!3)zmmmYh=(eFV$\"3dTi= ay$Rs\"FV7$$!3+,++v#\\N)\\FV$\"3s^w\")o\")yT7FV7$$!3commmCC(>%FV$\"3z \\]*>;A%3))!#>7$$!39*****\\FRXL$FV$\"39g(e#)3w&fbFhq7$$!3t*****\\#=/8D FV$\"3ik'*42'*odJFhq7$$!3=mmm;a*el\"FV$\"31O)oa\"[*4P\"Fhq7$$!3komm;Wn (o)Fhq$\"31F-]QVytP!#?7$$!3IqLLL$eV(>F]s$\"3!=Y,UTX!\\>!#B7$$\"3)Qjmm \"f`@')Fhq$\"3uQ'=\"ySa;PF]s7$$\"3%z****\\nZ)H;FV$\"3\\\\9&=s,#G8Fhq7$ $\"3ckmm;$y*eCFV$\"3q6%=4=(GBIFhq7$$\"3f)******R^bJ$FV$\"31v4-V0W'\\&F hq7$$\"3'e*****\\5a`TFV$\"3\\Lv,F;&fi)Fhq7$$\"3'o****\\7RV'\\FV$\"3y'G +u9LAB\"FV7$$\"3Y'*****\\@fkeFV$\"3.z?Ha?n>$)FV$\"3 '\\NJ![D)4Y$FV7$$\"3M*******pfa<*FV$\"32/iJNIX4UFV7$$\"39HLLeg`!)**FV$ \"3W*e%e+]b!)\\FV7$$\"3w****\\#G2A3\"F0$\"3Ir,[6I'e&eFV7$$\"3;LLL$)G[k 6F0$\"3im:)y#>5!y'FV7$$\"3#)****\\7yh]7F0$\"3CB9rkXA?yFV7$$\"3xmmm')fd L8F0$\"3[o)p5cC@*))FV7$$\"3bmmm,FT=9F0$\"30mY7'HZf+\"F07$$\"3FLL$e#pa- :F0$\"3A+c;KO#)G6F07$$\"3!*******Rv&)z:F0$\"3cDuMB\\(zC\"F07$$\"3ILLLG UYo;F0$\"3esRhSk)=R\"F07$$\"3_mmm1^rZ:P-&y;F07$$\"34++]2%)38>F0$\"35VaXFO&*H=F07$F+F+-%'COLOURG6& %$RGBG$\"*++++\"!\")$F*F*F_[l-F$6$7S7$F($\"3++++++++]FV7$F.Fd[l7$F4Fd[ l7$F9Fd[l7$F>Fd[l7$FCFd[l7$FHFd[l7$FMFd[l7$FRFd[l7$FXFd[l7$FgnFd[l7$F \\oFd[l7$FaoFd[l7$FfoFd[l7$F[pFd[l7$F`pFd[l7$FepFd[l7$FjpFd[l7$F_qFd[l 7$FdqFd[l7$FjqFd[l7$F_rFd[l7$FdrFd[l7$FirFd[l7$F_sFd[l7$FesFd[l7$FjsFd [l7$F_tFd[l7$FdtFd[l7$FitFd[l7$F^uFd[l7$FcuFd[l7$FhuFd[l7$F]vFd[l7$Fbv Fd[l7$FgvFd[l7$F\\wFd[l7$FawFd[l7$FfwFd[l7$F[xFd[l7$F`xFd[l7$FexFd[l7$ FjxFd[l7$F_yFd[l7$FdyFd[l7$FiyFd[l7$F^zFd[l7$FczFd[l7$F+Fd[l-Fiz6&F[[l F_[lF_[lF\\[l-F$6$7Y7$F($\"3vf/xE\\L.^FV7$F.$\"31.7A*eZ`9&FV7$F4$\"3iF uf^W<$>&FV7$F9$\"3'pWCyepBE&FV7$F>$\"3oWG\"=\"H'=N&FV7$FC$\"3uK\"z62tV Y&FV7$FH$\"3Y9il7T!Gf&FV7$FM$\"3s0$Qti'H`dFV7$FR$\"3))QfXmGa^fFV7$FX$ \"3M)Q1o\")pR='FV7$Fgn$\"3_B(yA3Y/Y'FV7$F\\o$\"3i'y@8#)z]t'FV7$Fao$\"3 9!fp]t:r2(FV7$Ffo$\"3M:*Q$3,U^uFV7$F[p$\"3MREr\\yENyFV7$F`p$\"3-)HZ3.J o>)FV7$Fep$\"3D_#Ghn$\\K')FV7$Fjp$\"3!=GetNal**)FV7$F_q$\"3P%[/\"Q_8,% *FV7$Fdq$\"3N4'4LS)fI(*FV7$Fjq$\"3(Gm$3F%=[+\"F07$F_r$\"3$[\\$[j1mH5F0 7$Fdr$\"3c#HI:tH*[5F07$$!3_mm;H9Li7FV$\"3=$\\%y\"eq_0\"F07$Fir$\"3Ee&> LNZ*f5F07$$!3$G++]7bDW%Fhq$\"3%>Z!Q!=xI1\"F07$F_s$\"3w,)*G%Q(=k5F07$$ \"3qbm;/rI2?Fhq$\"3/[tX,B'R1\"F07$$\"3V[mmT+07UFhq$\"391!p:_*=j5F07$$ \"3:Tm;zHz;kFhq$\"3)HY%3)\\b5F07$Fjs$\"3ik3mS(*R\\5F07$F_t$\"3NgIKKj3J5F07$Fdt$\"3V gpVpiX05F07$Fit$\"3CZxhRQ*yu*FV7$F^u$\"3AvL&y'Qb4%*FV7$Fcu$\"3%>$G^7b' ****)FV7$Fhu$\"3$*e28eA<9')FV7$F]v$\"3_$R9>cAn>)FV7$Fbv$\"3jF:9')4jByF V7$Fgv$\"3aeqcn_1JuFV7$F\\w$\"3Ok21o\\i$3(FV7$Faw$\"3#zc(>ie+\\nFV7$Ff w$\"3]nIdWv#QX'FV7$F[x$\"3A:&*3++y!='FV7$F`x$\"3Ed!R8kQH&fFV7$Fex$\"3% oz0UG>Xv&FV7$Fjx$\"3'eQ))GL@,f&FV7$F_y$\"3w8-B-X*\\Y&FV7$Fdy$\"3-;F>?n p[`FV7$Fiy$\"3;f03$H#*fE&FV7$F^z$\"3EmdQ\"egl>&FV7$Fcz$\"3_z\"eX^F V7$F+F\\_l-Fiz6&F[[lF_[lF\\[lF_[l-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEWG6 $;F(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" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "We see that the classical turning point for a particle of energy 1 /2 is at " }{TEXT 295 1 "x" }{TEXT -1 10 " = 1, and " }{TEXT 294 1 "x " }{TEXT -1 12 " = -1. For |" }{TEXT 296 1 "x" }{TEXT -1 546 "| > 1 a \+ classical particle is not to be found with this energy value, since th e kinetic energy is positive (by definition), and the potential energy exceeds the value of 1/2 in this range. Yet the quantum particle (whi ch does not follow any trajectory) has a finite probability to be foun d in the classically forbidden regime. This regime is called the quant um tunneling region. It is a feature of Schroedinger solutions of all \+ finite potentials that quantum particles penetrate beyond the classica l turning points for the given amount of energy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 11 "Exercise 1:" }} {PARA 0 "" 0 "" {TEXT -1 202 "Search for the next two acceptable value a of lambda. Hint: use the asymptotic behaviour of the solution to det ermine whether an energy (lambda) value can be found such that the sol ution merges with the " }{TEXT 270 1 "x" }{TEXT -1 83 "-axis asymptoti cally. Tabulate the lambda-values together with the one found above." }}{PARA 0 "" 0 "" {TEXT -1 78 "Graph the probability distributions for the properly normalized wavefunctions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "We proceed with the determin ation of antisymmetric solutions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICa:=psi(0)=0,D(psi)(0)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ICaG6$/-%$psiG6#\"\"!F*/--%\"DG6#F(F)\"\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol:=dsolve(\{subs(lambda=1/ 2,SE),ICa\},psi(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%$psi G6#%\"xG,$*&**-%%sqrtG6#%#PiG\"\"\"-%$expG6#,$*$)F)\"\"#F1#!\"\"F8F1F) F1-%$erfG6#*$-F.6#,$F6F:F1F1F1*$-F.6#FAF1F:#F1F8" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot(rhs(sol),x=-5..5,view=[-5..5,0..2]);" }} {PARA 13 "" 1 "" {GLPLOT2D 755 178 178 {PLOTDATA 2 "6%-%'CURVESG6$7_o7 $$!\"&\"\"!$!+S!Q1u#F)7$$!+dNvs\\!\"*$!+Q(yhS#F)7$$!+9r]X\\F0$!+pv<9@F )7$$!+r1E=\\F0$!+'\\c!f=F)7$$!+HU,\"*[F0$!+ZJ*fj\"F)7$$!+V8_O[F0$!+^6 \"*p7F)7$$!+e%G?y%F0$!+JEB)))*!\"'7$$!+A0?(o%F0$!+cdrXkFL7$$!+'esBf%F0 $!+3JnTUFL7$$!+`'Gd[%F0$!+HqpzEFL7$$!+@Z3zVF0$!+O*GMr\"FL7$$!+$)QtrUF0 $!+U/w06FL7$$!+YIQkTF0$!+0SHCs!\"(7$$!+HCadSF0$!+TMv(y%Feo7$$!+6=q]RF0 $!+(30A@$Feo7$$!+_>f_PF0$!+GP^$e\"Feo7$$!+p1YZNF0$!+GoYlz!\")7$$!+-OJN LF0$!+2k36TFjp7$$!+%*o%Q7$F0$!+X(\\rB#Fjp7$$!+#RFj!HF0$!+y^Pj7Fjp7$$!+ '4OZr#F0$!+hqr3!)F07$$!+v'\\!*\\#F0$!+-x!y1&F07$$!+DwZ#G#F0$!+h%zVS$F0 7$$!+D.xt?F0$!+D/iiCF07$$!+.TC%)=F0$!+\"f\"HF>F07$$!+#4z)e;F0$!+*p=)=: F07$$!+n`'zY\"F0$!+G!HYG\"F07$$!+>t)eC\"F0$!+gbdz5F07$$!+<1J\\5F0$!+:n LZ#*!#57$$!*>[jL)F0$!+cyI#f(Fgt7$$!*d/EG'F0$!+a8YCfFgt7$$!*bQ(RTF0$!+C 9[GSFgt7$$!*h=><#F0$!+*eB^:#Fgt7$$!((*e$\\F0$!+ep(e$\\!#77$$\"*(RQb@F0 $\"+ra'*Q@Fgt7$$\"*=>Y2%F0$\"+8aKoRFgt7$$\"*yXu9'F0$\"+pIz4eFgt7$$\"* \\y))G)F0$\"+u#>F07$$\"+(pe*z?F0$\"+lOJ%[#F07$$\"+C\\'QH#F0$\"+Od*3Z$F07$$\" +8S8&\\#F0$\"+!z#pG]F07$$\"+0#=bq#F0$\"+h_-WyF07$$\"+2s?6HF0$\"+48))y7 Fjp7$$\"+IXaEJF0$\"+1*\\QD#Fjp7$$\"+l*RRL$F0$\"+qK>%4%Fjp7$$\"+`<.YNF0 $\"+#)[sGzFjp7$$\"+8tOcPF0$\"+<8O/;Feo7$$\"+\\Qk\\RF0$\"+4Et*>$Feo7$$ \"+3ASgSF0$\"+t?**R[Feo7$$\"+p0;rTF0$\"+\"[p$=uFeo7$$\"+mTAqUF0$\"+Up1 *4\"FL7$$\"+lxGpVF0$\"+8%eak\"FL7$$\"+A-\"\\Z%F0$\"+hrVfDFL7$$\"+!oK0e %F0$\"+l.VGSFL7$$\"+[oi\"o%F0$\"+EWY(G'FL7$$\"+<5s#y%F0$\"+r(H&>**FL7$ $\"+i2/P[F0$\"+$GiHF\"F)7$$\"+30O\"*[F0$\"+S\\kQ;F)7$$\"+\"Q?&=\\F0$\" +X)H8'=F)7$$\"+a-oX\\F0$\"+*>6f6#F)7$$\"+F,%G(\\F0$\"+95<2CF)7$$\"\"&F *$\"+S!Q1u#F)-%'COLOURG6&%$RGBG$\"#5!\"\"$F*F*F^al-%+AXESLABELSG6$Q\"x 6\"Q!6\"-%%VIEWG6$;F(Fc`l;F^al$\"\"#F*" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 46 "sol:=dsolve(\{subs(lambda=5/2,SE),ICa\},psi(x));" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%$psiG6#%\"xG,$*&-%$expG6#,$ *$)F)\"\"#\"\"\"#!\"\"F2F3,(*&F)F3-F-6#F0F3!\"#*&-%%sqrtG6#%#PiGF3-%%e rfiGF(F3F5**F2F3F " 0 "" {MPLTEXT 1 0 42 "plot(rhs(sol),x=-5..5,view=[-5..5,-2..2]);" }}{PARA 13 "" 1 "" {GLPLOT2D 755 178 178 {PLOTDATA 2 "6%-%'CURVESG6$7_o7$$!\"& \"\"!$\"+I[.Jh!\"(7$$!+dNvs\\!\"*$\"(w&\\a!\"%7$$!+9r]X\\F1$\"(%4[[F47 $$!+r1E=\\F1$\"($o;VF47$$!+HU,\"*[F1$\"(#*o%QF47$$!+V8_O[F1$\"(BJ1$F47 $$!+e%G?y%F1$\"($fZCF47$$!+A0?(o%F1$\"(00n\"F47$$!+'esBf%F1$\")?Y_6F)7 $$!+`'Gd[%F1$\"(X1p(F)7$$!+@Z3zVF1$\"(%>/_F)7$$!+$)QtrUF1$\"(!4jNF)7$$ !+YIQkTF1$\"(%RvCF)7$$!+HCadSF1$\")iR[f_PF1$\"(>P0(F[p7$$!+p1YZNF1$\"))3#HTF-7$$!+-OJNLF1$\")$)GVDF-7$$ !+%*o%Q7$F1$\")L5!p\"F-7$$!+#RFj!HF1$\"*Qe'*>\"!\")7$$!+'4OZr#F1$\")]Z $R*Fjq7$$!+v'\\!*\\#F1$\")>PUuFjq7$$!+DwZ#G#F1$\")!\\a$fFjq7$$!+D.xt?F 1$\"*%\\/aXF17$$!+.TC%)=F1$\"*ES8>$F17$$!+#4z)e;F1$\"*:`hO\"F17$$!+n`' zY\"F1$!),A!*HF17$$!+>t)eC\"F1$!*27q<#F17$$!+<1J\\5F1$!+9YDSN!#57$$!*> [jL)F1$!+;x*)>WFht7$$!*d/EG'F1$!+w?W]WFht7$$!*bQ(RTF1$!+2M\"*yNFht7$$! *h=><#F1$!+1qx(3#Fht7$$!((*e$\\F1$!+)y'zN\\!#77$$\"*(RQb@F1$\"+#HJJ2#F ht7$$\"*=>Y2%F1$\"+#eq*QNFht7$$\"*yXu9'F1$\"+J,LAWFht7$$\"*\\y))G)F1$ \"+f+2IWFht7$$\"+i_QQ5F1$\"+&)4d-OFht7$$\"+!y%3T7F1$\"*$>y9AF17$$\"+O! 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" 0 "" {MPLTEXT 1 0 38 "int (rhs(sol)^2,x=-infinity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #,$*$-%%sqrtG6#%#PiG\"\"\"#F)\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "WF:=1/sqrt(%)*rhs(sol);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#WFG*&*(-%%sqrtG6#\"\"#\"\"\"%\"xGF+-%$expG6#,$*$)F,F*F+#!\"\" F*F+F+*$)%#PiG#F+\"\"%F+F4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "int(WF^2,x=-infinity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 259 "Now we graph the pro bability in such a way as to show that there is a finite probability t o find the quantum particle outside the classically allowed range. For this we plot the density superimposed on the potential while using th e energy value as a baseline:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "plot([V,3/2,WF^2+3/2],x=-2.5..2.5,color=[red,blue,green]);" }} {PARA 13 "" 1 "" {GLPLOT2D 694 299 299 {PLOTDATA 2 "6'-%'CURVESG6$7S7$ $!3++++++++D!#<$\"3+++++++DJF*7$$!3smm;HU,\"R#F*$\"3kM(Q?_u%eGF*7$$!3= L$3FH'='H#F*$\"3S>3TXdBOEF*7$$!3gmmTgBa*=#F*$\"3!Q\\HS(y/(R#F*7$$!3wmm \"H_\">#3#F*$\"319I0p2wn@F*7$$!3ML$3_!4Nv>F*$\"3+\"p`$*f05&>F*7$$!3km; /wfHw=F*$\"3Ev]&[HV-w\"F*7$$!3;+]PM.ttc\"F*$\"3DS*[pU-)>7F*7$$!3;LL$ epjJX\"F*$\"3'f*RWjB%e0\"F*7$$!3amm\"z/otN\"F*$\"3\\4b_(3SA@*!#=7$$!3) )****\\P[_\\7F*$\"3(f+Zwfhl!yF^o7$$!3#*****\\7)Q79\"F*$\"3O\\?yN,87lF^ o7$$!3'*****\\i^)o.\"F*$\"3'oDj5?acP&F^o7$$!3vlmT50A@%*F^o$\"3v#y%H&zp zV%F^o7$$!3OKLLeaR%H)F^o$\"3mI5'4!)\\)RMF^o7$$!3kJLLLo#)RtF^o$\"33+g;( *Gl$p#F^o7$$!3f***\\PfO%HiF^o$\"3#)zvx8SHS>F^o7$$!3/MLL$3`lC&F^o$\"3!e s6G'fJw8F^o7$$!3q'**\\P4u\"oTF^o$\"3/.V!zjPoo)!#>7$$!3*z**\\7G-89$F^o$ \"3f?K46+*Q$\\F\\r7$$!3%)GL$3Fp)p?F^o$\"3P[+<*Rz@9#F\\r7$$!3YKL3-$ff3 \"F^o$\"31$f1*G!Ql*e!#?7$$!3)Grmm\"z%zY#F\\s$\"3;%y*ofMQXI!#B7$$\"3!fL $e*)>px5F^o$\"3-X/1A,52eF\\s7$$\"3w++v$f4t.#F^o$\"3u'*GR!>:`2#F\\r7$$ \"3OPL$e*GstIF^o$\"3KQ]=?i)Qs%F\\r7$$\"3Y+++]#RW9%F^o$\"3c#Gq%[$)=)e)F \\r7$$\"3Y,+]7j#>>&F^o$\"3C$\\@<%\\!yM\"F^o7$$\"3t-+D1RU0iF^o$\"3=_H\" GHk`#>F^o7$$\"3+++](=S2L(F^o$\"3yw8$[e()po#F^o7$$\"3:jmm;p)=M)F^o$\"3] AF_mQNzMF^o7$$\"3O-++v=]@%*F^o$\"3*zUJ!z[BQWF^o7$$\"35L$e*[$z*R5F*$\"3 e\"**)HJ_y2aF^o7$$\"3e++]iC$p9\"F*$\"3!ed\"on.FxlF^o7$$\"3[m;H2qcZ7F*$ \"3[DMT)=<@y(F^o7$$\"3O+]7.\"fFN\"F*$\"3'pxPa&fy\\\"*F^o7$$\"3Ymm;/Ogb 9F*$\"3m%\\Ji#4Rf5F*7$$\"3w**\\ilAFj:F*$\"3)*fhB)35>A\"F*7$$\"3yLLL$)* ppm\"F*$\"3[nrmiWR*Q\"F*7$$\"3)RL$3xe,t]Ezr:F*7$$\"3Cn;HdO =y=F*$\"3[^(e_#pyj#[(>F*$\"3!QNIx\"3'*\\>F*7$$\"3SnmT&G !e&3#F*$\"35M$fMcA[<#F*7$$\"3#RLLL)Qk%=#F*$\"3cuG\\[WL'Q#F*7$$\"37+]iS jE!H#F*$\"3Q9**\\b*fEi#F*7$$\"3a+]P40O\"R#F*$\"3+J\"**Ga-$fGF*7$$\"3++ ++++++DF*F+-%'COLOURG6&%$RGBG$\"*++++\"!\")$\"\"!Fa[lF`[l-F$6$7S7$F($ \"3++++++++:F*7$F.Ff[l7$F3Ff[l7$F8Ff[l7$F=Ff[l7$FBFf[l7$FGFf[l7$FLFf[l 7$FQFf[l7$FVFf[l7$FenFf[l7$FjnFf[l7$F`oFf[l7$FeoFf[l7$FjoFf[l7$F_pFf[l 7$FdpFf[l7$FipFf[l7$F^qFf[l7$FcqFf[l7$FhqFf[l7$F^rFf[l7$FcrFf[l7$FhrFf [l7$F^sFf[l7$FdsFf[l7$FisFf[l7$F^tFf[l7$FctFf[l7$FhtFf[l7$F]uFf[l7$Fbu Ff[l7$FguFf[l7$F\\vFf[l7$FavFf[l7$FfvFf[l7$F[wFf[l7$F`wFf[l7$FewFf[l7$ FjwFf[l7$F_xFf[l7$FdxFf[l7$FixFf[l7$F^yFf[l7$FcyFf[l7$FhyFf[l7$F]zFf[l 7$FbzFf[l7$FgzFf[l-Fjz6&F\\[lF`[lF`[lF][l-F$6$7cq7$F($\"3[/(RkF9O^\"F* 7$F.$\"3s^Fd*o@7_\"F*7$F3$\"38=F1(eD0`\"F*7$F8$\"3KaCpAHyW:F*7$F=$\"3) *zW0=C1k:F*7$FB$\"3G*Rh.-X*)e\"F*7$FG$\"3scU?!*[_<;F*7$$!3_L$3_:8]#=F* $\"3LR7dk$HWj\"F*7$FL$\"3Qbu&[@A(p=F*7$Feo$\"3k\\uq40b**=F*7$$!3/++]() >1*3\"F*$\"3F*7$Fjo$\"3QK3X!*H*R\">F*7$$!3:3xct(R]-\"F*$\"30d ELY:f9>F*7$$!3c;aj%Q%>85F*$\"3F-D?=O'\\\">F*7$$!3(\\7.d**[8+\"F*$\"3ej bG()f5:>F*7$$!3mK$3x1O]*)*F^o$\"3QEpm;d,:>F*7$$!3m:aQy@ew(*F^o$\"3U1gM i,p9>F*7$$!3w*\\i!*GG\"e'*F^o$\"3]uy#R)p79>F*7$$!3w#eR(*Ru'R&*F^o$\"3_ F$=\\:CL\">F*7$F_p$\"3\\c)*es*zA\">F*7$$!31**\\P%)z!y&))F^o$\"37a!\\!p r(R!>F*7$Fdp$\"3;Za)3@g,*=F*7$$!3+KL$e96r\"yF^o$\"3W.(y+W]U(=F*7$Fip$ \"3E4OR(=)pa=F*7$$!3gl;a8'et\"F*7$$!3Pl;a )ejtq%F^o$\"34_)fS%HM+`bF\\r$\"3t/+K x*oM]\"F*7$$!3JmTgFTfEUF\\r$\"3*eOc)\\@,-:F*7$$!3kM$e*[V****GF\\r$\"3s ? J]\"F*7$$\"3S^$eRA\"*4-)F\\r$\"3?bAf=I@2:F*7$Fds$\"3/\"[(R&*Q&H^\"F*7$ $\"3Momm\"z+vb\"F^o$\"3L_P,4jrE:F*7$Fis$\"3fIIc(pI\\a\"F*7$$\"31p;zWi^ bDF^o$\"3c700M>.p:F*7$F^t$\"3%484$*)e*pf\"F*7$$\"3\"*om\"H2\"34OF^o$\" 3l?%[(fm-H;F*7$Fct$\"35l[**ekAj;F*7$$\"3'4+]7y#=oYF^o$\"3)f`du'ou(p\"F *7$Fht$\"3oI)e>\"oHKV$[#Ha=F *7$$\"3eJL3_NJOyF^o$\"3'pyx%)ei\\(=F*7$Fgu$\"3SVG$e?N:*=F*7$$\"3wKL$eR %p\")))F^o$\"33A6'))[TW!>F*7$F\\v$\"3%[Uc*RFG7>F*7$$\"3X=H#oK)yV&*F^o$ \"3h`l)R2cL\">F*7$$\"3aMekyZ2m'*F^o$\"3I\\K>VA<9>F*7$$\"3i](o/Bh$)y*F^ o$\"3h!H(GRJt9>F*7$$\"3qm;H#oZ1\"**F^o$\"3.:U)4-T]\">F*7$$\"3F*7$$\"3*)\\Pfe?_:5F*$\"3$=$z)3a3\\\">F*7$$\"3hTgx.2v F5F*$\"3ze$4XUuW\">F*7$Fav$\"3=F*7$$\"3%o;Hd!fX$4\"F*$\"3]D( fl!p73>F*7$Ffv$\"3azRuVQK)*=F*7$F[w$\"3wSI^v2Pq=F*7$$\"3UL$3_0j,I\"F*$ \"3eo1&Ge5=&=F*7$F`w$\"3IX&\\;%)\\7$=F*7$$\"3TLek`8=/9F*$\"3g_F^kit4=F *7$Few$\"3CLga=,K(y\"F*7$$\"36Le*[$zV4:F*$\"3_\"4la-$Qjc7t\"fZp\"F*7 $$\"3)QL3-$H**>@:F*7$FgzF^_l-Fjz6&F\\[lF`[lF][lF`[l-%+AXESLABELSG6$Q \"x6\"Q!6\"-%%VIEWG6$;$!#D!\"\"$\"#DF[jm%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" } }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 184 "We observe again the fact that the quantum particle penetrates into the classically forbidden region . We also observe that the particle has zero probability to be found a t the origin (" }{TEXT 297 1 "x" }{TEXT -1 220 "=0) when it is in an a ntisymmetric state. Given that there are no continuous trajectories fo r quantum particles this is not a problem, however: the interpretation of the quantum probability distribution is the following:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 467 "If repeated meas urements are made on an ensemble of particles prepared in a given eige nstate (the ensemble can be simultaneous, or there could be one partic le at a time with repeated measurements), then each time the probabili ty is recorded (a histogram is made of the position for the ensemble), the distribution will be an approximation to the above curve. There i s no meaning to to question as to how an individual quantum particle m oved in such a way as to cross " }{TEXT 298 1 "x" }{TEXT -1 3 "=0." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 271 11 " Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 69 "Find two more anti-symmetr ic solutions, and tabulate the eigenvalues." }}{PARA 0 "" 0 "" {TEXT -1 98 "Combine these eigenvalues with the ones found for the symmetric case, and observe how they behave." }}{PARA 0 "" 0 "" {TEXT -1 79 "Gr aph the probability distributions using the properly normalized wavefu nction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 272 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 40 "Change the value of th e spring constant " }{TEXT 273 1 "k" }{TEXT -1 7 ". Pick " }{TEXT 275 1 "k" }{TEXT -1 7 "=2 and " }{TEXT 274 1 "k" }{TEXT -1 118 "=4 respect ively, and determine the lowest 6 eigenvalues in each case. Draw your \+ conclusions about the energy spectrum." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 278 11 "Exercise 4:" }}{PARA 0 "" 0 "" {TEXT -1 38 "Keep the value of the spring constant " }{TEXT 277 1 "k" }{TEXT -1 27 "=1, but change the mass to " }{TEXT 276 1 "m" }{TEXT -1 110 "=2 (in Bohr units). Determine the 6 lowest energy values, and dra w your conclusions about the energy spectrum." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "Let us go back to the general case, and show what the exact eigenfunctions look like:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "V:=1/2*m*omega^2*x^2;" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "SE:=-h_^2/(2*m)*diff(psi(x),x$2)+V* psi(x)=lambda*psi(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG,$*&)%& omegaG\"\"#\"\"\")%\"xGF)F*#F*F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% #SEG/,&-%%diffG6$-%$psiG6#%\"xG-%\"$G6$F-\"\"##!\"\"F1**#\"\"\"F1F6)%& omegaGF1F6)F-F1F6F*F6F6*&%'lambdaGF6F*F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "alpha:=sqrt(m*omega/h_);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&alphaG*$-%%sqrtG6#%&omegaG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(orthopoly);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7(%\"GG%\"HG%\"LG%\"PG%\"TG%\"UG" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 43 "phi:=n->exp(-1/2*(alpha*x)^2)*H(n,alpha*x);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$phiGR6#%\"nG6\"6$%)operatorG%&arro wGF(*&-%$expG6#,$*&)%&alphaG\"\"#\"\"\")%\"xGF4F5#!\"\"F4F5-%\"HG6$9$* &F3F5F7F5F5F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "subs(p si(x)=phi(0),SE);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&-%%diffG6$-%$e xpG6#,$*&%&omegaG\"\"\")%\"xG\"\"#F.#!\"\"F1-%\"$G6$F0F1F2**#F.F1F.)F- F1F.F/F.F(F.F.*&%'lambdaGF.F(F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,$*&%&omegaG \"\"\"-%$expG6#,$*&F&F')%\"xG\"\"#F'#!\"\"F/F'#F'F/*&%'lambdaGF'F(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(%,lambda);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$%&omegaG#\"\"\"\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Clearly, the ground-state energy equals 1 /2 omega. Let's continue:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "subs(psi(x)=phi(1),SE);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simp lify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&-%%diffG6$,$*(-%$expG6# ,$*&%&omegaG\"\"\")%\"xG\"\"#F0#!\"\"F3F0-%%sqrtG6#F/F0F2F0F3-%\"$G6$F 2F3F4*()F/#\"\"&F3F0)F2\"\"$F0F*F0F0,$**%'lambdaGF0F*F0F6F0F2F0F3" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/,$*()%&omegaG#\"\"$\"\"#\"\"\"%\"xGF+ -%$expG6#,$*&F'F+)F,F*F+#!\"\"F*F+F),$**%'lambdaGF+F-F+-%%sqrtG6#F'F+F ,F+F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(%,lambda);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%&omegaG#\"\"$\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "For the first excited state we find that the energy equals 3/2 omega. Substitute different values of " }{TEXT 279 1 "n" }{TEXT -1 36 " and verify the eigenvalue spectrum." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 115 "The only remai ning point concerns the normalization of the eigenfunctions. The inter pretation of the functions phi(" }{TEXT 281 1 "n" }{TEXT -1 2 ", " } {TEXT 280 1 "x" }{TEXT -1 81 ") is that their square provides the prob ability to find the particle at location " }{TEXT 282 1 "x" }{TEXT -1 27 ". Therefore the sum of phi(" }{TEXT 285 1 "x" }{TEXT -1 2 ", " } {TEXT 284 1 "n" }{TEXT -1 23 ")^2 over all locations " }{TEXT 283 1 "x " }{TEXT -1 134 ", i.e., the integral is chosen to equal one. The func tions as defined do not meet that requirement, and one has to find an \+ omega- and " }{TEXT 286 1 "n" }{TEXT -1 52 "-dependent factor to achie ve this goal. For example:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assume(omega>0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "int (phi(5)^2,x=-infinity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#, $*&*$-%%sqrtG6#%#PiG\"\"\"F**$-F'6#%'omega|irGF*!\"\"\"%SQ" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "The normalization factor can be found to \+ be:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "c:=n->sqrt(sqrt(m*om ega/(h_*Pi))/(2^n*n!));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cGR6#% \"nG6\"6$%)operatorG%&arrowGF(-%%sqrtG6#*&-F-6#*&*&%\"mG\"\"\"%&omegaG F5F5*&%#h_GF5%#PiGF5!\"\"F5*&)\"\"#9$F5-%*factorialG6#F>F5F:F(F(F(" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "c(5)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$-%%sqrtG6#*&%'omega|irG\"\"\"%#PiG!\"\"F*#F*\"%SQ" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "Therefore, when the eigenfunct ion for the nth state is multiplied by c(n) it will lead to a normaliz ed probability distribution." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 305 22 "Anharmonic oscillators" }}{PARA 0 "" 0 "" {TEXT -1 316 "For other potentials (e.g., an x^4 anharmonic \+ oscillator) the eigenvalues and eigenfunctions cannot be found in term s of simple functions. Nevertheless, we can apply numerical techniques to find approximate eigenvalues and eigenfunctions using dsolve[numer ic]. We need to apply an iteration in the energy eigenvalue:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 7 "V:=x^4;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"VG*$)%\"xG\"\"%\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 " SE:=-1/2*diff(psi(x),x$2)+V*psi(x)=E_t*psi(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG/,&-%%diffG6$-%$psiG6#%\"xG-%\"$G6$F-\"\"##!\"\"F 1*&)F-\"\"%\"\"\"F*F7F7*&%$E_tGF7F*F7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "unprotect(Psi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICs:=psi(0)=1,D(psi)(0)=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ICsG6$/-%$psiG6#\"\"!\"\"\"/--%\"DG6#F(F)F*" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E_t:=0.668;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_tG$\"$o'!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(\{SE,ICs\},psi(x),numeric,output=listproc edure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,ps i(x)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "plot('Psi(x)',x=0 ..4,-0.5..1.5);" }}{PARA 13 "" 1 "" {GLPLOT2D 694 203 203 {PLOTDATA 2 "6%-%'CURVESG6$7_o7$$\"\"!F)$\"\"\"F)7$$\"+m;')=()!#6$\"+pGE\\**!#57$$ \"+e'40j\"F2$\"+2b%H#)*F27$$\"+<6m$[#F2$\"+7V\"4f*F27$$\"+(>%F2$\"+\"yD&\\))F27$$\"+\">K'*)\\F2$\"+V$R=R)F27$$\" +Dt:5eF2$\"+&y-4&yF27$$\"+\"fX(emF2$\"+3)*>IsF27$$\"+DCh/vF2$\"+T@#Gc' F27$$\"+L/pu$)F2$\"+cUIUeF27$$\"+;c0T\"*F2$\"+%H$=%>&F27$$\"+I,Q+5!\"* $\"+\"*3YnWF27$$\"+]*3q3\"Fbo$\"+Wf)*fPF27$$\"+q=\\q6Fbo$\"+:Vn;JF27$$ \"+fBIY7Fbo$\"+.#Hud#F27$$\"+j$[kL\"Fbo$\"+Z8q.?F27$$\"+`Q\"GT\"Fbo$\" +]0O\"e\"F27$$\"+s]k,:Fbo$\"+F&>n;\"F27$$\"+`dF!e\"Fbo$\"+-p\"Qn)F/7$$ \"+sgam;Fbo$\"+4E`sgF/7$$\"+= u#F/7$$\"+cK78>Fbo$\"+2'R^z\"F/7$$\"+Uc-)*>Fbo$\"+=k$**3\"F/7$$\"+f`@' 3#Fbo$\"+4Dy:h!#77$$\"+nZ)H;#Fbo$\"+P[W.MFfs7$$\"+Ky*eC#Fbo$\"+[_pB9Ff s7$$\"+S^bJBFbo$!+\\**o\"=\"!#87$$\"+0TN:CFbo$!+rZg(o\"Ffs7$$\"+7RV'\\ #Fbo$!+FF/7$$ \"+],s`FFbo$!+b-![j%F/7$$\"+zM)>$GFbo$!+!HVC1\"F27$$\"+qfa'RFbo$!+xa!pn)F_\\l7$$\"+`\"3u'RFbo$!+r# R`x*F_\\l7$$\"+F,%G(RFbo$!+_+l,6!\"#7$$\"+-@FyRFbo$!+)yR>C\"Fh^l7$$\"+ xSq$)RFbo$!+\\Ic+9Fh^l7$$\"+^g8*)RFbo$!+2](*z:Fh^l7$$\"+R?&=*RFbo$!+5[ Ly;Fh^l7$$\"+E!oX*RFbo$!+Ku'Hy\"Fh^l7$$\"+8SG(*RFbo$!+#=#G%*=Fh^l7$$\" \"%F)$!+Qdr7?Fh^l-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"x6 \"Q!6\"-%%VIEWG6$;F(Fh`l;$FjyFbal$\"#:Fbal" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 171 "The energy trial value was determined by trial and error . An energy value (and the corresponding eigenfunction) is considered \+ acceptable when the function is close to the " }{TEXT 287 1 "x" } {TEXT -1 95 "-axis for some range. This range can be extended by tunin g the energy value to higher accuracy." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 288 11 "Exercise 5:" }}{PARA 0 "" 0 "" {TEXT -1 314 "Find approximate eigenenergies for two more eigenvalues \+ (above the ground state). Observe the nodal structure of the wavefunct ions and compare with the exact harmonic oscillator result above. Calc ulate also several anti-symmetric eigenfunctions by changing the initi al conditions for the integration appropriately." }}{PARA 0 "" 0 "" {TEXT -1 106 "Compare the spacing between these eigenvalues with the s pacing property found for the harmonic oscillator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 292 11 "Exercise 6:" }}{PARA 0 " " 0 "" {TEXT -1 85 "Solve numerically for the lowest six eigenvalues a nd eigenfunctions of the potential " }{TEXT 291 1 "V" }{TEXT -1 1 "(" }{TEXT 290 1 "x" }{TEXT -1 3 ")=|" }{TEXT 289 1 "x" }{TEXT -1 99 "|. D iscuss the properties of the spacing of the eigenvalues as compared to the harmonic oscillator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 428 "Note that a systematic method to improve the trial val ues for the eigenenergies involves a bisection: one can systematically cut the energy range defined by two brackets (with a solution going t o positive infinity at large x for one bracketing value, and to negati ve infinity for the other) in half, and determine which half contains \+ the asymptotically acceptable solution to reduce the bracket systemati cally by factors of two." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Here we do the exercise for t he linear potential, and write a bisection algorithm." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 5 "V:=x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VG%\"xG " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "SE:=-1/2*diff(psi(x),x$ 2)+V*psi(x)=E_t*psi(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG/,&-% %diffG6$-%$psiG6#%\"xG-%\"$G6$F-\"\"##!\"\"F1*&F-\"\"\"F*F5F5*&%$E_tGF 5F*F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "unprotect(Psi);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICs:=psi(0)=1,D(psi)(0)=0; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ICsG6$/-%$psiG6#\"\"!\"\"\"/--% \"DG6#F(F)F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "By trial and erro r we find that the lowest eigenvalue is bracketed by 0.8 and 0.9:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E_t:=0.825;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_tG$\"$D)!\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(\{SE,ICs\},psi(x),numeric,output=listproc edure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,ps i(x)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "plot('Psi(x)',x=0 ..4,-0.5..1.5);" }}{PARA 13 "" 1 "" {GLPLOT2D 694 203 203 {PLOTDATA 2 "6%-%'CURVESG6$7V7$$\"\"!F)$\"\"\"F)7$$\"+m;')=()!#6$\"++TbR**!#57$$\" +e'40j\"F2$\"+3jz&z*F27$$\"+<6m$[#F2$\"+45[X&*F27$$\"+(>%F2$\"+hK'*)\\F2$\"+%ep%*R)F27$$\"+Dt:5 eF2$\"+-])H$zF27$$\"+\"fX(emF2$\"+Un;BuF27$$\"+DCh/vF2$\"+!pSz*oF27$$ \"+L/pu$)F2$\"+_#z*\\jF27$$\"+;c0T\"*F2$\"+jj'z'eF27$$\"+I,Q+5!\"*$\"+ S`#GL&F27$$\"+]*3q3\"Fbo$\"+*4#R4[F27$$\"+q=\\q6Fbo$\"+txrAVF27$$\"+fB IY7Fbo$\"+Xlm)*QF27$$\"+j$[kL\"Fbo$\"+W<.>MF27$$\"+`Q\"GT\"Fbo$\"+IP%[ .$F27$$\"+s]k,:Fbo$\"+\"*[19EF27$$\"+`dF!e\"Fbo$\"+$*Q'[E#F27$$\"+sgam ;Fbo$\"+%o&)f!>F27$$\"+Fbo$\"+&e#z.5F27$$\"+Uc-)*>Fbo$\"+?E#RF(F/7$$\"+f`@'3#Fbo$ \"+b1BKXF/7$$\"+nZ)H;#Fbo$\"+f_E>AF/7$$\"+Ky*eC#Fbo$!+\\mw3C!#77$$\"+S ^bJBFbo$!+B=Y)y#F/7$$\"+0TN:CFbo$!+csiS`F/7$$\"+7RV'\\#Fbo$!+$)R`BzF/7 $$\"+:#fke#Fbo$!+_[$**4\"F27$$\"+`4NnEFbo$!+`U!HS\"F27$$\"+],s`FFbo$!+ lL5k$GFbo$!+#3V]8#F27$$\"+qfa " 0 "" {MPLTEXT 1 0 5 "X:=4 ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"XG\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "SEroot:=proc(lambda) local SE,sol,Psi; global ICs,X,V;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "SE:=-1/2*diff(psi(x),x $2)+V*psi(x)=lambda*psi(x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "sol: =dsolve(\{SE,ICs\},psi(x),numeric,output=listprocedure): unprotect(Psi ):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x)):" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Psi(X); end:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 " fsolve('SEroot(E_t)',E_t,0.8..0.9);" }}{PARA 8 "" 1 "" {TEXT -1 37 "Er ror, (in fsolve) invalid arguments\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "Unfortunately, it is not as easy as that..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 146 "We first write a basic b isection step: given a map f and a bracketing interval, we come up wit h a new bracket for the root which is half the size:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 245 "Bisect:=proc(f,xL) local x1,x2,xh,f1,f2, fh; x1:=xL[1]: x2:=xL[2]: f1:=f(x1): f2:=f(x2): if f1*f2>0 then RETURN (\"Same sign on both ends of the bracket, cannot proceed\") fi: xh:=0. 5*(x1+x2): fh:=f(xh): if f1*fh<0 then [x1,xh] else [xh,x2] fi: end:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Bisect(SEroot,[0.8,0.9]); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$\"\")!\"\"$\"#&)!\"#" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Now we need a driver, to drive the trial energy bracket to some desired tolerance." }}{PARA 0 "" 0 "" {TEXT -1 100 "In this procedure we limit ourselves to a maximum of 100 iterations. We define an internal variable " }{TEXT 19 2 "rL" }{TEXT -1 105 " to represent the bracket as a list. We cannot overwrite the l ist that was passed down originally, i.e.. " }{TEXT 19 2 "xL" }{TEXT -1 71 " (Maple syntax). When successful we return the midpoint of the \+ bracket." }}{PARA 0 "" 0 "" {TEXT -1 33 "We also catch the incidence w hen " }{TEXT 19 6 "Bisect" }{TEXT -1 49 " returns with an error messag e and the result of " }{TEXT 19 2 "rL" }{TEXT -1 16 " is not of type \+ " }{TEXT 19 4 "list" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 268 "SEsolve:=proc(xL,tol) local i,rL; rL:=xL: for i from 1 to 100 do: rL:=Bisect(SEroot,rL): if not type(rL,list) then RETURN( rL) fi: if abs(rL[2]-rL[1]) < evalf(tol) then RETURN(0.5*(rL[1]+rL[2]) ) else fi: od: RETURN(\"Did not converge in 100 iterations in SEsolve \"); end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.8,0 .9],0.001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"*vV)*3)!\"*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "We should keep in mind that the nu mber is only accurate to the level determined by the variable " } {TEXT 19 3 "tol" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "SEsolve([0.8,0.9],0.00001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #$\"+$>ti3)!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "SEsolve([ 0.8,0.9],0.0000001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+^bD'3)!#5 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "E_t:=%;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_tG$\"+^bD'3)!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(\{SE,ICs\},psi(x),numeric,output=listproc edure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,ps i(x)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "plot('Psi(x)',x=0 ..4,-0.5..1.5);" }}{PARA 13 "" 1 "" {GLPLOT2D 694 203 203 {PLOTDATA 2 "6%-%'CURVESG6$7U7$$\"\"!F)$\"\"\"F)7$$\"+L3VfV!#6$\"+(H7\\)**!#57$$\" +m;')=()F/$\"+(R'zS**F27$$\"+7z>^7F2$\"+@Q(>% F2$\"+T/!G%))F27$$\"+\">K'*)\\F2$\"+0!y\"Q%)F27$$\"+Dt:5eF2$\"+2Ep%)zF 27$$\"+\"fX(emF2$\"+4?***[(F27$$\"+DCh/vF2$\"+eKW\")pF27$$\"+L/pu$)F2$ \"+cMD_kF27$$\"+;c0T\"*F2$\"+4'=\"))fF27$$\"+I,Q+5!\"*$\"+;(QYZ&F27$$ \"+]*3q3\"F\\p$\"+Klnu\\F27$$\"+q=\\q6F\\p$\"+DjR7XF27$$\"+fBIY7F\\p$ \"+N%z@6%F27$$\"+j$[kL\"F\\p$\"+mbEjOF27$$\"+`Q\"GT\"F\\p$\"+w,W2LF27$ $\"+s]k,:F\\p$\"+$HNG#HF27$$\"+`dF!e\"F\\p$\"+B<-4EF27$$\"+sgam;F\\p$ \"+V)4LH#F27$$\"+?F27$$\"+e/TM=F\\p$\"+I6(=w\"F27$$ \"+cK78>F\\p$\"+zvM[:F27$$\"+Uc-)*>F\\p$\"+i?uT8F27$$\"+f`@'3#F\\p$\"+ 2rY^6F27$$\"+nZ)H;#F\\p$\"+G)GX+\"F27$$\"+Ky*eC#F\\p$\"+i^0Q')F/7$$\"+ S^bJBF\\p$\"+i!eLO(F/7$$\"+0TN:CF\\p$\"+S2tviF/7$$\"+7RV'\\#F\\p$\"+[] Te`F/7$$\"+:#fke#F\\p$\"+L9syWF/7$$\"+`4NnEF\\p$\"+mx$*)z$F/7$$\"+],s` FF\\p$\"+Qz$\\<$F/7$$\"+zM)>$GF\\p$\"+n1\\*o#F/7$$\"+qfa " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.1,1.0],0.0 01);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+2*y$)3)!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.1,2.0],0.001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+_>D$3)!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.1,3.0],0.001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#QVSame~sign~on~both~ends~of~the~bracket,~cannot~proceed 6\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.1,4.0],0. 001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#QVSame~sign~on~both~ends~of~t he~bracket,~cannot~proceed6\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.1,5.0],0.001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\" +t>-&3)!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "SEsolve([4.,5 .0],0.001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+1k1=S!\"*" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SEsolve([0.1,6.0],0.001);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#QVSame~sign~on~both~ends~of~the~brack et,~cannot~proceed6\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "The pro gram still requires some attention, but the process of finding the eig envalues is automated to some extent." }}{PARA 0 "" 0 "" {TEXT -1 123 "Note that in the present form the program calculates the symmetric ei genfunctions (and the corresponding eigenvalues) only." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "How far does the pa rticle in the ground state explore the classically forbidden region in this case?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E0:=SEsolve( [0.1,2.0],0.00001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E0G$\"+S)fi3 )!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E_t:=E0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E_tG$\"+S)fi3)!#5" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(\{SE,ICs\},psi(x),numeric,output=l istprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=subs (sol,psi(x)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "P1:=plot(E 0+'Psi(x)^2',x=0..3,0..2,color=green):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "P2:=plot([x,E0],x=0..3,color=[red,blue]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plots[display](P1,P2);" }}{PARA 13 "" 1 "" {GLPLOT2D 707 299 299 {PLOTDATA 2 "6'-%'CURVESG6$7U7$$\"\"!F)$ \"+%)fi3=!\"*7$$\"+DJdpK!#6$\"+%R@p!=F,7$$\"+]i9RlF0$\"+$*>\">!=F,7$$ \"+XV)RQ*F0$\"+/w*\\z\"F,7$$\"+WA)GA\"!#5$\"+HW$ey\"F,7$$\"+Qeui=F@$\" +>,qd?fS*\\F@$\"+4dD?:F,7$$ 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$F^`lFeel7$Fa`lFeel7$Fd`lFeel7$Fg`lFeel7$Fj`lFeel7$F]alFeel7$F`alFeel7 $FcalFeel7$FfalFeel7$FialFeel7$F\\blFeel7$F_blFeel7$FbblFeel7$FeblFeel 7$FhblFeel7$F[clFeel7$F^clFeel7$FaclFeel7$FdclFeel7$FgclFeel7$FjclFeel 7$F]dlFeel7$F`dlFeel7$FcdlFeel7$FfdlFeel7$FidlFeel7$F\\elFeel7$F_[lFee l-Fd[l6&Ff[lF(F(Fg[l-%+AXESLABELSG6%Q\"x6\"Q!6\"%(DEFAULTG-%%VIEWG6$;F (F_[l;F($\"\"#F)" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 145 "Note that the green curve does not show a properly normalized pro bability distribution. 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global X,V,f_sym;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "SE:=-1/ 2*diff(psi(x),x$2)+V*psi(x)=lambda*psi(x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "if f_sym=S then IC:=psi(0)=1,D(psi)(0)=0; elif f_sym= A then IC:=psi(0)=0,D(psi)(0)=1; fi;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "sol:=dsolve(\{SE,IC\},psi(x),numeric,output=listprocedure): unpr otect(Psi):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x) ):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Psi(X); end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "f_sym:=S;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&f_symG%\"SG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E0:=SEsolve([-1.,-0.2],0.0001);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E0G$!+1Rv(p'!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "f_s ym:=A;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&f_symG%\"AG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "E1:=SEsolve([-1.,-0.2],0.0001);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E1G$!+%fevu#!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "ICa:=psi(0)=0,D(psi)(0)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ICaG6$/-%$psiG6#\"\"!F*/--%\"DG6#F(F)\"\"\"" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E_t:=E1;" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%$E_tG$!+%fevu#!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "SE:=-1/2*diff(psi(x),x$2)+V*psi(x)=E_t*psi(x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG/,&-%%diffG6$-%$psiG6#%\"xG-%\"$ G6$F-\"\"##!\"\"F1*&F*\"\"\"*$-%%sqrtG6#,&F5F5*$)F-F1F5F5F5F3F3,$F*$!+ %fevu#!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "sol:=dsolve(\{ SE,ICa\},psi(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 22 "Psi:=subs(sol,psi(x)):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 49 "P1:=plot(E1+'Psi(x)^2',x=0..8,-1..1,color=gree n):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "P2:=plot([V,E1],x=0. .8,color=[red,blue]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "pl ots[display](P1,P2);" }}{PARA 13 "" 1 "" {GLPLOT2D 707 299 299 {PLOTDATA 2 "6'-%'CURVESG6$7`o7$$\"\"!F)$!+%fevu#!#57$$\"+=arz@!#6$!+E &3Gu#F,7$$\"+N3VfVF0$!+19dGFF,7$$\"+`i9RlF0$!+9k)[q#F,7$$\"+q;')=()F0$ !+-'==n#F,7$$\"+]#HyI\"F,$!+I<#zd#F,7$$\"+MBxV5E$F,$!+!o'eOCF,7$$\"+^\\#o=*F,$\" +J*=o5$F,7$$\"+&Qk#z**F,$\"+u4xkPF,7$$\"+_*y*z5!\"*$\"+WTe/WF,7$$\"+l9 .i6F[q$\"+M(HJ*\\F,7$$\"+#H!*oC\"F[q$\"+!3![SbF,7$$\"+>\"\\?gF,7$$\"+-eL;9F[q$\"+.!**yU'F,7$$\"+&[A4]\"F[q$\"+^['Rw'F,7$$\"+'G Ize\"F[q$\"+DMGNqF,7$$\"+(3Q\\n\"F[q$\"+9d.LsF,7$$\"+1Yd^F[q$ \"+w.tDuF,7$$\"+wHid>F[q$\"+Y&G2T(F,7$$\"+g-w+?F[q$\"+;+8#Q(F,7$$\"+!3 *Q(3#F[q$\"+c^K'G(F,7$$\"++z,u@F[q$\"+'>kQ9(F,7$$\"+SP)4M#F[q$\"+3'Qzv 'F,7$$\"+>Zg#\\#F[q$\"+#HkqI'F,7$$\"+Fn*Gn#F[q$\"+,T.%o&F,7$$\"+2xiDGF [q$\"+(3T'4^F,7$$\"+Y,H.IF[q$\"+!H%p:WF,7$$\"+2:bgJF[q$\"+%Ryxz$F,7$$ \"+Y@4LLF[q$\"+n\"eI8$F,7$$\"+O;R(\\$F[q$\"+Q96DDF,7$$\"+<4#)oOF[q$\"+ $yHg#>F,7$$\"+7lCEQF[q$\"+Q%eGT\"F,7$$\"+%G^g*RF[q$\"*D,G-*F,7$$\"+>2V sTF[q$\"*kMb?%F,7$$\"+O&pfK%F[q$\")m$o8%F,7$$\"+kcz\"\\%F[q$!*iRTF$F,7 $$\"+\"G5Jm%F[q$!*/N5m'F,7$$\"+6#32$[F[q$!*/Fae*F,7$$\"+Ey'G*\\F[q$!+U !)o27F,7$$\"+J%=H<&F[q$!+'[P)[9F,7$$\"+3>qM`F[q$!+VRwO;F,7$$\"+,.W2bF[ q$!+F,7$$\"+T>4NeF[q$!+ey9v?F,7$$\"+8s5' *fF[q$!+[)H(y@F,7$$\"+mXTkhF[q$!+HJTrAF,7$$\"+od'*GjF[q$!+[(4'[BF,7$$ \"+EcB,lF[q$!+N'RrT#F,7$$\"+v>:nmF[q$!+W^!HZ#F,7$$\"+0a#o$oF[q$!+Y8.@D F,7$$\"+`Q40qF[q$!+Zr:hDF,7$$\"+\"3:(frF[q$!+x)>Bf#F,7$$\"+e%GpL(F[q$! +%o7Bi#F,7$$\"+:-V&\\(F[q$!+')ysWEF,7$$\"+ZhUkwF[q$!+awukEF,7$$\"+$!3!f&H_m_i(***!#=7$$\"39LLLL3 VfVFhal$!3gH!*4/7^!***F[bl7$$\"3s******\\i9RlFhal$!3Y*H#z1\")oy**F[bl7 $$\"3Hmmmm;')=()Fhal$!3u!HCw11A'**F[bl7$$\"3%*******\\#HyI\"F[bl$!3)*y C$))zgb\"**F[bl7$$\"3ELLLLBxV5E$F[bl$!3()ev-%y qF[bl7$$\"3))*****\\YJ?;\"!#<$!3z&*p@E1%G_'F[bl7$$\"3?LLL=\"\\(* ))yTF[bl7$$\"3)*******RP)4M#Ffel$!3O#\\)f3*3$GRF[bl7$$\"3ILLL=Zg#\\#Ff el$!3naUz_)*RBPF[bl7$$\"3cmmmEn*Gn#Ffel$!3:^W0Si0/NF[bl7$$\"3Tmmm1xiDG Ffel$!3$[E6!G$oiL$F[bl7$$\"3!)*****\\9!H.IFfel$!3K&)=th#f\"fJF[bl7$$\" 3Immm1:bgJFfel$!3n=$oRN5m,$F[bl7$$\"3<+++X@4LLFfel$!3Y^jM:'pO(GF[bl7$$ \"31+++N;R(\\$Ffel$!3E87(\\O0\"\\FF[bl7$$\"3wmmm;4#)oOFfel$!3JsmjiftHE F[bl7$$\"3jmmm6lCEQFfel$!3)zðbfGDF[bl7$$\"3ELLL$G^g*RFfel$!3.Jsk-F[bl7$$\"3G******H%=H<&Ffel$!3iJ_dA`+)* =F[bl7$$\"35mmm1>qM`Ffel$!38.BAC'GC%=F[bl7$$\"3%)*******HSu]&Ffel$!39m w-.\\^'y\"F[bl7$$\"3'HLL$ep'Rm&Ffel$!3!e0=fkc'Q4N eFfel$!3/,Xx^M9*o\"F[bl7$$\"3#emm;@2h*fFfel$!3'zU/%4$G]k\"F[bl7$$\"3]* ****\\c9W;'Ffel$!3iL**z&R\"G,;F[bl7$$\"3Lmmmmd'*GjFfel$!3)>+SJ\"eng:F[ bl7$$\"3j*****\\iN7]'Ffel$!3uGl$G_*G?:F[bl7$$\"3aLLLt>:nmFfel$!3I@#Qq# *)H$[\"F[bl7$$\"35LLL.a#o$oFfel$!3'z0/0qnsW\"F[bl7$$\"3ammm^Q40qFfel$! 3`IZeUd?89F[bl7$$\"3y******z]rfrFfel$!3_wdWWkF$Q\"F[bl7$$\"3gmmmc%GpL( Ffel$!3%eho\\1#[]8F[bl7$$\"3/LLL8-V&\\(Ffel$!3Z9si%))GCK\"F[bl7$$\"3=+ ++XhUkwFfel$!3]mr\"ogjPH\"F[bl7$$\"3=+++:o0 solutions are not quantized." }}{PARA 0 "" 0 "" {TEXT -1 60 "there is an infinite series of bound states (Rydberg series)" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "63 0 0" 22 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }