{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 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 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 "Warning" -1 7 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 2 2 2 2 2 1 1 1 3 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 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 "Map le 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 256 12 "Atomic Model" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 290 "A simple model fo r the structure of few-electron atoms is presented. A model potential \+ for the helium atom is introduced (it represents a good approximation \+ to the so-called Hartee-Fock potential for the helium atom). The eigen energies and eigenfunctions of this potential are determined by" }} {PARA 0 "" 0 "" {TEXT -1 96 "a) the variational method for the lowest- lying eigenstate for a given angular momentum symmetry;" }}{PARA 0 "" 0 "" {TEXT -1 62 "b) the numerical solution of the radial Schroedinger equation;" }}{PARA 0 "" 0 "" {TEXT -1 62 "c) a matrix diagonalization using a Slater-type orbital basis." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 304 "A simple model potential for the ground state of helium can be obtained as follows: assume that in the ground state the wavefunction corresponds to an 1s^2 configuration (one spin -up, one spin-down electron to form parahelium with total electron spi n zero). What potential should the electron experience?" }}{PARA 0 "" 0 "" {TEXT -1 96 "The nuclear attraction -2/r and a repulsive potentia l due to the presence of the other electron." }}{PARA 0 "" 0 "" {TEXT -1 133 "What is the repulsive potential due to the other electron (whi ch has the same 1s-wavefunction, it has just opposite spin projection) ?" }}{PARA 0 "" 0 "" {TEXT -1 426 "We have an chicken-and-egg problem \+ here: given a potential, we can determine the |1s> state, and calculat e the repulsive part of the potential correctly. This is the objective of a self-consistent field calculation. Then we would need to repeat \+ the calculation of the new |1s> state using the new potential. Repeati ng this procedure one can come up with a |1s> state and corresponding \+ potential that provide the lowest energy." }}{PARA 0 "" 0 "" {TEXT -1 89 "We break the closed circuit by stating an approximate answer for t he potential in helium:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 " restart: Digits:=14:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "V:= r->-2/r+1/r*(1-exp(-3.36*r)*(1+1.665*r));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VGf*6#%\"rG6\"6$%)operatorG%&arrowGF(,&*&\"\"#\"\" \"9$!\"\"F1*&F0F1,&F/F/*&-%$expG6#,$*&$\"$O$!\"#F/F0F/F1F/,&F/F/*&$\"% l;!\"$F/F0F/F/F/F1F/F/F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 200 " This is the potential experienced by an 1s-electron in the helium atom : at short distances the electron just feels the full nuclear attracti on (-2/r) as the second expression approaches a constant as " }{TEXT 272 1 "r" }{TEXT -1 76 " goes to zero. At large distances the electron ic repulsion term goes like 1/" }{TEXT 273 1 "r" }{TEXT -1 80 " and sc reens the nucleus by one unit so that the overall potential goes like \+ -1/" }{TEXT 274 1 "r" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "We can check for the consistency by a s imple variational calculation:" }}{PARA 0 "" 0 "" {TEXT -1 112 "We sta rt with an unnormalized 1s state that depends on a 'charge' parameter \+ (cf. the hydrogen-like wavefunction)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "chi:=r*exp(-beta*r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$chiG*&%\"rG\"\"\"-%$expG6#,$*&%%betaGF'F&F'!\"\"F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "We are using the radial wavefunction (i.e . " }{TEXT 19 9 "r*R_nl(r)" }{TEXT -1 82 ") so that the radial kinetic energy is just proportional to the second derivative." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "assume(beta>0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "E1sT:=Int(chi*(-1/2*Diff(chi,r$2)+V(r)*chi) ,r=0..infinity)/Int(chi^2,r=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E1sTG*&-%$IntG6$*(%\"rG\"\"\"-%$expG6#,$*&%&beta|irG F+F*F+!\"\"F+,&*&#F+\"\"#F+-%%DiffG6$*&F*F+F,F+-%\"$G6$F*F6F+F2*(,&*&F 6F+F*F2F2*&F*F2,&F+F+*&-F-6#,$*&$\"$O$!\"#F+F*F+F2F+,&F+F+*&$\"%l;!\"$ F+F*F+F+F+F2F+F+F+F*F+F,F+F+F+/F*;\"\"!%)infinityGF+-F'6$*&)F*F6F+)F,F 6F+FPF2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "E1sT:=value(E1sT );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E1sTG,$**$\"/+++++]7!#9\"\"\" ,,*&$\"'[?w\"\"!F*%&beta|irGF*!\"\"*&$\"'D*=&F/F*)F0\"\"#F*F1$\"'/FfF/ F1*&$\"&+D'F/F*)F0\"\"%F*F**&$\"&+]'F/F*)F0\"\"$F*F*F*F0F*,**&$\"&Dc\" F/F*FAF*F**&$\"&](yF/F*F5F*F**&$\"'+B8F/F*F0F*F*$\"&)3uF/F*F1F*" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot(E1sT,beta=1.55..1.8,thi ckness=3);" }}{PARA 13 "" 1 "" {GLPLOT2D 406 124 124 {PLOTDATA 2 "6&-% 'CURVESG6$7S7$$\"3/++++++]:!#<$!3[dk:#[*y'*))!#=7$$\"3km;a)G\\ab\"F*$! 3%yPhZ#HS-*)F-7$$\"3G$ek`o!>g:F*$!3]WTChe52*)F-7$$\"3mm\"z>)G_l:F*$!3C cW(p&H>7*)F-7$$\"3ymT&QU!*3d\"F*$!3*yQz\"Gw4<*)F-7$$\"3S$eRZXKid\"F*$! 3h-9fLLw@*)F-7$$\"3i;z>,_=\"e\"F*$!3.#*eW]l*e#*)F-7$$\"3+]7G$[8je\"F*$ !3)**Q&4#z!)*H*)F-7$$\"3u;z%*frh\"f\"F*$!3W%>iNA&*R$*)F-7$$\"3%*\\ilFQ !pf\"F*$!3N![shY%yP*)F-7$$\"3IL$3_\"=M-;F*$!3))y\"3)p1YT*)F-7$$\"3omTg (fJrg\"F*$!3+&>6GM7X%*)F-7$$\"3/+]7eP_7;F*$!3]48Kv'Qx%*)F-7$$\"38+]Pf! 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The value of beta=2 would be obtained if there was only one 1 s-electron (the hydrogen-like solution for the ground state of the He+ ion). The fact that the electron wants to have a slightly more diffus e wavefunction reflects the so-called inner screening: the electrostat ic repulsion when combined with the simple model of two identical elec trons (apart from the spin projection) leads to two electrons which re pel each other on average (independent electron model). Both electrons experience the same common central potential and are bound by the sam e eigenenergy." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Now we would li ke to check two things:" }}{PARA 0 "" 0 "" {TEXT -1 45 "1) how accurat e is this variational solution?" }}{PARA 0 "" 0 "" {TEXT -1 85 "2) how close is the used potential to the potential produced by this 1s-wave function?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 98 "Part 1 can be answered by using dsolve[numeric]. For part 2 we nee d to solve the Poisson equation." }}{PARA 0 "" 0 "" {TEXT -1 119 "Let \+ us start with the first question: To solve the SE we start the integra tion not at zero, but at some small value of " }{TEXT 257 5 "r=eta" } {TEXT -1 26 " to avoid the singularity." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eta:=10^(-8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$e taG#\"\"\"\"*++++\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "IC:= phi(eta)=eta,D(phi)(eta)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6 $/-%$phiG6##\"\"\"\"*++++\"F*/--%\"DG6#F(F)F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Et:=-0.9042;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#EtG$!%U!*!\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "SE:=-1/ 2*diff(phi(r),r$2)+(V(r)-Et)*phi(r);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%#SEG,&*&#\"\"\"\"\"#F(-%%diffG6$-%$phiG6#%\"rG-%\"$G6$F0F)F(!\"\"* &,(*&F)F(F0F4F4*&F0F4,&F(F(*&-%$expG6#,$*&$\"$O$!\"#F(F0F(F4F(,&F(F(*& $\"%l;!\"$F(F0F(F(F(F4F(F($\"%U!*!\"%F(F(F-F(F(" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 57 "sol:=dsolve(\{SE,IC\},phi(r),numeric,output=li stprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phir:=subs (sol,phi(r)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot(phir, 0..10,thickness=3);" }}{PARA 13 "" 1 "" {GLPLOT2D 404 186 186 {PLOTDATA 2 "6&-%'CURVESG6$7fo7$\"\"!$!/QIF&p***>!#H7$$\"/mTN@Ki8!#:$ \"/quXXvD8F/7$$\"/K$3FWYs#F/$\"/Q?'oK0e#F/7$$\"/)\\iSmp3%F/$\"/s4#Rwuw $F/7$$\"/lmT&)G\\aF/$\"/QW!f,'*)[F/7$$\"/I3x1h6oF/$\"/tTb?!)\\fF/7$$\" /+]7G$R<)F/$\"/`6*[`3&pF/7$$\"/l\"z%\\DO&*F/$\"/XgvdT&*yF/7$$\"/LL3x&) *3\"!#9$\"/$Q$R'Rgy)F/7$$F.FS$\"/o1'p8:/\"FS7$$\"/+]ilyM;FS$\"/x)p'ok& =\"FS7$$\"/Le*)4D2>FS$\"/@^NBo78FS7$$\"/mm;arz@FS$\"/mJ0p:C9FS7$$\"/$e *)4bQl#FS$\"/3.LX:&e\"FS7$$\"/+D\"y%*z7$FS$\"/B>_my4#>FS7$$\"/+voMrU^FS$\"/Z(zf;&\\>FS7$$\"/!z%*HB$4aFS$\"/rkll1c>FS7$$\" /&3-8Lfn&FS$\"/H%z:+%e>FS7$$\"/v$4'HaUfFS$\"//Urh#p&>FS7$$\"/mm\"z_\"4 iFS$\"/e#)H*=?&>FS7$$\"/lmm6m#G(FS$\"/1(3wUS!>FS7$$\"/lmT&phN)FS$\"/q^ W8]B=FS7$$\"/+v=ddC%*FS$\"/7SdTgB/x\"Fgs$\" /3y&33=())F/7$$\"/+D1J:w=Fgs$\"/ft.![,-)F/7$$\"/LL3En$4#Fgs$\"/Ky>K'HZ 'F/7$$\"/m;/RE&G#Fgs$\"/K``!ReK&F/7$$\"/++D.&4]#Fgs$\"/7li++^UF/7$$\"/ ++vB_taaFsLF/7$$\"/++v'Hi#HFgs$\"/%GpESko#F/7$$\"/m\"z*ev:JF gs$\"/mZwb]y@F/7$$\"/LL347TLFgs$\"/v@0EF#p\"F/7$$\"/LLLY.KNFgs$\"/eK68 (GO\"F/7$$\"/*\\7o7Tv$Fgs$\"/Z@j\"fn0\"F/7$$\"/LL$Q*o]RFgs$\"/``Y_c>%) !#;7$$\"/*\\7=lj;%Fgs$\"/:.)e<([lFdy7$$\"/*\\PaRY2aFgs$\"//by()o*\\\"Fdy7$$\"/m;zXu9cFgs$\"/$yEEZ.<\"Fdy7$$\"/****\\ y))GeFgs$\"/O9(=\"zp!*!#<7$$\"/**\\i_QQgFgs$\"/r3v11(3(Fb\\l7$$\"/*\\7 y%3TiFgs$\"/!*evqA4cFb\\l7$$\"/**\\P![hY'Fgs$\"/!*))fM1pVFb\\l7$$\"/KL $Qx$omFgs$\"/8q3!=9a$Fb\\l7$$\"/***\\P+V)oFgs$\"/.fyeA**GFb\\l7$$\"/m \"zpe*zqFgs$\"/s/d6K'\\#Fb\\l7$$\"/***\\#\\'QH(Fgs$\"/\\gps-BAFb\\l7$$ \"/Ke9S8&\\(Fgs$\"/n&Q&eH4@Fb\\l7$$\"/*\\i?=bq(Fgs$\"/Mm,7-K@Fb\\l7$$ \"/KL3s?6zFgs$\"/Q\"=rciH#Fb\\l7$$\"/*\\7`Wl7)Fgs$\"/GnY_IJEFb\\l7$$\" /lmm*RRL)Fgs$\"/u^?U)\\8$Fb\\l7$$\"/l;a<.Y&)Fgs$\"/4**exEqQFb\\l7$$\"/ Ke9tOc()Fgs$\"/&GD4u;([Fb\\l7$$\"/****\\Qk\\*)Fgs$\"/`kZ,z\"4'Fb\\l7$$ \"/K$3dg6<*Fgs$\"/eN&Qu6%zFb\\l7$$\"/lmmxGp$*Fgs$\"/Ny/7'=,\"Fdy7$$\"/ )\\7oK0e*Fgs$\"/%\\a9]XJ\"Fdy7$$\"/)\\(=5s#y*Fgs$\"/r6zGN#p\"Fdy7$\"#5 $\"/bu,.cBAFdy-%'COLOURG6&%$RGBG$Fcbl!\"\"$F(F(F\\cl-%*THICKNESSG6#\" \"$-%+AXESLABELSG6$Q!6\"Fdcl-%%VIEWG6$;F\\cl$FcblF(%(DEFAULTG" 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 "" {MPLTEXT 1 0 8 "E1s:=Et;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E1sG$!%U!*!\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 272 "The numerically exact solution for the radial SE yields a slightl y lower eigenvalue of E_1s = -0.9042 a.u. (1 a.u. = 27.21 eV) compared to the variational result of E_1s_v =-0.897 a.u.. This means that the true eigenfunction is somewhat different from the hydrogenic form." } }{PARA 0 "" 0 "" {TEXT -1 243 "The numerical eigenfunction is not norm alized properly. To compare the graphs we simply change the normalizat ion of the variational answer. In fact, the variational state was norm alized such that the derivative of the function equals unity at " } {TEXT 258 1 "r" }{TEXT -1 3 "=0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "eval(subs(r=0,diff(subs(beta=beta0,chi),r)));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"\"\"\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}{PARA 7 "" 1 "" {TEXT -1 50 "War ning, the name changecoords has been redefined\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "P1:=plot(phir,0..6,color=red,thickness=3):" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "P2:=plot(subs(beta=beta0,c hi),r=0..6,color=blue,thickness=3):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display([P1,P2]);" }}{PARA 13 "" 1 "" {GLPLOT2D 384 174 174 {PLOTDATA 2 "6&-%'CURVESG6%7ao7$\"\"!$!/QIF&p***>!#H7$$\"/+]il yM;!#:$\"/$>m*eG#e\"F/7$$\"/++DJdpKF/$\"/)=MgyK1$F/7$$\"/+](ofV!\\F/$ \"/%45vT$[WF/7$$\"/++]i9RlF/$\"/!o\"o5eUdF/7$$\"/+]7G$R<)F/$\"/`6*[`3& pF/7$$\"/++v$>(3)*F/$\"/O_Pfxx!)F/7$$\"/+v$f]V9\"!#9$\"/]u?ntF\"*F/7$$ \"/++]#HyI\"FN$\"/K7,-\\55FN7$$\"/]PfIJ#f\"FN$\"/jJ+UMk6FN7$$\"/+voozw =FN$\"/$**\\w#G*H\"FN7$$\"/]7y1Gh@FN$\"//f(>wqT\"FN7$$\"/+]([kdW#FN$\" /If`uL>:FN7$$\"/+D\"3Gc3$FN$\"/?g*Hy**p\"FN7$$\"/++v;\\DPFN$\"/]\\vrkB =FN7$$\"/+++nfpVFN$\"/CA]XO->FN7$$\"/++DFN7$$\"/Dc^E 'R<&FN$\"/HDAm^]>FN7$$\"/]7yNAM`FN$\"/t%)=nma>FN7$$\"/vo/X[%\\&FN$\"/Y Kr*\\s&>FN7$$\"/+DJauacFN$\"/T>D\"f$e>FN7$$\"/D\"yN1]\"eFN$\"/Moy=3e>F N7$$\"/]P%Gn_(fFN$\"/&G+\\,l&>FN7$$\"/v$4@Gb8'FN$\"/l&H(pp`>FN7$$\"/+] P\"*y&H'FN$\"/I^*=V(\\>FN7$$\"/+]7(=,*oFN$\"/a\"Q+$=E>FN7$$\"/+](G[W[( FN$\"/*QKvT4*=FN7$$\"/+]()fB:()FN$\"/CBzFi\"z\"FN7$$\"/+](Q=\"))**FN$ \"/%4h.ygm\"FN7$$\"/+vj=pD6!#8$\"/j\"y'fNI:FN7$$\"/++lN?c7Fgt$\"/H18:Q )Q\"FN7$$\"/+]U$e6P\"Fgt$\"/u#4kqeE\"FN7$$\"/++&>q0]\"Fgt$\"/s@\\.*R8 \"FN7$$\"/++DM^I;Fgt$\"/mkY&)**45FN7$$\"/++0ytb@ Fgt$\"/[o0ru3jF/7$$\"/+v3wY_AFgt$\"/N(o18(3bF/7$$\"/++IOTqBFgt$\"/3P!z zc([F/7$$\"/+v3\">)*\\#Fgt$\"/qB@(*3cUF/7$$\"/+DEP/BEFgt$\"/7I6G%Ht$F/ 7$$\"/+](o:;v#Fgt$\"/U>9J=]KF/7$$\"/+v$)[opGFgt$\"/'[4T+#eGF/7$$\"/+]i %Qq*HFgt$\"/f>r:#\\[#F/7$$\"/+vQIKHJFgt$\"/vXlV%e9#F/7$$\"/+D^rZWKFgt$ \"/D)4QUm)=F/7$$\"/+]Zn%)oLFgt$\"/[,m37S;F/7$$\"/++5FL(\\$Fgt$\"/8e%[$ z<9F/7$$\"/+]d6.BOFgt$\"/jo$\\`$G7F/7$$\"/+vo3lWPFgt$\"/iO$pI$o5F/7$$ \"/+]A))ozQFgt$\"/L+b>kT\"*!#;7$$\"/++Ik-,SFgt$\"/oIepPTzFh[l7$$\"/++D -eITFgt$\"/`-X8WGoFh[l7$$\"/+v=_(zC%Fgt$\"/U)yL%z^fFh[l7$$\"/++b*=jP%F gt$\"/Nxo=f=^Fh[l7$$\"/+v3/3(\\%Fgt$\"/!G9/5\"RWFh[l7$$\"/+vB4JBYFgt$ \"/!*[DL;BQFh[l7$$\"/++DVsYZFgt$\"/!3v(e?-LFh[l7$$\"/+v=n#f([Fgt$\"/yq \\q^JGFh[l7$$\"/++!)RO+]Fgt$\"/#*o@$z3W#Fh[l7$$\"/+]_!>w7&Fgt$\"/.ds7 \\'4#Fh[l7$$\"/+v)Q?QD&Fgt$\"/\")>Q2g-=Fh[l7$$\"/++5jyp`Fgt$\"/j%3P$*) o:Fh[l7$$\"/+]Ujp-bFgt$\"/$*)[y6\"Q8Fh[l7$$\"/++gEd@cFgt$\"/su!zU3;\"F h[l7$$\"/+v3'>$[dFgt$\"/F&[!)p0)**!#<7$$\"/+D6EjpeFgt$\"/ZaC!fDk)Fd`l7 $\"\"'$\"/`W@+_7uFd`l-%'COLOURG6&%$RGBG$\"*++++\"!\")$F(F(Feal-%*THICK NESSG6#\"\"$-F$6%7_o7$FealFeal7$$\"3$*****\\ilyM;!#>$\"3e[J#[C/.f\"Fab l7$$\"3')*****\\7t&pKFabl$\"3umIzpX1%4$Fabl7$$\"3z****\\(ofV!\\Fabl$\" 3quj:ZK\"[^%Fabl7$$\"3s******\\i9RlFabl$\"3]**)p*)Raf&eFabl7$$\"3m**** \\7G$R<)Fabl$\"3)*zpyb$o27(Fabl7$$\"3e*****\\P>(3)*Fabl$\"3F&oCI(eT7$) Fabl7$$\"3&****\\Pf]V9\"!#=$\"3]4)H&4A%RV*Fabl7$$\"3%*******\\#HyI\"F` dl$\"3?0@F2\"G)[5F`dl7$$\"37+]PfIJ#f\"F`dl$\"3@$R?=J=r@\"F`dl7$$\"3-++ voozw=F`dl$\"3_`Z;3zKn8F`dl7$$\"3#***\\7y1Gh@F`dl$\"35hx-@@y+:F`dl7$$ \"33++]([kdW#F`dl$\"3pu[ul;s=;F`dl7$$\"3K++D\"3Gc3$F`dl$\"3CcPK\\,>L=F `dl7$$\"3++++v;\\DPF`dl$\"3Q'\\N!o*)z')>F`dl7$$\"3A++++nfpVF`dl$\"3;3e qDLH!4#F`dl7$$\"3W+++D0)\\g*z@F`dl7$$ \"3y*\\P4@Gb8'F`dl$\"3%fWS,&Rqy@F`dl7$$\"3o****\\P\"*y&H'F`dl$\"3m*o7d `if<#F`dl7$$\"3i****\\7(=,*oF`dl$\"3o3Y7)QFT:#F`dl7$$\"3e****\\(G[W[(F `dl$\"3)G'Q1IWk;@F`dl7$$\"3i****\\()fB:()F`dl$\"3\\#Qi,-tC+#F`dl7$$\"3 9++](Q=\"))**F`dl$\"3kfPk%fP8&=F`dl7$$\"3(****\\P'=pD6Fd`l$\"3Oq5v50O% o\"F`dl7$$\"33+++lN?c7Fd`l$\"3Z?:CQ643:F`dl7$$\"3-++]U$e6P\"Fd`l$\"3!p Ef(z)ReN\"F`dl7$$\"36+++&>q0]\"Fd`l$\"3gnF#Q#4r#>\"F`dl7$$\"3'******\\ U80j\"Fd`l$\"3:A&*y/6\"3/\"F`dl7$$\"35+++0ytb@Fd`l$\"3^Kn6\"fX,$fFabl7$$\"3'****\\(3wY_AFd`l $\"3cJZS?>zL]Fabl7$$\"3#)******HOTqBFd`l$\"3!zP2Mz?8M%Fabl7$$\"37++v3 \">)*\\#Fd`l$\"3mC()[--=!o$Fabl7$$\"3:++DEP/BEFd`l$\"3t\\4/ddfOJFabl7$ $\"3=++](o:;v#Fd`l$\"3)>qzf/)e[EFabl7$$\"3=++v$)[opGFd`l$\"3/!)z&QBXKE #Fabl7$$\"3%*****\\i%Qq*HFd`l$\"3c\"\\4gN&e1>Fabl7$$\"3&****\\(QIKHJFd `l$\"3_**oZw7&Fd`l$\"3')y\\$eQYX&*)!#@7$$\"3O++v)Q?QD&Fd`l$\"3mrnj_ \"\\]T(Fhdm7$$\"3G+++5jyp`Fd`l$\"3>==E&3@$[dF d`l$\"39)y`b/(*=_$Fhdm7$$\"37++D6EjpeFd`l$\"3wRyCp$*[IHFhdm7$$F[alF($ \"3A!\\]?;@SS#Fhdm-F_al6&FaalFealFealFbalFfal-%+AXESLABELSG6%Q!6\"F`gm -%%FONTG6#%(DEFAULTG-%%VIEWG6$;FealFhfmFegm" 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 -1 152 "Note that the agreement for the energy was at \+ the level of 7/900 , i.e. in the 1 % range. The deviation between the wavefunctions is in the 10 % range." }}{PARA 0 "" 0 "" {TEXT -1 154 " Nevertheless, we can state that the simple hydrogenic wavefunction cat ches the main feature of the numerical solution, namely the most likel y location in " }{TEXT 259 1 "r" }{TEXT -1 21 " for the 1s electron." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "The result for the binding ene rgy for one of the two helium electrons is the first quantity that can be compared with experiment:" }}{PARA 0 "" 0 "" {TEXT -1 4 "The " } {TEXT 260 20 "ionization potential" }{TEXT -1 82 " of helium is measur ed to be 24.481 eV (cf. R. Liboff, table 12.2). Our answer is:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "-Et*27.21*_eV;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#,$*&$\")#G.Y#!\"'\"\"\"%$_eVGF(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "For any atom the eigenenergy of the high est occupied orbital should equal the negative of the ionization poten tial. Our result is indeed quite close." }}{PARA 0 "" 0 "" {TEXT -1 328 "There is another quantity that can be measured, namely the total \+ energy of the atom (equal to the sum of ionization energies for both e lectrons). This is not simply twice the eigenenergy, since after ioniz ing on of the two He-electrons, the other is left in a hydrogen-like s tate (energy of -2 a.u. due to Z=2, and E_1s=-Z^2/2)." }}{PARA 0 "" 0 "" {TEXT -1 97 "Combining this answer with the calculated 1s binding e nergy we have for the total binding energy:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "(Et-2)*27.21*_eV;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&$\")#GB!z!\"'\"\"\"%$_eVGF(!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 261 28 "The electrostatic repulsion:" }}{PARA 0 "" 0 "" {TEXT -1 256 "Now we look at the question as to what potential is associated with the approximate wavefunction. For this we need a solution to the Poisson equation for a spherically symmetric charge distribution base d on the properly normalized variational wavefunction." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "A1s:=1/sqrt(int(chi^2,r=0..infinity ));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$A1sG,$*&\"\"#\"\"\")%&beta|i rG#\"\"$F'F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "rho:=subs (beta=beta,(A1s*chi)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$rhoG,$* *\"\"%\"\"\")%&beta|irG\"\"$F()%\"rG\"\"#F()-%$expG6#,$*&F*F(F-F(!\"\" F.F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "int(rho,r=0..infi nity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 116 "Note that the 4Pi from the integration over theta and phi are cancelled by the square of the spherical harmonic Y00!" } }{PARA 0 "" 0 "" {TEXT -1 91 "The solution to Poisson's equation in mu ltipole expansion leads to the monopole expression:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "V0:=unapply(simplify(int(rho,r=0..R)/R+in t(rho/r,r=R..infinity)),R);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#V0Gf *6#%\"RG6\"6$%)operatorG%&arrowGF(,$*&,(*(%&beta|irG\"\"\"9$F1-%$expG6 #,$*(\"\"#F1F0F1F2F1!\"\"F1F1F3F1F1F9F1F2F9F9F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "plot([V(r)+2/r,subs(beta=beta0,V0(r))],r= 0..5,color=[red,blue],view=[0..5,0..2],thickness=3);" }}{PARA 13 "" 1 "" {GLPLOT2D 436 152 152 {PLOTDATA 2 "6'-%'CURVESG6$7fn7$$\"3-+?aQ`!eS $!#?$\"3UY&Qy'Gz%p\"!#<7$$\"3/+S3x1h6oF*$\"3#f@#G#f:Xp\"F-7$$\"30+Ec,; 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The abov e potential shows how asymptotically the charge distribution of electr on 1 screens one of the protons for electron 2 (located at a large " } {TEXT 262 1 "r" }{TEXT -1 8 "-value)." }}{PARA 0 "" 0 "" {TEXT -1 216 "A sophisticated central-field or Hartree-Fock calculation take the el ectrostatic repulsion due to the numerically obtained charge density a nd re-calculates the eigenenergy/eigenfunction until convergence is ac hieved." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 263 22 "Electronic excitations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 443 "We proceed to calculate the energy lev els for the 2s and 2p states. We simply assume that we can use the pot ential obtained for the ground state, and calculate the energy spectru m for this potential. For the 2s-state we will not carry out a variati onal calculation as the energy will not be guranteed to be above the e xact eigenenergy for the given potential. We repeat our trial-and-erro r procedure to find a radial function with one node at " }{TEXT 264 1 "r" }{TEXT -1 3 ">0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "IC: =phi(eta)=eta,D(phi)(eta)=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG 6$/-%$phiG6##\"\"\"\"*++++\"F*/--%\"DG6#F(F)F+" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 13 "Et:=-0.15768;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#EtG$!&od\"!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "SE: =-1/2*diff(phi(r),r$2)+(V(r)-Et)*phi(r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG,&*&#\"\"\"\"\"#F(-%%diffG6$-%$phiG6#%\"rG-%\"$G6$F0F)F(! \"\"*&,(*&F)F(F0F4F4*&F0F4,&F(F(*&-%$expG6#,$*&$\"$O$!\"#F(F0F(F4F(,&F (F(*&$\"%l;!\"$F(F0F(F(F(F4F(F($\"&od\"!\"&F(F(F-F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "sol:=dsolve(\{SE,IC\},phi(r),numeri c,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phir:=subs(sol,phi(r)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "plot(phir,0..20,-0.5..0.5,thickness=3);" }}{PARA 13 "" 1 "" {GLPLOT2D 478 222 222 {PLOTDATA 2 "6&-%'CURVESG6$7bo7$\"\"!$!/B\\]'o** *>!#H7$$\"/qmT&)G\\a!#:$\"/Fc;Y&e)[F/7$$\"/ML3x&)*3\"!#9$\"/jsG#3\"e() F/7$$\"/+]ilyM;F5$\"/a-14&o<\"F57$$\"/nm;arz@F5$\"/.<+jn/9F57$$\"/++DJ dpKF5$\"/>00f=$o\"F57$$\"/MLL3VfVF5$\"/%otF/7$$\"/L$eR\"=\\8Fhn$!/ciC%=T:\"F/7$$\"/LLL A`c9Fhn$!/,4Q;_=UF/7$$\"/L$32$)Qc\"Fhn$!/w&3(GVIsF/7$$\"/LL3RBr;Fhn$!/ <$QTdg,\"F57$$\"/+vV^\"\\)=Fhn$!/6`%)pRm:F57$$\"/n;zjf)4#Fhn$!/9;0xJj? 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The National Institute of Standards (NIST) website in the US \+ lists spectroscopic data using wavenumbers in cm^(-1). Relative to the ground state (1s^2) it lists for parahelium:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dE1s2s:=166277.4*icm;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'dE1s2sG,$*&$\"(uFm\"!\"\"\"\"\"%$icmGF*F*" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dE1s2p:=171134.9*icm;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%'dE1s2pG,$*&$\"(\\8r\"!\"\"\"\"\"%$i cmGF*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dE1s3s:=184864.8 *icm;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'dE1s3sG,$*&$\"(['[=!\"\"\" \"\"%$icmGF*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "The conversion \+ factor is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "eV:=8065.541* icm;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eVG,$*&$\"(Tb1)!\"$\"\"\"%$ icmGF*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "dE1s2s/eV;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"/6Xmxdh?!#7" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 10 "dE1s2p/eV;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# $\"/Kx7J!=7#!#7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "Our 2s excitat ion energy in eV:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "(E2s-E 1s)*27.21;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"*#4GJ?!\"(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "The experimental difference betwee n parahelium 1s2s and 1s2p states:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "(dE1s2p-dE1s2s)/eV;" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#$\"/)3AjMD-'!#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 116 "N ow let us carry out the calculation for the 2p state. First we carry o ut the hydrogen-like variational calculation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "l:=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"lG\" \"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 146 "The trial function is hy drogen-like: the radial function picks up a factor of r^l, and we add \+ the centrifugal potential to the radial Hamiltonian:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "chi:=r^(1+l)*exp(-beta*r);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%$chiG*&)%\"rG\"\"#\"\"\"-%$expG6#,$*&%&beta|ir GF)F'F)!\"\"F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "E2pT:=In t(chi*(-1/2*Diff(chi,r$2)+(V(r)+1/2*l*(l+1)/r^2)*chi),r=0..infinity)/I nt(chi^2,r=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E2pTG*& -%$IntG6$*()%\"rG\"\"#\"\"\"-%$expG6#,$*&%&beta|irGF-F+F-!\"\"F-,&*&#F -F,F--%%DiffG6$*&F*F-F.F--%\"$G6$F+F,F-F4*(,(*&F,F-F+F4F4*&F+F4,&F-F-* &-F/6#,$*&$\"$O$!\"#F-F+F-F4F-,&F-F-*&$\"%l;!\"$F-F+F-F-F-F4F-F-*&F-F- *$F*F-F4F-F-F*F-F.F-F-F-/F+;\"\"!%)infinityGF--F'6$*&)F+\"\"%F-)F.F,F- FSF4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "E2pT:=value(E2pT); " }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%E2pTG,$**$\"/+++++]7!#9\"\"\",0 *&$\"+sI3L5\"\"!F*%&beta|irGF*!\"\"*&$\"*++q\\(F/F*)F0\"\"$F*F**&$\"*v =ny&F/F*)F0\"\"%F*F**&$\"*+?N'HF/F*)F0\"\"#F*F1*&$\")+D1RF/F*)F0\"\"'F *F**&$\"*+++]#F/F*)F0\"\"&F*F*$\"*G\\wA&F/F1F*F0F*,.*&$\"(Dcw*F/F*FIF* F**&$\")]7.#)F/F*F:F*F**&$\"*+]iv#F/F*F5F*F**&$\"*++0j%F/F*F?F*F**&$\" *+?'*)QF/F*F0F*F*$\"*K7pI\"F/F*F1F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot(E2pT,beta=0.45..0.6,thickness=3);" }}{PARA 13 " " 1 "" {GLPLOT2D 466 188 188 {PLOTDATA 2 "6&-%'CURVESG6$7S7$$\"35+++++ ++X!#=$!3'Q1(p*=!**[7F*7$$\"3#****\\7t&pKXF*$!3)o\"f!*=6\"4D\"F*7$$\"3 ;+v=7T9hXF*$!3/=SsM7]_7F*7$$\"3=+](=HPJf%F*$!3sy2a+#*>a7F*7$$\"3/+]7VD MDYF*$!3s&omz&>\"eD\"F*7$$\"3-+vVGZRdYF*$!3u4o$)G5Kd7F*7$$\"3')*\\(=27 6(o%F*$!3OUI4++D\"*3 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***********fF*$!3Nhd@Z4SN7F*-%'COLOURG6&%$RGBG$\"#5!\"\"$\"\"!F_[lF^[l -%*THICKNESSG6#\"\"$-%+AXESLABELSG6$Q&beta|ir6\"Q!Fh[l-%%VIEWG6$;$\"#X !\"#$\"\"'F][l%(DEFAULTG" 1 2 0 1 10 3 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " beta0:=fsolve(diff(E2pT,beta),beta=0..2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "E2pV:=subs(beta=beta0,E2pT);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&beta0G$\"/*H5Oi%\\^!#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%E2pVG$!/YT8h^o7!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "No w we verify the variational calculation by a numerical solution:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "IC:=phi(eta)=eta^(l+1),D(phi )(eta)=(l+1)*eta^l;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ICG6$/-%$phi G6##\"\"\"\"*++++\"#F+\"2++++++++\"/--%\"DG6#F(F)#F+\")+++]" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Et:=-0.12699;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%#EtG$!&*p7!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "SE:=-1/2*diff(phi(r),r$2)+(V(r)+1/2*l*(l+1)/r^2-Et)*p hi(r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#SEG,&*&#\"\"\"\"\"#F(-%%d iffG6$-%$phiG6#%\"rG-%\"$G6$F0F)F(!\"\"*&,**&F)F(F0F4F4*&F0F4,&F(F(*&- %$expG6#,$*&$\"$O$!\"#F(F0F(F4F(,&F(F(*&$\"%l;!\"$F(F0F(F(F(F4F(F(*&F( F(*$)F0F)F(F4F($\"&*p7!\"&F(F(F-F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "sol:=dsolve(\{SE,IC\},phi(r),numeric,output=listproce dure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "phir:=subs(sol,ph i(r)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot(phir,0..25,t hickness=3);" }}{PARA 13 "" 1 "" {GLPLOT2D 436 174 174 {PLOTDATA 2 "6& -%'CURVESG6$7bo7$\"\"!$\"/\"zA%zEeI!#J7$$\"/L$3FWYs#!#9$\"/Jqrx&*Ha!#: 7$$\"/mmT&)G\\aF/$\"/<,.kQ[yF/$\"/MWZ3@ZIF/7$$\"/S&)ehL0!*F/$\"/>xt[bQPF/7$$\"/$ek`o!>5!#8$\"/g :Z\"H@W%F/7$$\"/H#=XtB:\"FJ$\"/=U(o]kB&F/7$$\"/v=n$ycG\"FJ$\"/WksARBgF /7$$\"/@b#G$)*=9FJ$\"/21$\\%p$z'F/7$$\"/n\"z>)G_:FJ$\"/&za)pmRvF/7$$\" /FJ$\"/,wxcp'e*F/7$$\"/nT&QU!*3#FJ$\"/Ywlqm=5FJ7$$\"/voHR9cBFJ$\"/^ 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n;Fdx$\"/c\"=:wQ*QF27$$\"/+v$4v5s\"Fdx$\"/q)R'*oF:$F27$$\"/#zWn*)*pFdx$\"/0MM2ak8F27$$\"/L3-=!y(>Fdx$\"/F>([*)44\"F27$$ \"/D\"G8O;.#Fdx$\"/m`jl=M&)!#;7$$\"/nm\"*\\[$3#Fdx$\"/MNL2PBmFf^l7$$\" /Qa&\\!p\\Ff^l7$$\"/ekG=4*=#Fdx$\"/4+WiicNFf^l7$$\"/+]i 4TPAFdx$\"/)oN)**>0CFf^l7$$\"/$3F9!z#H#Fdx$\"/;Q$*)e(*>\"Ff^l7$$\"/nmT >KUBFdx$\"/By3jymFf^l7$\"#D$!/eDX()[WKFf^l-%'COLOURG6&%$RGBG$\"#5!\"\"$F(F (F[bl-%*THICKNESSG6#\"\"$-%+AXESLABELSG6$Q!6\"Fcbl-%%VIEWG6$;F[bl$Faal F(%(DEFAULTG" 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 "" {MPLTEXT 1 0 8 "E2p:=Et;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$E2pG$!&*p7!\"&" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 115 "The answer is just slightly below the value obtai ned by the variational calculation with a hydrogenic wavefunction." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "(E2p-E1s)*27.21*_eV;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&$\"*T)y9@!\"(\"\"\"%$_eVGF(F(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "We learn that the model predicts \+ an excitation energy of slightly more than 21 eV. What does this have \+ to do with reality?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 83 "This compares well with the experimental observation of parahelium (about 21.2 eV)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "One important piece of information that does em erge from this calculation is the lifting of the (" }{TEXT 265 2 "nl" }{TEXT -1 196 ")-degeneracy observed in the hydrogen atom spectrum: th e 2s and 2p states are no longer degenerate, and thus, there is the po ssibility of observing photons that correspond to 2p-2s de-excitations ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "How \+ well do we get the splitting?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "(E2p-E2s)*27.21;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"(\\2N)!\" (" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "This should be compared with the experimental result of 0.6 eV!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT 266 21 "Matrix representation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 210 "We proceed wit h the calculation of approximate spectra in an angular momentum symmet ry sector. For this purpose we first define ourselves a suitable basis . We defined so-called Slater-type orbitals for a given " }{TEXT 267 1 "l" }{TEXT -1 66 "-symmetry, and use a Gram-Schmidt procedure to ort hogonalize them." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "Our radial functions are real-valued, and therefore we h ave the simple inner product and normalization:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 34 "IP:=(f,g)->int(f*g,r=0..infinity):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "NO:=xi->xi/sqrt(IP(xi,xi));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#NOGf*6#%#xiG6\"6$%)operatorG%&arrow GF(*&9$\"\"\"-%%sqrtG6#-%#IPG6$F-F-!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 38 "STO:=(n,l,beta)->r^(n+l)*exp(-beta*r);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$STOGf*6%%\"nG%\"lG%%betaG6\"6$%)ope ratorG%&arrowGF**&)%\"rG,&9$\"\"\"9%F3F3-%$expG6#,$*&9&F3F0F3!\"\"F3F* F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "L:=0;" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%\"LG\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "N:=8;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"NG\"\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "B1:=[seq(STO(n,L,17/10),n =1..N)];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#B1G7**&%\"rG\"\"\"-%$ex pG6#,$*(\"# " 0 "" {MPLTEXT 1 0 16 "IP(B1[1],B1[2]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6## \"%]P\"&@N)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 119 "The Gram-Schmidt \+ procedure takes a list of functions (state vectors) and orthonormalize s them using the two procedures." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "GS:=proc(vecs) local i,n,j,res,xi;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "n:=nops(vecs); res:=[NO(vecs[1])];" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 21 "for i from 2 to n do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "xi:=vecs[i]-add(IP(vecs[i],res[j])*res[j],j=1..i-1); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "res:=[op(res),NO(xi)]; od: end: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "B1ON:=GS(B1):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "IP(B1ON[5],B1ON[5]);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "IP(B1ON[4],B1ON[5]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "B1ON[2];" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&#\"$*G\"$](\"\"\"*&,&*&)%\"rG\"\"# F(-%$expG6#,$*(\"# " 0 "" {MPLTEXT 1 0 77 "plot([seq(B1 ON[i],i=1..4)],r=0..10,color=[red,blue,green,black],thickness=3);" }} {PARA 13 "" 1 "" {GLPLOT2D 422 168 168 {PLOTDATA 2 "6)-%'CURVESG6$7do7 $$\"\"!F)F(7$$\"3gmmTN@Ki8!#>$\"3$yBJ!)z$*4!fF-7$$\"3ALL$3FWYs#F-$\"3% 3m.;0!=`6!#=7$$\"3%)***\\iSmp3%F-$\"3_$Hs'Q)p,p\"F57$$\"3WmmmT&)G\\aF- $\"3`\\U'HCo>?#F57$$\"3m****\\7G$R<)F-$\"3gp4(\\9_M:$F57$$\"3GLLL3x&)* 3\"F5$\"3&4eUEf\"H9SF57$$\"3YmmTN@Ki8F5$\"3)*ocn&)Gu!z%F57$$\"3$***** \\ilyM;F5$\"3#p;PXn\"o)[&F57$$\"3RLLe*)4D2>F5$\"3miF^,Yi8hF57$$\"3emmm ;arz@F5$\"3%4CKDw\\2n'F57$$\"3;L$e*)4bQl#F5$\"3G)[')[APG\\(F57$$\"3v** *\\7y%*z7$F5$\"3'GCeqU$fZ\")F57$$\"3Lm;ajW8-OF5$\"3u'[jGy/gl)F57$$\"3[ 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calculate the kinetic and poten tial energy matrix elements:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "Tkin:=(phi,psi)->-1/2*int(expand(phi*diff(psi,r$2)),r=0..infinit y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%TkinGf*6$%$phiG%$psiG6\"6$%) operatorG%&arrowGF),$*&#\"\"\"\"\"#F0-%$intG6$-%'expandG6#*&9$F0-%%dif fG6$9%-%\"$G6$%\"rGF1F0/FA;\"\"!%)infinityGF0!\"\"F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Vpot:=(phi,psi)->int(expand(phi*psi *(V(r)+L*(L+1)/(2*r^2))),r=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%VpotGf*6$%$phiG%$psiG6\"6$%)operatorG%&arrowGF)-%$intG6$-%'ex pandG6#*(9$\"\"\"9%F5,&-%\"VG6#%\"rGF5**\"\"#!\"\"%\"LGF5,&F?F5F5F5F5F ;!\"#F5F5/F;;\"\"!%)infinityGF)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "with(LinearAlgebra): Digits:=15:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 30 "HM:=Matrix(N,shape=symmetric):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "We make use of the symmetry of the hamil tonian matrix: it allows to save almost a factor of 2 in computation t ime." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "for i from 1 to N \+ do: for j from 1 to i do: HM[i,j]:=Tkin(B1ON[i],B1ON[j])+Vpot(B1ON[i], B1ON[j]); od: od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "#for \+ i from 1 to N do: for j from i+1 to N do: HM[i,j]:=HM[j,i]: od: od:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "HMf:=Matrix(evalf(HM),shap e=symmetric,datatype=float[8]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "evals:=Eigenvalues(HMf, output='list'):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "ev_s:=evalf(evals,8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%ev_sG7*$!)u*>/*!\")$!)Eh^:F($\")'>\"pG!\"*$\")&[/!QF($\")-8'4 \"!\"($\")0B3FF2$\")UyUtF2$\")G%>K$!\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "(-.90419974+.15516126)*27.21;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!-3/P8Q?!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "W e see that the chosen basis set can reproduce the 1s and 2s states, bu t that it does not have a prediction for a bound 3s eigenstate." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 233 "The 1s-2 s energy difference can be compared with the experimental result for p arahelium. We calculate 20.38 eV as compared to 20.61 eV in experiment . From the numerical solution we had 20.31 eV. Is the matrix size suff iciently large?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 121 "Increase the matr ix size and observe the stability of the 1s and 2s eigenvalues. Do you find an acceptable 3s eigenvalue?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 269 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 158 "Change the value of the parameter that controls the Slater type o rbital (STO) basis from the chosen value of 17/10. Find the best basis for the 3s eigenvalue." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 248 "Now the eigenfunctions. We use a symmetr ic matrix diagonalization now, and so the eigenvalues come out as real s and properly sorted. The eigenvectors are stored as a matrix, as a s econd entry in the sequence of expressions returned by Eigenvectors:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "VEs:=Eigenvectors(HMf):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "VEs[1],VEs[2];" }}{PARA 12 "" 1 "" {XPPMATH 20 "6$-%'RTABLEG6%\")ctRF-%'MATRIXG6#7*7#$!0NySW(* >/*!#:7#$!0c,Ac7;b\"F.7#$\"0lg4g>\"pG!#;7#$\"0AKh`[/!QF.7#$\"0L]b>Ih4 \"!#97#$\"0E[^]I#3FF<7#$\"0LrAA%yUtF<7#$\"0h3!*yU>K$!#8&%'VectorG6#%'c olumnG-F$6%\")'pEi#-F(6#7*7*$\"0#3H,0$4)**F.$\"0:)*[MhI<%F5$!0\\g\\%Q- ZKF5$\"0QRU+:^dCyJ\"F.7*$!0[5fDC9#oFfn$!0aoZ!zNB]F.$!0$*e$=;G+[F5 $\"0eOm`$p#p%F.$!0`^H'y)\\S$F.$!0Q\\Tc+6[#F.$!0pbfOn=N&F.$\"0(zOEWGtCF .7*$\"0$)H^_^_r$Ffn$!0%\\Xk&y$!0Gd/tJed%F5$!0^;&z-L)z\"F.$!0L)o^S]8HF.$!0u; jgE^!QF.$!0S,y!e-!\\%F.$\"0PwB*e%=+&F.$\"0^ku$[8I`F.%'MatrixG" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 252 "Suppose that we would like to see the eigenfunctions corresponding to the four lowest-lying eigenvalues . The eigenvector for a given eigenvalue contains the expansion coeffi cients for the expansion of the eigenstate in terms of the chosen basi s states." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "The expansion coefficients are stored in columns. We pick the colu mn index for a given state " }{TEXT 19 1 "i" }{TEXT -1 38 ", and then \+ step through the row index " }{TEXT 19 1 "j" }{TEXT -1 24 " to assembl e the states." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 149 "We have incorporated an explicit normalization of the as sembled eigenfunctions even though this isn't necessary, as shown by t he calculated value of " }{TEXT 19 2 "No" }{TEXT -1 317 " (it turns ou t to be practically equal to one). For an orthonormal basis (created b y the Gram-Schmidt), and matrix eigenvectors normalized according to t he Euclidean norm (usually adhered to by numerical diagonalization rou tines) this is true that normalized eigenfunctions are created automat ically by the procedure." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "for i from 1 to 4 do:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "psi0:=add (VEs[2][j,i]*B1ON[j],j=1..N):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "No :=1/sqrt(int(expand(psi0^2),r=0..infinity));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "phi_a[i]:=No*add(VEs[2][j,i]*B1ON[j],j=1..N): od:" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "plot([seq(phi_a[i],i=1..4)] ,r=0..15,color=[red,blue,green,black],thickness=3);" }}{PARA 13 "" 1 " " {GLPLOT2D 446 208 208 {PLOTDATA 2 "6)-%'CURVESG6$7eo7$$\"\"!F)F(7$$ \"0]i:gT<-\"!#;$\"0zD70B[u%F-7$$\"0+DJ?$[V?F-$\"0y7=s\\uH*F-7$$\"0](o/ [AlIF-$\"0oRjkWkO\"!#:7$$\"0+]iSmp3%F-$\"0VcA1;_y\"F:7$$\"0+v$4'\\/8'F -$\"0@Wr$ySrDF:7$$\"0++D\"G$R<)F-$\"0U9B'p)HH$F:7$$F,F:$\"0vHof8V&RF:7 $$\"0+v=#**3E7F:$\"0jPJTh%fXF:7$$\"0](=U#Q/V\"F:$\"0#\\&yKOA6&F:7$$\"0 ++Dc'yM;F:$\"0cRX-%>;cF:7$$F2F:$\"0#y3p%F:$\"0(zg[S3K\"*F:7$$ \"0v$fLpfZ]F:$\"0<5w&*G(>#*F:7$$\"0]7`p,KS&F:$\"0qwIVhlE*F:7$$\"0DJqX1 )edF:$\"0AAL$fqx#*F:7$$\"0+](=7T9hF:$\"0iFyO:xD*F:7$$\"0+DJ?qSr(F:$\"0 wPY_46)))F:7$$\"0++v=HPJ*F:$\"0Vy\\$ol;#)F:7$$\"0++]<*R#4\"!#9$\"0m0q] \\-U(F:7$$\"0+]7VDMD\"Fer$\"0KF(yeD\"f'F:7$$\"0]7yN'o89Fer$\"0NxgQ()yy &F:7$$\"0+vVGZRd\"Fer$\"0oPdIKb.&F:7$$\"0+D\"y'HDs\"Fer$\"0E(y1[R'R%F: 7$$\"0+v=276(=Fer$\"03a%edA=QF:7$$\"0+vV`g\\-#Fer$\"0`-\\8)=%G$F:7$$\" 0+vo**3)y@Fer$\"0$>=$\\VM\"GF:7$$\"0+voH>zL#Fer$\"0uSxIJ()Q#F:7$$\"0+v ofHq\\#Fer$\"0F#*ot`;-#F:7$$\"0+v$f'HU\"GFer$\"0F(H-#3\"Q9F:7$$\"0++D \"*309$Fer$\"0r&pY0)R+\"F:7$$\"0+]i&e*yU$Fer$\"0sk@O>1F(F-7$$\"0++v[D9 v$Fer$\"0fuJ3gd-&F-7$$\"0++Dc$GwSFer$\"0[Lh^G!\\MF-7$$\"0++D^W$*Q%Fer$ \"0!\\`Puo(Q#F-7$$\"0+vo%QjtYFer$\"0N!>4g=. 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reside. A smaller value of beta should improve matters." }}{PARA 0 " " 0 "" {TEXT -1 133 "For this reason one mixes usually STO's with diff erent values of beta. The Gram-Schmidt process takes care of the ortho normalization." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 271 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 300 "Calculate t he eigenfunctions for a few low-lying eigenstates in the L=2 and L=3 s ymmetry sectors. Observe how the eigenenergies approach the hydrogenic result in this case of E_n=-1/(2n^2). Can you explain this behaviour? 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