{VERSION 4 0 "IBM INTEL NT" "4.0" } {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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{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 "" 1 16 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 1 14 0 0 0 0 0 0 1 0 0 0 0 0 0 1 } {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 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 1 14 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Tim es" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 259 19 "Perturbation Theory" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "In this w orksheet we provide two examples of perturbation calculations:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 417 "1) to pr ovide a motivation for perturbative calculations we consider the first relativistic correction to the kinetic energy in the hydrogen-like gr ound state. It represents an example where a natural smallness paramet er is present (1/c^2), and where we really are only interested in the \+ first-order correction to the energy as the higher-order contributions do compete with corrections to the energy operator itself." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 365 "2) to show in \+ a model problem how one can calculate higher-order corrections, and th at there are examples where the series expansion is only semi-converge nt we calculate the energy corrections to high, but finite order for a harmonic oscillator perturbed by a power-law (quartic) anharmonic pot ential. The Dalgarno-Lewis method is used to carry out the calculation ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 263 38 "A first-order perturbation calculation" }}{PARA 0 "" 0 "" {TEXT -1 204 "Suppose we are interested in a calculation to account fo r the correction to the Schroedinger kinetic energy due to special rel ativity. We just consider the hydrogenic ground state which is non-deg enerate." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assume(c>0,m>0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "taylor (sqrt(m^2*c^4+p^2*c^2),p=0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "T he rest energy in atomic units is given as" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 22 "subs(m=1,c=137,m*c^2);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 12 "%*27.12*_eV;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "This is slightly off from the accepted electron mass of 512 keV d ue to our approximate value of the fine structure constant as 1/137." }}{PARA 0 "" 0 "" {TEXT -1 247 "We know from the virial theorem that t he kinetic energy in the hydrogen atom ground state has the same magni tude as the binding energy (13.6 eV). We recognize that the inverse po wers of c will generate converging series. Note that in atomic units \+ " }{TEXT 256 1 "m" }{TEXT -1 3 "=1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 280 "To first order we need to calcu late the expectation value of the fourth derivative for the H0-eigenst ate, i.e., the hydrogenic 1s-state. The angular momentum is zero, and \+ therefore only a radial derivative is required when considering the mo mentum operator as it acts on s-states." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "psi:=2*r*exp(-r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "int(psi^2,r=0..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 125 "It is most convenient to perform the calculation in mome ntum space, where the momentum operator is a multiplicative operator. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "phi:=simplify(sqrt(2/Pi )*int(psi*sin(k*r),r=0..infinity));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "int(phi^2,k=0..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "If we choose units in which h_bar equals unity, we do no t distinguish between p and k (momentum and wavenumber)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "int(phi^2*k^2/2,k=0..infinity);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "We recognize that the non-relati vistic kinetic energy is calculated correctly. Let us now obtain the \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "int(phi^2*(-k^4/(8*c^2) ),k=0..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "evalf( subs(c=137,%));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "This is a ver y small correction when compared to the non-relativistic answer of 0.5 . Now we repeat the calculation for arbitrary Z (hydrogen-like situati on)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assume(Z>0);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "psi:=2*Z^(3/2)*r*exp(-Z*r); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "int(psi^2,r=0..infinity );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "phi:=simplify(sqrt(2/ Pi)*int(psi*sin(k*r),r=0..infinity));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "No:=int(phi^2,k=0..infinity);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 37 "T_nr:=int(phi^2*k^2/2,k=0..infinity);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "T_r1:=int(phi^2*(-k^4/(8*c^2 )),k=0..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "subs( Z=20,c=137,evalf([1*c^2,T_nr,T_r1])); # Calcium" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 49 "subs(Z=79,c=137,evalf([1*c^2,T_nr,T_r1])); # G old" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "subs(Z=92,c=137,eval f([1*c^2,T_nr,T_r1])); # Uranium" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "It is evident that:" }}{PARA 0 "" 0 "" {TEXT -1 64 "a) the kinetic energy becomes comparable to the rest energy for " }{TEXT 257 1 "Z" } {TEXT -1 73 " approaching 100 (heavy-ion collisions allow one to explo re this regime)." }}{PARA 0 "" 0 "" {TEXT -1 78 "b) the relativistic k inetic energy becomes appreciable for moderate values of " }{TEXT 258 1 "Z" }{TEXT -1 50 ". The Schroedinger equation is no longer adequate. " }}{PARA 0 "" 0 "" {TEXT -1 68 "c) there are two expansions that do c ome up in the present context: " }}{PARA 0 "" 0 "" {TEXT -1 76 "(i) th e expansion of the relativistic kinetic energy as a power series in p; " }}{PARA 0 "" 0 "" {TEXT -1 146 "(ii) the energy correction from the \+ first term in this expansion beyond the non-relativistic expression ha s been evaluated only in first-order PT." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 979 "It is important to distinguish th e two sources of expansions. The expansion of the Hamiltonian itself h as as its origin the model-like character of the approach; a proper ap proach to special relativity would start from a different wave equatio n, and would be fully compliant with the modern world after Einstein. \+ This approach for spin-1/2 particles such as electrons was introduced \+ by Dirac. It leads not only to an equation where the relativistic kine tic energy expression is quantized, but one finds that all the magneti c interactions resulting from the particle's spin and orbital angular \+ momenta is automatically included. One has to include perturbatively, \+ however corrections to the Hamiltonian that arise from interactions wi th the nuclear magnetic moment (hyperfine structure), and further deta ils such as nuclear finite size effects (the nucleus is not a point pa rticle and one should take into account at short distances the extent \+ of the nuclear charge distribution)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 260 49 "The Dalgarno-Lewis method for \+ perturbation theory" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 144 "In 1955 A. Dalgarno and J.T. Lewis published a method \+ that allows to calculate the perturbation series to high orders for no n-degenerate states." }}{PARA 0 "" 0 "" {TEXT -1 380 "The method is ex plained in the classic textbook by L.I. Schiff (QM, McGraw-Hill, 3rd e d., 1968, chapter 33). The method is based on the conversion of an eig envalue problem into a series of inhomogeneous differential equations. These equations determine successively the corrections to the eigenfu nctions, and the energy corrections are obtained by a simple expectati on value. For " }{XPPEDIT 18 0 "H = H0+lambda*W;" "6#/%\"HG,&%#H0G\"\" \"*&%'lambdaGF'%\"WGF'F'" }{TEXT -1 20 " the equations read:" }}{PARA 0 "" 0 "" {TEXT -1 20 "(H0-E[0]) psi[0] = 0" }}{PARA 0 "" 0 "" {TEXT -1 34 "(H0-E[0]) psi[1] = (E[1]-W) psi[0]" }}{PARA 0 "" 0 "" {TEXT -1 48 "(H0-E[0]) psi[2] = (E[1]-W) psi[1] + E[2] psi[0]" }}{PARA 0 "" 0 " " {TEXT -1 62 "(H0-E[0]) psi[3] = (E[1]-W) psi[2] + E[2] psi[1] + E[3] psi[0]" }}{PARA 0 "" 0 "" {TEXT -1 3 "..." }}{PARA 0 "" 0 "" {TEXT -1 5 "Here " }{XPPEDIT 18 0 "E = E[0]+lambda*E[1]+lambda^2*E[2]+lambda ^3*E[3]+O(lambda^4);" "6#/%\"EG,,&F$6#\"\"!\"\"\"*&%'lambdaGF)&F$6#F)F )F)*&F+\"\"#&F$6#F/F)F)*&F+\"\"$&F$6#F3F)F)-%\"OG6#*$F+\"\"%F)" } {TEXT -1 175 ", and a similar expansion holds for the wavefunction. Th e energy corrections are calculated from the expectation value E[i] = \+ . psi[0] has to be normalized." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 738 "The equations to be solv ed for psi[i] are inhomogeneous, as they are driven by a known right-h and side, namely the wavefunction correction obtained in the previous \+ step. The method is started with the known solution psi[0], E[0], and \+ E[1] is calculated directly. At every stage one has to calculate a new wavefunction from solving a distinct inhomogeneous problem, although \+ the operator on the LHS remains the same. The energy correction is the n obtained by a simple matrix element. This is in contrast with the us ual approach where the second-order energy already involves an infinit e sum. Note that the differential equations are of a structure where t hey can be solved by a Green's function that involves a sum over all H 0 eigenstates." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 183 "In this worksheet we consider cases of power-law potenti als that allow exact solutions to the Dalgarno-Lewis equations. The H0 problem is assumed to be the simple harmonic oscillator." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 7 "W:=x^4;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "u00:=exp(-x^2/2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "No:=1/sqrt (int(u00^2,x=-infinity..infinity)):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "u0:=No*u00;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "E0:=1/2; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "E1:=int(u0*W*u0,x=-infi nity..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "Now that we \+ have the first-order energy correction we can proceed with the calcula tion of the first-order wavefunction calculation:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 50 "SE1:=-1/2*diff(f(x),x$2)+(x^2-1)/2*f(x)=u0*( E1-W);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 262 "We can solve SE1 by a \+ matrix method. The left-hand side will be common to all subsequent equ ations. Inhomogeneous differential equations in matrix representations become matrix inversion problems. We use a harmonic oscillator basis \+ for the matrix representation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(orthopoly);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "xi:=n->exp(-x^2/2)*H(n,x)/(Pi^(1/4)*sqrt(n!*2^n));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "int(xi(4)^2,x=-infinity..infinity); # an \+ example of normalization." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "subs(f(x)=xi(n),lhs(SE1));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "xi(m)*simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "ME:=unapply(Int(expand(%),x=-infinity..infinity),m,n);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 236 "Now we show that the structure of the ma trix is very simple indeed: only diagonal matrix elements occur, and t he structure is that =n. This diagonal matrix is trivial \+ to invert! Check it for yourself using matrix inversion." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "value(ME(1,2));" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 15 "value(ME(2,2));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "value(ME(2,4));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "value(ME(2,6 ));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "value(ME(0,0));" }}{PARA 0 " > " 0 "" {MPLTEXT 1 0 15 "value(ME(4,4));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "value(ME(6,6)); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "value(ME(1,1));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "rhs(S E1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "RH1:=unapply(Int(ex pand(xi(m)*rhs(SE1)),x=-infinity..infinity),m);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 189 "The solution is given as a superposition of the bas is functions, where the coefficients follow from the matrix inversion \+ mentioned above in combination with the right-hand-side calculation." }}{PARA 0 "" 0 "" {TEXT -1 257 "The first two calculations state that \+ we do not need to admix the psi[0] state (c0=0), which is good since w e can't invert the corresponding matrix coefficient; also all states w ith odd symmetry do not contribute (the integrals vanish for symmetry \+ reasons):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "c0:=simplify(v alue(RH1(0)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c1:=simpl ify(value(RH1(1))/1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "Now we c ould get an infinite sequence of non-vanishing right-hand-side calcula tions:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c2:=simplify(valu e(RH1(2))/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c4:=simpli fy(value(RH1(4))/4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c6: =simplify(value(RH1(6))/6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "In fact, all higher coefficients vanish!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c8:=simplify(value(RH1(8))/8);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "The solution to the equations SE1 which determines \+ the first-order wave-function correction is then given exactly as:" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "sol:=c2*xi(2)+c4*xi(4);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "We now verify that it does solve \+ SE1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "subs(f(x)=sol,SE1): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalb(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 96 "So we found an exact solution! Now we calculate t he second-order energy correction using psi[1]:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 42 "E2:=int(u0*sol*(W),x=-infinity..infinity);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "int(u0*sol*(W-E1),x=-infini ty..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "That is the ex act answer! The two above calculations agree since u0 is orthogonal to sol, i.e., to psi[1]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "Now do the same for the next equation: (H0-W0)*psi[2 ]=(E[1]-W)*psi[1]+E[2]*psi[0]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 54 "Note that it is only the right-hand side \+ that changes!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "E1;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "RH2:=unapply(Int(expand(xi(m )*((E1-W)*sol+E2*u0)),x=-infinity..infinity),m):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "value(RH2(0));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c2:=simplify(value(RH2(2))/2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c4:=simplify(value(RH2(4))/4);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c6:=simplify(value(RH2(6))/6);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c8:=simplify(value(RH2(8))/8);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "value(RH2(10));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 305 "All higher coefficients vanish. Note that we produced two more co efficients than before: this is due to the x^4 interaction, which can \+ couple two adjacent columns or rows in the matrix that represents this interaction in the HO basis (generate the matrix representation and o bserve). There are four terms." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "sol2:=c2*xi(2)+c4*xi(4)+c6*xi(6)+c8*xi(8):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "SE2:=lhs(SE1)=((E1-W)*sol+E2*u0):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "subs(f(x)=sol2,SE2):" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 22 "simplify(%): evalb(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "Let us check the orthogonality of the successive solution corrections:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "int(sol2*u 0,x=-infinity..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "int(sol2*sol,x=-infinity..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 249 "Note that orthogonality to ground state is preserved. Th e solution psi[1] is not orthogonal to psi[2]. The orthogonality to th e ground state means that the energy correction calculation can be mad e with W or with W displaced by some energy constant:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E3:=int(u0*W*sol2,x=-infinity..infi nity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "int(u0*sol2*(W-E1 ),x=-infinity..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Now \+ the fourth order:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "RH3:=u napply(Int(expand(xi(m)*((E1-W)*sol2+E2*sol+E3*u0)),x=-infinity..infin ity),m):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "value(RH3(0)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c2:=simplify(value(RH3( 2))/2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c4:=simplify(value(RH3(4 ))/4);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c6:=simplify(value(RH3(6) )/6);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c8:=simplify(value(RH3(8)) /8);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c10:=simplify(value(RH3(10) )/10);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c12:=simplify(value(RH3(1 2))/12);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "value(RH3(14)); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "All subsequent coefficients v anish. There are four terms in the solution:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "sol3:=c2*xi(2)+c4*xi(4)+c6*xi(6)+c8*xi(8)+c10*xi (10)+c12*xi(12):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "SE3:=lh s(SE1)=((E1-W)*sol2+E2*sol+E3*u0):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "subs(f(x)=sol3,SE3):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "simplif y(%): evalb(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "We continue t o find exact solutions to the Dalgarno-Lewis equations, and we preserv e the orthogonality to the ground state:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "int(sol3*u0,x=-infinity..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "int(sol2*sol3,x=-infinity..infinity);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Orthogonality to ground state is \+ again preserved." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E4:=int (u0*W*sol3,x=-infinity..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "int(u0*sol3*(W-E1),x=-infinity..infinity);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "This is the fourth order. Now wan t the fifth! Our series expansion for the ground-state energy so far r eads as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "E0+lambda*E1+la mbda^2*E2+lambda^3*E3+lambda^4*E4;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "It is somewhat worrisome that the coefficients are growing..." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "RH4:=unapply(Int(expand(x i(m)*((E1-W)*sol3+E2*sol2+E3*sol+E4*u0)),x=-infinity..infinity),m):" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "value(RH4(0));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c2:=simplify(value(RH4(2))/2);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c4:=simplify(value(RH4(4))/4);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c6:=simplify(value(RH4(6))/6);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c8:=simplify(value(RH4(8))/8);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c10:=simplify(value(RH4(10))/10);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c12:=simplify(value(RH4(12))/12); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c14:=simplify(value(RH4(14))/14 );" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c16:=simplify(value(RH4(16))/ 16);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c18:=simplify(value(RH4(18) )/18);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "sol4:=c2*xi(2)+c4 *xi(4)+c6*xi(6)+c8*xi(8)+c10*xi(10)+c12*xi(12)+c14*xi(14)+c16*xi(16): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "SE4:=lhs(SE1)=((E1-W)*s ol3+E2*sol2+E3*sol+E4*u0):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "subs( f(x)=sol4,SE4):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "simplify(%): eva lb(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E5:=int(u0*W*sol4 ,x=-infinity..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "int(u0*sol4*(W-E1),x=-infinity..infinity);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 23 "Do one more, the sixth:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "RH5:=unapply(Int(expand(xi(m)*((E1-W)*sol4+E2*sol3+E 3*sol2+E4*sol+E5*u0)),x=-infinity..infinity),m):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "value(RH5(0));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c2:=simplify(value(RH5(2))/2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c4:=simplify(value(RH5(4))/4);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c6:=simplify(value(RH5(6))/6);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c8:=simplify(value(RH5(8))/8);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c10:=simplify(value(RH5(10))/10);" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 33 "c12:=simplify(value(RH5(12))/12);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c14:=simplify(value(RH5(14))/14);" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 33 "c16:=simplify(value(RH5(16))/16);" }}{PARA 0 " > " 0 "" {MPLTEXT 1 0 33 "c18:=simplify(value(RH5(18))/18);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c20:=simplify(value(RH5(20))/20);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c22:=simplify(value(RH5(22))/22);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c24:=simplify(value(RH5(24))/24); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 108 "sol5:=c2*xi(2)+c4*xi(4 )+c6*xi(6)+c8*xi(8)+c10*xi(10)+c12*xi(12)+c14*xi(14)+c16*xi(16)+c18*xi (18)+c20*xi(20):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "SE5:=lh s(SE1)=((E1-W)*sol4+E2*sol3+E3*sol2+E4*sol+E5*u0):" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 20 "subs(f(x)=sol5,SE5):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "simplify(%): evalb(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "E6:=int(u0*W*sol5,x=-infinity..infinity);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 14 "And, one more:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "RH6:=unapply(Int(xi(m)*((E1-W)*sol5+E2*sol4+E3*sol3+ E4*sol2+E5*sol+E6*u0),x=-infinity..infinity),m):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "value(RH6(0));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c2:=simplify(value(RH6(2))/2);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c4:=simplify(value(RH6(4))/4);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c6:=simplify(value(RH6(6))/6);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "c8:=simplify(value(RH6(8))/8);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c10:=simplify(value(RH6(10))/10);" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 33 "c12:=simplify(value(RH6(12))/12);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c14:=simplify(value(RH6(14))/14);" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 33 "c16:=simplify(value(RH6(16))/16);" }}{PARA 0 " > " 0 "" {MPLTEXT 1 0 33 "c18:=simplify(value(RH6(18))/18);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c20:=simplify(value(RH6(20))/20);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c22:=simplify(value(RH6(22))/22);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c24:=simplify(value(RH6(24))/24); " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c26:=simplify(value(RH6(26))/26 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "sol6:=c2*xi(2)+c4*xi (4)+c6*xi(6)+c8*xi(8)+c10*xi(10)+c12*xi(12)+c14*xi(14)+c16*xi(16)+c18* xi(18)+c20*xi(20)+c22*xi(22)+c24*xi(24):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "SE6:=lhs(SE1)=((E1-W)*sol5+E2*sol4+E3*sol3+E4*sol2+E5 *sol+E6*u0):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "subs(f(x)=sol6,SE6) :" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "simplify(%): evalb(%);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "E7:=int(u0*sol6*(W-W1),x=-in finity..infinity);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "This worris ome trend of growing coefficients has continued:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "E1,E2,E3,E4,E5,E6,E7;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 17 "lambda:='lambda':" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "EO2:=E0+E1*lambda+E2*lambda^2;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 21 "EO3:=EO2+E3*lambda^3;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 21 "EO4:=EO3+E4*lambda^4;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "EO5:=EO4+E5*lambda^5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "EO6:=EO5+E6*lambda^6;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "EO7:=EO6+E7*lambda^7;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "plot([EO2,EO3,EO4,EO5,EO6,EO7],lambda=0..0.5,-2..2,co lor=[red,blue,green,black,grey,magenta]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "Let us find the exact eigenenergy at lambda=0.1 and compa re the various approximations:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "E_num:=0.559145;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "SE:=-1/2*diff(u(x),x$2)+(1/2*x^2+1/10*x^4-E_num)*u(x)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "solSE:=dsolve(\{SE,u(0)=1,D(u)(0)=0 \},u(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "ux:=subs(solSE,u(x)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot('ux(x)',x=0..4,view=[0..4,-0.2..1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 170 "Pt:=[[1,E0+0.1*E1],[2,subs(lambda= 0.1,EO2)],[3,subs(lambda=0.1,EO3)],[4,subs(lambda=0.1,EO4)],[5,subs(la mbda=0.1,EO5)],[6,subs(lambda=0.1,EO6)],[7,subs(lambda=0.1,EO7)]];" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot([Pt,E_num],1..8);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 206 "Clearly this is an example where \+ perturbation theory does not work! For the x^4 potential even a coupli ng of lambda=0.1 leads to a useless asymptotic series. For smaller lam bda is should be useful, however." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "E_num:=0.514085;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "SE:=-1/2*diff(u(x),x$2)+(1/2*x^2+1/50*x^4-E_num)*u(x) =0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "solSE:=dsolve(\{SE,u (0)=1,D(u)(0)=0\},u(x),numeric,output=listprocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "ux:=subs(solSE,u(x)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot('ux(x)',x=0..4,view=[0..4,-0.2 ..1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 177 "Pt:=[[1,E0+1/50* E1],[2,subs(lambda=1/50,EO2)],[3,subs(lambda=1/50,EO3)],[4,subs(lambda =1/50,EO4)],[5,subs(lambda=1/50,EO5)],[6,subs(lambda=1/50,EO6)],[7,sub s(lambda=1/50,EO7)]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "pl ot([Pt,E_num],1..8);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 306 "Eventual ly the expansion does diverge! The message is: for small coupling pert urbation theory (even though it can result in a semi-convergent series ) can be extremely useful. This is particularly true for problems wher e many terms of the expansion can be calculated symbolically or by num erical techniques." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 142 "Pick a diff erent power-law potential (possibly a linear combination). Carry out t he calculations and compare the results with exact solutions." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 196 " We can also look at the wavefunctions: one can see that they fail at m oderate lambda, and then fail at large lambda [we show the one used to calculate the enegy correction E3, and the one for E5]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "u2:=u0+lambda*sol+lambda^2*sol2:" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "u5:=u2+lambda^3*sol3+lambd a^4*sol4+lambda^5*sol5:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 " plot([subs(lambda=1/50,u2),subs(lambda=1/50,u5)],x=0..3,view=[0..3,-0. 2..1],color=[red,blue]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "plot([subs(lambda=1/10,u2),subs(lambda=1/10,u5)],x=0..3,view=[0..3,-0 .2..1],color=[red,blue]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 143 "We \+ see that the higher wavefunction corrections are very small in the fir st case, and that they are destroying the state in the second example. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "Can P ade approximants rescue the situation of the diverging series for the \+ energy eigenvalue?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with( numapprox):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Pade:=pade(E O7,lambda,[3,4]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot(P ade,lambda=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "subs(l ambda=0.1,Pade);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "This agrees r ather well with the numerical result. Let us check a value at strong c oupling:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "subs(lambda=1., Pade);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "E_num:=0.80375;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "SE:=-1/2*diff(u(x),x$2)+( 1/2*x^2+1*x^4-E_num)*u(x)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "solSE:=dsolve(\{SE,u(0)=1,D(u)(0)=0\},u(x),numeric,output=listpr ocedure):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "ux:=subs(solSE ,u(x)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "plot('ux(x)',x=0 ..4,view=[0..4,-0.2..1]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "We \+ see that the Pade result based on the perturbation expansion to sevent h order does not cure everything." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 262 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 185 "Carry out the perturbaton expansion to tenth order, construct a b etter Pade approximant and observe whether the ground-state eigenvalue can be obtained within 1% of the numerical value." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "167" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }