Physics 195aProblem set number 7 – Solution to Problem 40Due 2 PM, Thursday, November 21, 2002READING: Read sections 1-5 of the “Solving the Schr¨odinger Equation:Resolvents” course note.PROBLEMS:35. Harmonic oscillator in three dimensions: Exercise 2 of the HarmonicOscillator course note.36. Resolvent mathematics: Exercise 1 of the Resolvent course note.37. More resolvent mathematics: Exercise 2 of the Resolvent course note.38. Still more resolvent mathematics: Exercise 3 of the Resolvent coursenote.39. Green’s function solution of the infinite square well: Exercise 4 of theResolvent course note.40. The one-electron atom (review?): We have had a couple of examplesof looking at the qualitative features of wave functions. Now apply thesame reasoning to the one-electron atom. Thus, sketch the effectivepotential and the lowest three radial wave functions (do both R(r)andu(r)=rR(r)) for the 1-electron atom for  = 0. Now do the same forthe qualitative solutions for  = 1. Pay attention to the turning points,and to the dependence at r = 0. Since you have already computed theactual wave functions, you may produce graphs of the functions youobtained. If you do this, however, you should look carefully at yourgraphs and make sure you understand at a physically intuitive level thequalitative features.Solution: Since we have solved for the wave functions, we’ll plot them.The effective equivalent one-dimensional potential is:Veff(r)=−Ze2r+( +1)2mr2. (12)17We found the solutions:Rnp(r)=−2Znpa03/2 (np−  − 1)!2np[(np+ )!]3(ρ/np)e−ρ/2npL2+1np+(ρ/np),(13)whereρ =2Za0r, (14)a0=1me2=1mα, is the Bohr radius, (15)np= nr+  +1. (16)The Associated Laguerre polynomials are given by:L2+1n+(x)=n−−1k=0(−)k+1[(n + )!]2(n −  − 1 − k)!(2 +1+k)!k!xk. (17)The first few of these polynomials are:L11(x)=−1 (18)L12(x)=2(−2+x) (19)L13(x)=6(−3+3x −12x2) (20)L33(x)=−6 (21)L34(x) = 24(−4+x) (22)L35(x) = 120(−10 + 5x −12x2). (23)Hence, the first few radial wave functions are:R10(ρ)=2Za03/2e−ρ/2(24)R20(ρ)=12√2Za03/2(2 − ρ/2)e−ρ/4(25)R30(ρ)=29√3Za03/2(3 − ρ + ρ2/18)e−ρ/6(26)18R21(ρ)=14√6Za03/2ρe−ρ/4(27)R31(ρ)=127√6Za03/2ρ(4 − ρ/3)e−ρ/6(28)R41(ρ)=164√15Za03/2ρ10 −54ρ +ρ232e−ρ/8. (29)In terms of ρ, we may express the effective potential as:Ueff(ρ) ≡a0Veff(r(ρ))2Z2e2= −1ρ+( +1)ρ2. (30)The bond state energies are:Enp= −Z2e4m21n2p. (31)In terms of our scaled Ueff(ρ) potential, they appear at:Enp= −14n2p. (32)We’ll make graphs of Ueff(ρ)andofRnp(ρ)(a0/Z)3/2(as well as