Physics 221B: Solution to HW #5 The Lithium-like Atom
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1 Physics B: Solution to HW #5 The Lithium-like Atom The gound state will be composed of two electons in s with spins up and down and a thid electon in an n = state. When choosing a state fo this thid electon we needn t woy about its spin, since the Hamiltonian does not act is spin-space. So lets just say it has spin up. If we choose the p state we won t specify m, the eigenvalue L z, since the Hamiltonian is otationally invaiant (i.e. choosing an m will not change anything in ou esults). a) Lets stat with notation. s fo example, means a paticle in the state s which is a function of x, with spin. We can fom a Slate deteminant and likewise fo s p. b) s s = s s s 3 s s s 3, 3! s s s 3 The unpetubed Hamiltonian is a sum of thee single paticle Hamiltonians. Theefoe ou single paticle states ae eigenstates of H 0 with thei usual enegies. Evey tem in s s contains the thee diffeent single paticle states. So, s s H 0 s s = (E n= + E n= ) s s s = (E n= + E n= ), and likewise fo s p. c) Some moe notation ψ ψ ψ 3 3 ψ ψ ψ 3. The expectation value H potentially contains 08 tems because ou wave function consists of pemutations, and H is the sum of thee /ijs. We can educe this numbe by using the anti-symmety of the wavefunction and the othogonality of single paticle states as follows. Fo example ψ ψ ψ 3 ψ ψ ψ 3 = ψ ψ 3 ψ ( ψ ψ 3 ψ ) = ψ ψ ψ 3 3 ψ ψ ψ 3.
2 In the last step we simply switched the labels of paticles and 3. Following this we can simplify H to H = e + e 3 + e 3 = 3 e. We ae down to 3 tems. The opeato / acts only on the paticles at x and x. In othe wods, ψ i ψ j ψ k ψ l ψ m ψ n = ψ i ψ j ψ l ψ m ψ k ψ n. By othogonality, only tems with the same thid state in the ba and the ket will suvive. This leaves us with the following twelve tems. H = 3 e = 3 ( s s e s s s s e s s + s s e s s s s e s s + s s e s s s s e s s + s s e s s s s e s s + s s e s s s s e s s ) + s s e s s s s e s s. Since = thee ae exactly two of evey tem. H = s s e s s s s e s s and likewise fo the p states. + s s e s s s s e s s + s s e s s s s e s s, Sounds a bit like a beauty contest, doesn t it? we ae down to the finalists...
3 d) The nd and th tems vanish by othogonality of spin wavefunctions. The 3d and 5th tems ae equal. Thus H = ss e ss + ss e ss ss e ss, and likewise fo the p states. e) Petubation Theoy Calculation In pat (f), we will have to distinguish between a vaiational paamete λ and the chage Z in the Hamiltonian which isn t vaied, so I will use λ in the wavefunctions in this calculation. I use atomic units (in these units e /a. To get an enegy in ev, multiply the esults by 7. ev). We need to calculate the 3 tems in pat (d) fo both s and p cases. I ll do one example and quote esults fo the othes. Using the expession in the lectue notes = l,m l + < l > l+ Y lm ()Y lm(), whee the agument in the spheical hamonics means (θ, φ ), sp sp = d 3 d 3 l,m l+ < l > l+ Y lm ()Y lm() (λ 3/ e λ Y00 ()) ( λ3/ λ e λ/ Ym ()) (λ 3/ e λ Y 00 ()) ( λ3/ λ e λ/ Y m ()). We want to evaluate this by using the othogonality elations fo spheical hamonics, but they don t hold if thee is othe angula dependence (e.g. a thid spheical hamonic) in the integal. Howeve, note that Y00 = is actually independent of angle, so we can pull it outside the integal. Then we can evaluate the emaining θ, φ angula dependence, d cos θ dφ Ylm ()Y 00() = δ l0 δ m0. We then use the delta functions to cancel the sum and fix l = 0, m = 0 elsewhee. That is good because then Y lm () Y 00 () =, and then I thank Ed Boyda fo TEXing the following sections. 3
4 thee ae only two emaining θ, φ spheical hamonics: d cos θ dφ Ym ()Y m () =. We ae left with sp sp = d d = λ8 (λ 3/ e λ ) ( > (λ 3/ e λ ) ( λ3/ λ e λ / ) d d > 4 e λ e λ. Because of the / > we need to split the integal into two pats, = λ8 { d e λ 8λ 7. Following analo- Mathematica does these integals nicely, giving λ8 gous pocedues, I find 0 λ3/ λ e λ / ) d 4 e λ } + d 4 e λ. ss ss = 5λ 8 ss ss = 7λ 3 4 ss ss = 4 λ 3 sp sp = 59λ 3 5 sp ps = 7 4 λ 3 8. Setting λ Z, and then using Z = 3, 8 E s s = 5Z 8 + 7Z Z 3.0Z This contibution aises the enegy, as one would expect fo electon epulsion, and is a significant offset to the zeoth-ode esult E 0 = 9Z 8 = 8 8. When we have a p electon instead, E s p = 5Z Z Z Z
5 In total, (E 0 + E) s s (E 0 + E) s p.843. Confiming ou intuition, the total E 0 + E has smalle magnitude fo the p than the s case: The p electon is in a moe cicula obit, so it sees less of the nuclea chage (i.e. it is sceened moe by the inne electons). f) Vaiational Calculation In ou tial wavefunctions we eplace Z with λ as above. The zeoth-ode single-paticle contibutions to the enegy with this wavefunction ae s p λ p p λ s =, s s = p p = 8, s Z s = Zλ, as you can easily compute. Thus Z Z Zλ s s = p p = 4, ψ va (s s) H ψ va (s s) = λ + λ Zλ Zλ λ, the last being the E contibution. taking Z=3) gives Minimizing with espect to λ (and λ Z , E va 7.89 (s s). We find λ < Z, popely eflecting the sceening effect of the electons. As with Helium, the vaiational enegy countes the ove-coection fom petubation theoy. Repeating fo the s p case, λ.54, E va 7.0 (s p). 5
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