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Universität Heidelberg SS 2009 Thirteenth quantum mechanics sheet To be handed in on 07.07. All points in this sheet are awarded as bonus points. Q 34: Hydrogen atom, fine structure (16 points) ~ S ~ and J~ are angular momentum operators. The notationˆis omitted Remark: In this question L, for these operators. We consider the Hamiltonian of the Hydrogen atom ~2 P̂ Ze2 Ĥ0 = − + Ŵ , 2m R̂ Z=1 with the spin-orbit interaction (see the lectures) Ŵ = Ze2 1 ~ ~ S·L 2m2e c2 R̂3 The aim of this exercise is to treat Ŵ as a perturbation from Ĥ0 and from this calculate the 2 2 (0) Z 2 e4 1 = − me2c α2 Zn2 (α = finestructure constant) at corrections to the energy levels En = − me2h̄ 2 n2 first order in perturbation theory. The energy En 0 depends only on the main quantum number n and not on the angular momentum quantum numbers l and m. Therefore we must use the ~ 2 and L3 with |n, l, ml i. degenerate perturbation theory. We describe the joint Eigenstates Ĥ0 , L It follows: Ĥ0 |n, l, ml i = En(0) |n, l, ml i ~ 2 |n, l, ml i = h̄2 l(l + 1)|n, l, ml i L L3 |n, l, ml i = h̄ml |n, l, ml i ψnlml (~r) = h~r|n, l, ml i = Rnl (r)Ylml (Ω) ~ is given by S ~ = h̄ ~τ with the Pauli matrix τi . The state space of Ĥ = Ĥ0 + Ŵ The spin operator S 2 ~ 2 , L3 : is spanned by the product of Eigenstates of Ŝ3 and the Eigenstates |n, l, ml i of Ĥ0 , L |n, l, ml , ms i := |n, l, ml i|ms i ~ ·S ~ commutes with L ~ 2 and S ~ 2 , but not with L3 and S3 . (4 points) a) Show that Ŵ ∼ L This means that the product state |n, l, ml , ms i does not build an appropriate basis for the perturbed theory, because of the energy level degeneracy of l, ml and ms . One must build for a fixed l sub vector space, a new basis through linear combinations of |n, l, ml , ms i, in which ~ ·S ~ is diagonal (= angular momentum coupling). To do this one uses L ~ and S ~ to build a Ŵ ∼ L ~ new operator J of the total angular momentum: ~ +S ~ J~ = L b) Show that the components Ji satisfy the angular momentum commutation relation, i.e. that [Ji , Jj ] = ih̄ X k ǫijk Jk . show furthermore that h i h i h i i h i h i ~ 2 = J3 , J~2 = 0 ~ 2 = J3 , S J3 , L h ~ 2 = J~2 , S ~2 = L ~ 2, S ~2 = 0 J~2 , L (5 points) ~ 2 and S ~ 2 . We describe From b) it follows that it is possible to form joint Eigenvectors of J~2 , J3 , L these Eigenvectors |j, mj ; l, si with J~2 |j, mj ; l, si = h̄2 j(j + 1)|j, mj ; l, si J3 |j, mj ; l, si = h̄mj |j, mj ; l, si ~ 2 |j, mj ; l, si = h̄2 l(l + 1)|j, mj ; l, si L ~ 2 |j, mj ; l, si = h̄2 s(s + 1)|j, mj ; l, si . S (1) Here s = 1/2. ~ ·S ~ and hence Ŵ is diagonal c) Show that the states |j, mj ; l, si are also Eigenstates of L ~ ·S ~ (in terms of j, l and s). (3 in this basis. Determine the corresponding Eigenvalues L points) d) Calculate now the energy corrections at first order in perturbation theory En(1) = hn; j, mj ; l, s|Ŵ |n; j, mj ; l, si . For this use without proof that hn; j, mj ; l, s| 1 R̂3 |n; j, mj ; l, si = Z3 , a30 n3 l(l + 1)(l + 21 ) (0) a0 = h̄2 me e2 (1) The degeneracy of the Energy Eigenvalues En are broken by the energy corrections En . How large is the splitting ? Consider (1) En (0) En . (4 points)