Homework Solution Set # 3. Thursday, September 22, Textbook: Claude Cohen Tannoudji, Bernard Diu and Franck Lalo, Second Volume Complement G X

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1 Deparmen of Physics Quanum Mechanics II, 570 Temple Universiy Insrucor: Z.-E. Meziani Homework Soluion Se # 3 Thursday, Sepember, 06 Texbook: Claude Cohen Tannoudji, Bernard Diu and Franck Lalo, Second Volume Complemen G X Exercise, page 086 Consider a deuerium aom (composed of a nucleus of spin I and an elecron). The elecronic angular momenum is J L + S, where L is he orbial angular momenum of he elecron and S is is spin. The oal angular momenum of he aom is F J + I, where I is he nuclear spin. The eigenvalues of J and F are J(J + ) and F (F + ) respecively. a. Wha are he possible values of he quanum number J and F for a deuerium aom in he s ground sae? In he s ground sae of he deuerium aom he values of n and l are n, l 0 and s /. Therefore he possible values of J are such ha L / L + / / J / leaving he only value possible for J o be J /. Wih his value of J we can find he possible values of F assuming ha I, J I F J + I / F / +. Thus F / or F 3/. b. Same quesion for he deuerium in he p excied sae. In he case of he p sae of deuerium we have This implies J values such ha n, l () L / J L + / / J + / () resuling in possible values of J 3/ and J /. Now for he case of J / we have leading o F / and F 3/ For he case where J 3/ we find leading o F /, F 3/ and F 5/ I / F I + / / F + / (3) I 3/ F I + 3/ 3/ F + 3/ (4) Exercise, page 086 The hydrogen aom nucleus is a proon of spin I /. a) In he noaion of he preceding exercise, wha are he possible values of he quanum numbers J and F for a hydrogen aom in he p level? In he case of he p sae for hydrogen we have n and l and he elecron spin s /. This implies ha L s J L + s L / J L + / / J + / (5)

2 Thus he wo possible values of J are J 3/ and J /. Now for he case of J / we have leading o F 0 and F For he case where J 3/ we find leading o F and F I / F I + / / / F / + / (6) I 3/ F I + 3/ / 3/ F / + 3/ (7) b) Le { n, l, s, J, M J } be he basis obained by adding L and S o form J (M J is he eigenvalue of J z ); and le { n, l, s, J, I, F, M F } be he basis obained by adding J and I o form F (M F is he eigenvalue of F z ) The magneic operaor of he elecron is: M µ B ( L + S)/ In each of he subspaces E(n, l, s /, J, I /, F ) arising from he p level and subended by he F + vecors n, l, s, J, I, F, M F (8) corresponding o he fixed values of J and F, he projecion heorem enables us o wrie M g JF µ B F / (9) Calculae he various possible Landé facors g JF corresponding o he p level. Using he projecion heorem Complemen D X we can wrie L S L J J(J + ) J S J J(J + ) J Using he resuls of Complemen D X of equaions 4-a and 4-b afer adding he wo erms we can wrie: L J E0,l,s,J S J E0,l,s,J [J(J + ) + L(L + ) S(S + )] [J(J + ) L(L + ) + S(S + )] Therefore we obain ha { µ B M J(J + ) 3 [J(J + ) + L(L + ) S(S + )] + [ µ B 3 J(J + ) J(J + ) L(L + ) + ] S(S + ) J } [J(J + ) L(L + ) + S(S + )] J Now armed wih he relaion beween Mand J we use he projecion heorem again o find he relaion beween M and F and deduce he Landé facor.

3 Firs les remember ha F I + J, herefore J J F F (F + ) F J F E0,l,s,J,I,F Thus J F (F + ) We now replace he expression of J in ha of M o obain: F M M µ B [3J(J + ) L(L + ) + S(S + )] F J(J + ) F (F + ) µ B 4J(J + )F (F + ) {3J(J + ) L(L + ) + S(S + )} {F (F + ) + J(J + ) I(I + ))} F Since s / and I /, he above expression becomes M µ B We can wrie he Landé facor as : g JF 4J(J + )F (F + ) {3J(J + ) 3)} {F (F + ) + J(J + ) 3/4)} F (0) {3J(J + ) 3)} {F (F + ) + J(J + ) 3/4)} () 4J(J + )F (F + ) We can summarize he resul for all values of J and F l, s, J, and F 0, g 0 no defined l, s, J, and F, g 6 l, s, J 3, and F, g l, s, J 3, and F, g () Exercise 3, page Consider a sysem composed of wo spins / paricles whose orbial variables are ignored. The Hamilonian of he sysem is: Ĥ ω Ŝ z + ω Ŝ z (3) where S z and S z are he projecions of he spin S and S of he wo paricles ono Oz, and ω and ω are real consans. a. The iniial sae of he sysem, a ime 0 is: ψ(0) > [ + > + + >] (4) where + >,, +, > and + >,,, + > are wo of he common eigenvecors of Ŝ, Ŝ Ŝz and Ŝz. A ime, Ŝ ( S + S ) is measured, wha resuls can be found and wih wha probabiliies? 3

4 Firs, when a measuremen is performed on Ŝ he only possible resuls of he measuremen are eigenvalues of Ŝ. In his case, he possible eigenvalues are S(S + ) wih S and S 0. Therefore he possible resuls are or 0. Wrien in he basis of common vecors of Ŝ, Ŝ, Ŝ z and Ŝz hus also a common basis of Ĥ ω Ŝ z + ω Ŝ z ψ(0) [ ] 0 ψ() [ ] e i E e i E + + ψ() ] [e i(ω ω) + + e i(ω ω) + where E + (ω ω ) and E + ( ω + ω ). The probabiliy o find when measuring S a ime is given by P( ) + ψ() + 0 ψ() + ψ() In he nex expression we swich from he basis SM S o he basis m S, m S o find P( ) + + ψ() + ( )ψ() + ψ() 0 + [ e i(ω ω)/] + ( ) ω ω cos while he probabiliy o find 0 is given by swiching basis like before we have [ e i(ω ω)/] + 0 ) ( cos ω ω P(0 ) 00 ψ() P(0 ) ( + + )ψ() [ e i(ω ω)/] ( ) ω ω i sin [e i(ω ω)/] ( sin ω ω b If he iniial sae of he sysem is arbirary, wha Bohr frequencies can appear in he ime evoluion of < S ˆ >? Using he Ehrenfes Theorem we can wrie in general ha Since Ĥ and Ŝ commue hen d S ˆ [Ĥ, d i Ŝ ] [Ĥ, Ŝ ] 0 d S ˆ consan d Thus < ˆ S > does no evolve in ime and hus no frequencies will be needed. c Same quesion for < Ŝx >< Ŝx + Ŝx > If he iniial sae is arbirary we can wrie i in he basis of eigensaes of Ĥ, S, S, S z and S z as ) ψ(0) > α(0) β(0) + + γ(0) + + δ(0) 4

5 since his sae is an eigensae of Ĥ he ime evoluion is wrien as ψ() > α(0)e ie++/ β(0)e ie+ / + + γ(0)e ie +/ + + δ(0)e ie / where we have E ++ / ω + ω E + / ω ω E + / ω ω E / ω + ω (5) he corresponding bra is: ψ() α() β() + + γ() + + δ() + + for convenience we leave ou in he noaion of he coefficiens he ime dependence in wha follows. Ŝx Ŝx + Ŝx (α β + + γ + + δ ) S x (α β + + γ + + δ ) + (α β + + γ + + δ ) S x (α β + + γ + + δ ) (α β + + γ + + δ ) (S + + S ) (α β + + γ + + δ ) + (α β + + γ + + δ ) (S + + S ) (α β + + γ + + δ ) (α β + ) S + (γ + + δ ) + (δ + γ + ) S (α β + ) + (α γ + ) S + (β + + δ ) + (δ + β + ) S (α γ + ) (α γ + β δ) + (δ β + γ α) + (α β + γ δ) + (δ γ + β α) Boh frequencies will appear in he ime dependence of < ˆ S x >. 5