dx x ψ, we should find a similar expression for rθφ L ψ. From L = R P and our knowledge of momentum operators, it follows that + e y z d

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1 PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 11 Topics Coveed: Obital angula momentum, cente-of-mass coodinates Some Key Concepts: angula degees of feedom, spheical hamonics 1. [20 pts] In ode to deive the popeties of the spheical hamonics, we need to detemine the action of the angula momentum opeato in spheical coodinates. Just as we have x P x ψ = i d dx x ψ, we should find a simila expession fo θφ L ψ. Fom L = R P and ou knowledge of momentum opeatos, it follows that ( θφ L ψ = ı ( e x y d dz z d ) ( + e y z d dy dx x d ) ( + e z x d dz dy y d )) θφ ψ. dx Catesian coodinates ae elated to spheical coodinates via the tansfomations x = sin θ cos φ y = sin θ sinφ z = cos θ and the invese tansfomations = x 2 + y 2 + z 2 x θ = actan( 2 + y 2 ) z φ = actan( y x ). Thei deivatives can be elated via expansions such as x = x + θ x θ + φ x φ. Using these elations, and simila expessions fo y and z, find expessions fo θφ L x ψ, θφ L y ψ, and θφ L z ψ, involving only spheical coodinates and thei deivatives. x = x = sin θ cos φ x θ = z2 x cos θ cos φ = 2 z x 2 +y2 x φ = x2 csc θ sinφ = x 2 So d dx = sinθ cos φ + x 2 +y 2 y cos θ cos φ y = y = sin θ sin φ y θ = z2 y cos θ sinφ = 2 z x 2 +y2 y φ = x2 1 x 2 +y 2 x csc θ cos φ = So d dy = sin θ sin φ + z = z = cos θ θ cos θ sin φ θ + csc θ sinφ φ csc θ cos φ φ 1

2 x 2 +y 2 2 z 2 = sinθ z θ = z2 z φ = 0 So d dz = cos θ sinθ θ Now θφ L x ψ = i (y d dz z d dy ) θφ ψ So we can say ( L x = i y d dz z d ) dy = i ( sin θ cos θ sin φ sin 2 θ sinφ θ sinθ cos θ sin φ cos 2 ) θ sin φ θ + cot θ cos φ φ Which means θφ L x ψ = i ( sinφ θ cot θ cos φ φ ) θφ ψ Similaly we can say L y = i (z d dx x d dz ) = i ( sin θ cos θ cos φ + cos 2 θ cos φ θ cot θ sinφ φ sinθ cos θ cos φ + sin 2 ) θ cos φ θ so that θφ L y ψ = i (cos φ θ cot θ sinφ φ ) θφ ψ Lastly, we have L z = i (x d dy y d dx ) = i ( sin 2 θ sin φcos φ + sin θ cos θ sin φcos φ θ + cos 2 φ φ + sin 2 θ sin φcos φ sin θ cos θ sin φcos φ θ + sin 2 φ φ ) so that θφ L z ψ = i φ θφ ψ 2

3 2. [15pts] Fom you pevious answe and the definition L 2 = L 2 x + L2 y + L2 z, pove that θφ L 2 ψ = 2 ( 1 sin θ θ sinθ θ + 1 sin 2 θ 2 ) 2 φ 2 θφ ψ. L 2 = 2 [(sin φ θ + cot θ cos φ φ )(sin φ θ + cot θ cos φ φ ) + (cos φ θ cot θ sinφ φ )(cos φ θ cot θ sin φ φ ) + φ 2 ] = 2 [ sin 2 φ θ 2 + cot θ sinφcos φ θ φ csc 2 θ sin φcos φ φ + cot θ sin φcos φ φ θ + cot θ cos 2 φ θ + cot 2 θ cos 2 φ φ 2 cot2 θ cos φsin φ φ + cos 2 φ θ 2 cot θ cos φsin φ θ φ + csc 2 θ sinφcos φ φ cot θ sin φcos φ φ θ + cot θ sin 2 φ θ + cot 2 θ sin 2 θ φ 2 + cot2 θ sin φcos φ φ + φ 2 ] = 2 [ θ 2 + cot θ θ + (1 + cot 2 θ) φ 2 ] Noting that and the poof is complete. 1 sin θ θ sin θ θ = 2 θ + cot θ θ 1 + cot 2 θ = 1 sin 2 θ 3

4 3. [10 pts] We can factoize the Hilbet space of a 3-D paticle into adial and angula Hilbet spaces, H (3) = H () H (Ω). Two altenate basis sets that both span H (Ω) ae { θφ } and { lm }. As the angula momentum opeato lives entiely in H (Ω), we can use ou esults fom poblem 11.1 to deive an expession fo θφ L z lm. Combine this with the fomula L z lm = m lm, to deive and then solve a diffeential equation fo the φ-dependence of θφ lm. You solution should give θφ lm in tems of the as of yet unspecified initial condition θ lm θ,φ lm. What estictions does this φ=0 solution impose on the quantum numbe m, which descibes the z-component of the obital angula momentum? Since m max = l, what estictions ae then placed on the total angula momentum quantum numbe l? and Thus θφ L z lm = i φ θφ lm θφ L z lm = m θφ lm i θφ lm = m θφ lm φ The solution to this simple fist-ode diffeential equation is θφ lm = θ0 lm e imφ Since the wave-function must be single valued, we equie m to be a whole intege. As m max = l, this implies that l must be a whole intege also. 4

5 4. [10 pts] Using L ± = L x ± il y we can use the elation L + l,l = 0 and the expessions fom poblem 11.1 to wite a diffeential equation fo θφ ll. Plug in you solution fom 11.3 fo the φ-dependence, and show that the solution is θφ ll = c l e ilφ sin l (θ). Detemine the value of the nomalization coefficient c l by pefoming the necessay integal. We have θφ L + ll = 0 This implies θφ L x ll + i θφ L y ll = 0 Using the expessions fom 11.1 gives ( sin θ θ cot θ cos φ φ +icos φ θ icot θ cos φ φ ) θφ ll = 0 This simplifies to (i(cos φ + isin φ) θ cot θ(cos φ + isin φ) φ ) θφ ll = 0 Factoing out the e iφ gives (i θ cot θ φ ) θφ ll = 0 Plugging in the solution fom 11.3 gives (i θ ilcot θ) θ0 ll e ilφ = 0 which educes to θ θ0 ll = lcot θ θ0 ll Now θ sin l θ = lsin l 1 θ cos θ = lcot θ sin l θ So the solution is θφ ll = c l e ilφ sin l θ The nomalization integal is π 0 sin θdθ 2π 0 dφ θφ ll 2 = 1 With ou solution this becomes c l 2 π 0 sin θdθ 2π 0 dφ sin 2l θ = 1 Pefoming the phi integal gives 2π c l 2 π 0 sin θdθ sin2l θ = 1 u-substitution with u = cos θ gives 2π c l du(1 u2 ) l = 1 Since the integand is even, this educes to 4π c l du(1 u2 ) l = 1 Fom Mathematic we get 2π c l 2 Γ[1/2]Γ[l+1] Γ[l+3/2] = 1 which gives c l = Γ(l+3/2) 2πΓ[1/2]Γ[l+1] Thus we have Γ(l + 3/2) θφ ll = 2πΓ[1/2]Γ[l + 1] sinl θe ilφ 35 Fo the special case l = 3 this gives θφ 33 = 1 8 e3iφ π sin3 θ, which agees with the spheical hamonic Y3 3 (θ,φ) up to a non-physical phase facto. 5

6 5. [10 pts] Using L lm = l(l + 1) m(m 1) l,m 1 togethe with you pevious answes to deive an expession fo θφ l,m 1 in tems of θφ lm. Explain how in pinciple you can now ecusively calculate the value of the spheical hamonic Yl m (θφ) θφ lm fo any θ and φ and fo any l and m. Follow you pocedue to deive popely nomalized expessions fo spheical hamonics fo the case l = 1, m = 1,0,1. To constuct the othe m states fo the same l, we can begin fom the expession Now θφ L lm = (l + m)(l m + 1) θφ l,m 1 θφ L lm = θφ L x lm i θφ L y lm = i ( sin φ θ cot θ cos φ φ ) θφ lm (cos φ θ cot θ sinφ φ ) θφ lm = e iφ ( θ + icot θ φ ) θφ lm Putting the pieces togethe gives θφ l,m 1 = e iφ ( θ + icot θ φ ) (l + m)(l m + 1) θφ lm Stating fom ou expession fo θφ ll, we can find θφ l, l 1 by applying the above diffeential fomula. Successive iteations will then geneate all the emaining θφ lm states. 6

7 6. [10 pts] A paticle of mass M is constained to move on a spheical suface of adius a. Does the system live in H (3), H (), o H (Ω)? What is the Hamiltonian? What ae the enegy levels and degeneacies? What ae the wavefunctions of the enegy eigenstates? Because the adial motion is constained to a fixed value, it is only necessay to conside the dynamics in H (Ω). The Hamiltonian is then H = L2 2Ma 2 Choosing simultaneous eigenstates of L 2 and L z, we have H l,m = 2 l(l + 1) 2Ma 2 l,m so that and E l = 2 l(l + 1) 2Ma 2 d l = 2l + 1 The wavefunctions ae the spheical hamonics θφ l,m = Y l m (θ,φ) 7

8 7. [10 pts] Two paticles of mass M 1 and M 2 ae attached to a massless igid od of length d. The od is attached to an axle at its cente-of-mass, and is fee to otate without fiction in the x-y plane. Descibe the Hilbet space of the system and then wite the Hamiltonian. What ae the enegy levels and degeneacies? What ae the wavefunctions of the enegy eigenstates? Only a single angle, φ is equied to specify the state of the system, whee φ is the azimuthal angle, thus the Hilbet space is H (φ). The Hamiltonian is then whee is the moment of inetia. This gives The enegy levels ae then I = 2M H = L2 z 2I ( ) d 2 = Md2 2 2 H = L2 z Md 2 E m = 2 m 2 Md 2 whee m = 0, ±1, ±2, ±3,... The enegy levels all have a degeneacy of 2, except fo E 0, which is not degeneate. The wavefunctions ae given by φ m = eimφ 2π 8

9 8. [10 pts] Fo a two-paticle system, the tansfomation to elative and cente-of-mass coodinates is defined by R = R 1 R 2 The coesponding momenta ae defined by R CM = m 1 R 1 + m 2 R2 m 1 + m 2 P = µ d dt R P CM = M d dt R CM whee µ = m 1 m 2 /M is the educed mass, and M = m 1 + m 2 is the total mass. Invet these expessions to wite R 1, R 2, P 1, and P 2 in tems of R, R CM, P, and P CM. The solutions ae R 1 = R CM + m 2 M R R 2 = R CM m 1 M R Witing P and P CM in tems of P 1 and P 2 gives P = m 2 P 1 m 1 P2 m 1 + m 2 P CM = P 1 + P 2 Inveting this gives P 1 = m 1 M P CM + P P 2 = m 2 M P CM P 9