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PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 11 Topics Covered: Orbital angular momentum, center-of-mass coordinates Some Key Concepts: angular degrees of freedom, spherical harmonics 1. [20 pts] In order to derive the properties of the spherical harmonics, we need to determine the action d of the angular momentum operator in spherical coordinates. Just as we have hx|Px |ψi = −i~ dx hx|ψi, ~ ~ =R ~ × P~ and our knowledge of momentum we should find a similar expression for hrθφ|L|ψi. From L operators, it follows that d d d d d d ~ −z −x + ~ey z + ~ez x − y hrθφ|ψi. hrθφ|L|ψi = −ı~ ~ex y dz dy dx dz dy dx Cartesian coordinates are related to spherical coordinates via the transformations x = r sin θ cos φ y = r sin θ sin φ z = r cos θ and the inverse transformations p x2 + y 2 + z 2 p x2 + y 2 θ = arctan( ) z y φ = arctan( ). x Their derivatives can be related via expansions such as r= ∂x = ∂θ ∂φ ∂r ∂r + ∂θ + ∂φ . ∂x ∂x ∂x Using these relations, and similar expressions for ∂y and ∂z , find expressions for hrθφ|Lx |ψi, hrθφ|Ly |ψi, and hrθφ|Lz |ψi, involving only spherical coordinates and their derivatives. ∂x r = ∂x θ = x r = sin θ cos φ z2 √ x = cos θrcos φ r 2 z x2 +y 2 2 ∂x φ = − x2x+y2 xy2 = − csc θrsin φ So d dx = sin θ cos φ ∂r + cos θ cos φ ∂θ r ∂y r = ∂y θ = y r = sin θ sin φ z2 √ y = cos θrsin φ r 2 z x2 +y 2 ∂y φ = x2 1 x2 +y 2 x So d dy ∂z r = = csc θ cos φ r sin θ sin φ∂r + cos θrsin φ ∂θ z r = cos θ − csc θ sin φ ∂φ r = + csc θ cos φ ∂φ r 1 √ 2 x2 +y 2 = − sinr θ ∂z θ = − zr2 z 2 ∂z φ = 0 d So dz = cos θ∂r − sinr θ ∂θ Now hrθφ|Lx |ψi = −i~(y d d − z )hrθφ|ψi dz dy So we can say d d −z Lx = i~ y dz dy = −i~ r sin θ cos θ sin φ∂r − sin2 θ sin φ∂θ − r sin θ cos θ sin φ∂r − cos2 θ sin φ∂θ + cot θ cos φ∂φ Which means hrθφ|Lx |ψi = −i~ (− sin φ∂θ − cot θ cos φ∂φ ) hrθφ|ψi Similarly we can say d d −x ) dx dz = −i~ r sin θ cos θ cos φ∂r + cos2 θ cos φ∂θ − cot θ sin φ∂φ − r sin θ cos θ cos φ∂r + sin2 θ cos φ∂θ Ly = −i~(z so that hrθφ|Ly |ψi = −i~ (cos φ∂θ − cot θ sin φ∂φ ) hrθφ|ψi Lastly, we have d d −y ) dy dx 2 = −i~ r sin θ sin φ cos φ∂r + sin θ cos θ sin φ cos φ∂θ + cos2 φ∂φ +r sin2 θ sin φ cos φ∂r − sin θ cos θ sin φ cos φ∂θ + sin2 φ∂φ Lz = −i~(x so that hrθφ|Lz |ψi = −i~∂φ hrθφ|ψi 2 2. [15pts] From your previous answer and the definition L2 = L2x + L2y + L2z , prove that hrθφ|L2 |ψi = −~2 ∂2 1 ∂ ∂ 1 sin θ + 2 sin θ ∂θ ∂θ sin θ ∂ 2 φ2 hrθφ|ψi. L2 = −~2 [(sin φ∂θ + cot θ cos φ∂φ )(sin φ∂θ + cot θ cos φ∂φ ) + (cos φ∂θ − cot θ sin φ∂φ )(cos φ∂θ − cot θ sin φ∂φ ) + ∂φ2 = −~2 sin2 φ∂θ2 + cot θ sin φ cos φ∂θ ∂φ − csc2 θ sin φ cos φ∂φ + cot θ sin φ cos φ∂φ ∂θ + cot θ cos2 φ∂θ + cot2 θ cos2 φ∂φ2 − cot2 θ cos φ sin φ∂φ + cos2 φ∂θ2 − cot θ cos φ sin φ∂θ ∂φ + csc2 θ sin φ cos φ∂φ − cot θ sin φ cos φ∂φ ∂θ + cot θ sin2 φ∂θ + cot2 θ sin2 θ∂φ2 + cot2 θ sin φ cos φ∂φ + ∂φ2 = −~2 ∂θ2 + cot θ∂θ + (1 + cot2 θ)∂φ2 Noting that 1 ∂θ sin θ∂θ = ∂θ2 + cot θ∂θ sin θ and 1 + cot2 θ = the proof is complete. 3 1 sin2 θ 3. [10 pts] We can factorize the Hilbert space of a 3-D particle into radial and angular Hilbert spaces, H(3) = H(r) ⊗ H(Ω) . Two alternate basis sets that both span H(Ω) are {|θφi} and {|ℓmi}. As the angular momentum operator lives entirely in H(Ω) , we can use our results from problem 11.1 to derive an expression for hθφ|Lz |ℓmi. Combine this with the formula Lz |ℓmi = ~m|ℓmi, to derive and then solve a differential equation for the φ-dependence of hθφ|ℓmi. Your solution should give hθφ|ℓmi in terms of the as of yet unspecified initial condition hθ|ℓmi ≡ hθ, φ|ℓmi . What restrictions does this φ=0 solution impose on the quantum number m, which describes the z-component of the orbital angular momentum? Since mmax = ℓ, what restrictions are then placed on the total angular momentum quantum number ℓ? hθφ|Lz |ℓmi = −i~ ∂ hθφ|ℓmi ∂φ and hθφ|Lz |ℓmi = ~mhθφ|ℓmi Thus −i~ ∂ hθφ|ℓmi = ~mhθφ|ℓmi ∂φ The solution to this simple first-order differential equation is hθφ|ℓmi = hθ0|ℓmieimφ Since the wave-function must be single valued, we require m to be a whole integer. As mmax = ℓ, this implies that ℓ must be a whole integer also. 4 4. [10 pts] Using L± = Lx ± iLy we can use the relation L+ |ℓ, ℓi = 0 and the expressions from problem 11.1 to write a differential equation for hθφ|ℓℓi. Plug in your solution from 11.3 for the φ-dependence, and show that the solution is hθφ|ℓℓi = cℓ eiℓφ sinℓ (θ). Determine the value of the normalization coefficient cℓ by performing the necessary integral. We have hθφ|L+ |ℓℓi = 0 This implies hθφ|Lx |ℓℓi + ihθφ|Ly |ℓℓi = 0 Using the expressions from 11.1 gives (− sin θ∂θ − cot θ cos φ∂φ + i cos φ∂θ − i cot θ cos φ∂φ )hθφ|ℓℓi = 0 This simplifies to (i(cos φ + i sin φ)∂θ − cot θ(cos φ + i sin φ)∂φ )hθφ|ℓℓi = 0 Factoring out the eiφ gives (i∂θ − cot θ∂φ )hθφ|ℓℓi = 0 Plugging in the solution from 11.3 gives (i∂θ − iℓ cot θ)hθ0|ℓℓie−iℓφ = 0 which reduces to ∂θ hθ0|ℓℓi = ℓ cot θhθ0|ℓℓi Now ∂θ sinℓ θ = ℓ sinℓ−1 θ cos θ = ℓ cot θ sinℓ θ So the solution is hθφ|ℓℓi = cℓ eiℓφ sinℓ θ Rπ R 2π The normalization integral is 0 sin θdθ 0 dφ |hθφ|ℓℓi|2 = 1 Rπ R 2π 2ℓ With our solution this becomes |cℓ |2 0 sin θdθ 0 dφ sin θ = 1 R π Performing the phi integral gives 2π|cℓ |2 0 sin θdθ sin2ℓ θ = 1 R1 u-substitution with u = cos θ gives 2π|cℓ |2 −1 du (1 − u2 )ℓ = 1 R1 Since the integrand is even, this reduces to 4π|cℓ |2 0 du (1 − u2 )ℓ = 1 From Mathematic we get 2π|cℓ |2 Γ[1/2]Γ[ℓ+1] Γ[ℓ+3/2] = 1 q Γ(ℓ+3/2) which gives cℓ = 2πΓ[1/2]Γ[ℓ+1] Thus we have s Γ(ℓ + 3/2) sinℓ θeiℓφ 2πΓ[1/2]Γ[ℓ + 1] q 1 3iφ 35 3 For the special case ℓ = 3 this gives hθφ|33i = 8 e π sin θ, which agrees with the spherical hθφ|ℓℓi = harmonic Y33 (θ, φ) up to a non-physical phase factor. 5 p 5. [10 pts] Using L− |ℓmi = ~ ℓ(ℓ + 1) − m(m − 1)|ℓ, m − 1i together with your previous answers to derive an expression for hθφ|ℓ, m − 1i in terms of hθφ|ℓmi. Explain how in principle you can now recursively calculate the value of the spherical harmonic Yℓm (θφ) ≡ hθφ|ℓmi for any θ and φ and for any ℓ and m. Follow your procedure to derive properly normalized expressions for spherical harmonics for the case ℓ = 1, m = −1, 0, 1. To construct the other m states for the same ℓ, we can begin from the expression p hθφ|L− |ℓmi = ~ (ℓ + m)(ℓ − m + 1)hθφ|ℓ, m − 1i Now hθφ|L− |ℓmi = hθφ|Lx |ℓmi − ihθφ|Ly |ℓmi = −i~(− sin φ∂θ − cot θ cos φ∂φ )hθφ|ℓmi − ~(cos φ∂θ − cot θ sin φ∂φ )hθφ|ℓmi = ~e−iφ (−∂θ + i cot θ∂φ )hθφ|ℓmi Putting the pieces together gives e−iφ (−∂θ + i cot θ∂φ ) hθφ|ℓmi hθφ|ℓ, m − 1i = p (ℓ + m)(ℓ − m + 1) Starting from our expression for hθφ|ℓℓi, we can find hθφ|ℓ, ℓ − 1i by applying the above differential formula. Successive iterations will then generate all the remaining hθφ|ℓmi states. 6 6. [10 pts] A particle of mass M is constrained to move on a spherical surface of radius a. Does the system live in H(3) , H(r) , or H(Ω) ? What is the Hamiltonian? What are the energy levels and degeneracies? What are the wavefunctions of the energy eigenstates? Because the radial motion is constrained to a fixed value, it is only necessary to consider the dynamics in H(Ω) . The Hamiltonian is then H= L2 2M a2 Choosing simultaneous eigenstates of L2 and Lz , we have H|ℓ, mi = ~2 ℓ(ℓ + 1) |ℓ, mi 2M a2 so that Eℓ = ~2 ℓ(ℓ + 1) 2M a2 and dℓ = 2ℓ + 1 The wavefunctions are the spherical harmonics hθφ|ℓ, mi = Ymℓ (θ, φ) 7 7. [10 pts] Two particles of mass M1 and M2 are attached to a massless rigid rod of length d. The rod is attached to an axle at its center-of-mass, and is free to rotate without friction in the x-y plane. Describe the Hilbert space of the system and then write the Hamiltonian. What are the energy levels and degeneracies? What are the wavefunctions of the energy eigenstates? Only a single angle, φ is required to specify the state of the system, where φ is the azimuthal angle, thus the Hilbert space is H(φ) . The Hamiltonian is then H= where L2z 2I 2 M d2 d = I = 2M 2 2 is the moment of inertia. This gives H= L2z M d2 Em = ~2 m2 M d2 The energy levels are then where m = 0, ±1, ±2, ±3, . . . The energy levels all have a degeneracy of 2, except for E0 , which is not degenerate. The wavefunctions are given by eimφ hφ|mi = √ 2π 8 8. [10 pts] For a two-particle system, the transformation to relative and center-of-mass coordinates is defined by ~ =R ~1 − R ~2 R ~ ~ ~ CM = m1 R1 + m2 R2 R m1 + m2 The corresponding momenta are defined by d ~ P~ = µ R dt d ~ P~CM = M R CM dt where µ = m1 m2 /M is the reduced mass, and M = m1 + m2 is the total mass. Invert these expres~ 1, R ~ 2 , P~1 , and P~2 in terms of R, ~ R ~ CM , P~ , and P~CM . sions to write R The solutions are ~1 = R ~ CM + m2 R ~ R M ~2 = R ~ CM − m1 R ~ R M Writing P~ and P~CM in terms of P~1 and P~2 gives m2 P~1 − m1 P~2 P~ = m1 + m2 P~CM = P~1 + P~2 Inverting this gives m1 ~ PCM + P~ P~1 = M m2 ~ P~2 = PCM − P~ M 9