Homework 6 Problems Solutions - Quantum Mechanics I | PHYS 5250, Assignments of Quantum Mechanics

Download Homework 6 Problems Solutions - Quantum Mechanics I | PHYS 5250 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics - 1: HW 6 Solutions Leo Radzihovsky Paul Martens November 12, 2007 1 Problem 1 Show that H ( ~E1 · ~E2 ) commutes with ~L, where ~E1,2 are vector operators. The definition of a vector operator.[ E (1 or 2) i , Lj ] = ı~ijkEk (1) (transforms like a vector under rotations)[ E (1) i E (2) i , Lj ] = E(1)i [ E (2) i , Lj ] + [ E (1) i , Lj ] E (2) i (2) = ı~ijk ( E (1) i E (2) k + E (1) k E (2) i ) (3) = 0 (4) Because ~E1 · ~E2 commute with ~L all functions of ~E1 · ~E2 commute with ~L[ H ( ~E1 · ~E2 ) , ~L ] = 0 (5) 2 Problem 2 Show that Lz = −ı~ (x∂y − y∂x) = −ı~∂φ relating coordinates x = r sin θ cosφ (6) y = r sin θ sinφ (7) z = r cos θ (8) and derivatives ∂y = ∂r ∂y ∂r + ∂θ ∂y ∂θ + ∂φ ∂y ∂φ (9) ∂x = ∂r ∂x ∂r + ∂θ ∂x ∂θ + ∂φ ∂x ∂φ (10) 1 r = √ x2 + y2 + z2 (11) z r = cos θ (12) calculating derivatives ∂r ∂x = x r = sin θ cosφ (13) ∂r ∂y = y r = sin θ sinφ (14) θ derivatives can be taken easily by making use of the chain rule and equation (12) ∂ ∂x z r = − z r2 ∂r ∂x = − sin θ ∂θ ∂x = − sin θ ∂θ ∂x (15) solving for ∂θ∂x . ∂θ ∂x = 1 r cos θ cosφ (16) the calculation for ∂θ∂y is almost identical ∂θ ∂y = 1 r cos θ sinφ (17) finding φ derivatives y x = tanφ (18) taking derivatives of both sides − tanφ x = sec2 φ ∂φ ∂x (19) solving for ∂φ∂x . ∂φ ∂x = − sinφ r sin θ (20) a similar argument can be used for y ∂φ ∂y = cosφ r sin θ (21) Putting all of it together Lz −ı~ = x∂y − y∂x = r sin θ cosφ (∂yr∂r + ∂yθ∂θ + ∂yφ∂φ) −r sin θ sinφ (∂xr∂r + ∂xθ∂θ + ∂xφ∂φ) (22) Lz = −ı~∂φ (23) 2 5.2 b: ψ (x, y, z) = A (xy + yz + xz) e− r r0 (46) use the result from problem 4: xy = −ı 4 √ 32π 15 ( Y 22 − Y −22 ) r2 (47) yz = ı 2 √ 8π 15 ( Y 12 + Y −1 2 ) r2 (48) xz = −1 2 √ 8π 15 ( Y 12 − Y −12 ) r2 (49) (50) therefore ψ (x, y, z) = −A 2 e− r r0 √ 8π 15 [ ıY 22 + (1− ı)Y 12 − (1 + ı)Y −12 − ıY −2 2 ] r2 (51) via orthogonality of the Y m` ’s C2,2 = Nı ⇒ P2,2 = N2 = 1 6 (52) C2,1 = N (1− ı) ⇒ P2,1 = 2N2 = 1 3 (53) P2,0 = 0 (54) C2,−1 = N (1 + ı) ⇒ P2−1 = 2N2 = 1 3 (55) C2,−2 = −ıN ⇒ P2,−2 = N2 = 1 6 (56) all other coefficients are 0. 6 Problem 6 Rotation of ` = 1 (vector) eigenstates. θxx̂ rotation: x→ x (57) y → y cos θx − z sin θx (58) z → z cos θx + y sin θx (59) (60) ψ (x, y, z) → ψ (x, y cos θx − z sin θx, z cos θx + y sin θx) (61) check on ψ = Aze− r2 a2 → A (z cos θx − y sin θx) e− r2 a2 by working in Y m` repre- sentation. ψ = Ae −r2 a2 z = A √ 4π 3 re− r2 a2 Y 01 (62) 5 first we write this as a vector ψ = 01 0  (63) then we rotate using the rotation matrix e−ıθxLx/~ = (cos θx − 1) ( Lx ~ )2 − ı sin θx ( Lx ~ ) + I (64) because L3x = Lx. In matrix form e−ıθxLx/~ = (cos θx − 1)  0 1√ 2 0 1√ 2 0 1√ 2 0 1√ 2 0  2 − ı sin θx  0 1√ 2 0 1√ 2 0 1√ 2 0 1√ 2 0  (65) As a check you should convince yourself that x→ x here. eıθxLx/~ 01 0  = 01 0 − ı sin θx  1√20 1√ 2 + (cos θx − 1) 01 0  (66) = cos θxY 01 − ı sin θx 1√ 2 ( Y 11 + Y − 1 1 ) (67) = 1 r √ 3 4π (cos θxz − sin θxy) (68) therefore ψ → A √ 4π 3 re− r2 a2 1 r √ 3 4π (cos θxz − sin θxy) = A (cos θxz − sin θxy) e− r2 a2 (69) 7 Problem 7 Harmonic Oscillator 7.1 a: 2D in polar H = p2 2µ + 1 2 µω2ρ2 (70) 7.1.1 i: [H,Lz] One can just use answer to problem 1. WE do it here explicitly.[ p2x + p 2 y, xpy − ypx ] = [ p2x, x ] py + [ p2y, x ] py − [ p2x, y ] px − [ p2y, y ] px (71) = 2px [px, x] py − 2py [py, y] px (72) = −2ı~ (pxpy − pypx) = 0 (73) 6 therefore [H,Lz] = 0 (74) So eigenfunctions of H are eigenfunctions of Lz. Keep this in mind as we look at the HO Hamiltonian. Hψ = Eψ (75)[ − ~ 2 2µ ∇2 + V (ρ) ] ψ = Eψ (76)[ − ~ 2 2µ ( ∂2 ∂ρ2 + 1 ρ ∂ ∂ρ + 1 ρ2 ∂2 ∂φ2 ) + V (ρ) ] R (ρ) Φ (φ) = ER (ρ) Φ (φ) (77) We extract the angular dependence by taking Φ = Φm, where LzΦm = ~mΦm, with m integers. H in terms of Lz. Note that Φm = eımφ, forms a complete set.[ − ~ 2 2µ ( ∂2ρ + 1 ρ ∂ρ ) + L2z 2µρ2 + V (ρ) ] R (ρ) Φm (φ) = ER (ρ) Φm (φ) (78)[ − ~ 2 2µ ( ∂2ρ + 1 ρ ∂ρ ) + ( 1 2 µω2ρ2 + ~2m2 2µρ2 )] R (ρ) = ER (ρ) (79) 7.1.2 ii: ρ→ 0 limit In the ρ → 0 the effective potential is dominated by a centrifugal barrier, due to angular momentum conservation. (Veff → m 2 ρ2 ) − ~ 2 2µ ( ∂2ρ + 1 ρ ∂ρ − m2 ρ2 ) Rm (ρ) = ERE,m (ρ) ≈ 0 (80) suppose Rm (ρ) = ρα (81) we get a relation for α using eqn (80). α (α− 1) + α−m2 = 0 ⇒ α = ±m (82) We expect our wavefunction to be well behaved at ρ = 0, so we’ll take α = |m|. lim ρ→0 RE,m = Cρ|m| (83) 7.1.3 iii: ρ→∞ limit In this limit we drop the m 2 ρ2 , E, and 1 ρ∂ρ terms.( − ~ 2 2µ ∂2ρ + 1 2 µω2ρ2 ) Rm (ρ) = 0 (84) solution to this is e − ρ 2 2ρ20 , where ρ0 = √ ~ µω (85) 7 m is even/odd when n is even/odd |m| = { n, n− 2, n− 4, · · · , 0 n ∈ even n, n− 2, n− 4, · · · , 1 n ∈ odd (104) This means the degeneracy of En is n + 1. This agrees with our treatment in Cartesian coordinates. e = ~ω nx + ny︸ ︷︷ ︸ n +1  (105) 7.1.9 ix: Eigenfunctions n = 0 ⇒ m = 0, D0 = 1 φ0,0 (ρ, φ) = N0,0e− ρ2 2ρ (106) n = 1 ⇒ m ∈ {1, 0,−1}, D1 = 2 u1 = c0 + c2x2, c2 = 0 (107) φ1,±1 (ρ, φ) = N1,±1e − ρ 2 2ρ0 ( ρ ρ0 ) e±ıφ (108) n = 2 ⇒ m ∈ {2, 0,−2}, D2 = 3 u2,2 (x) = c (2) 0 + c (2) 2 x 2 (109) c (2) 2 = 2 (−3 + 2 + 1 + 0) c (2) 0 = 0 (110) u2,0 (x) = c (0) 0 + c (0) 2 x 2 (111) c (0) 2 = 2 −3 + 1 + 0 2 ∗ 2 c (0) 0 = −c (0) 0 (112) φ2,±2 (ρ, φ) = N2,±2e − ρ2ρ0 ( ρ ρ0 )2 e±2ıφ (113) φ2,0 (ρφ) = N2,0e − ρ 2 2ρ0 ( 1− ρ 2 ρ20 ) (114) 7.1.10 x: relating Cartesian and polar coordinate solutions n = 0 state is non-degenerate so it must match the Cartesian form exactly φ0,0 ∝ e− ρ 2ρ0 = e− x2 2ρ0 e− y2 2ρ0 (115) 10 n = 1 states are two-fold degenerate so the a Cartesian state must just be some linear combination of the old states φ1,±1 ∝ ρe − ρ 2 2ρ20 e±ıφ (116) ∝ (x± ıy) e − ρ 2 2ρ20 (117) ∝ ( H1 (x) e−x 2 2ρ0 )( H0 (y) e − y 2 2ρ0 ) ± ı ( H0 (x) e − x2 2ρ20 )( H1 (y) e − y 2 2ρ20 ) (118) φ2,±2 ∝ ρ2 cos 2φ︸ ︷︷ ︸ x2−y2 ±ı ρ2 sin 2φ︸ ︷︷ ︸ 2xy  e− ρ22ρ20 (119) ∝ H2 (x) e − x2 2ρ20H0 (y) e − y 2 2ρ20 −H0 (x) e − x2 2ρ20H2 (y) e − y 2 2ρ20 ±ı2H1 (x) e − x2 2ρ20H1 (y) e − y 2 2ρ20 (120) φ2,0 ∝ ( 1− x 2 + y2 ρ20 ) e − ρ 2 2ρ20 (121) ∝ H2 (x) e − x2 2ρ20H0 (y) e − y 2 2ρ20 +H0 (x) e − x2 ρ20 H2 (y) e − y 2 2ρ20 (122) Parity from before was (−1)n. Eince odd/even m corresponds to odd/even n, parity in the polar representation is (−1)m. 7.2 b: Rotating system If we rotate the system with frequency Ω, then H → HΩ = p2 2µ + 1 2 µω2ρ2 − ΩLz (123) En,m = ~ω (n+ 1)− ~Ωm (124) For small Ω < ω, we get a small splitting of previously degenerate energy levels. For Ω ≥ ω, the system becomes unstable as a result of centrifugal kinetic energy dominating the potential. Veff = 1 2 mω2r2 − 1 2 mΩ2r2 (125) 11 7.3 c: 3D Oscillator H = p2 2µ + 1 2 µω2r2 (126) The eigenfunctions have the form φE,`,m = UE,` r Y m` (127) En = ~ω ( n+ 3 2 ) (128) These states are degenerate. Each n, ` has 2` + 1 m-states. For a given n the allowed values of ` are given by ` = n− 2k, where k ∈ Z (129) 7.3.1 i: degeneracy Dn = n/2∑ k=0 n∑ `=0 ∑̀ m=−`︸ ︷︷ ︸ 2`+1 δ`,n−2k (130) = n/2∑ k=0 [2 (n− 2k) + 2] (131) = n/2∑ k=0 (2n+ 1)− 4 n/2∑ k=0 k (132) = (2n+ 1) (n 2 + 1 ) − n (n 2 + 1 ) = (n+ 1) (n 2 + 1 ) (133) Dn = (n+ 1) (n+ 2) 2 (134) 7.3.2 ii: eigenfunctions ψ0,0,0 (r, θ, φ) = N0,0e − r2 2r20 = e − x2 2r20 e − y 2 2r20 e − z2 2r20 (135) 12