Answers to Homework #7 - Introduction to Quantum Mechanics 1 | PHYS 5250, Assignments of Quantum Mechanics

Download Answers to Homework #7 - Introduction to Quantum Mechanics 1 | PHYS 5250 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics - 1: HW 7 Solutions Leo Radzihovsky Paul Martens November 28, 2007 1 Problem 1 3D Spherical square-well potential. Suppose V (r) = { −V0 r < d 0 r > d (1) 1.1 a: Boundstates For these states E < 0. Define: Rn` (r) = Un` (r) r (2) The radial equation for U − ~ 2 2m d2Un` (r) dr2 + ( V (r) + ~ 2` (` + 1) 2mr2 ) Un` = En`Un` (3) 1.1.1 i: spectrum for general ` Solve for internal and external separately then match boundary conditions. 0 < r < d ⇒ V (r) = −V0 ⇒ free schr. solutions with E → E + V0 d2R dr2 + 2 r dR dr + ( κ2 − ` (` + 1) r2 ) R = 0 (4) where κ ≡ √ 2m (V0 − |E|) ~2 (5) we require a regular solution at r = 0; therefore Rn` (r) = Aj` (κr) (6) for the external solution d2R dr2 + 2 r dR dr + ( −α2 − ` (` + 1) r2 ) R = 0 (7) 1 where α ≡ √ 2m |E| ~2 (8) outer solution is R (r) = Bj` (ıαr) + Cn` (ıαr) (9) we need a decaying solution, which is a spherical Hankel function of the first kind. h (1) ` ≡ j` + ın`. Matching boundary conditions leads to a discrete spec- trum. κ j′` (κd) j` (κd) = ıα h (1)′ ` (ıαd) h` (ıαd) (10) 1.1.2 ii: ` = 0 case { U ′′ + κ2U = 0 0 < r ≤ d U ′′ − α2U = 0 d < r (11) Solution is U (r) = { A sin κr 0 < r ≤ d Be−αr d < r (12) BC B A = eαd sinκd (13) The wavefunction is ΨE0 (r) = { A r sin κr 0 < r ≤ d B r e −αr d < r (14) spectrum is found through the relation κ cosκd sinκd = −α = − √ 2m ~2 V0 − κ2 (15) κ2 = ( κ2 + α2 ) ︸ ︷︷ ︸ 2m ~2 V0 sin2 κd (16) sinκd = ± √ ~2 2md2V0 κd (17) 2 therefore C B = − tan δ` (k) (32) 1.2.3 iii For ` = 0 R0 (r) = { 1 r (B sin kr − C cos kr) r > d A r sin kr 0 ≤ r ≤ d (33) ¿From Continuity BC’s A sinκd = B sin kd − C cos kd (34) Aκ cosκd = Bk cos kd + Ck sin kd (35) κ cotκd = k B cos kd + C sin kd B sin kd − C cos kd (36) κ k cotκd = cot kd − CB 1 − CB cot kd (37) κ k cotκd − cot kd = C B (κ k cot κd cot kd − 1 ) (38) C B = κ k cot κd − cot kd κ k cotκd cot kd − 1 (39) C B = − tan δ0 (k) = tan kd − kκ tan κd 1 − kκ tan kd tanκd (40) define κ0 ≡ √ emV0~ 2 (41) x ≡ kd (42) x0 ≡ κ0d (43) in terms of these new parameters κ = κ0 √ 1 + ( k κ0 )2 cot δ0 = x tan x tan " x0 r 1+ “ x x0 ”2 # x0 r 1+ “ x x0 ”2 − 1 x   1 + 13x 2 − 1x0 tan " x0 r 1− “ x x0 ”2 # r 1+ “ x x0 ”2    (44) one can expand this in terms of k k cot δ0 ≈ − 1 a + 1 2 r0k 2 (45) 5 a ≡ ( 1 − tan κ0d κ0d ) d (46) r0 ≡ 2d ( 1 κ0d tanκ0d + 1/3− 1 2κ20d 2 + tan κ0d 2(κ0d) 3/2 − tan 2 κ0d 2κ20d 2 1− tan κ0dκ0d ) 1 − tan κ0dκ0d (47) a = d ( 1 − tanκ0d κ0d ) (48) r0 = 2d    1 κ0d tan κ0d 1 − tanκ0d + 1 3 − 12κ20d2 + tan κ0d 2(κ0d) 3/2 − tan 2 κ0d 2κ20d 2 ( 1 − tan κ0dκ0d )2    (49) 0.5 1 1.5 2 2.5 3 Κ0d -10 -7.5 -5 -2.5 2.5 5 7.5 10 a €€€€ d Diverging Scattering Length Note that: (1): scattering length a diverges (changing sign from negative to positive) at κ0d = π 2 , which implies 2mV ∗0 d 2 ~2 = π2 4 ⇒ V ∗0 = π2~2 8md2 = V0C (50) ie. a → ∞ when V0 → V −0C as a bound state comes in (2): r0 is smooth and finite near V0C . 1.2.4 iv for V0 → ∞, κ0d → ∞. a = d ( 1 − tan κ0d κ0d ) → d (51) 6 1 1.25 1.5 1.75 2 2.25 Κ0d -20 -15 -10 -5 r0€€€€€€€ d Well Behaved Effective Range 1.2.5 v κ j′` (κd) j` (κd) = k j′ell (kd) + C B n ′ ` (kd) j` (kd) + C B n` (kd) (52) κ j′` (κd) j` (κd) j` (kd) + κ j′` (κd) j` (κd) n` (kd) C B = k ( j′` (kd) + C B n′` (kd) ) (53) ⇒ C B ( κ j` (κd) j` (κd) n` (kd) − kn′` (kd) ) = kj′` (kd) − κ j′` (κd) j` (κd) j` (kd) (54) in the limit kd → ∞ n` (kd) → − (kd) ` (2` + 1)!! (55) j` (kd) → (kd) ` (2` + 1)!! (56) ⇒ C B ( −κ0 j` (κ0d) j` (κ0d) (kd)−(`+1) (2` + 1)!! − k (` + 1) (kd) −(`+2) (2` + 1)!! ) (57) ≈ ` (2` + 1)!! k (kd) `−1 − κ0 j′` (κ0d) j` (κd) (kd)` (2` + 1)!! (58) solving the above expression for CB . C B = − (kd)2`+1 ` d − κ0 j′`(κ0d) j`(κ0d) `+1 d + κ0 j′ ` (κ0d) j`(κ0d) (59) 7 consider a spin- 12 particle. (j = 1 2 and m ∈ { − 12 , 12 } ). Then 〈jm |Jx| jm′〉 ≡ Jxmm′ (81) = ~ 2 (√ 3 4 + 1 4 δm,m−1 + √ 3 4 + 1 4 δm,m+1 ) (82) 〈jm |Jx| jm′〉 ≡ ~ 2 σx (83) likewise Jym,m′ = ~ 2ı (δm,m−1 − δm,m+1) ≡ ~ 2 σy (84) Jzm,m′ = ~mδm,m′ ≡ ~ 2 σz (85) Therefore σx = ( 0 1 1 0 ) σy = ( 0 −ı ı 0 ) (86) σz = ( 1 0 0 −1 ) 3.2 b [σi, σj ] = 2ıijkσk (87) {σi, σj} = 2δij (88) 3.2.1 i 2σiσj = [σi, σj ] + {σi, σj} (89) = 2δij + 2ıijkσk (90) Tr [σiσj ] = 2δij (91) 3.2.2 ii (~a · ~σ) ( ~b · ~σ ) = aibjσiσj (92) = aibj (δij + ıijkσk) (93) (~a · ~σ) ( ~b · ~σ ) = ~a ·~b + ı ( ~a ×~b ) · ~σ (94) 3.3 c n̂ · ~σ = nxσx + nyσy + nzσz (95) n̂ · ~σ = ( cos θ sin θe−ıφ sin θeıφ − cos θ ) (96) 10 3.3.1 i: eigen values and vectors ∣ ∣ ∣ ∣ cos θ − λ sin θe−ıφ sin θeıφ − (cos θ + λ) ∣ ∣ ∣ ∣ = 0 (97) ⇒ λ = ±1 (98) Finding eigen vectors (cos θ − 1) α+ + sin θe−ıφβ+ = 0 (99) ( −2 sin2 θ 2 ) α+ + 2 sin θ 2 cos θ 2 e−ıφβ+ = 0 (100) ⇒ Ψn̂+ = ( cos θ2 eıφ sin θ2 ) (101) (cos θ + 1) α− + sin θe −φβ− = 0 (102) ( 2 cos2 θ 2 ) α− + 2 sin θ 2 cos θ 2 e−ıφβ− = 0 (103) ⇒ Ψn̂− = ( sin θ2 −eıφ cos θ2 ) (104) 3.3.2 ii: rotating e−ı θ 2 θ̂·~σ = cos θ 2 − ıθ̂ · ~σ sin θ 2 (105) =   cos θ2 − ıθ̂z sin θ2 ( −ıθ̂x − θ̂y ) sin θ2 ( −ıθ̂x + θ̂y ) sin θ2 cos θ 2 + ıθ̂z sin θ 2   (106) θ̂ = ẑ × n̂ |ẑ × n̂| = 1 √ n2x + n 2 y (−nyx̂ + nyŷ) (107) where n̂ ≡ x̂ sin θ cosφ + ŷ sin θ sin φ + ẑ cos θ. θ̂ = − sin φx̂ + cosφŷ (108) so e−ı θ 2 θ̂·~σ = ( cos θ2 −e−ıφ sin θ2 eıφ sin θ2 cos θ 2 ) (109) Eigenvectors through rotation Ψn̂+ = e −ı θ2 θ̂·~σΨẑ+ = ( cos θ2 eıθ sin θ2 ) (110) 11 Ψn̂− = e −ı θ2 θ̂·~σ = ( − sin θ2e−ıφ cos θ2 ) (111) In operator language (n̂ · ~σ) Un̂U †n̂Ψn̂± = λ±Ψn̂± (112) U †n̂ (n̂ · ~σ) Un̂ ︸ ︷︷ ︸ σz ( U †n̂Ψ n̂ ± ) = λ± ( U †n̂Ψ n̂ ± ) (113) σz ( U †n̂Ψ n̂ ± ) = ± ( U †n̂Ψ n̂ ± ) (114) eigenvectors Ψn̂+ = Un̂ ( 1 0 ) (115) Ψn̂− = Un̂ ( 0 1 ) (116) 3.4 d: Eigenvalues and vectors of S x + S z for S = ~ 2 suppose O = Sx + Sz = ~ 2 (σx + σz) = ~ 2 ( 1 1 1 −1 ) (117) Eigenvalues from the characteristic equation −1 + λ2 − 1 = 0 ⇒ λ± = ± √ 2 (118) where n̂ = 1√ 2 (x̂ + ẑ) (119) or φ = 0, θ = π4 . We calculate the eigenvectors directly ( 1 − √ 2 ) α + β = 0 (120) Ψ+ = N ( −1 1 − √ 2 ) (121) Ψ− = N ( 1 − √ 2 1 ) (122) where N = 1√ 1+1+2−2 √ 2 = 1√ 4−2 √ 2 . This agrees with the result we get by rotating the eigenvectors of σz . Ψ+ = ( cos π8 sin π8 ) = (√ (cosπ4 + 1) /2 √ ( 1 − cos π2 ) /2 ) = 1 2 1 √ 1 − 1/ √ 2 ( 1 − ( 1 − √ 2 ) ) (123) 12 and by using cos θ = Bz √ B2x + B 2 z (150) sin θ = Bx √ B2x + B 2 z (151) Ψ+ (t) = ( cos2 θ2e −ıω0t/2 + sin2 θ2e ıω0t/2 cos θ2 sin θ 2 ( e−ıω0t/2 − eıω0t/2 ) ) (152) = ( 1 2 (cos θ + 1) (−2ı) sin ω0t2 + eıω0t/2 −ı sin θ sin ω0t/2 ) (153) Ψ+ (t) =   cos ω0t2 − ı Bz√ B2x+B 2 z sin ω0t2 −ı Bx√ B2x+B 2 z sin ω0t2   (154) 3.6 f The Maximum eigen value of S2tot = ~ 2Stot (Stot + 1). Where Stot = 1 2 + 1 2 + · · · + 1 2 ︸ ︷︷ ︸ N = N2 . In ket form ∣ ∣ ∣ ∣ N 2 , N 2 〉 = ∣ ∣ ∣ ∣ 1 2 1 2 〉 ∣ ∣ ∣ ∣ 1 2 1 2 〉 · · · ∣ ∣ ∣ ∣ 1 2 1 2 〉 (155) which is symmetric. The lowering operator S−tot = S (x) tot − ıS (y) tot (156) = ( S−1 + S − 2 + S − 3 + · · · + S−N ) (157) is also symmetric, which means that ( S−tot )n ∣ ∣ ∣ ∣ N 2 N 2 〉 = ∣ ∣ ∣ ∣ N 2 , N 2 − n 〉 (158) must also be symmetric. 3.7 g Hd = 1 r3 [~µ1 · ~µ2 − 3 (~µ1 · r̂) (~µ2 · r̂)] (159) we choose ẑ to be in the same direction as r̂. Hd = − ( gµPB )2 ~2a3 [ ~S1 · ~S2 − 3S1zS2z ] (160) 15 ( ~S1 + ~S2 )2 = S21 + S 2 2 + 2 ~S1 · ~S22 (161) S2 = ~2S (S + 1) = ~2S1 (S1 + 1) + ~ 2S2 (S2 + 1) + 2~S1~S2 (162) (S1z + S2z) 2 = S21z + S 2 2z + 2S1zS2z = S 2 z (163) so Hd = − ( gµPB )2 a3 [ 1 2 S (S + 1) − 1 2 S1 (S1 + 1) − 1 2 S2 (S2 + 1) − 3 2 ( S2z − S21z − S22z ) ] (164) Hd = − ( gµPB )2 2a3 [ S (S + 1) − 3 2 − 3 ( S2z − 1 2 )] (165) Hd |S, Sz; S1, S2〉ES,Sz |S, Sz; S1, S2〉 (166) S = 0; 1 Sz = 0; 0,±1 { ⇒ ES,Sz = −2 µ (P )2 B a3 ( S (S + 1) − s2z ) (167) Energy levels are E0,0 = 0 (168) E1,0 = − 4µ (P )2 B a3 (169) E1,±1 = 2µ (P )2 B a3 (170) 4 Problem 4: 3D charge particle in constant B- field 4.1 a H = ( ~p − q ~A c )2 /2m (171) ~A = 1 2 ~B × ~r =   −By2 Bx 2 0   (172) we use the the radiation gauge ~∇ · A = 0, in which case ~p · ~A = ~A · ~p. H = p2 2m − q 2mc ( ~B × ~r) · ~p ︸ ︷︷ ︸ ~r×~p·~B + q2B2 8mc2 ( x2 + y2 ) (173) = P 2perp 2m − q 2mc BLz + q2B2 8mc2 r2perp − p2z 2m (174) 16 H |kz〉 = ( p2perp 2m + q2B2 8mc2 r2perp − qB 2mc Lz + ~ 2k2z 2m ) |kz〉 = E |kz〉 (175) This is an effective 2D H.O. with an extra − qB2mcLz term. H2D |E2D〉 = E2D |E2D〉 (176) E2D + ~ 2k2z 2m = E (177) H2D = p2perp 2m + mω20 2 r2perp − ΩLz (178) with ω0 = qB 2mc (179) Ω = qB 2mc (180) 4.2 b This has energy spectrum En,m = ~ω0 (n + 1) (181) note that when 4ω0 = Ω = 1 2ωB that the Landau levels are spaced ~ωB apart. This agrees with our result done in Cartesian coordinates. 4.3 c Compare − qB2mcLz and diamagnetic q2B2 8mc2 r 2 perp for an e − in an atom. R = q2B2 8mc2 〈 r2perp 〉 qB 2mc 〈LZ〉 = qBa20 4~c (182) become comparable for R ≈ 1. B∗ = ~c ea20 (183) R = e2 4~c B e/a20 = B ∗ ( 0.5 ∗ 10−8cm )2 4 ∗ 137 ∗ 4.8 ∗ 10−10esu (184) R ≈ B 9 ∗ 109Guass (185) B∗ ∼ 9 × 109Gauss (186) under typical conditions B  B∗, which means diamagnetic term is negligible in atomic physics. 17