Homework 3 with Solutions - Quantum Mechanics 1 | PHYS 5260, Assignments of Quantum Mechanics

Download Homework 3 with Solutions - Quantum Mechanics 1 | PHYS 5260 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics II: HW 3 Solutions Leo Radzihovsky February 28, 2008 1 Gaussian electric field pulse on a hydrogen atom We start with the following definitions: i~∂t |Ψ〉 = (H0 + H1 (t)) |Ψ〉 , (1) |Ψ (t)〉 = ∑ n Cne − i ~ E0nt |n〉 , (2) ⇒ ∂tCn = − i ~ ∑ m eiω 0 nmt 〈n|H1 (t) |m〉Cm (t) . (3) Our unperturbed Hamiltonian and eigenenergies are: H0 = p2 2m − e 2 r , E0n = − ERy n2 , (4) and the eigenstates are |n〉 → |n l m〉. The perturbing Hamiltonian is: H1 (t) = e~r · ~E (t) = eEz e−t 2/τ2 . (5) Since we start out in the ground state, our coefficient at t = −∞ is: Cnlm (−∞) = δnlm, 100, (6) and our final states are |2 l m〉, with l = 0, 1 and m running from −l to +l. From the dipole selection rule, we know that the only final state that doesn’t vanish is |2 1 0〉. So we can write for our time-dependent coefficient: C210 = − i ~ eE 〈2 1 0| z |1 0 0〉 ∫ ∞ −∞ dt′ e− t′2 τ2 +iω0t ′ ︸ ︷︷ ︸ √ πτe− 1 4 ω2 0 τ2 , (7) 1 V(x) x α −E cl 0 t<0 t>0 Figure 1: A sketch of the potential and shifted potential. with ω0 = (E210 − E100) /~ = 34~ERy. We find the matrix element in a straightforward manner: 〈2 1 0| z |1 0 0〉 = ∫ ∞ 0 dr r2 ( 1 πa30 ) 1√ 32 r2 a0 e− 3r 2a0 ∫ π 0 dθ sin θ cos2 θ ∫ 2π 0 dφ ︸ ︷︷ ︸ 2π (8) = 2a0√ 32 ∫ ∞ 0 dx x4e− 3 2 x ︸ ︷︷ ︸ ( 23 ) 5 4! ∫ 1 −1 dµ µ2 ︸ ︷︷ ︸ 2 3 (9) = √ 215a20 310 . (10) Finally, we can write our transition probability: P100→210 = |C210|2 = ( eEa0 ~ )2 ( 215 310 ) πτ2e−ω 2 0τ 2/2. (11) Since H1 (t) commutes with ~S, Our answer is independent of whether spin is included, which implies that S and Sz are conserved. So the probability is unchanged. 2 Shifted harmonic oscillator We examine the following Hamiltonian: H = p2 2m + 1 2 mω2x2 − qEx, t < 0. (12) i.e., a harmonic oscillator with a uniform ~E field switched off at t = 0. 2 (b) Adiabatic decay If we shutoff E adiabatically, then our final state is |Ψ (t = 0+)〉 = |0>〉, i.e., |0<〉 evolves slowly into |0>〉. This implies E> adiabatic = ~ω2 . So our energy change in this case is: ∆Eadiabatic = E cl 0 = 1 2 mω2α2. (37) This is half as large as the sudden change, because the adiabatic change leaves the system in a ground state. (c) Classical treatment The classical change is just Ecl0 , i.e., the same as ∆Eadiabatic. 3 Perturbed harmonic oscillator transitions (a) Oscillating electric field We now consider the following Hamiltonian: H = H0 + H1 = p2 2m + 1 2 mω2x2 − qEx cos ωet Θ (t) . (38) The system starts out in the ground state for t = 0−, so our transition coefficient is given by: Cn (t) = − i ~ ∫ t 0 dt′ 〈n|H1 (t′) |0〉 eiωt ′ , (39) with 〈n|H1 (t) |0〉 = −qE cosωet 〈n|x |0〉 ︸ ︷︷ ︸ x0√ 2 〈n|a†+a|0〉 (40) = −qEx0√ 2 δ1, n cosωet. (41) So Cn (t) = iqEx0 2 √ 2~ δ1,n ∫ t 0 dt′ ( ei(ω+ωe)t ′ + ei(ω−ωe)t ′ ) (42) = iqEx0 2 √ 2~ δ1,n [ ei(ω+ωe)t−1 i (ω + ωe) + ei(ω−ωe)t − 1 i (ω − ωe) ] . (43) Thus, our transition probability for a given energy level n is: Pn (t) = |Cn (t)|2 (44) = q2E2x20 8~2 δ1,n [( sin ((ω + ωe) t/2) (ω + ωe) /2 )2 + ( sin ((ω − ωe) t/2) (ω − ωe) /2 )2 + 2 sin ((ω + ωe) t/2) sin ((ω − ωe) t/2) cos 2ωet (ω2 − ω2c ) /4 ] .(45) 5 The probability of leaving the ground state is thus: Pout (t) = ∞∑ n=1 Pn (t) (46) = q2E2x20 8~2 [( sin ((ω + ωe) t/2) (ω + ωe) /2 )2 + ( sin ((ω − ωe) t/2) (ω − ωe) /2 )2 + 8 sin ((ω + ωe) t/2) sin ((ω − ωe) t/2) cos 2ωet (ω2 − ω2c ) ] .(47) As t → ∞, the ( sin ω̃t ω̃t )2 function develops a narrower and narrower peak (of height 1) around the origin with width ∆ω̃ = 2~t , which goes to zero as t → ∞. So we can take the limit of these functions: lim t→∞ ( sin ω̃t/2 ω̃t/2 )2 t2 = 2πtδ (ω̃) , (48) and get the transition probability out of the ground state: Pout (t → ∞) ≈ q2E2x20π 4~2 tδ (ω − ωe) , (49) which then leads immediately to the transition rate at long times: Rout (t → ∞) ≈ q2E2x20 4~2 δ (ω − ωe) . (50) (b) Modulated oscillator frequency Now we consider the Hamiltonian: H = p2 2m + 1 2 mω2 (t) x2 = p2 2m + 1 2 mω2x2 ︸ ︷︷ ︸ H0 + mωω0x 2 cosωet + 1 2 mω20x 2 cos2 ωet ︸ ︷︷ ︸ H1(t) . (51) Since ω0  ω, we neglect the rightmost term. So our matrix element is: 〈n|H1 (t) |0〉 = mωω0 cosωet 〈n|x2 |0〉 = mωω0 cosωet x20 2 〈n| ( a† + a )2 ︸ ︷︷ ︸ ((a†)2+a2+a†a+aa†) |0〉 (52) = 1 2 mωω0x 2 0 cosωet ( δn,0 + √ 2δn,2 ) . (53) Thus, the asymptotic transition rate is: Rout (t → ∞) ≈ π ~2 ( mωω0x 2 0 )2 δ (ω − ωe) . (54) 4 Spin decay We have an initial state of ∣ ∣+ 12 , ẑ 〉 . 6 (a) Response to a weak magnetic field pulse We now apply a weak B field pulse and compute dσ : the amplitudes of finding the spin in state σ = ± 12 . i. B parallel to initial spin configuration Our magnetic field and Hamiltonian are: B (t) = ẑB0τδ (t) ⇒ H1 (t) = −µB~σ · B (t) , (55) which leads to and expression for the amplitudes: dσ (t) = − i ~ ∫ t 0 dt′ 〈σẑ|H1 (t) |↑ ẑ〉 + δσ,↑ (56) = i ~ µBB0τ 〈σẑ|σz |↑ ẑ〉 ︸ ︷︷ ︸ δσ,↑ ∫ t 0 dt′ δ (t′) + δσ,↑ (57) dσ (t) = δσ,↑ ( 1 + i µBB0τ ~ ) . (58) ii. B transverse to initial spin configuration Now our magnetic field and Hamiltonian are: B (t) = ŷB0τδ (t) ⇒ H1 (t) = −µBB0τδ (t) σy, (59) which leads to and expression for the amplitudes: dσ (t) = i ~ µBB0τ 〈σẑ|σy |↑ ẑ〉 ︸ ︷︷ ︸ i|↓ẑ〉 +δσ,↑ (60) dσ (t) = δσ,↑ − µBB0τ ~ δσ,↓. (61) (b) Response to a weak magnetic field of finite duration We now repeat the analysis for a finite pulse (see figure 2). Since our dσ (t) is independent of t, the expressions must be the same: dσ (t) = δσ, ↑ ( 1 + i µBB0τ ~ ) (62) dσ (t) = δσ, ↑ − µBB0τ ~ δσ, ↓, (63) where equations 62 and 63 correspond to parts (a.i) and (a.ii), respectively. 7 z x B(θ,φ)S y Figure 4: Great circle in the x-z plane. Inset: How S and B are related. 5 Berry’s phase We study Berry’s phase: γn = i ∫ t 0 dt′ 〈Ψn (α (t′))| ∂t′ |Ψn (α (t′))〉 (75) = i ∫ ~α1 ~α0 d~α · 〈Ψn (α)| ~∇α |Ψn (α)〉 , (76) with the standard magnetic Hamiltonian and eigenstates: H = −B · ~σ, ∣ ∣ ∣↑ B̂ (t) 〉 = ( cos θ2 eiφ sin θ2 ) . (77) (a) Berry’s phase with rotating B i. Great circle in the x-z plane Since we are in the x-z plane, B is changing due to a changing 0 ≤ θ < 2π (see figure 4): γ↑ = i ∫ 2π 0 dθ 〈 ↑ B̂ ∣ ∣ ∣∂θ ∣ ∣ ∣↑ B̂ 〉 ︸ ︷︷ ︸ 0 , (78) so γ↑ = 0 in this case. 10 z x y Figure 5: Great circle in the equatorial plane. ii. Equatorial rotation γ↑ = i ∫ 2π 0 dφ 〈 ↑ B̂ ∣ ∣ ∣∂φ ∣ ∣ ∣↑ B̂ 〉 ︸ ︷︷ ︸ i 2 , (79) so γ↑ = −π in this case (see figure 5). iii. Horizontal rotation at latitude θ Now we look at a non-equatorial azimuthal circle (see figure 6): γ↑ (θ) = i ∫ 2π 0 dφ ( cos θ2 e −iφ sin θ2 ) · ( 0 ieiφ sin θ2 ) (80) = − ∫ 2π 0 dφ sin2 θ 2 ︸ ︷︷ ︸ 1 2 (1−cos θ) (81) γ↑ (θ) = −2π sin2 θ 2 = −π (1 − cos θ) . (82) 11 z x y Figure 6: Circle at latitude θ. (b) Ordinary phase contribution After one complete revolution, we have: (i) ∣ ∣ ∣↑ B̂ 〉 → − ∣ ∣ ∣↑ B̂ 〉 (83) (ii) ∣ ∣ ∣↑ B̂ 〉 → ∣ ∣ ∣↑ B̂ 〉 , (84) (85) But with Berry’s phase factor both operations give the same eiπ = −1 factor. (c) Gauge transformation properties AB = i~ 〈n ~α| ~∇α |n ~α〉 . (86) Our gauge transformation rotates |n ~α〉 into eiχ |n ~α〉, so AB becomes: AB → i~ 〈n ~α| ~∇α |n ~α〉 − ~~∇χ, (87) and Berry’s phase becomes: γ = 1 ~ ∮ AB · d~α → 1 ~ ∮ AB · d~α − (χ (α1) − χ (α0)) , (88) with the last part cancelled by eiχ(α). 12