Quantum Mechanics 1 - Homework 4 Solutions | PHYS 5260, Assignments of Quantum Mechanics

Download Quantum Mechanics 1 - Homework 4 Solutions | PHYS 5260 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics II: HW 4 Solutions Leo Radzihovsky March 17, 2008 1 NMR and Rabi oscillations We shall study the interaction of a spin-1/2 particle with a magnetic field of the following form (see figure 1 ): B = B0 + B⊥ = B0 + B⊥n̂ (t) = B0ẑ + B⊥ (x̂ cosωt + ŷ sin ωt) . (1) (a) Parameterization of the Hamiltonian The Hamiltonian has the standard form, H = −~µ ·B. Since we are concerned with an electron, we have the following expression for ~µ: ~µ = −gµB ~ S = −µB~σ. (2) So our Hamiltonian becomes: H = µBB0 ︸ ︷︷ ︸ 1 2  σz + µBB⊥ ︸ ︷︷ ︸ g n̂ (t) · ~σ⊥ (3) = 1 2  σz ︸︷︷︸ 0 @ 1 0 0 −1 1 A +g cosωt σx ︸︷︷︸ 0 @ 0 1 1 0 1 A +g sin ωt σy ︸︷︷︸ 0 @ 0 −i i 0 1 A (4) = 1 2 σz + g ( e−iωtσ+ + e iωtσ− ) (5) = ( 1 2 ge −iωt geiωt − 12 ) . (6) Thus, we see that g = µBB⊥ and  = 2µBB0. (b) Spin dynamics We need to compute Pσz (t) for the initial state: |Ψ (0)〉 = |↓〉 ≡ |2〉 note: |↑〉 ≡ |1〉 . (7) So we need to find |Ψ (t)〉: |Ψ (t)〉 = ∑ i=1, 2 Ci |σz〉 = C1 (t) |↑〉 + C2 (t) |↓〉 , (8) 1 S B0 B Figure 1: A sketch showing the components of B, as well as the spin orientation S. from which we can easily find Pσz (t): Pσz (t) = |〈σz |Ψ (t)〉|2 = |Ci (t)|2 . (9) An exact solution is possible by solving the Schrödinger equation, H |Ψ〉 = i~∂t |Ψ〉. We perform a unitary time-dependent transformation to put the system into the rotating frame of B⊥ (t). In that frame B⊥ is time-independent, and therefore so is H . It can then be diagonalized. We first write: |Ψ (t)〉 = U (t) ∣ ∣ ∣Ψ̃ (t) 〉 , (10) So: ( U †HU ) ∣ ∣ ∣Ψ̃ (t) 〉 = i~U †∂t ( U (t) ∣ ∣ ∣Ψ̃ (t) 〉) (11) = i~U †∂tU (t) ∣ ∣ ∣Ψ̃ (t) 〉 + i~∂t ∣ ∣ ∣Ψ̃ (t) 〉 . (12) Now we just pick a U (t) so that H is time-independent: U †HU = 1 2 U †σzU + gU †B̂⊥ · ~σ⊥U. (13) Recall: U †θ B̂ · ~σUθ = ~~Rθ · B̂, (14) where ~~Rθ is the O(3) rotation matrix. Thus, we simply rotate around ẑ by ωt: U (t) = e−i ωt 2 σz . (15) As an aside, we demonstrate that this indeed is the right transformation. Uθ = e −iθn̂·S/~ = e−iθn̂·~σ/2, (16) 2 B’ ε/~ 2 θ g x y z Figure 3: The relation between the orientation of B′ and the parameter g and ̃. So ∣ ∣ ∣ ˜̃Ψ 〉 = |σz〉 with Eσz = σz b̃ = ±b̃. Recalling ∣ ∣ ∣Ψ̃ 〉 = U ∣ ∣ ∣ ˜̃Ψ 〉 , we can write our states in component form: 〈 σz |Ψ̃ 〉 = U ( 1 0 ) , U ( 0 1 ) , (41) or: ê+ = ( cos θ/2 sin θ/2 ) , ê− = ( − sin θ/2 cos θ/2 ) , (42) and our eigenenergies become: E± = ± √ 1 4 ( − ~ω)2 + g2 = ±Eω. (43) From figure 3 , we can see that the angle θ is defined by cos θ = ̃/2 √ 1 4 ̃ 2 + g2 and sin θ = g √ 1 4 ̃ 2 + g2 . (44) As a check, we note that for g = B⊥ = 0, B′ = ̃ 2 ẑ. So our eigenstates and eigenenergies become: ê+ = ( 1 0 ) , ê− = ( 0 2 ) , E± = ± ̃ 2 , (45) As they should. 5 Now, the initial condition is such that Ci(0) = δi2, so C̃(0) = ( 0 1 ) = C+(0)ê+ + C−(0)ê− (46) C(0) = sin θ 2 ê+ + cos θ 2 ê− (47) = sin θ 2 ( cos θ/2 sin θ/2 ) + cos θ 2 ( − sin θ/2 cos θ/2 ) = ( 0 1 ) . (48) So, C̃(t) = sin θ 2 ê+e − i ~ Eωt + cos θ 2 ê−e i ~ Eωt (49) =   i sin θ sin ( Eω ~ t ) sin2 θ2e − i ~ Eωt + cos2 θ2e i ~ Eωt   . (50) We need to transform this back to the non-rotating frame. Using |Ψ〉 = U ∣ ∣ ∣Ψ̃ 〉 = e−i ωt 2 σz ∣ ∣ ∣Ψ̃ 〉 , we can write: C(t) =    i sin θ sin ( Eω ~ t ) e−iωt/2 ( sin2 θ2e − i ~ Eωt + cos2 θ2e i ~ Eωt ) eiωt/2    , (51) So our probabilities are: |C1(t)|2 = sin2 θ sin2 ( Eω ~ t ) = 1 2 sin2 θ ( 1 − cos ( 2Eω ~ t )) , (52) |C2(t)|2 = sin4 θ 2 + cos4 θ 2 + sin2 θ 2 cos2 θ 2 2 cos ( 2Eω ~ t ) = 1 − 1 2 sin2 θ ( 1 − cos ( 2Eωt ~ )) . (53) We note here that |C1(t)|2 + |C2(t)|2 = 1. See figure 4 for a plot as a function of time. Calculating 〈Sz〉 is now straightforward: 〈Ψ(t)|Sz |Ψ(t)〉 = ~ 2 〈σz〉 = ~ 2 ( |C1(t)|2 − |C2(t)|2 ) (54) = −~ 2 + ~ sin2 θ sin2 ( Eω ~ t ) . (55) i. Frequency and amplitude of oscillations It is easy from the final form of |C2(t)|2 that the frequency of oscillations is: ωR = 2Eω ~ = 2 ~ √ 1 4 ( − ~ω)2 + g2 = 2g ~ (56) and the amplitude of oscillations is: AR = 1 2 sin2 θ = g2/2 1 4 ( − ~ω) 2 + g2 = 1 2 , (57) where the last equality for equations 56 and 57 apply at resonance. 6 P(t)σz P(t) AR t 1 Figure 4: A plot of P↓(t) with time. The oscillations in P↓(t) are a result of the system transitioning between |↑〉 and |↓〉, which is due to B⊥. ii. Parameter dependence of amplitude See figure 5. iii. Oscillating field directed along x̂ We can write the oscillating field as: B⊥ = B⊥x̂ cosωt = 1 2 ( B+⊥ + B − ⊥ ) (58) = 1 2 B⊥ [(x̂ cosωt + ŷ sinωt) + (x̂ cosωt − ŷ sin ωt)] , (59) where B±⊥ are counter-rotating fields. When written this way, we can immediately go to rotating reference frames as before. For the ωt reference frame, we have: Hω = 1 2 ( − ~ω)σz + gσx + g (cos (2ωt)σx − cos (2ωt)σy) , (60) where the last term is a quickly oscillating off resonance perturbation, which can be treated with time- dependent perturbation theory. Going to the other frame, the −ωt frame, we obtain: Hω = 1 2 ( + ~ω)σz + gσx + g (cos (2ωt)σx + cos (2ωt)σy) . (61) Examination of the first term reveals that we are off-resonance in this frame. Most of the dynamics are governed by the on-resonance component of B, so our answer does indeed look similar to the original case. iv. Conservation of probability We examined this earlier. See equations 52 and 53, and the note immediately following. 7 hω0 |0 > spin−up ladderspin−down ladder ε |0 > |1 > |2 > |3 > |4 > |5 > |5 > |4 > |3 > |2 > |1 > Figure 6: The g = 0 spectrum. Note the shift in energy  for a given n value and opposite spins. (b) Off-resonance pertubation theory Our zeroth order and perturbing hamiltonians are: H0 = 1 2 σz + ~ω0a †a, (70) H1 = g ( a†σ− + aσ+ ) . (71) The first order correction is given by: E(1)n σz = g 〈n σz| ( a†σ− + aσ+ ) |n σz〉 = 0. (72) So we must go to second order: E(2)n σz = ∑ m, σ′z 6=n, σz g2 ∣ ∣〈m σ′z | ( a†σ− + aσ+ ) |n σz〉 ∣ ∣ 2 ( ~ω0n + 1 2σz ) − ( ~ω0m + 1 2σ ′ z ) (73) = ∑ m, σ′z 6=n, σz g2 ( δm,n+1 √ n + 1δσz↑δσ′z↓ + δm,n−1 √ nδσz↓δσ′z↑ )2 ~ω0 (n − m) + 12 (σz − σ′z) (74) = g2 [ (n + 1) δσz↑  − ~ω0 − nδσz↓  − ~ω0 ] (75) = g2 ∆     n (δσz↑ − δσz↓) ︸ ︷︷ ︸ σz + δσz↑ ︸︷︷︸ 1 2 (σz+1)     , (76) 10 where ∆ ≡  − ~ω0 is the detuning. So our spectrum, to second order in g, is: En σz = ~ω0n + 1 2 σz + g2 ∆ ( n + 1 2 ) σz + g2 2∆ . (77) The term in g2 involving nσz is a result of the AC Stark effect, while the one with 1 2σz is the Lamb shift. Equivalently, we can derive the second order shift in the following manner: E(2)n σz = ∑ m, σ′z 6=n, σz g2 〈n σz| ( a†σ− + aσ+ )† |m σ′z〉 〈m σ′z | ( a†σ− + aσ+ ) |n σz〉 ∆σz (78) = g2 ∆ 〈n σz| ( aσ+ + a †σ− ) ( a†σ− + aσ+ ) ︸ ︷︷ ︸ aa†σ+σ−+a†aσ−σ+ σz |n σz〉 (79) = g2 ∆ 〈n σz|   a †a {σ+, σ−} ︸ ︷︷ ︸ 1 + σ+σ− ︸ ︷︷ ︸ 1 2 (σz+1)   σz |n σz〉 (80) = g2 ∆ 〈n σz| ( a†aσz + 1 2 (σz + 1) ) |n σz〉 . (81) Hence to second order in g we have Heff ≈ ~ω0a †a + 1 2 σ̂z + g2 ∆ ( a†a + 1 2 ) σ̂z + g2 2∆ . (82) So we note the following: there is a spin-dependent oscillator frequency ~ω↑↓ = ~ω0± g 2 ∆ , and an n-dependent shift in ↑↓ levels Espm = ( 1 2 + g2 ∆ n ) σz. Our first order states can readily be found: ∣ ∣ ∣n(1) 〉 = ∑ m6=n C(1)m ∣ ∣ ∣m(0) 〉 (83) C(1)m = 〈 m0|n(1) 〉 = 〈 m0 ∣ ∣H1 ∣ ∣n(0) 〉 E0n − E0m (84) ⇒ ∣ ∣ ∣n(1), σz 〉 = ∣ ∣ ∣n(0), σz 〉 + g ∆ √ n + 1δσz↑ ∣ ∣ ∣(n + 1)(0), −σz 〉 − g ∆ √ nδσz↓ ∣ ∣ ∣(n − 1)(0), −σz 〉 .(85) (c) Exact spectrum and eigenstates To find the exact solution, we first rewrite our eigenstates as an expansion in the zeroth order ones: |Eα〉 = ∑ n σz Cαn σz |n σz〉 = ∑ n ( Cαn↑ |n ↑〉 + Cαn↓ |n ↓〉 ) . (86) If we act on a zeroth order eigenstate with the complete Hamiltonian, we get: H |n σz〉 = ~ω0n |n σz〉 + 1 2 σz |n σz〉 + g (√ n + 1δσz↑ |n + 1 ↓〉 + √ nδσz↓ |n − 1 ↑〉 ) , (87) 11 so our matrix elements are: 〈m σ′z |H |n σz〉 = ( ~ω0n + 1 2 σz ) δnmδσ′zσz + g √ n + 1δσz↑δσ′z↓δm n+1 + g √ nδσz↓δσ′z↑δm n−1 (88) Or, writing it out in matrix form:            ~ω0n − 2 g √ n 0 0 0 0 · · · g √ n ~ω0 (n − 1) + 2 0 0 0 0 · · · 0 0 ~ω0 (n − 1) − 2 g √ n − 1 0 0 · · · 0 0 g √ n − 1 ~ω0 (n − 2) + 2 0 0 · · · 0 0 0 0 ~ω0 (n − 2) − 2 g √ n − 2 · · · 0 0 0 0 g √ n − 2 ~ω0 (n − 3) + 2 · · · ... ... ... ... ... ... . . .            . (89) Written like this, it is easy to see that the full Hilbert space decouples into 2 × 2 subspaces, labeled by n. So we can write the eigenvectors in a closed form and attach physical meaning to the terms: |Eαn 〉 = Cαn↓ |n ↓〉 + Cαn−1↑ |n − 1 ↑〉 , (90) where the absorption of a photon (n → n − 1) and flipping of spin takes us from the left term to the right one, exciting the state, and emission of a photon (n − 1 → n) with a spin flip, de-exciting it. Equivalently, we can show the decoupling more directly: H |Eα〉 = ∑ n σz Cαn σz [ ( 1 2 σz + ~ω0n ) |n, σz〉 + g √ n + 1δσz↑ |n + 1, −σz〉 + g √ nδσz↓ |n − 1, −σz〉], (91) 〈m σ′z |H |Eα〉 = ( 1 2 σz + ~ω0m ) Cαm σ′z +g √ mCαm−1 ↑δσ′z↓ + g √ m + 1Cαm+1 ↓δσ′z↑ = E αCαm σ′z . (92) So we have pairs of coupled equations: ( 1 2 + ~ω0m ) Cαm↑ + g √ m + 1Cαm+1 ↓ = E αCαm↑( 1 2 + ~ω0 (m + 1) ) Cαm+1 ↓ + g √ m + 1Cαm ↑ = E αCαm+1 ↓ } , (93) ( 1 2 + ~ω0 (m − 1) ) Cαm−1 ↑ + g √ mCαm↓ = E αCαm−1 ↑( 1 2 + ~ω0m ) Cαm↓ + g √ mCαm−1 ↑ = E αCαm↓ } . (94) Now we can take a look at the “nth” Hamiltonian, a 2 × 2 array: Hn = ( ~ω0n − /2 g √ n g √ n ~ω0 (n − 1) + /2 ) (95) = ~ω0 ( n − 1 2 ) ︸ ︷︷ ︸ 1 2 (En↓+En−1↑)≡Ēn 1 + 1 2 (~ω0 − ) ︸ ︷︷ ︸ bz= 1 2 (En↑−En↓)≡∆2 σx + bx ︷︸︸︷ g √ n σx, (96) 12 i. absorption Cn−1, ↑(t) = − i ~ ∫ t 0 dt′ e i ~ −~ω0=∆ ︷ ︸︸ ︷ ( E0n−1, ↑ − E0n, ↓ ) t′ 〈n − 1, ↑| g ( a†σ− + aσ+ ) |n, ↓〉 ︸ ︷︷ ︸ g √ n (116) = −g √ n ∆ ( e i ~ ∆t − 1 ) , (117) So the probability of absorption is: Pn, ↓→n−1, ↑ = |Cn−1, ↑(t)|2 = g2n ∆2/4 sin2 [ ∆ ~ t ] . (118) ii. emission Cn+1, ↓(t) = − i ~ ∫ t 0 dt′ e i ~ −∆ ︷ ︸︸ ︷ ( E0n+1, ↓ − E0n, ↑ ) t′ 〈n + 1, ↓| g ( a†σ− + aσ+ ) |n, ↑〉 ︸ ︷︷ ︸ g √ n+1 (119) = g √ n + 1 ∆ ( e− i ~ ∆t − 1 ) , (120) So the probability of emmision is: Pn, ↑→n+1, ↓ = |Cn+1, ↓(t)|2 = g2 (n + 1) ∆2/4 sin2 [ ∆ ~ t ] (121) = Pn, ↓→n−1, ↑ + g2 ∆2/4 sin2 [ ∆ ~ t ] , (122) where we can identify the leftmost term on the right with stimulated emission and the other term as spon- taneous emission. (f) Average value of spin and field on resonance Our initial coherent state can be written: |Ψ(0)〉 = ∑ n αn√ n! e− |α|2 2 |n ↓〉 (123) = ∑ n αn√ n! e− |α|2 2 [ cos θn 2 |n +〉 − sin θn 2 |n−〉 ] , (124) So our wavefunction is: |Ψ(t)〉 = ∑ n αn√ n! e− |α|2 2 [ cos θn 2 e− i ~ E+n t |n +〉 − sin θn 2 e− i ~ E−n t |n−〉 ] . (125) 15 Note that |0 ↓〉 is completely decoupled so cos θn=0 = 1 and sin θn=0 = 0. We can now immedately write down the spin expectation value: 〈Ψ(t)|σz |Ψ(t)〉 = ∑ n1, n2 (α∗)n1 αn2√ n1!n2! e−|α| 2 [ cos θn1 2 e i ~ E+n1t 〈n1 +| − sin θn1 2 e i ~ E−n1t 〈n1 −| ] σz × [ cos θn2 2 e− i ~ E+n2t |n2 +〉 − sin θn2 2 e− i ~ E−n2 t |n2 −〉 ] (126) = ∑ n1, n2 (α∗)n1 αn2√ n1!n2! e−|α| 2 [ cos θn1 2 cos θn2 2 e− i ~ (E − n1 −E−n2)t 〈n1 +|σz |n2 +〉 + · · · ] ,(127) where the other terms are straightforward to find. At this point, we turn our attention to a slightly faster and cleaner approach. We first find the probability coefficients Cm↑↓. Cm↑ = 〈m ↑ |Ψ(t)〉 (128) = ∑ n=1 αn√ n! e− |α|2 2     cos θn 2 e− i ~ E+n t sin θn 2 δm, n−1 ︷ ︸︸ ︷ 〈m ↑ |n +〉 − sin θn 2 e− i ~ E−n t cos θn 2 δm, n−1 ︷ ︸︸ ︷ 〈m ↑ |n−〉     ︸ ︷︷ ︸ δm, n−1 1 2 sin θne − i 2~ (E + n +E − n )t „ e − i 2~ (E + n −E − n )t−e i 2~ (E + n +E − n )t « (129) = αm+1 √ (m + 1)! e− |α|2 2 (−i) sin θm+1e− i ~ Ēm+1t sin [ 1 ~ √ ∆2/4 + g2 (m + 1)t ] , (130) Cm↓ = ∑ n=0 δm, n αn√ n! e− |α|2 2 [ cos2 θn 2 e− i ~ E+n t + sin2 θn 2 e− i ~ E−n t ] , (131) so the probabilities associated with these amplitudes are: |Cm↑|2 = ( |α|2 )m+1 (m + 1)! e|α| 2 sin2 θm+1 sin 2 [ 1 ~ √ ∆2/4 + g2 (m + 1)t ] , (132) |Cm↓|2 = ( |α|2 )m m! e|α| 2       cos4 θm 2 + sin4 θm 2 + cos2 θm 2 sin2 θm 2 ( e i ~ (E + m−E−m)t + e− i ~ (E + m−E−m)t ) ︸ ︷︷ ︸ 2 cos h 1 ~ 2 √ ∆2/4+g2mt i       (133) = ( |α|2 )m m! e|α| 2 [ 1 − 1 2 sin2θm ( 1 − cos [ 1 ~ 2 √ ∆2/4 + g2mt ])] . (134) We now recall the relation between these and the spin expectation value: 〈σz〉 = ∑ m=0 ( |Cm↑|2 − |Cm↓|2 ) , (135) 16 to write the spin expectation value in our “pseudo” spin system: 〈σz〉 = − |C0↓|2 + ∑ m=1 ( |Cm−1, ↑|2 − |Cm↓|2 ) (136) = e|α| 2  −1 + sin2 θ0 ︸ ︷︷ ︸ 0 sin2 [ ∆ 2~ t ]   + ∑ m=1 ( |α|2 )m m! e−|α| 2 ( 2 sin2 θm sin 2 [ 1 ~ √ ∆2/4 + g2mt ] − 1 ) (137) 〈σz〉 = −1 + 2e−|α| 2 ∑ m=1 ( |α|2 )m m! g2m ∆2/4 + g2m sin2 [ 1 ~ √ ∆2/4 + g2mt ] . (138) We note that at t = 0, 〈σz〉 = −1, as it should. On resonance, ∆ = 0, so our expression reduces to: 〈σz〉 (t) = −1 + 2e−|α| 2 ∑ m=0 ( |α|2 )m m! sin2 [g ~ √ mt ] ︸ ︷︷ ︸ 1 2 (1−cos [ 2g ~ √ mt]) (139) = −e−|α|2 ∑ m=0 ( |α|2 )m m! cos [ 2g ~ √ mt ] . (140) Thus, the spin expectation value exhibits “collapse and revival” (see figure 8). Next, we calculate the expectation value of the oscillator field 〈Ψ(t)| a |Ψ(t)〉 ≡ 〈a〉. We first look at a acting on the wavefunction: a |Ψ(t)〉 = ∑ n αn√ n! e− |α|2 2 [ cos θn 2 e− i ~ E+n t a |n +〉 ︸ ︷︷ ︸ √ n cos θn 2 |n−1 ↓〉+ √ n−1 sin θn 2 |n−2 ↑〉 − sin θn 2 e− i ~ E−n t a |n−〉 ︸ ︷︷ ︸ −√n sin θn 2 |n−1 ↓〉+ √ n−1 cos θn 2 |n−2 ↑〉 ]. (141) The bra vector and the state vectors are given by 〈Ψ(t)| = ∑ m α∗m√ m! e− |α|2 2 [ cos θm 2 e i ~ E+mt 〈m +| − sin θm 2 e i ~ E−mt 〈m−| ] , (142) 〈m +| = cos θm 2 |m ↓〉 + sin θm 2 |m − 1 ↑〉 , (143) 〈m−| = − sin θm 2 |m ↓〉 + cos θm 2 |m − 1 ↑〉 , (144) 17 where we have defined the following quantities: E++n−1, n = −~ω0 + √ ∆2/4 + g2 (n − 1) − √ ∆2/4 + g2n, (153) E−−n−1, n = −~ω0 − √ ∆2/4 + g2 (n − 1) + √ ∆2/4 + g2n, (154) E+−n−1, n = −~ω0 + √ ∆2/4 + g2 (n − 1) + √ ∆2/4 + g2n, (155) E−+n−1, n = −~ω0 − √ ∆2/4 + g2 (n − 1) − √ ∆2/4 + g2n. (156) (157) Now we continue simplifying the expectation value: 〈a〉 = α 2 ∑ n=1 |α|2(n−1) (n − 1)! e −|α|2e−iω0t[ √ n − 1 n sin θn−1 sin θn × ( cos [(√ ∆2/4 + g2 (n − 1) − √ ∆2/4 + g2n ) t ] − cos [(√ ∆2/4 + g2 (n − 1) + √ ∆2/4 + g2n ) t ]) + ( cos [(√ ∆2/4 + g2 (n − 1) − √ ∆2/4 + g2n ) t ] + cos [(√ ∆2/4 + g2 (n − 1) + √ ∆2/4 + g2n ) t ]) + i cos θn−1 ( sin [(√ −√ ) t ] + sin [(√ + √ ) t ]) + i cos θn ( sin [(√ −√ ) t ] − sin [(√ + √ ) t ]) + cos θn−1 cos θn ( cos [(√ −√ ) t ] − cos [(√ + √ ) t ]) , (158) where the blank spaces under the radicals indicate the same contents as previous terms. So our “final” expression becomes: 〈a〉 = α 2 ∑∞ n=1 |α|2(n−1) (n − 1)! e −|α|2e−iω0t × [cos [(√ −√ ) t ] (√ n − 1 n sin θn−1 sin θn + cos θn−1 cos θn + 1 ) + cos [(√ + √ ) t ] ( − √ n − 1 n sin θn−1 sin θn − cos θn−1 cos θn + 1 ) + sin [(√ −√ ) t ] (i cos θn−1 + i cos θn) + sin [(√ + √ ) t ] (i cos θn−1 − i cos θn)]. (159) Note that for g = 0, 〈a〉 = αe−iω0t. On resonance, we have drastic simplifications: E±n = ~ω0 ( n − 1 2 ) ± g √ n, (160) cos θn = 0, sin θn = 1, (161) 20 so the expectation value on resonance is: 〈a〉 = α 2 ∑∞ n=1 |α|2(n−1) (n − 1)! e −|α|2e−iω0t × [cos [ g (√ n − 1 − √ n ) t ] ( 1 + √ n − 1 n ) + cos [ g (√ n − 1 + √ n ) t ] ( 1 − √ n − 1 n ) ]. (162) 3 Dyson Expansion (a) Path integral derivation The evolution operator can be written: U (xN , x0; tN , t0) = ∫ x(tN )=xN x(t0)=x0 Dx[t]e i~ R tN t0 dt[L0+L1], (163) with L0 the free Lagrangian and L1 the interaction term. Explicity, these are: L0 = pẋ − H0 (x(t)) L0 = 1 2 mẋ2, (164) = e.g. (165) L1 = −H1 (x(t)) L1 = −V (x). (166) (167) Now we Taylor expand U in powers of S1 ≡ ∫ tN t0 dt L1: U = ∫ xN x0 ∏ t dx(t)e i ~ S0 [ 1 + (−i ~ )∫ tN t0 dt1H1 + · · · ] = ∫ xN x0 Dx e i~ S0 + (−i ~ )∫ tN t0 dt1 ∫ ∞ −∞ dx1 ∫ x(tN )=xN x(t1)=x1 t=tN∏ t=t1 dx(t)e i ~ R tN t1 dt′ L0 ︸ ︷︷ ︸ U0(xN ,x1;tN ,t1) ×H1 (x1) × ∫ x(t1)=x1 x(t0)=x0 t=t1∏ t=t0 dx(t)e i ~ R t1 t0 dt′ L0 ︸ ︷︷ ︸ U0(x1,x0;t1,t0) + · · · , (168) 21 so we can rewrite the evolution operator in terms of the free particle evolution operator: U (xN , x0; tN , t0) = U0 (xN , x0; tN , t0) + (−i ~ )∫ tN t0 dt1 ∫ ∞ −∞ dx1U0 (xN , x1; tN , t1) H1 (x1) U0 (x1, x0; t1, t0) + (−i ~ )2 ∫ tN t0 dt2 ∫ t2 t0 dt1 ∫ ∞ −∞ dx2 ∫ ∞ −∞ dx1U0 (xN , x2; tN , t2) H1 (x2) U0 (x2, x1; t2, t1) × H1 (x1) U0 (x1, x0; t1, t0) + · · · . (169) The above is just the coordinate representation of the time-ordered evolution operator, i.e., the Dyson equation (x-integrals are “matrix products”): Û (tN , t0) = e − i ~ R tN t0 dt′H0T ( e− i ~ R tN t0 dt′ĤI(x̂,t′) ) . (170) (b) Dyson series for a time-independent Hamiltonian For a time-independent Hamiltonian, we have a considerable simplification: Û (tN − t0) = Û0 (tN − t0) + (−i ~ )∫ tN t0 dt1Û0 (tN − t1) Ĥ1U0 (t1 − t0) + (−i ~ )2 ∫ tN t0 dt2 ∫ t2 t0 dt1Û0 (tN − t2) Ĥ1Û0 (t2 − t1) Ĥ1Û0 (t1 − t0) + · · · . (171) The above are just a convolutions of Ĥ1Û0 over intermediate times. So we define a Fourier transform of Û (tN − t0): Û(E) = ∫ ∞ 0 dt e i ~ EtÛ(t), (172) with E ≡ ER + i, where  is a convergence factor. So our relation becomes: Û (E) = Û0 (E) + (−i ~ ) Û0 (E) Ĥ1U0 (E) + (−i ~ )2 Û0 (E) Ĥ1Û0 (E) Ĥ1Û0 (E) + · · · (173) = Û0 (E) ( 1 + (−i ~ ) Ĥ1U0 (E) + (−i ~ )2 Ĥ1Û0 (E) Ĥ1Û0 (E) + · · · ) (174) =               1 + (−i ~ ) Ĥ1U0 (E) + (−i ~ )2 Ĥ1Û0 (E) Ĥ1Û0 (E) + · · · ︸ ︷︷ ︸ 1 1+ i ~ Ĥ1Û0        −1 Û−10        −1 (175) = [( 1 + i ~ Ĥ1Û0 ) Û−10 ]−1 (176) = [ Û−10 + i ~ Ĥ1 ]−1 . (177) 22 In the continuum limit, we take xi = ci, with c → 0 as the lattice constant: H = 1 c Nc≡L∑ xi=0 c [ 1 2m Π (xi) 2 + 1 2 mω20u (xi) 2 ] (203) = ∫ L 0 dx c [ 1 2m Π (x)2 + 1 2 mω20u (x) 2 ] . (204) We can absorb c through a redefinition of Π and u: Π̃(x) = 1√ c Π(x), ũ(x) = 1√ c u(x), (205) ⇒ [ ũ(x), Π̃ (x′) ] = i~ δx, x′ c ︸ ︷︷ ︸ δ(x−x′) . (206) (b) Generalized Einstein model This is a trivial generalization, leading to: H = N∑ i=1 [ 1 2m Π2i + 1 2 mω2i u 2 i ] , (207) H = N∑ i=1 ~ωi ( a†i ai + 1 2 ) , (208) ∣ ∣E{n} 〉 = |{ni}〉 = |n1〉 ⊗ |n2〉 ⊗ |n3〉 ⊗ · · · ⊗ |nN 〉 , (209) E{n} = N∑ i=1 ~ωi ( ni + 1 2 ) . (210) (c) Debye model H = Nc∑ xi=0 [ 1 2m Π (xi) 2 + 1 2 B (u (xi+1) − u (xi))2 ] , (211) Hcont. = ∫ L 0 dx c [ 1 2m Π (x) 2 + 1 2 Bc2 (∂xu) 2 ] . (212) i. Fourier series decoupling We first exploit a discrete Fourier transform to decouple the system: u (xi) = 1 L π/c ∑ kn=−π/c eiknxi ũ (kn) , (213) 25 2(B/m)2 1 2π/ccπ/ π/c +ε−π/c −π/c +ε−2π/c ωk k Figure 9: The first Brillouin zone, indicated by the solid line, is the only range of values of kn = k that are physical. It is easy to see from the figure that other values of k give the same ωk, i.e., the same mode of vibration. with a similar expression for Π (xi). Note that since u (xi) is real, ũ ∗ (kn) = ũ (−kn). Now we rewrite the Hamiltonian in terms of these transformed quantities: H = 1 L2 ∑ knkm ∑ xi eiknxi+ikmxi 1 2m Π̃ (km) Π̃ (kn) + 1 L2 ∑ knkm ∑ xi 1 2 B ( eiknxi+1 − eiknxi ) ũkn ( eikmxi+1 − eikmxi ) ũkm (214) H = 1 L2 ∑ knkm Nδkn,−km 1 2m Π̃ (km) Π̃ (kn) + 1 L2 ∑ knkm 1 2 BNδkn,−km ( eiknc − 1 ) ( eikmc − 1 ) ũkn ũkm (215) H = 1 Lc ∑ kn      1 2m Π̃ (−kn) Π̃ (kn) + 1 2 B · 4 sin2 knc 2 ︸ ︷︷ ︸ ≡ 1 2 mω2kn ũkn ũ−kn      (216) H = 1 Lc ∑ kn [ 1 2m ∣ ∣ ∣Π̃ (kn) ∣ ∣ ∣ 2 + 1 2 mω2kn |ũkn | 2 ] , (217) ωkn = 2 √ B m |sin (knc/2)| . (218) See figure 9 for a plot of ωkn . The first Brillouin zone is the only physical range of kn because for a kn outside it, we do not have a new physical mode of vibration (displacement u (xi)). This is due to the discreteness of xi = ci, or aliasing. Graphically (see figure 10), one can see that the distortion of wavevector |kn| > πc is identical to k′n “folded” back in the first Brillouin zone. For example, kn = 2π c means u (xi) = 0, which is already included by kn = 0. The two modes kn = π c +  and k ′ n =  − πc are identical. 26 c Figure 10: Distortion of the lattice by an amount greater than c, i.e. distorting the wavevector by an amount |kn| > πc , is “folded back.” Mathematically, we can easily see: ukn (xi) = ũkne i(πc +) xi︸︷︷︸ ci = ũkne i(−πc +)ci, (219) since eiπ = e−iπ. Also note that limiting the range to the first Brillouin zone produces Nm normal modes: Nm = 2π/c 2π/L = L c = N, (220) as it must, since we have N independent atoms and therefore N linearly independent combinations of their displacements u (xi); these linearly independent combinations are precisely the N ũkn ’s for N kn’s in the first Brillouin zone. A final note: requiring kn to have the form kn = 2π L n ensures that periodic boundary conditions are satisfied, i.e.: u (xi + L) = u (xi) (221) u (xi + L) = 1 L ∑ kn eikn(xi+L)ũkn (222) = 1 L ∑ kn eiknxi ũkn (223) ⇒ eiknL = 1 ⇒ kn = 2π L n. (224) ii. Second quantization form We have the Hamiltonian: H = 1 Lc ∑ kn [ 1 2m ∣ ∣ ∣Π̃ (kn) ∣ ∣ ∣ 2 + 1 2 mω2kn |ũkn | 2 ] . (225) As in a single harmonic oscillator we want to construct the operators ak and a † k. A slight complica- tion/modification is that kn and −kn couple as shown above: H = 1 Lc ∑ kn [ 1 2m Π̃ (−kn) Π̃ (kn) + 1 2 mω2knũ−kn ũkn ] . (226) 27 The advantage of keeping the previous form is that it allows a cleaner transition to the continuum limit: u (xi) = 1 L ∑ kn eiknxi ũ (kn) (247) ⇒ u (x) = ∫ dk 2π ũ (k) eikx, (248) ũ (kn) = c ∑ xi e−iknxiu (xi) (249) ⇒ ũ (k) = ∫ dx u (x) e−ikx, (250) H = ∑ i [ 1 2m Π (xi) 2 + 1 2 B (u (xi+1) − u (xi))2 ] (251) ⇒ H = ∫ dx   1 2m ( Π̂√ c )2 + 1 2 Bc2 ︸︷︷︸ B̃ ( ∂xũ√ c )2   (252) = ∫ dx [ 1 2m ˆ̂ Π2 + 1 2 B̃ ( ∂x ˆ̂u )2 ] , (253) Where the new operators have the following commutation relation: [ ˆ̂u (x) , ˆ̂ Π (x′) ] = i~ 1 c δxixj = i~δ (x − x′) . (254) Turning to the standard form Hamiltonian (equation 239), we see that it takes the following continuum form: H = ∫ dk 2π ~ωk ( ˆ̂a†k ˆ̂ak + 1 2 (2π)δ(k = 0) ) , (255) where 2πδ(k = 0) = 2π 1 “k = 0” = 2π 2π/L = L, (256) ˆ̂ak = 1√ c âk, (257) ˆ̂a†k = 1√ c â†k (258) ⇒ [ ˆ̂ak, ˆ̂a † k′ ] = Lc 1 c δk, k′ ︸ ︷︷ ︸ 2πδ(k−k′) . (259) iii. Quantum numbers and the Debye spectrum |E〉 = ∣ ∣ ∣ ∣ ∣ ∣ ∣ n k1 ︸︷︷︸ −π/c , nk1 , nk1 , · · · n kN ︸︷︷︸ π/c 〉 , (260) 30 so there are N quantum numbers, corresponding to phonon occupation numbers for each mode. The energy is given by: E = ∫ π/c −π/c dk 2π ~ωk ( nk + 1 2 ) . (261) 5 Electromagnetic field We are concerned with the electromagnetic field of a single photon, with the following vector field: A (r, t) = 2A0̂ cos (k · r − ωt), (262) where we work in cgs units. Maxwell’s equations give us expressions for the electric and magnetic fields: E = −1 c ∂tA, B = ~∇×A. (263) We now look at the total electromagnetic energy: HEM = 1 8π ∫ d3r [ |E|2 + |B|2 ] (264) = 4A20 8π ∫ d3r [ ω2 c2 sin2 (k · r− ωt) + k2 sin2 (k · r − ωt) ] . (265) Now the time average value of sin2 (ωt) is 1/2, and we also know that ω = ck. so the time averaged total energy becomes: HEM = 4 8π A20 ω2 c2 · 2 · 1 2 V = ~ω, (266) Since we are dealing with a single photon. So our amplitude becomes: A0 = √ 2πc2~ V ω . (267) 31