Solutions - Chapter 4

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Καλλιστώ Καλάρης
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1 Solutions - Chapter Kevin S. Huang Problem.1 Unitary: Ût = 1 ī hĥt Û tût = 1 Neglect t term: 1 + hĥ ī t 1 īhĥt = 1 + hĥ ī t ī hĥt = 1 Ĥ = Ĥ Problem. Ût = lim 1 ī ] n hĥ1t 1 ī ] hĥt... 1 ī ] hĥnt 1 ī ] hĥ1t 1 ī ] hĥt = 1 ī i h Ĥ1 + Ĥt + Ĥ1 tĥt] 1 ī ] hĥ1t 1 ī ] hĥt 1 ī ] hĥ3t = 1 ī h Ĥ1 + Ĥ + Ĥ3t + i Ĥ1 ĤtĤ3t] 3 i tĥ + Ĥ3t + Ĥ3 tĥtĥ1t] Ût = 1 + ī t Ĥt 1 t 1 + ī t t1 Ĥt 1 t 1 Ĥt t +... h 0 h 0 0 Û n = ī n t t1 tn 1 Ĥt 1 t 1 Ĥt t... Ĥt n t n h Ĥt 1, Ĥt ] = 0: 0 Û n = 1 n! 0 ī n t n Ĥt t h 0 1 0

2 Problem.3 Time-inepenent observable: Ût = exp ī h t 0 ] t Ĥt t A = ī ψt Ĥ, Â] ψt h Ĥ E = E E E Ĥ, Â] E = E ĤÂ ÂĤ E = E ĤÂ E E ÂĤ E A = E EÂ E E ÂE E = E E Â E E E Â E = 0 t Problem. Ĥ = ˆµ B = ˆµ = gq mcŝ B = B 0 î ge mcŝxb 0 = ω 0 Ŝ x Ût = e iĥt/ = e iω 0 ˆ S xt/ = e i ˆ S xφ/ = ˆRφî z cos z ˆRφî + z = 1 φ +z i sin sin φ = 1 ] φ z = 1 φ = π 3 t = l 0 v 0 ω 0 l 0 v 0 = π 3 l 0 = πv 0 3ω 0

3 Problem.5 B = B 0 sin θî + B 0 cos θẑ Below we use φ as efine by ω 0 t: θ = 0: Ĥ = ω 0 B 0 Ŝ B = ω 0 Ŝx sin θ + Ŝz cos θ = ω 0 Ŝ n φ = 0 Ût = e iφ S ˆ n/ = 1 + iφ Ŝ n + 1 iφ Ŝn +...! = 1 1 φ φ iσ n 1! 3! cos θ sin θ σ n Sz sin θ cos θ 3 φ +...] = cos φ iσ n sin cosφ/ i sinφ/ cos θ i sinφ/ sin θ Ût Sz i sinφ/ sin θ cosφ/ + i sinφ/ cos θ Ût +z = cos +y = 1 +z i z φ i sin 1 +y Ût + z = cos P φ φ i sin ] cos θ +z i sin ] φ cos θ 1 sin φ sin θ z S y = = +y ÛT + z = 1 sinω 0T sin θ P = 1 φ sin θ φ θ = π/: Problem.6 P = 1 sinω 0T = π = 0 ψt = e iω 0t/ +z + eiω 0t/ z S z = 0 t S z = ī h ψt Ĥ, Ŝz] ψt + ψt Ŝz t ψt 3

4 Ĥ = ω 0 Ŝ z : Ŝz t = 0 Ĥ, Ŝz] = 0 S x = cos ω 0t t S x = ī h ψt Ĥ, Ŝx] ψt + ψt Ŝx t ψt Ĥ = ω 0 Ŝ z : Ŝx t = 0 Ĥ, Ŝx] = ω 0 σ z, σ x ] = ω 0 iσ y = ω 0iŜy t S x = ω 0 ψt Ŝy ψt Ŝ y ψt = i e iω 0 t/ z eiω 0t/ +z ω 0 ψt Ŝy ψt = ω 0i = ω 0i e iω 0 t + e iω 0t e iω 0 t/ +z + e iω 0t/ z eiω 0t/ +z + e iω 0t/ z = ω 0i cos ω0 t i sin ω 0 t + cos ω 0 t i sin ω 0 t t Ŝx = t cos ω 0t = ω 0 sin ω 0t S y = sin ω 0t = ω 0 sin ω 0t t S y = ī h ψt Ĥ, Ŝy] ψt + ψt Ŝy t ψt Ĥ = ω 0 Ŝ z : Ŝy t = 0

5 Ĥ, Ŝx] = ω 0 σ z, σ y ] = ω 0 iσ x = ω 0iŜx t S y = ω 0 ψt Ŝx ψt Ŝ x ψt = e iω 0 t/ z + eiω 0t/ +z ω 0 ψt Ŝx ψt = ω 0 = ω 0 e iω 0 t + e iω 0t Problem.7 - Skippe Problem.8 e iω 0 t/ +z + e iω 0t/ e iω 0 t/ z +z + e iω 0t/ z = ω 0 cos ω0 t + i sin ω 0 t + cos ω 0 t i sin ω 0 t t Ŝy = t sin ω 0t = ω 0 cos ω 0t ψ0 = cos θ +z + sin θ z = ω 0 cos ω 0t ψt = e iω 0t/ cos θ +z + eiω 0t/ sin θ z S x = ψt Ŝx ψt = e iω0t/ cos θ +z + e iω 0t/ sin θ z Ŝ x e iω0t/ cos θ +z + eiω 0t/ sin θ z = e iω 0t/ cos θ +z + e iω 0t/ sin θ z e iω 0t/ cos θ z + eiω 0t/ sin θ +z = e iω0t sin θ cos θ + e iω 0t sin θ cos θ = sin θ cos θ cos ω 0t S x = sin θ cos ω 0t S y = ψt Ŝy ψt 5

6 = e iω0t/ cos θ +z + e iω 0t/ sin θ z Ŝ y e iω0t/ cos θ +z + e iω 0t/ sin θ z = i e iω 0t/ cos θ +z + e iω 0t/ sin θ z e iω 0t/ cos θ z eiω 0t/ sin θ +z = i e iω0t sin θ cos θ eiω 0t sin θ cos θ = i sin θ cos θ i sin ω 0t S y = sin θ sin ω 0t S z = ψt Ŝz ψt = e iω0t/ cos θ +z + e iω 0t/ sin θ z Ŝ z e iω0t/ cos θ +z + e iω 0t/ sin θ z = e iω 0t/ cos θ +z + e iω 0t/ sin θ z e iω 0t/ cos θ +z eiω 0t/ sin θ z = cos θ θ sin = cos θ Problem.9 S z = cos θ Ĥ = ω 0 Ŝ z + ω 1 cos ωtŝx ψ0 = +z Ĥ ψt = i ψt t ω 0 ω 1 cos ωt at ω 1 cos ωt ω 0 bt ȧt = i ḃt Approximation: B 1 B 0, ω 1 ω 0 at bt cte iω 0 t/ = te iω 0t/ 6

7 Approximation: ω ω 0 i ct = ω 1 iċt = ω 1 eiω 0 ωt t i t = ω 1 eiω ω 0t ct i ct = ω 1 iω 0 ωe iω 0 ωt t + e iω 0 ωt t] i t = ω 1 iω ω 0e iω ω0t ct + e iω ω 0tċt] iω 0 ωe iω 0 ωt i e iω ω 0tċt e iω 0 ωt iω ] 1 ω 1 eiω ω 0t ct ct = ω 1 ω 0 ω i ċt ω ] 1 ω 1 ct ct = iω 0 ωċt ω1 ct Characteristic equation: c iω 0 ωċ + ω1 c = 0 r + iω ω 0 ]r + ω1 = 0 r = iω 0 ω ± i ω 0 ω + ω 1 / ct = e iω ω0 ω 0 ωt/ A sin + ω 1 / ω0 ω t + B cos + ω 1 / t c0 = 1, 0 = 0: A = iω ω 0 ω0 ω + ω 1 / B = 1 ct = e iω iω ω 0 ωt/ 0 ω0 ω + ω 1 / sin ω0 ω + ω 1 / ω0 ω t + cos + ω 1 / t ċt = e iω ω 0 ωt/ 1 / ω 0 ω + ω 1 / sin ω0 ω + ω 1 / t 7

8 t = ie iω ω 0 ωt/ 1 / ω0 ω + ω 1 / sin ω0 ω + ω 1 / t z ψt = b tbt = tt = Problem.10 ω 1 / ω0 ω + ω 1 / ω 0 ω + ω 1 / sin t I = II = I = 1, I = 1 1 II = 1, II = 1 I Ĥ I = 1, I a a Ĥ a a I = 1 Ĥ Ĥ + Ĥ 1 + Ĥ = E 0 A Compare with: Analogous Hamiltonian: 1 Ĥ 1 1 Ĥ + Ĥ 1 Ĥ I Ĥ II = 1 Ĥ Ĥ Ĥ 1 Ĥ II Ĥ I = 1 Ĥ 1 1 Ĥ Ĥ 1 + Ĥ II Ĥ II = E Ĥ 0 A µ e E 0 cos ωt I,II µ e E 0 cos ωt E 0 + A Ĥ +z, z ω 0 ω 1 cos ωt ω 1 cos ωt ω 0 E + = E 0 + A E = E 0 A = µ e E = µ e E = E 0 + A 8

9 E Ĥ + µ e E 0 cos ωt II,I µ e E 0 cos ωt E ċt i t = µ e E 0 te ie + E t/ cos ωt cte ie + E t/ +z II, z I E + = ω 0 E 0 + A E = ω 0 E 0 A E + E = ω 0 A Analogue of Rabi s formula: ω 1 µ e E 0 I ψt = Problem.11 ψ0 = II µ e E 0 / A/ ω + µ e E 0 / sin µ = gq S mc B = B 0ˆk Ĥ = µ B = gq mc B 0Ŝz = ω 0 Ŝ z Ĥ 1, 1 = ω 0 Ŝ z 1, 1 = ω 0 1, 1 = E 1 1, 1 Ĥ 1, 0 = ω 0 Ŝ z 1, 0 = 0ω 0 1, 0 = E 0 1, 0 A/ ω + µ e E 0 / t Ĥ 1, 1 = ω 0 Ŝ z 1, 1 = ω 0 1, 1 = E 1 1, 1 ψ0 = 1, 1 y = 1 1, 1 + i 1, 0 1 1, 1 ψt = e iĥt/ 1 i 1, 1 + 1, 0 1 1, 1 9

10 = e ie 1t/ 1, 1 + i e ie 0t/ 1, 0 e ie 1t/ 1, 1 ψt = e iω 0t 1, 1 + i 1, 0 eiω0t 1, 1 x = 1 1, 1 + 1, , 1 1, 0 x = 1, 1 1, 1 1, 1 1, 1 x = 1 1, 1 1, , 1 1, 1 y = 1 1, 1 + i 1, 0 1 1, 1 1, 0 y = 1, 1 + 1, 1 1, 1 y = 1 1, 1 i 1, 0 1 1, 1 1, 1 x ψt = e iω0t e iω 0t + i = 1 sin ω 0t 1, 0 x ψt e iω0t + e iω0t = = cos ω 0 t 1, 1 x ψt = e iω0t e iω 0t i = 1 + sin ω 0t S x = 1 sin ω 0t sin ω 0t = sin ω 0 t 1, 1 y ψt = e iω0t + e iω 0t + 1 = 1 + cos ω 0t 1, 0 y ψt e iω0t e iω0t = = sin ω 0 t 1, 1 y ψt = e iω0t + e iω 0t 1 = 1 cos ω 0t S y = 1 + cos ω 0t cos ω 0t = cos ω 0 t 10

11 S z = = 0 Problem.1 Ĥ = ω 0 Ŝ x Problem.13 ψ0 = 1, 1 = 1 1, 1 x + 1, 0 x + 1 1, 1 x ψt = e ie 1t/ e ie 0 t/ 1, 1 x + 1, 0 x + e ie 1t/ 1, 1 x = e iω 0t 1, 1 x + 1, 0 x + eiω0t 1, 1 x 1, 1 = 1 1, 1 x 1, 0 x + 1 1, 1 x 1, 1 ψt = e iω0t + e iω 0t 1 = 1 cos ω 0t Ĥ 1,,3 E 0 0 A 0 E 1 0 A 0 E 0 Fin eigenstates: E 0 λ 0 A 0 E 1 λ 0 A 0 E 0 λ = E 0 λe 1 λe 0 λ A E 1 λ = 0 λ E 0 A = 0 λ = E 1, E 0 ± A E 0 λ 0 A 0 E 1 λ 0 A 0 E 0 λ a b c = 0 λ = E 1 : E 0 E 1 a + Ac = 0 Aa + E 0 E 1 c = 0 a = c = 0, b = 1 E 1 = 11

12 λ = E 0 + A: Aa + Ac = 0 E 1 E 0 Ab = 0 Aa Ac = 0 a = c = 1, b = 0 E 0 + A = λ = E 0 A: Aa + Ac = 0 E 1 E 0 + Ab = 0 Aa + Ac = 0 a = c = 1, b = 0 E 0 A = a ψ0 = b ψt = e iĥt/ ψ0 = e ie 1t/ E 1 ψ0 = 3 = 1 E 0 + A 1 E 0 A ψt = e iĥt/ ψ0 = e ie 0+At/ E 0 + A e ie 0 At/ E 0 A Problem.1 0 ie0 Ĥ x,y ie 0 0 ω 0 0 i Ĥ x,y i 0 +ω 0 / = +y = 1 +z + i z ω 0 / = y = 1 +z i z a By analogy, the eigenstates an eigenvalues are: 1

13 +E 0 = 1 x + i y E 0 = 1 x i y b ψ0 = x = 1 +E E 0 ψt = e iĥt/ ψ0 = e ie 0t/ x ψt = e ie0t/ + e ie 0t/ y = i +E 0 + i E 0 y ψt = ie ie0t/ + ie ie 0t/ +E 0 + eie 0t/ E 0 = cos E 0t = sin E 0t The photon polarization is oscillating between the x an y states. Problem.15 Â, ˆB] = iĉ A B C Â, ˆB] = i Ĉ Ĉ = Â, ˆB] A t = ī h ψt Ĥ, Â]ψt = ī Ĥ, Â] h Ĥ, Â] = A /t Ĥ, Â] E A A E A /t t is time neee for expecte value of observable to change significantly. Problem.16 - Skippe 13

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