Exam (elaborations) TEST BANK FOR Introductory Quantum Optics 1st Edition By Gerry C and Knight P [Cambridge Solution Manual]

Eq. (2.5) has the form Ex(z; t) = s 2!2 V "0 q(t) sin(kz); (2.1.1) and Eq. (2.2) r £ B = ¹0"0 @E @t : (2.1.2) Both equations lead to ¡@zBy = ¹0"0 s 2!2 V "0 q_(t) sin(kz); (2.1.3) which itself leads to Eq. (2.6) By(z; t) = ¹0"0 k s 2!2 V "0 q_(t) cos(kz): (2.1.4) 2.2 problem 2.2 H = 1 2 Z dV · "0E2 x(z; t) + 1 ¹0 B2 y (z; t) ¸ : (2.2.1) 9 10 CHAPTER 2. FIELD QUANTIZATION From the previous problem Ex(z; t) = s 2!2 V "0 q(t) sin(kz); (2.2.2) so "0E2 x(z; t) = 2!2 V q2(t) sin2(kz): (2.2.3) Also By(z; t) = ¹0"0 k s 2!2 V "0 q_(t) cos(kz); (2.2.4) and 1 ¹0 B2 y (z; t) = 2 V p2(t) cos2(kz); (2.2.5) where we have used that c2 = (¹0"0)¡1, p(t) = q_(t), and ck = !. Eq. 2.2.1 becomes then H = 1 V Z dV £ !2q2(t) sin2(kz) + p2(t) cos2(kz) ¤ : (2.2.6) Using these simple trigonometric identities cos2 x = 1+cos 2x 2 and sin2 x = 1¡cos 2x 2 , we can simplify equation 2.2.6 further to: H = 1 2V Z dV £ !2q2(t)(1 + cos 2kz) + p2(t)(1 ¡ cos 2kz) ¤ : (2.2.7) Because of the periodic boundaries both cosine terms drop out, also 1 V R dV = 1 and we end up by H = 1 2 ¡ p2 + w2q2¢ : (2.2.8) It is easy to see that this Hamiltonian has the form of a simple harmonic oscillator. 2.3 problem 2.3 Let f be a function de¯ned as: f(¸) = ei¸ ^ A ^B e¡i¸A^: (2.3.1) 2.4. PROBLEM 2.4 11 If we expand f as f(¸) = c0 + c1(i¸) + c2 (i¸)2 2! + :::; (2.3.2) where c0 = f(0) c1 = f0(0) c2 = f00(0) ¢ ¢ ¢ Also c0 = f(0) = ^B c1 = f0(0) = h ^ Aei¸ ^ A ^B e¡i¸ ^ A ¡ ei¸ ^ A ^B A^e¡i¸A^ i¯¯¯ ¸=0 = h ^ A; ^B i c2 = h ^B ; h ^ A; ^B ii : The same way we can determine the other coe±cients. 2.4 problem 2.4 Let f(x) = e A^xe ^B x (2.4.1) df(x) dx = A^e A^xe ^B x + e ^ Ax ^B e ^B x = ³ A^ + e ^ Ax ^B e¡A^x ´ f(x) It is easy to prove that h ^B ;A^n i = nA^n¡1 h ^B ;A^ i (2.4.2) 12 CHAPTER 2. FIELD QUANTIZATION h ^B ; e¡A^x i = X 2 4^B ; ³ ¡A^x ´n n! 3 5 = X (¡1)n xn n! h ^B ;A^n i = X (¡1)n xn (n ¡ 1)! A^n¡1 h ^B ;A^ i = ¡e¡A^x h ^B ;A^ i x So ^B e¡A^x ¡ e¡ ^ Ax ^B = ¡e¡A^x h ^B ;A^ i x e¡ ^ Ax ^B e ^ Ax = ^B ¡ e¡A^x h ^B ;A^ i x (2.4.3) e ^ Ax ^B e¡ ^ Ax = ^B + e A^x h ^ A; ^B i x (2.4.4) Equation 4.1.1 becomes df(x) dx = ³ ^ A + ^B + h ^ A; ^B i´ f(x): (2.4.5) Since h ^ A; ^B i commutes with ^ A and ^B , we can solve equation 2.4.5 as an ordinary equation. The solution is simply f(x) = exp h³ ^ A + ^B ´ x i exp µ 1 2 h ^ A; ^B i x2 ¶ (2.4.6) If we take x = 1 we will have e ^ A+^B = e A^e ^B e¡1 2 [ ^ A;^B ] (2.4.7) 2.5 problem 2.5 jª(0)i = 1 p 2 ¡ jni + ei'jn + 1i ¢ : (2.5.1) 2.5. PROBLEM 2.5 13 jª(t)i = e¡i ^H t ~ jª(0)i = 1 p 2 ³ e¡i ^H t ~ jni + e¡i ^H t ~ jn + 1i ´ = 1 p 2 ¡ e¡in!tjni + ei'e¡i(n+1)!tjn + 1i ¢ ; where we have used E ~ = ! ^njª(t)i = ^ay^ajª(t)i = 1 p 2 ¡ e¡in!tnjni + ei'e¡i(n+1)!t(n + 1)jn + 1i ¢ h^ni = hª(t)j^njª(t)i = 1 2 (n + n + 1) = n + 1 2 the same way ­ ^n2® = hª(t)j^n^njª(t)i = 1 2 ¡ n2 + (n + 1)2¢ = n2 + n + 1 2 ­ (¢^n)2® = ­ ^n2® ¡ h^ni2 = 1 4 ^E jª(t)i = E0 sin(kz) ¡ ^ay + ^a ¢ jª(t)i = 1 p 2 E0 sin(kz) ¡ ^ay + ^a ¢ ¡ e¡in!tjni + ei'e¡i(n+1)!tjn + 1i ¢ = 1 p 2 E0 sin(kz) h e¡in!t ³p n + 1jn + 1i + p njn ¡ 1i ´ + ei'e¡i(n+1)!t ³p n + 2jn + 2i + p n + 1jni ´i 14