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); (...