1. In the lecture, we have derived the expression of second order susceptibility as χ (2) (ω1 + ω2, ω1, ω2) = Ne3a ε0m2D (ω1 + ω2) D (ω1) D (ω2) (a) Based on this result, show Miller’s rule: the quantity χ (2) (ω1 + ω2, ω1, ω2) χ(1) (ω1 + ω2) χ(1) (ω1) χ(1) (ω2) is almost a constant for all noncentrosymmetric crystals. (b) By writing ω3 = ω1 + ω2, show that χ (2) (ω3, ω1, ω2) = χ (2) (ω1, ω3, ω−2) = χ (2) (ω2, ω3, ω−1)
2. We have derived the third order susceptibility as χ (3) ijkl (ωq, ωm, ωn, ωp) = N be4 [δij δkl + δikδjl + δilδjk] 3ε0m3D (ωq) D (ωm) D (ωn) D (ωp) Show that the nonlinear susceptibility possesses the following tensor properties: χ1122 = χ1212 = χ1221 = χ1133 = χ1313 = χ1331 = χ2233 = χ2323 = χ2332 = χ2211 = χ2121 = χ2112 = χ3311 = χ3131 = χ3113 = χ3322 = χ3232 = χ3223 = 1 3 χ1111 = 1 3 χ2222 = 1 3 χ3333 with all other elements vanishing.
3. Consider the Difference-Frequency Generation. Suppose the ω3 wave is a strong wave (i.e., it is undepleted by the nonlinear interaction, so that A3 can be treated essentially as a constant), and no wave at frequency ω2 is incident on the medium. The coupled amplitude equations are: dA1 dz = 8πideff ω 2 1 k1c 2 A3A2 ∗ e i∆kz dA2 dz = 8πideff ω 2 2 k2c 2 A3A1 ∗ e i∆kz ∆k = k3 − k1 − k2 If ∆k = 0, the solution for A1 and A2 that meet the boundary conditions (A2(0) = 0, A1(0) specified) can be found as, ( A1(z) = A1(0) cosh κz A2(z) = i n1ω2 n2ω1 1/2 A3 |A3|A∗ 1 (0) sinh kz Show that if ∆k 6= 0, the solution for A1 and A2 can be, A1(z) = A1(0) cosh gz − i∆k 2g sinh gz + K1 g A ∗ 2 (0) sinh gz e (1/2)i∆kz A2(z) = A2(0) cosh gz − i∆k 2g sinh gz + K2 g A ∗ 1 (0) sinh gz e (1/2)i∆kz where, g = " κ1κ ∗ 2 − ∆k 2 2 #1/2 ; κj = 8πiω2 j def fA3 kj c 2