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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)

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)

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  1. In the lecture, we have derived the expression of second order susceptibility as
χ ( 2 ) ( ω 1 + ω 2 , ω 1 , ω 2 ) = N e 3 a ε 0 m 2 D ( ω 1 + ω 2 ) D ( ω 1 ) D ( ω 2 )
(a) Based on this result, show Miller's rule: the quantity
χ ( 2 ) ( ω 1 + ω 2 , ω 1 , ω 2 ) χ ( 1 ) ( ω 1 + ω 2 ) χ ( 1 ) ( ω 1 ) χ ( 1 ) ( ω 2 )
is almost a constant for all noncentrosymmetric crystals.
(b) By writing ω 3 = ω 1 + ω 2 , show that
χ ( 2 ) ( ω 3 , ω 1 , ω 2 ) = χ ( 2 ) ( ω 1 , ω 3 , ω − 2 ) = χ ( 2 ) ( ω 2 , ω 3 , ω − 1 )

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