1.3 Solutions to Problems
7
1.3 Solutions to Problems
1.3.1 Inversion Symmetry of χ (2)
[Problem 1.1] If we suppose that material properties such as χ (2) are invariant by
inversion, show χ (2) = 0. This indicates that the second-order optical processes in
Eq. (1.1) are forbidden for a centrosymmetric material.
We impose electric field on the material and consider the induced polarization.
The second term in Eq. (1.1) is denoted by P
(2)
p ,
P
(2)
p =
q,r
χ
(2)
pqr E q E r ,
(1.5)
which is the source of the second-order optical process. If we operate inversion
upon the coordinate system, vector quantities such as P
(2)
p , E q or E r change their
signs. However, material properties such as χ (2) are assumed to be invariant by the
inversion. Therefore, the inversion operation transforms Eq. (1.5) to
− P
(2)
p =
q,r
χ
(2)
pqr
−E q
(−E r ) .
(1.6)
Both Eqs. (1.5) and (1.6) should hold simultaneously, which necessarily leads to
χ (2) = 0. This means that P (2) = 0 in Eq. (1.5) for a centrosymmetric material.
1.3.2 Time and Frequency Domains
[Problem 1.2] Derive Eq. (1.4) from Eqs. (1.2) and (1.3).
During the derivation, make use of the fact that the coefficient χ (2) (t, t , t ) in
Eq. (1.2) does not depend on the origin of time, i.e. χ (2) (t, t , t ) = χ (2) (t + t +
, t + ) with an arbitrary time shift by . (In other words, χ (2) (t, t , t ) is a
function of the time intervals, τ ≡ t − t and τ ≡ t − t .)
Explain that only the component of sum frequency, = ω 1 + ω 2 , appears in
the left-hand side of Eq. (1.4), when the right-hand side is composed of E(ω 1 ) and
E(ω 2 ).
The time-dependent polarization of Eq. (1.2) is represented using the Fourier
series of the electric fields as