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ID: past-exam-of-the-laser-science-and-quantum-information-processing-course-of-the-university-of-oxford/2015/4/b/solution

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Past exam of the laser science and quantum information processing course of the University of Oxford / 2015 / 4 / b / Solution by

Codex  0 2026-10-25

For a fundamental electric field in the

x y

E_{z} = 0

. The specified tensor then gives

P_{x} = P_{y} = 0, P_{z} (2 ω) = 2 ϵ_{0} d_{36} E_{x} E_{y} .

E_{x} = E_{0} cos ϕ

and

E_{y} = E_{0} sin ϕ

yields

P_{z} = ϵ_{0} d_{36} E_{0}^{2} sin 2 ϕ, I_{2 ω} \propto sin^{2} 2 ϕ, ϕ = 4 5^{\circ} (mod 9 0^{\circ}) .

The source electric polarization is therefore along

z

and is largest along the in-plane diagonals. A radiated wave also needs a transverse component of this source: propagation along

z

would not generate a propagating

z

-polarized output. Propagation in the

x y

plane provides the simple strictly transverse example. For a tilted extraordinary wave, the electric displacement, rather than generally the entire electric field, is perpendicular to the wave vector; the tensor source must be projected onto that wave's allowed polarization. This distinction matters when using the type-I geometry below.

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