Direction of nonlinear polarization

  • #1

Homework Statement



A linearly-polarized electromagnetic wave with a frequency w and with an
intensity of 1MW/cm2
is propagating in x-direction in a nonlinear crystal with a
refractive index n=1.5. Assume that a second-order nonlinear optical susceptibility tensor
for second-harmonic generation in the nonlinear crystal has only one nonzero component
[itex]\chi^{(2)}_{zzz}(2w,w,w)=10 pm/V (10-11 m/V)[/itex]. The electromagnetic wave is polarized along [itex]\frac{1}{\sqrt{2}}(\hat{y}+\hat{z})[/itex]-direction.

Calculate the amplitude and direction of a nonlinear polarization
oscillating at frequency 2w in the crystal. (Use the expression for the Poynting
vector to deduce the value of E-field in the EM wave and be careful with the
geometry and various factors of 2 when doing calculations.)



Homework Equations



Poynting vector: [itex]\vec{S}=<\vec{E} \times \vec{H}>[/itex]

[itex]P_z^{(2)}(2w)=\epsilon_0 \chi^{(2)}_{zzz}(2w;w,w)E_z^2(w)[/itex]




The Attempt at a Solution



I don't need the amplitude explained, just the direction. (The direction is along z axis, why is that?) The linear polarization is along the same direction as the EM wave. But how do I find the direction for the nonlinear case? I tried crossing [itex]\hat{x}\ with\ \frac{1}{\sqrt{2}}(\hat{y}+\hat{z})[/itex] but that gives [itex]-\frac{1}{\sqrt{2}}(\hat{y}+\hat{z})[/itex] which is wrong.
 

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