Direction of nonlinear polarization

[snickersnee]

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.

 

[mark726]

The direction of the nonlinear polarization can be found by considering the Poynting vector. Since the EM wave is propagating in the x-direction, the Poynting vector will also be in the x-direction, as it is given by the cross product of the electric and magnetic fields.

Using the expression for the Poynting vector, we can write:

\vec{S} = <\vec{E} \times \vec{H}> = <E_xH_z - E_zH_x, E_yH_x - E_xH_y, E_zH_y - E_yH_z>

Since the EM wave is polarized along the \frac{1}{\sqrt{2}}(\hat{y}+\hat{z}) direction, we can write the electric field as:

\vec{E} = <0, E_y, E_z>

Plugging this into the Poynting vector equation, we get:

\vec{S} = <0, E_zH_x - E_xH_y, E_yH_x - E_xH_y>

Since we are only interested in the z-component of the Poynting vector, we can ignore the first two terms. The z-component of the Poynting vector is given by:

S_z = E_yH_x - E_xH_y

To find the direction of the nonlinear polarization, we need to find the direction of the electric field at the frequency 2w. Since the electric field at this frequency is given by:

E_z(2w) = \sqrt{2}E_z(w)

We can write the z-component of the Poynting vector as:

S_z = \sqrt{2}E_z(w)H_x - E_xH_y

Since the electric field and magnetic field at the frequency 2w are perpendicular to each other, we can write:

H_x = \frac{1}{\sqrt{2}}H_z

Plugging this into the equation for S_z, we get:

S_z = \sqrt{2}E_z(w)\frac{1}{\sqrt{2}}H_z - E_xH_y

Simplifying, we get:

S_z = E_z(w)H_z - E_xH_y

Now, using the expression for the nonlinear polarization, we can write:

P_z^{(2)}(2w) = \epsilon_0 \chi^{