How does this type of pump-probe spectroscopy work?

Question

In optical Kerr effect spectroscopy, a pump beam causes a temporary birefringence in a sample, while a probe beam measures the return of the sample to equilibrium.

Here’s a schematic:

The pump and probe are polarized linearly, 45˚ relative to one another, with the probe coming in horizontally. The birefringence causes a small elliptical polarization of the probe beam exiting the sample. A polarizer after the sample is adjusted to let vertically polarized light pass through.

I have some questions about this set up:

-Why must the polarizations of the pump and probe be different by 45˚?

-How does anything get through, namely the elliptically polarized light that constitutes the signal?

-And how does the intensity of the signal change due to the polarization-sensitive components?

Answer 1

It is not necessary that the polarization of the pump is $45^\circ$ but the effect is strongest for this configuration and will be zero for either $0$ or $90^\circ$ Look at how lambda/4 or lambda/2 plates work if you want to know more.

The reason that something goes through is that the birefringence of the crystal introduces other electric field components than $0^\circ$, which can pass the polarizer. The signal changes according to the nonlinear kerr effect, which is $n_e = n_o + n_2 I_p$.