Papers reporting on devices such as this Mach-Zehnder modulator:
High-performance hybrid silicon and lithium niobate Mach–Zehnder modulators for 100 Gbit s−1 and beyond | Nature Photonics
Apply electric fields that are well beyond the coercive field of lithium niobate: ~210 kV/cm.
Coercive fields in ferroelectrics: A case study in lithium niobate and lithium tantalate (scitation.org)
How do these modulators still work?
These modulators do not work by the ferroelectric effect. They work by the electrooptic effect. The ferroelectric polarization messes up the electrooptic effect. Opposite domains result in opposite phase retardation effectively cancelling each other. In the regime of low field, the hysteresis of the polarization makes the electrooptic response nonlinear. Working with fields well above the coercive force ensures the domains are aligned so that the electrooptic effect is linear and predictable.
The crystals also have a piezoelectric response which also messes up the electrooptic effect. On acoustic time scales the physical movement of the crystal cancels some of the field reducing the electrooptic response. This makes the electrooptic response frequency dependent.
The point is you never get just the behavior you want. Nonlinear crystals have every nonlinear behavior. They have nonisotropic coefficients of thermal conductivity, complicated stress tensors, etc. all for the same reason: the crystal symmetry. If you want to exploit one effect, you usually have to work around the others. Thus, in this case, you want an electrooptic modulator, but you wind up having to think about ferroelectrics, piezoelectrics, walk-off from the nonisotropic dielectric tensor, etc., etc.
ok, 10 kV/cm isn‘t going to change any domains, so the operation of the modulator would depend on what domains were manufactured into the film. I don’t know if they purposely align the domains once the device is made. I found a couple of papers indicating that would be helpful/effective, but I couldn’t find any proof it is actually done.
I see. I think people just fabricate their waveguides perpendicular to the built-in polarization.
And anyway for my application I don't have to apply the electric field opposite to the polarization. If I understand correctly, I suppose I can apply an E field in the same direction as the intrinsic polarization up to LN's breaking strength of 65 V/um and still be fine.
Again, thank you for spending time to write your answers !