# Nonlinear polarization (second and third order)

Why does second order nonlinear polarization occur only in crystal materials with a non-centrosymmetric crystal structure? (Nonlinear effects at crystal surfaces are an exception). Why does third order polarization occur in basically all media? The answer, found here http://www.rp-photonics.com/nonlinear_polarization.html did not help (at least for now).

-

## 1 Answer

It's because centrosymmetric crystals prohibit a nonzero second-order nonlinear polarization by the centrosymmetry (I will call it parity, sorry), $$P: (x,y,z) \to (-x,-y,-z).$$ Work with the formula $$P_i = A_{ij} E_{j} + B_{ijk} E_{j} E_{k}$$ where I renamed the coefficients $A,B$ relatively to the usual conventions. Under the parity symmetry $P$, the vectors and tensors map as $$P_i\to -P_i, \qquad E_i\to -E_i$$ because they have an odd number of indices and they're polar vectors, not axial vectors. Centrosymmetric crystals have the internal constants $A,B$ constant under the parity symmetry $$A_{ij} \to A_{ij}, \qquad B_{ijk} \to B_{ijk}.$$ It's important that there's no sign in the transformation of $B$, either (despite the odd number of three indices): we literally get the same crystal back by the parity symmetry so there can't be any changes of its material constants, not even the tensor-valued ones.

When the transformation laws are applied, the original equation transforms to $\to$ $$-P_i = -A_{ij} E_{j} + B_{ijk} E_{j} E_{k}$$ which contradicts the original formula unless $B_{ijk}=0$. So the nonlinear term must vanish because of the transformation rules under parity.

Only odd powers of $E_i$ might appear on the right hand side with a nonzero coefficient. In particular, terms that are cubic in $E_i$ are exactly as allowed by the symmetry as the terms that are linear in $E_i$; for example, $E^2\cdot E_i$ has pretty much identical transformation rules under any parity-like symmetry as $E_i$, so if $E_i$ is allowed, so must be the $E^2\cdot E_i$ terms. In fact, the lowest order (linear) terms might be set to zero by some special tuning but it's virtually impossible to get rid of the cubic terms which have a higher number of independent tensor components.

-