# Parity conservation in second harmonic generation?

The second harmonic arises from susceptibility of third rank tensor $X^{(2)}$ which have (-1) parity.

Let say two photons are absorbed and one is emitted, so the total change in parity is $(-1)^{(2+1)}$. The initial state equals the final state so $(-1)^0=1$.

Where is the mistake here and how to conserve parity?

The requirement is that $\chi^{(2)}$ be non-centrosymmetric. That's a bit different than having a particular parity. The states involved must be neither odd nor even; the parity must be mixed. That way the dipole matrix element exists between all three intermediate states involved in calculation of the susceptibility.
• $P_\mu$ has negative parity, and so does the product $\chi_{\mu\nu\lambda}^{(2)}E_\nu E_\lambda$. Doesn't that work out ? – garyp May 25 '16 at 18:34