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

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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?

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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.

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  • $\begingroup$ But isnt the polarization P and the E field with (-1) parity as a vectors. And this brings definite parity for all χ(n) ? This guy at least says so books.google.com/… $\endgroup$ – Anonymous May 25 '16 at 18:12
  • $\begingroup$ $P_\mu$ has negative parity, and so does the product $\chi_{\mu\nu\lambda}^{(2)}E_\nu E_\lambda$. Doesn't that work out ? $\endgroup$ – garyp May 25 '16 at 18:34
  • $\begingroup$ Its perfect. But I do not understand how the parity is conserved for SHG, seems to work fine for χ(n) odd, but not for χ(n) even. My question basically how one conserves parity through out the process. $\endgroup$ – Anonymous May 25 '16 at 19:16

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