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I am current studying nonlinear optics, the second order polarization reads as $$ \begin{equation} P^{(2)}_\mu = \epsilon_0\chi^{(2)}_{\mu\alpha\beta}\vdots E_\alpha E_\beta \end{equation} $$ And the energy can be wrote as $H = \frac{1}{2}\vec{D}\cdot\vec{E}$, so I think the quantum mechanical Hamiltonian should look like $$ \begin{align} H = &\frac{1}{2}\epsilon_0\chi^{(2)}_{\mu\alpha\beta}E_\mu E_\alpha E_\beta\\ =& g (\hat{a}^\dagger \hat{a} \hat{a} + H.c.) \end{align} $$ But instead, the Kerr nonlinearity term is $\hat{a}^\dagger \hat{a}^\dagger \hat{a} \hat{a}$. Could anyone tell me why? Thanks a lot.

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    $\begingroup$ I may be wrong, but isn't the process you are describing the Pockel's effect? The Kerr effect being the third-order polarization? This would explain why the Kerr nonlinearity term is $\hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\hat{a}$ rather than the Hamiltonian that you got? $\endgroup$ – Thomas Russell Jun 12 '17 at 13:30
  • $\begingroup$ @Thomas yes it is. I made a mistake here. I remembered that in some literature the Kerr term is called second order nonlinearity. Maybe my memory was wrong. Thanks a lot.^_^ $\endgroup$ – yangcs11 Jun 12 '17 at 13:49

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