# Why the Kerr non-linearity is $a^\dagger a^\dagger a a$

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.

• 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? – Thomas Russell Jun 12 '17 at 13:30
• @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.^_^ – yangcs11 Jun 12 '17 at 13:49