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The non-linear optical Hamiltonian of the third order is written as: $$ H = H_l - \frac{1}{3\varepsilon_0}\int d\mathbf{r} D^i(\mathbf{r})\Gamma_2^{ijk} D^j(\mathbf{r}) D^k(\mathbf{r}) - \frac{1}{4\varepsilon_0}\int d\mathbf{r} D^i(\mathbf{r})\Gamma_3^{ijkl} D^j(\mathbf{r}) D^k(\mathbf{r})D^l(\mathbf{r}) $$ where $H_l$ is the linear part.

I have some problems getting this result; my approach is to take the polarization $$ P_i = \Gamma^{ij}_1 D^j(\mathbf{r}) + \Gamma^{ijk}_2 D^j(\mathbf{r})D^k(\mathbf{r})+ \Gamma^{ijkl}_3 D^j(\mathbf{r})D^k(\mathbf{r})D^l(\mathbf{r}) $$ where the sum is implicit over repeated indexes and write the standard Hamiltonian as follow: $$ H = \frac{1}{2}\int d\mathbf{r} \mathbf{E}\cdot\mathbf{D} = \frac{1}{2\varepsilon_0}\int d\mathbf{r} (\mathbf{D-P})\cdot\mathbf{D} = \frac{1}{2\varepsilon_0}\int d\mathbf{r}\mathbf{D}\cdot\mathbf{D} - \frac{1}{2\varepsilon_0}\int d\mathbf{r}\mathbf{P}\cdot\mathbf{D}\, . $$ (Taken from section VII of this paper.)

Focusing on the last term: $$\int d\mathbf{r}\mathbf{P}\cdot\mathbf{D} = \int d\mathbf{r}P^iD^i =\int d\mathbf{r}D^i(\mathbf{r})\left(\Gamma^{ij}_1 D^j(\mathbf{r}) + \Gamma^{ijk}_2 D^j(\mathbf{r})D^k(\mathbf{r})+ \Gamma^{ijkl}_3 D^j(\mathbf{r})D^k(\mathbf{r})D^l(\mathbf{r})\right). $$ This leads to a very similar Hamiltonian apart from the factors $\frac{1}{3},\frac{1}{4}$ in front of the integral. What am I doing wrong? Is there an explanation for those factors?

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Your expression for the hamiltonian, $H = \frac12\int\mathbf E\cdot \mathbf D \, \mathrm d\mathbf r$, is incorrect. Following up the reference trail of the paper you quote quickly leads to a related paper that makes it clear that the relevant quantity in that formalism is $$ H = \int\left[ \int \mathbf E\cdot \mathrm d\mathbf D\right]\mathrm d\mathbf r, $$ and further pointing to chapter 4 of J.D. Jackson's Classical Electrodynamics as a reference for that.

In principle, the factors of $1/3$ and $1/4$ come out of the integration with respect to $\mathbf D$. Thus, for the linear case, the relevant integral is of the form $$ \int D^j\mathrm dD^i=\frac{2-\delta^{ij}}{2}D^jD^i, $$ with a prefactor to handle the behaviour where $i=j$ and therefore the integral should read e.g. $\int D^1\mathrm dD^1=\frac{1}{2}(D^1)^2$. As you can imagine, the higher-order terms get messier on a combinatorically fast scale, because of the many ways the polynomial could integrate; as an example, the second-order nonlinearity gives a term of the form $$ \int D^jD^k\mathrm dD^i=\left(1-\frac{\delta^{ji}+\delta^{ki}}{2}+\frac{\delta^{ji}\delta^{ki}}{3}\right)D^jD^kD^i $$ to account for the cases where $j=i$ or $k=i$ separately, and where $i=j=k$.

Finally, you need to contract this with the $\Gamma_2^{ijk}$, using the fact that they are symmetric in the Cartesian components (because the line integral $\int \mathbf E\cdot \mathrm d\mathbf D$ needs to be independent of the path of integration). This should give you the results that Sipe et al. claim in that second paper, but to be honest with you I'm not sure this actually works, even in the linear case: the integral still gives $$ \int \Gamma_1^{ij}D^j\mathrm dD^i=\frac{2-\delta^{ij}}{2}\Gamma_1^{ij}D^jD^i=\frac12\Gamma_1^{xx}D^xD^x+(\Gamma_1^{xy}+\Gamma_1^{yx})D^xD^y+\frac12\Gamma_1^{yy}D^yD^y $$ when only the $x$ and $y$ components are nonzero, and this does not reduce to $\Gamma_1^{ij}D^jD^i$ as claimed. Ultimately this can be fixed by refactoring those combinatoric terms into the $\Gamma$s, but they seem to be explicitly not doing that, so I would tread with extreme caution with those results.

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  • $\begingroup$ Thanks for the answer, but it's not completely satisfactory. I don't think the terms could be refactored in the $\Gamma$'s, since later in the paper he expresses the $\Gamma$'s as a function of the usual $\chi$ and those terms don't appear. So there's still a discrepancy in the results. Anyway, it could be the right way, I'll try something around this line and maybe I will figure something out. Thanks! $\endgroup$ – user85231 Mar 30 '17 at 12:03
  • $\begingroup$ Indeed, I agree that there's probably some discrepancies in those papers. I would recommend looking for how other papers in the literature deal with this. In any case, the $\int \mathbf E\cdot \mathbf D \,\mathrm d\mathbf r$ hamiltonian in your question is definitely not correct - you need the $\int\mathbf E\cdot\mathrm d\mathbf D$ formulation to account for how energy is stored in the polarization as the field strength increases (see Jackson for details), and this obviously needs to change if the polarization's dependence on $E$ is no longer linear. $\endgroup$ – Emilio Pisanty Mar 30 '17 at 12:43

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