Theory of electric polarization induced by inhomogeneity in crystals

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Strictly speaking, there is still an ambiguity in the definition of 𝑷 \bm{P} given here because 𝒋 = ∂ 𝑷 / ∂ t + ∇ × 𝑴 \bm{j}=\partial\bm{P}/\partial t+\bm{\nabla}\times\bm{M} also contains a contribution from the magnetization current. As a result, 𝑷 \bm{P} is only defined up to a divergence-free field. One can of course fix the gauge by imposing that 𝑷 \bm{P} only have a longitudinal component. However, this is not a critical issue because the ambiguity can be removed by spatial averaging

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The integral of terms that do not contain λ ˙ \dot{\lambda} is given by − e ∫ BZ d 𝒌 ∫ 0 T d t [ ∇ α k ε + ( Ω β ​ β k ​ r ∇ α k ε − Ω α ​ β k ​ r ∇ β k ε + Ω α ​ β k ​ k ∇ β r ε ) ] -e\int_{\text{BZ}}d\bm{k}\int_{0}^{T}dt[\nabla^{k}_{\alpha}\varepsilon+(\Omega^{kr}_{\beta\beta}\nabla^{k}_{\alpha}\varepsilon-\Omega^{kr}_{\alpha\beta}\nabla^{k}_{\beta}\varepsilon+\Omega^{kk}_{\alpha\beta}\nabla^{r}_{\beta}\varepsilon)] . The integral of ∇ α k ​ ε \nabla^{k}_{\alpha}\varepsilon over the entire Brillouin zone obviously vanishes. After integration by parts and making use of the Bianchi identity ∇ α k Ω β ​ β r ​ k + ∇ β k Ω α ​ β k ​ r + ∇ β r Ω β ​ α k ​ k = 0 \nabla^{k}_{\alpha}\Omega^{rk}_{\beta\beta}+\nabla^{k}_{\beta}\Omega^{kr}_{\alpha\beta}+\nabla^{r}_{\beta}\Omega^{kk}_{\beta\alpha}=0 , one can show that the integral of the last three terms contributes a divergence-free part, which can be discarded

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The formal derivation of Eq. ( 9

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