Electric Polarization by Berry Phase: Ver. 1.1

The polarization coming from the electric contribution is given by

$𝐏={\displaystyle \sum _{k=1}^3}{P}_i{𝐑}_i.$ (1)

$P_i$ can be evaluated by the following Berry phase formula [1, 2]:

$2\pi P_i$ $=$ ${𝐆}_i\cdot 𝐏$ (2)

$=$ $-{\displaystyle \frac{e}{{(2\pi )}^3}}{\displaystyle \sum _{\sigma }}{\displaystyle {\int }_{\mathrm{B}}}𝑑k^3{𝐆}_i\cdot {\left({\displaystyle \frac{\partial }{\partial {𝐤}^{\prime }}}{\eta }_{\sigma }(𝐤,{𝐤}^{\prime })\right)}_{{𝐤}^{\prime }=𝐤},$

where ${\int }_{\mathrm{B}}$ means that the integral over the first Brillouin zone of which volume is $V_{\mathrm{B}}$. The quantum phase ${\eta }_{\sigma }(𝐤,{𝐤}^{\prime })$ is given by

${\eta }_{\sigma }(𝐤,{𝐤}^{\prime })=\mathrm{Im}\left\{\ln \left(\det ⟨u_{\sigma \mu }^{(𝐤)}|u_{\sigma \nu }^{({𝐤}^{\prime })}⟩\right)\right\},$ (3)

where $\mu$ and $\nu$ run over the occupied states. The integration and derivative in Eq. (2) are approximated by a discretization:

${𝐆}_i\cdot 𝐏\approx -{\displaystyle \frac{e}{V_{\mathrm{B}}{N}_2{N}_3}}{\displaystyle \sum _{\sigma }}{\displaystyle \sum _{i_2=0,i{3}_{=}0}^{N_2-1,N_3-1}}{\displaystyle \sum _{i_1=0}^{N_1-1}}{\eta }_{\sigma }({𝐤}_{i_1{i}_2{i}_3},{𝐤}_{i_1+1i_2{i}_3}^{\prime }).$ (4)

Noting that

${\psi }_{\sigma \mu }^{(𝐤)}(𝐫)$ $=$ ${\mathrm{e}}^{\mathrm{i}𝐤\cdot 𝐫}{u}_{\sigma \mu }^{(𝐤)}(𝐫),$ (5)

$=$ ${\displaystyle \frac{1}{\sqrt{N}}}{\displaystyle \sum _{\mathrm{n}}^N}{\mathrm{e}}^{\mathrm{i}{𝐑}_{\mathrm{n}}\cdot 𝐤}{\displaystyle \sum _{i\alpha }}{c}_{\sigma \mu ,i\alpha }^{(𝐤)}{\varphi }_{i\alpha }(𝐫-{\tau }_i-{𝐑}_{\mathrm{n}}),$

the overlap matrix $⟨u_{\sigma \mu }^{(𝐤)}|u_{\sigma \nu }^{(𝐤+\mathrm{\Delta }𝐤)}⟩$ in Eq. (3) is evaluated as

$⟨u_{\sigma \mu }^{(𝐤)}|u_{\sigma \nu }^{(𝐤+\mathrm{\Delta }𝐤)}⟩$ $=$ $⟨{\psi }_{\sigma \mu }^{(𝐤)}|{\mathrm{e}}^{\mathrm{i}𝐤\cdot 𝐫}{\mathrm{e}}^{-\mathrm{i}𝐤\cdot 𝐫}{\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot 𝐫}|{\psi }_{\sigma \nu }^{(𝐤+𝚫𝐤)}⟩,$ (6)

$=$ $⟨{\psi }_{\sigma \mu }^{(𝐤)}|{\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot 𝐫}|{\psi }_{\sigma \nu }^{(𝐤+𝚫𝐤)}⟩,$

$=$ ${\displaystyle \frac{1}{N}}{\displaystyle \sum _{\mathrm{n},{\mathrm{n}}^{\prime }}}{\displaystyle \sum _{i\alpha ,j\beta }}{c}_{\sigma \mu ,i\alpha }^{(𝐤)*}{c}_{\sigma \nu ,j\beta }^{(𝐤+𝚫𝐤)}{\mathrm{e}}^{-\mathrm{i}𝐤\cdot ({𝐑}_{\mathrm{n}}-{𝐑}_{{\mathrm{n}}^{\prime }})}\times$

$⟨{\varphi }_{i\alpha }(𝐫-{\tau }_i-{𝐑}_{\mathrm{n}})|{\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot (𝐫-{𝐑}_{{\mathrm{n}}^{\prime }})}|{\varphi }_{j\beta }(𝐫-{\tau }_j-{𝐑}_{{\mathrm{n}}^{\prime }})⟩.$

Defining that

${𝐫}^{\prime }=𝐫-{\tau }_i-{𝐑}_{\mathrm{n}},$ (7)

we have

$⟨u_{\sigma \mu }^{(𝐤)}|u_{\sigma \nu }^{(𝐤+\mathrm{\Delta }𝐤)}⟩$ $=$ ${\displaystyle \frac{1}{N}}{\displaystyle \sum _{\mathrm{n},{\mathrm{n}}^{\prime }}}{\displaystyle \sum _{i\alpha ,j\beta }}{c}_{\sigma \mu ,i\alpha }^{(𝐤)*}{c}_{\sigma \nu ,j\beta }^{(𝐤+𝚫𝐤)}{\mathrm{e}}^{-\mathrm{i}𝐤\cdot ({𝐑}_{\mathrm{n}}-{𝐑}_{{\mathrm{n}}^{\prime }})}\times$ (8)

$⟨{\varphi }_{i\alpha }({𝐫}^{\prime })|{\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot ({𝐫}^{\prime }+{\tau }_i+{𝐑}_{\mathrm{n}}-{𝐑}_{{\mathrm{n}}^{\prime }})}|{\varphi }_{j\beta }({𝐫}^{\prime }+{\tau }_i-{\tau }_j+{𝐑}_{\mathrm{n}}-{𝐑}_{{\mathrm{n}}^{\prime }})⟩.$

Since each term depends on only the relative position ${𝐑}_{\mathrm{n}}-{𝐑}_{{\mathrm{n}}^{\prime }}$, Eq. (8) becomes

$⟨u_{\sigma \mu }^{(𝐤)}|u_{\sigma \nu }^{(𝐤+\mathrm{\Delta }𝐤)}⟩$ $=$ ${\displaystyle \sum _{\mathrm{n}}}{\displaystyle \sum _{i\alpha ,j\beta }}{c}_{\sigma \mu ,i\alpha }^{(𝐤)*}{c}_{\sigma \nu ,j\beta }^{(𝐤+𝚫𝐤)}{\mathrm{e}}^{\mathrm{i}𝐤\cdot {𝐑}_{\mathrm{n}}}\times ⟨{\varphi }_{i\alpha }({𝐫}^{\prime })|{\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot ({𝐫}^{\prime }+{\tau }_i-{𝐑}_{\mathrm{n}})}|{\varphi }_{j\beta }({𝐫}^{\prime }+{\tau }_i-{\tau }_j-{𝐑}_{\mathrm{n}})⟩,$ (9)

$=$ ${\displaystyle \sum _{\mathrm{n}}}{\displaystyle \sum _{i\alpha ,j\beta }}{c}_{\sigma \mu ,i\alpha }^{(𝐤)*}{c}_{\sigma \nu ,j\beta }^{(𝐤+𝚫𝐤)}{\mathrm{e}}^{\mathrm{i}𝐤\cdot {𝐑}_{\mathrm{n}}}{\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot ({\tau }_i-{𝐑}_{\mathrm{n}})}⟨{\varphi }_{i\alpha }({𝐫}^{\prime })|{\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot {𝐫}^{\prime }}|{\varphi }_{j\beta }({𝐫}^{\prime }+{\tau }_i-{\tau }_j-{𝐑}_{\mathrm{n}})⟩,$

The exponential function in Eq. (9) can be approximated by

${\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot {𝐫}^{\prime }}\approx 1-i\mathrm{\Delta }𝐤\cdot {𝐫}^{\prime }.$ (10)

Thus, Eq. (9) becomes

$⟨u_{\sigma \mu }^{(𝐤)}|u_{\sigma \nu }^{(𝐤+\mathrm{\Delta }𝐤)}⟩$ $=$ ${\displaystyle \sum _{\mathrm{n}}}{\displaystyle \sum _{i\alpha ,j\beta }}{c}_{\sigma \mu ,i\alpha }^{(𝐤)*}{c}_{\sigma \nu ,j\beta }^{(𝐤+𝚫𝐤)}{\mathrm{e}}^{\mathrm{i}𝐤\cdot {𝐑}_{\mathrm{n}}}{\mathrm{e}}^{-\mathrm{i}\mathrm{\Delta }𝐤\cdot ({\tau }_i-{𝐑}_{\mathrm{n}})}\times$ (11)

$\left\{⟨{\varphi }_{i\alpha }({𝐫}^{\prime })|{\varphi }_{j\beta }({𝐫}^{\prime }+{\tau }_i-{\tau }_j-{𝐑}_{\mathrm{n}})⟩-i\mathrm{\Delta }𝐤\cdot ⟨{\varphi }_{i\alpha }({𝐫}^{\prime })|{𝐫}^{\prime }|{\varphi }_{j\beta }({𝐫}^{\prime }+{\tau }_i-{\tau }_j-{𝐑}_{\mathrm{n}})⟩\right\},$

where the overlap integral is evaluated in momentum space, and the expectation value for the position operator is evaluated using the same real space mesh as for the solution of Poisson’s equation in OpenMX.