As discussed
Unlike the Bloch functions which are delocalized in space, the Wannier functions are localized.
Its center (Wannier Charge Center or WCC) can be calculated as:
r ˉ n , R = ∫ w n , R ∗ ( r ) r w n , R ( r ) d 3 r (1) \bar{\mathbf{r}}_{n,\mathbf{R}}=\int w_{n,\mathbf{R}}^{*}(\mathbf{r}) \mathbf{r} w_{n,\mathbf{R}}(\mathbf{r}) d^{3} \mathbf{r}
\tag{1} r ˉ n , R = ∫ w n , R ∗ ( r ) r w n , R ( r ) d 3 r ( 1 ) The use of
r ˉ n \bar{\mathbf{r}}_{n} r ˉ n
p = 1 a ( ∑ i ( q i x i ) i o n s + ∑ n o c c ( q n r ‾ n , R ) W F s ) (2) p=\frac{1}{a}\left(\sum_{i}\left(q_{i} x_{i}\right)^{i o n s}+\sum_{n}^{o c c}\left(q_{n} \overline{\mathbf{r}}_{n,\mathbf{R}}\right)^{W F s}\right)
\tag{2} p = a 1 ( i ∑ ( q i x i ) i o n s + n ∑ occ ( q n r n , R ) W F s ) ( 2 ) Since Wannier transformation is a basis transformation (unitary), we can also express Eq. 1 using Bloch functions:
⟨ w n 0 ∣ r ∣ w n R ⟩ = V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k (3) \begin{aligned}
\left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle
&=\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} | i \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\
\end{aligned}
\tag{3} ⟨ w n 0 ∣ r ∣ w n R ⟩ = ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k ( 3 ) However, the derivation of this equity turns out to be not very straight forward.
Now, lets first consider a 1D system. For this system, we can write the following relation:
( x − a ) ∣ w n R ⟩ = a 2 π ∫ − π / a π / a ( x − R ) e i k ( x − R ) ∣ u n k ⟩ d k = a 2 π ∫ − π / a π / a ( − i ∂ k e i k ( x − R ) ) ∣ u n k ⟩ d k = − i a 2 π e i k ( x − R ) ∣ u n k ⟩ ∣ − π a π a − ( − i ) a 2 π ∫ − π / a π / a e i k ( x − R ) ( ∂ k ∣ u n k ⟩ ) d k = 0 + a 2 π ∫ − π / a π / a e i k ( x − R ) ( i ∂ k ∣ u n k ⟩ ) d k , (4) \begin{aligned}
(x-a)\left|w_{n R}\right\rangle &=\frac{a}{2 \pi} \int_{-\pi/a}^{\pi / a}(x-R) e^{i k(x-R)}\left|u_{n k}\right\rangle d k \\
&=\frac{a}{2 \pi} \int_{-\pi/a}^{\pi / a}\left(-i \partial_{k} e^{i k(x-R)}\right)\left|u_{n k}\right\rangle d k \\
&=-\frac{ia}{2 \pi} e^{i k(x-R)} \left|u_{n k}\right\rangle \bigg |_{\frac{-\pi}{a}}^{\frac{\pi}{a}} - (-i)
\frac{a}{2 \pi} \int_{-\pi/a}^{\pi / a} e^{i k(x-R)}\left( \partial_{k}\left|u_{n k}\right\rangle\right) d k\\
&=0+\frac{a}{2 \pi} \int_{-\pi/a}^{\pi / a} e^{i k(x-R)}\left(i \partial_{k}\left|u_{n k}\right\rangle\right) d k,
\end{aligned}
\tag{4} ( x − a ) ∣ w n R ⟩ = 2 π a ∫ − π / a π / a ( x − R ) e ik ( x − R ) ∣ u nk ⟩ d k = 2 π a ∫ − π / a π / a ( − i ∂ k e ik ( x − R ) ) ∣ u nk ⟩ d k = − 2 π ia e ik ( x − R ) ∣ u nk ⟩ a − π a π − ( − i ) 2 π a ∫ − π / a π / a e ik ( x − R ) ( ∂ k ∣ u nk ⟩ ) d k = 0 + 2 π a ∫ − π / a π / a e ik ( x − R ) ( i ∂ k ∣ u nk ⟩ ) d k , ( 4 ) here, we have use of the fact that $e^{ik(x−a)} u_{nk}(x) = \psi_{nk}(x−a)$ has the same value at $k=-\pi/a$ and $\pi/a$ (Bloch functions are periodic in $k$ space, for details, check out
Generalizing this result to 3D and moving $a$ (in 3D case $\mathbf{R}$) from the left hand side to the right hand side:
r ∣ w n R ⟩ = V c e l l ( 2 π ) 3 ∫ B Z e − i k ⋅ R [ e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ ] d 3 k . (5) \mathbf{r}\left|w_{n \mathbf{R}}\right\rangle=\frac{V_\mathrm{cell}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left[e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle\right] d^{3} k.
\tag{5} r ∣ w n R ⟩ = ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R [ e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ ] d 3 k . ( 5 ) Multiply
⟨ w n 0 ∣ \left\langle w_{n \mathbf{0}}\right| ⟨ w n 0 ∣
⟨ w n 0 ∣ r ∣ w n R ⟩ = [ V cell ( 2 π ) 3 ∫ B Z e i k ′ ⋅ 0 e − i k ′ ⋅ r ⟨ u n k ′ ∣ d 3 k ′ ] ⋅ { V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R [ e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ ] d k } (6) \begin{aligned}
\left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle&= \left[ \frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{i \mathbf{k'} \cdot \mathbf{0}} e^{-i \mathbf{k'} \cdot \mathbf{r}} \left\langle u_{n \mathbf{k'}}\right| d^{3} k' \right] \\
& \quad \cdot \left\{
\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left[e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle\right] dk \right \}\\
\end{aligned}
\tag{6} ⟨ w n 0 ∣ r ∣ w n R ⟩ = [ ( 2 π ) 3 V cell ∫ BZ e i k ′ ⋅ 0 e − i k ′ ⋅ r ⟨ u n k ′ ∣ d 3 k ′ ] ⋅ { ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R [ e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ ] d k } ( 6 ) Taking a closer look at
e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩
The first term (
R ⋅ e i k ⋅ r ∣ u n k ⟩ \mathbf{R} \cdot e^{i \mathbf{k} \cdot \mathbf{r}}\left|u_{n \mathbf{k}}\right\rangle R ⋅ e i k ⋅ r ∣ u n k ⟩ R \mathbf{R} R
In the second term (
i e i k ⋅ r ∇ k ∣ u n k ⟩ ie^{i \mathbf{k} \cdot \mathbf{r}}\nabla_{\mathbf{k}}\left|u_{n \mathbf{k}}\right\rangle i e i k ⋅ r ∇ k ∣ u n k ⟩ ∣ u n k ⟩ \left|u_{n \mathbf{k}}\right\rangle ∣ u n k ⟩ r \mathbf{r} r
So that
e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩
For two Bloch functions, we have the following relation:
⟨ ψ k ∣ χ k ′ ⟩ = ( 2 π ) 3 V cell ⟨ u k ∣ v k ′ ⟩ δ 3 ( k − k ′ ) (7) \left\langle\psi_{\mathbf{k}} | \chi_{\mathbf{k}^{\prime}}\right\rangle=\frac{(2 \pi)^{3}}{V_{\text {cell }}}\left\langle u_{\mathbf{k}}|v_{\mathbf{k}^{\prime}}\right\rangle \delta^{3}\left(\mathbf{k}-\mathbf{k}^{\prime}\right)
\tag{7} ⟨ ψ k ∣ χ k ′ ⟩ = V cell ( 2 π ) 3 ⟨ u k ∣ v k ′ ⟩ δ 3 ( k − k ′ ) ( 7 ) we can rewrite Eq. 6 as:
⟨ w n 0 ∣ r ∣ w n R ⟩ = [ V cell ( 2 π ) 3 ∫ B Z e i k ′ ⋅ 0 e − i k ′ ⋅ r ⟨ u n k ′ ∣ d 3 k ′ ] ⋅ { V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R [ e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ ] d k } = V cell 2 ( 2 π ) 6 ∫ B Z ∫ B Z e i k ′ ⋅ 0 e − i k ⋅ R ( e − i k ′ ⋅ r ⟨ u n k ′ ∣ ) [ e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ ] d 3 k d 3 k ′ = V cell 2 ( 2 π ) 6 ∫ B Z ∫ B Z e i k ′ ⋅ 0 e i − k ⋅ R ⟨ ψ n k ′ ∣ χ n k ⟩ d 3 k d 3 k ′ = V cell ( 2 π ) 3 ∫ B Z ∫ B Z e i k ′ ⋅ 0 e − i k ⋅ R ⟨ u n k ′ ∣ v n k ⟩ δ 3 ( k − k ′ ) d 3 k d 3 k ′ = V cell ( 2 π ) 3 ∫ B Z e i k ( 0 − R ) ⟨ u n k ′ ∣ ( R + i ∇ k ) ∣ u n k ⟩ d 3 k = V cell ( 2 π ) 3 R ∫ BZ e i k ( 0 − R ) d 3 k + V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k = V cell ( 2 π ) 3 R δ 0 , R + V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k = V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k (8) \begin{aligned}
\left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle&= \left[ \frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{i \mathbf{k'} \cdot \mathbf{0}} e^{-i \mathbf{k'} \cdot \mathbf{r}} \left\langle u_{n \mathbf{k'}}\right| d^{3} k' \right] \\
& \quad \cdot \left\{
\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left[e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle\right] dk \right \}\\
&=\frac{V^2_{\text {cell }}}{(2 \pi)^{6}} \int_{\mathrm{BZ}}\int_{\mathrm{BZ}} e^{i\mathbf{k'}\cdot\bf{0}} e^{-i\mathbf{k}\cdot\bf{R}}
\left (e^{-i\mathbf{k'}\cdot\bf{r}} \bra{u_{n\mathbf{k'}}}\right )
\left[ e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle \right] d^3 k d^3 k' \\
&= \frac{V^2_{\text {cell }}}{(2 \pi)^{6}} \int_{\mathrm{BZ}}\int_{\mathrm{BZ}} e^{i\mathbf{k'}\cdot\bf{0}} e^{i\mathbf{-k}\cdot\bf{R}}
\braket{\psi_{n\mathbf{k'}} | \chi_{n\mathbf{k}}}
d^3 k d^3 k'\\
&= \frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}}\int_{\mathrm{BZ}} e^{i\mathbf{k'}\cdot\bf{0}} e^{-i\mathbf{k}\cdot\bf{R}}
\left\langle u_{n\mathbf{k'}}|v_{n\mathbf{k}}\right\rangle \delta^{3}\left(\mathbf{k}-\mathbf{k}^{\prime}\right)
d^3 k d^3 k' \\
&= \frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{i\mathbf{k}\bf{(0-R)}}
\langle u_{n\mathbf{k'}}| \left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle
d^3 k \\
&= \frac{V_{\text{cell}}}{(2 \pi)^{3}} \mathbf{R} \int_\text{BZ} e^{i\mathbf{k}(\bf{0} - \bf{R})} d^3 k +
\frac{V_{\text{cell}}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} \mid i \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\
&= \frac{V_{\text{cell}}}{(2 \pi)^{3}} \mathbf{R} \delta_{0,\mathbf{R}} +
\frac{V_{\text{cell}}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} \mid i \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\
&=\frac{V_{\text{cell}}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} \mid i \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k
\end{aligned}
\tag{8} ⟨ w n 0 ∣ r ∣ w n R ⟩ = [ ( 2 π ) 3 V cell ∫ BZ e i k ′ ⋅ 0 e − i k ′ ⋅ r ⟨ u n k ′ ∣ d 3 k ′ ] ⋅ { ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R [ e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ ] d k } = ( 2 π ) 6 V cell 2 ∫ BZ ∫ BZ e i k ′ ⋅ 0 e − i k ⋅ R ( e − i k ′ ⋅ r ⟨ u n k ′ ∣ ) [ e i k ⋅ r ( R + i ∇ k ) ∣ u n k ⟩ ] d 3 k d 3 k ′ = ( 2 π ) 6 V cell 2 ∫ BZ ∫ BZ e i k ′ ⋅ 0 e i − k ⋅ R ⟨ ψ n k ′ ∣ χ n k ⟩ d 3 k d 3 k ′ = ( 2 π ) 3 V cell ∫ BZ ∫ BZ e i k ′ ⋅ 0 e − i k ⋅ R ⟨ u n k ′ ∣ v n k ⟩ δ 3 ( k − k ′ ) d 3 k d 3 k ′ = ( 2 π ) 3 V cell ∫ BZ e i k ( 0 − R ) ⟨ u n k ′ ∣ ( R + i ∇ k ) ∣ u n k ⟩ d 3 k = ( 2 π ) 3 V cell R ∫ BZ e i k ( 0 − R ) d 3 k + ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k = ( 2 π ) 3 V cell R δ 0 , R + ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k = ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k ( 8 ) Phase factor
One can add an arbitrary phase factor $\phi(k)$ to the Bloch function:
ψ n k ( x ) = e i ϕ ( k ) ψ n k ( x ) , \psi_{nk}(x) = e^{i\phi(k)} \psi_{nk}(x), ψ nk ( x ) = e i ϕ ( k ) ψ nk ( x ) , and this won’t change the charge density:
∣ ψ n k ( x ) ∣ 2 = e i ϕ ( k ) e − i ϕ ( k ) ∣ ψ n k ( x ) ∣ 2 = ∣ ψ n k ( x ) ∣ 2 . |\psi_{nk}(x)|^2 = e^{i\phi(k)} e^{-i\phi(k)} |\psi_{nk}(x)|^2 = |\psi_{nk}(x)|^2. ∣ ψ nk ( x ) ∣ 2 = e i ϕ ( k ) e − i ϕ ( k ) ∣ ψ nk ( x ) ∣ 2 = ∣ ψ nk ( x ) ∣ 2 . The only constraint on
ϕ ( k ) \phi(\mathbf{k}) ϕ ( k ) e i k ⋅ R e^{i\mathbf{k} \cdot \mathbf{R}} e i k ⋅ R
In the following, we are going to combine this phase factor into the cell-periodic part of the Bloch function:
e i ϕ ( k ) ψ n k ( x ) = e i k x e i ϕ ( k ) ∣ u n k ⟩ = e i k x ∣ u n k ′ ⟩ e^{i\phi(k)} \psi_{nk}(x) = e^{ikx} e^{i\phi(k)} \ket{u_{nk}} = e^{ikx} \ket{u_{nk}'} e i ϕ ( k ) ψ nk ( x ) = e ik x e i ϕ ( k ) ∣ u nk ⟩ = e ik x ∣ u nk ′ ⟩ The effect of this phase factor can be found by replacing this new $\ket{u_{nk}’}$ in the final expression of Eq. 8:
⟨ w n 0 ∣ r ∣ w n R ⟩ = V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k ′ ∣ i ∇ k u n k ′ ⟩ d 3 k = i V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k e − i ϕ ( k ) ∣ ∇ k e i ϕ ( k ) u n k ⟩ d 3 k = i V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k e − i ϕ ( k ) ∣ i ϕ ′ ( k ) e i ϕ ( k ) u n k + e i ϕ ( k ) ∇ k u n k ⟩ d 3 k = i V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k e − i ϕ ( k ) ∣ i ϕ ′ ( k ) e i ϕ ( k ) u n k ⟩ d 3 k + i V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k ∣ ∇ k u n k ⟩ d 3 k = − V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ϕ ′ ( k ) d 3 k + i V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k ∣ ∇ k u n k ⟩ d 3 k (9) \begin{aligned}
\left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle
&= \frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}}' \mid i \nabla_{\mathbf{k}} u_{n \mathbf{k}}'\right\rangle d^{3} k\\
&= i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} e^{-i\phi(\mathbf{k})} \bigg | \nabla_{\mathbf{k}} e^{i\phi(\mathbf{k})} u_{n \mathbf{k}}\right\rangle d^{3} k\\
&= i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} e^{-i\phi(\mathbf{k})} \bigg |
i\phi' (\mathbf{k}) e^{i\phi(\mathbf{k})} u_{n \mathbf{k}} +e^{i\phi(\mathbf{k})} \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\
&= i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} e^{-i\phi(\mathbf{k})} \bigg |
i\phi' (\mathbf{k}) e^{i\phi(\mathbf{k})} u_{n \mathbf{k}}\right\rangle d^{3} k +i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} | \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\
&= -\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}
\phi' (\mathbf{k})
d^{3} k +i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} | \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\
\end{aligned}
\tag{9} ⟨ w n 0 ∣ r ∣ w n R ⟩ = ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k ′ ∣ i ∇ k u n k ′ ⟩ d 3 k = i ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k e − i ϕ ( k ) ∇ k e i ϕ ( k ) u n k ⟩ d 3 k = i ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k e − i ϕ ( k ) i ϕ ′ ( k ) e i ϕ ( k ) u n k + e i ϕ ( k ) ∇ k u n k ⟩ d 3 k = i ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k e − i ϕ ( k ) i ϕ ′ ( k ) e i ϕ ( k ) u n k ⟩ d 3 k + i ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k ∣ ∇ k u n k ⟩ d 3 k = − ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ϕ ′ ( k ) d 3 k + i ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k ∣ ∇ k u n k ⟩ d 3 k ( 9 ) Because like $\phi(\mathbf{k})$, $\phi’ (\mathbf{k})$ also has the same periodicity as $e^{-i \mathbf{k} \cdot \mathbf{R}}$, the first term is zero. We again have:
⟨ w n 0 ∣ r ∣ w n R ⟩ = V cell ( 2 π ) 3 ∫ B Z e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k . \begin{aligned}
\left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle
&=\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} | i \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\
\end{aligned}. ⟨ w n 0 ∣ r ∣ w n R ⟩ = ( 2 π ) 3 V cell ∫ BZ e − i k ⋅ R ⟨ u n k ∣ i ∇ k u n k ⟩ d 3 k . Now we see that adding a phase factor does not affect the charge center.
Note that, up to now, all we’ve considered is a single band system.
Expanding this into multi-band systems, we have to replace the phase factor with a unitary matrix:
∑ m U m n ( k ) ψ m k ( r ) . (10) \sum_{m} U_{m n}^{(\mathbf{k})} \psi_{m \mathbf{k}}(\mathbf{r}).
\tag{10} m ∑ U mn ( k ) ψ m k ( r ) . ( 10 ) The result of this band mixing in Eq. 10 is that the total WCC is well invariant, not individual charge centers.
before, Wannier functions can be used to calculate the poalrization of a system.