Wannier Charge Center in Bloch representation

DFT  Wannier  Polarization 

As discussed :link: before, Wannier functions can be used to calculate the poalrization of a system.

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:

\[\bar{\mathbf{r}}_{n,\mathbf{R}}=\int w_{n,\mathbf{R}}^{*}(\mathbf{r}) \mathbf{r} w_{n,\mathbf{R}}(\mathbf{r}) d^{3} \mathbf{r} \tag{1}\]

The use of \(\bar{\mathbf{r}}_{n}\) eliminated ambiguities introduced by the charge density and unit cell so that both cores and charges can be treated as point charges. The polarization of the system can be calculated as:

\[p=\frac{1}{a}\left(\sum_{i}\left(q_{i} x_{i}\right)^{i o n s}+\sum_{n}^{o c c}\left(q_{n} \overline{\mathbf{r}}_{n,\mathbf{R}}\right)^{W F s}\right) \tag{2}\]

Since Wannier transformation is a basis transformation (unitary), we can also express Eq. 1 using Bloch functions:

\[\begin{aligned} \left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle &=\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} | i \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\ \end{aligned} \tag{3}\]

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:

\[\begin{aligned} (x-a)\left|w_{n R}\right\rangle &=\frac{a}{2 \pi} \int_{-\pi/a}^{\pi / a}(x-R) e^{i k(x-R)}\left|u_{n k}\right\rangle d k \\ &=\frac{a}{2 \pi} \int_{-\pi/a}^{\pi / a}\left(-i \partial_{k} e^{i k(x-R)}\right)\left|u_{n k}\right\rangle d k \\ &=-\frac{ia}{2 \pi} e^{i k(x-R)} \left|u_{n k}\right\rangle \bigg |_{\frac{-\pi}{a}}^{\frac{\pi}{a}} - (-i) \frac{a}{2 \pi} \int_{-\pi/a}^{\pi / a} e^{i k(x-R)}\left( \partial_{k}\left|u_{n k}\right\rangle\right) d k\\ &=0+\frac{a}{2 \pi} \int_{-\pi/a}^{\pi / a} e^{i k(x-R)}\left(i \partial_{k}\left|u_{n k}\right\rangle\right) d k, \end{aligned} \tag{4}\]

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 :link: this and :link: this) so that the first term is zero.

Generalizing this result to 3D and moving $a$ (in 3D case $\mathbf{R}$) from the left hand side to the right hand side:

\[\mathbf{r}\left|w_{n \mathbf{R}}\right\rangle=\frac{V_\mathrm{cell}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left[e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle\right] d^{3} k. \tag{5}\]

Multiply \(\left\langle w_{n \mathbf{0}}\right|\) to the left of Eq. 5, we get

\[\begin{aligned} \left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle&= \left[ \frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{i \mathbf{k'} \cdot \mathbf{0}} e^{-i \mathbf{k'} \cdot \mathbf{r}} \left\langle u_{n \mathbf{k'}}\right| d^{3} k' \right] \\ & \quad \cdot \left\{ \frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left[e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle\right] dk \right \}\\ \end{aligned} \tag{6}\]

Taking a closer look at \(e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle\) , we see that:

  • The first term ( \(\mathbf{R} \cdot e^{i \mathbf{k} \cdot \mathbf{r}}\left|u_{n \mathbf{k}}\right\rangle\) ) can still be treated as a Bloch function but with a prefactor \(\mathbf{R}\).
  • In the second term ( \(ie^{i \mathbf{k} \cdot \mathbf{r}}\nabla_{\mathbf{k}}\left|u_{n \mathbf{k}}\right\rangle\)) , \(\left|u_{n \mathbf{k}}\right\rangle\) is a cell-periodic function, taking a derivative against $\mathbf{k}$ won’t change it’s periodicity (in \(\mathbf{r}\) ). This means that we can treat this term as a Bloch function.

So that \(e^{i \mathbf{k} \cdot \mathbf{r}}\left(\mathbf{R}+i \nabla_{\mathbf{k}}\right)\left|u_{n \mathbf{k}}\right\rangle\) can be treated as another Bloch function with the same cell periodicity and k dependency.

For two Bloch functions, we have the following relation:

\[\left\langle\psi_{\mathbf{k}} | \chi_{\mathbf{k}^{\prime}}\right\rangle=\frac{(2 \pi)^{3}}{V_{\text {cell }}}\left\langle u_{\mathbf{k}}|v_{\mathbf{k}^{\prime}}\right\rangle \delta^{3}\left(\mathbf{k}-\mathbf{k}^{\prime}\right) \tag{7}\]

we can rewrite Eq. 6 as:

\[\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}\]

Phase factor

One can add an arbitrary phase factor $\phi(k)$ to the Bloch function:

\[\psi_{nk}(x) = e^{i\phi(k)} \psi_{nk}(x),\]

and this won’t change the charge density:

\[|\psi_{nk}(x)|^2 = e^{i\phi(k)} e^{-i\phi(k)} |\psi_{nk}(x)|^2 = |\psi_{nk}(x)|^2.\]

The only constraint on \(\phi(\mathbf{k})\) is that it needs to have the same periodicity (which is $\mathbf{G}$) as \(e^{i\mathbf{k} \cdot \mathbf{R}}\). This is because of the Bloch functions are $\mathbf{k}$ translational invariant (i.e. $\psi_{\mathbf{k}} = \psi_{\mathbf{k+G}}$, for details, again check out :link: this and :link: this)

In the following, we are going to combine this phase factor into the cell-periodic part of the Bloch function:

\[e^{i\phi(k)} \psi_{nk}(x) = e^{ikx} e^{i\phi(k)} \ket{u_{nk}} = e^{ikx} \ket{u_{nk}'}\]

The effect of this phase factor can be found by replacing this new $\ket{u_{nk}’}$ in the final expression of Eq. 8:

\[\begin{aligned} \left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle &= \frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}}' \mid i \nabla_{\mathbf{k}} u_{n \mathbf{k}}'\right\rangle d^{3} k\\ &= i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} e^{-i\phi(\mathbf{k})} \bigg | \nabla_{\mathbf{k}} e^{i\phi(\mathbf{k})} u_{n \mathbf{k}}\right\rangle d^{3} k\\ &= i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} e^{-i\phi(\mathbf{k})} \bigg | i\phi' (\mathbf{k}) e^{i\phi(\mathbf{k})} u_{n \mathbf{k}} +e^{i\phi(\mathbf{k})} \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\ &= i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} e^{-i\phi(\mathbf{k})} \bigg | i\phi' (\mathbf{k}) e^{i\phi(\mathbf{k})} u_{n \mathbf{k}}\right\rangle d^{3} k +i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} | \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\ &= -\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}} \phi' (\mathbf{k}) d^{3} k +i\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} | \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\ \end{aligned} \tag{9}\]

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:

\[\begin{aligned} \left\langle w_{n \mathbf{0}}|\mathbf{r}| w_{n \mathbf{R}}\right\rangle &=\frac{V_{\text {cell }}}{(2 \pi)^{3}} \int_{\mathrm{BZ}} e^{-i \mathbf{k} \cdot \mathbf{R}}\left\langle u_{n \mathbf{k}} | i \nabla_{\mathbf{k}} u_{n \mathbf{k}}\right\rangle d^{3} k\\ \end{aligned}.\]

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:

\[\sum_{m} U_{m n}^{(\mathbf{k})} \psi_{m \mathbf{k}}(\mathbf{r}). \tag{10}\]

The result of this band mixing in Eq. 10 is that the total WCC is well invariant, not individual charge centers.



Author | Chengcheng Xiao

Currently a PhD student at Imperial College London. Predicting electron behaviour since 2016.