Berry curvature flux around a Weyl node

Question

How can I formally show (or at least argue) that, given the crystal Hamiltonian expansion around a Weyl node in a three-dimensional Brillouin Zone located at $\vec{k}_{0}$,

$\hat{H}=f_{0}(\vec{k}_{0})\mathbb{I}+\vec{v}_{0}\cdot\vec{q}\mathbb{I}+\sum_{a=x,y,z}\vec{v}_{a}\cdot\vec{q}\sigma^{a}$

with $\vec{k}=\vec{k}_{0}+\vec{q}$, the Berry flux through a sphere surrounding $\vec{k}_{0}$ corresponds to the one of a sink or source of Berry curvature, depending on the chirality of the Weyl node, defined as $\chi=\vec{v}_{x}\cdot(\vec{v}_y\times\vec{v}_z)$?

Answer 1

At Weyl points ($K$, $q = 0$), we can approximate a tight-binding Hamiltonian as \begin{equation} H(\vec{K}+\vec{q})= v \vec{q} \cdot \vec{\sigma}. \end{equation} As this is a general $2x2$ Hamiltonian, we know that the eigenvalues will be of the form $ v|q|$. \ \end{definition} The corresponding eigenvectors are given as \begin{equation} |- \rangle = \biggl(\begin{array}{c} e^{-i \varphi} \sin \theta /2 \\ - \cos \theta /2 \end{array}\biggr), \ |+ \rangle = \biggl( \begin{array}{c} e^{-i \varphi} \cos \theta /2 \\ \sin \theta /2 \end{array}\biggr) \end{equation}

Note that these are not well-defined (single valued) at $\theta = \pi$. We can define a $U(1)$gauge transformation so that the eigenvectors are \begin{equation} |- \rangle = \biggl(\begin{array}{c} \sin \theta /2 \\ - e^{i \varphi}\cos \theta /2 \end{array}\biggr), \ |+ \rangle = \biggl( \begin{array}{c} \cos \theta /2 \\ e^{i \varphi}\sin \theta /2 \end{array}\biggr) \end{equation} Now these are single valued except at the north-pole $\theta = 0$. \ We can compute the Berry curvature in both gauges \begin{gather} A_\theta = i \langle - | \partial_\theta | - \rangle = 0 \\ A_\phi = i \langle - | \partial_\phi | - \rangle = (\sin \theta /2)^2 \\ \end{gather} and \begin{gather} A_\theta = i \langle - | \partial_\theta | - \rangle = 0 \\ A_\phi = i \langle - | \partial_\phi | - \rangle = - (\cos \theta /2)^2 \\ \end{gather}

From here you can see that this is no longer single valued at $\theta = 0$. This requires that we define a new gauge so that \begin{gather} A_\theta = i \langle - | \partial_\theta | - \rangle = 0 \\ A_\phi = i \langle - | \partial_\phi | - \rangle = - (\cos \theta /2)^2 \\ \end{gather} However, both of these give the gauge independent \begin{equation} F_{ \theta \phi} = \partial_\theta A_\phi - \partial_\phi A_\theta = \frac{\sin(\theta)}{2} \end{equation} Since we just have $q\cdot \sigma$, our angles are just spherical coordinate parameterization. Now we can write the Berry curvature in terms of the inverse Jacobian \begin{equation} F_{q_i,q_j} = F_{\theta, \phi} \frac{\partial(\theta ,\phi)}{\partial(q_i,q_j)} = \frac{\sin}{2}\frac{\partial(\theta, \phi)}{\partial(q_i,q_j)} = \frac{1}{2|q|^2} \end{equation} Now the field strength pseudovector is given by \begin{equation} \mathcal{F}_i = \epsilon_{ijk}F_{jk} \end{equation} and so we have \begin{equation} \mathcal{F} = \frac{-\vec{q}}{2 |q|^3} \end{equation} From here it is clear that we can use Gauss's theorem \begin{equation} \int_{\text {sphere }} \mathcal{F}(q) \cdot d \mathbf{S}= 2 \pi \int d \theta \ d \phi \ \sin (\theta) =- 2 \pi (1) \end{equation} which gives a quantized Berry curvature given by the Chern number $C = \frac{1}{2 \pi} \gamma$

Answer 2

To simplify the maths, let us set the zero of energy at the crossing point and make the velocity isotropic (essentially re-scale your definition of $\mathbf{q}$), so that the Hamiltonian around the Weyl point becomes: $$ \hat{H}(\mathbf{q})=v_{\mathrm{F}}\mathbf{q}\cdot\mathbf{\sigma}=v_{\mathrm{F}}(q_1\sigma_1+q_2\sigma_2+q_3\sigma_3), $$ for $\mathbf{q}=(q_1,q_2,q_3)$. The $i$th component of the Berry curvature is: $$ \Omega_i(\mathbf{q})=-\frac{1}{8\pi}\frac{1}{|\mathbf{q}|^3}\varepsilon_{ijk}\mathbf{q}\cdot\partial_{q_j}\mathbf{q}\times\partial_{q_k}\mathbf{q}. $$ If you carry out the calculation for all three components, and combine them into a vector $\Omega=(\Omega_1,\Omega_2,\Omega_3)$, you get: $$ \Omega(\mathbf{q})=-\frac{1}{4\pi}\frac{\mathbf{q}}{|\mathbf{q}|^3}. $$ The flux over a sphere S of fixed radius $|\mathbf{q}|$ is $\Omega(\mathbf{q})\cdot d\mathbf{S}=\Omega(\mathbf{q})\cdot\hat{\mathbf{q}}dS$. The surface element is $dS=q^2\sin(\theta)d\theta d\phi$ (where I use $(q,\theta,\phi)$ as my spherical coordinates for $\mathbf{q})$, so that $\Omega(\mathbf{q})\cdot d\mathbf{S}=-\frac{1}{4\pi}\sin(\theta)d\theta d\phi$, and the integral now becomes very simple: $$ \int_{\mathrm{sphere}}\Omega(\mathbf{q})\cdot d\mathbf{S}=-\frac{1}{4\pi}\int\sin(\theta)d\theta\int d\phi=-1, $$ and this corresponds to a left-handed Weyl point. If you repeat the same analysis for the Hamiltonian $\hat{H}(\mathbf{q})=-v_{\mathrm{F}}\mathbf{q}\cdot\sigma$, then you get the Berry curvature as: $$ \Omega(\mathbf{q})=+\frac{1}{4\pi}\frac{\mathbf{q}}{|\mathbf{q}|^3}, $$ and the integral over the flux gives you $+1$, corresponding to a right-handed Weyl point.