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

Andrew Hardy · Accepted Answer

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 $qcdot sigma$, 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 $C = frac{1}{2 pi} gamma$

ProfM · Answer

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:
$$
hat{H}(mathbf{q})=v_{mathrm{F}}mathbf{q}cdotmathbf{sigma}=v_{mathrm{F}}(q_1sigma_1+q_2sigma_2+q_3sigma_3),
$$
for $mathbf{q}=(q_1,q_2,q_3)$. The $i$th component of the Berry curvature is:
$$
Omega_i(mathbf{q})=-frac{1}{8pi}frac{1}{|mathbf{q}|^3}varepsilon_{ijk}mathbf{q}cdotpartial_{q_j}mathbf{q}timespartial_{q_k}mathbf{q}.
$$
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:
$$
Omega(mathbf{q})=-frac{1}{4pi}frac{mathbf{q}}{|mathbf{q}|^3}.
$$
The flux over a sphere S of fixed radius $|mathbf{q}|$ is $Omega(mathbf{q})cdot dmathbf{S}=Omega(mathbf{q})cdothat{mathbf{q}}dS$. The surface element is $dS=q^2sin(theta)dtheta dphi$ (where I use $(q,theta,phi)$ as my spherical coordinates for $mathbf{q})$, so that $Omega(mathbf{q})cdot dmathbf{S}=-frac{1}{4pi}sin(theta)dtheta dphi$, and the integral now becomes very simple:
$$
int_{mathrm{sphere}}Omega(mathbf{q})cdot dmathbf{S}=-frac{1}{4pi}intsin(theta)dthetaint dphi=-1,
$$
and this corresponds to a left-handed Weyl point. If you repeat the same analysis for the Hamiltonian $hat{H}(mathbf{q})=-v_{mathrm{F}}mathbf{q}cdotsigma$, then you get the Berry curvature as:
$$
Omega(mathbf{q})=+frac{1}{4pi}frac{mathbf{q}}{|mathbf{q}|^3},
$$
and the integral over the flux gives you $+1$, corresponding to a right-handed Weyl point.

Berry curvature flux around a Weyl node

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