I've hit a wall with the Berry's phase.
So here's the thing:
- I know Berry's phase is defined as \begin{equation} \gamma_C = \int_S \mathcal F d^2\mathbf k \end{equation}
- I'm considering the 2D $H$ of \begin{equation} H = \left(\begin{matrix}d_z & d_x - id_y \\ d_x + id_y & -d_z\end{matrix}\right) = \hat {\mathbf d}(\mathbf k) \cdot \vec\sigma \end{equation} where $\hat{\mathbf d}$ is essentially a point on $S^2$.
What I don't understand is why it now automatically follows that \begin{equation} \mathcal F = \frac12 \epsilon_{ij}\hat{\mathbf d} \cdot (\partial_{i}\hat{\mathbf d} \times \partial_j\hat{\mathbf d}). \end{equation}
Hopefully somebody can clarify, in particular a geometric argument would be nice. The method I'm reviewing now states,
The Berry curvature is given by the solid angle per unit area in $\mathbf k$ space, which is simply half the solid angle element for the mapping $\mathbf {\hat d}(\mathbf k)$.
Edit: Maybe this additional information is useful: $\mathcal F = \nabla \times \mathbf A$ with \begin{equation} \mathbf A = -i\left< u(\mathbf k)\right| \nabla \left| u(\mathbf k)\right>. \end{equation} with $\left|u(\mathbf k)\right>$ the cell periodic eigenstate of the Bloch Hamiltonian