Derivation of expression for Berry curvature
I will write my response as detailed as possible.
We can write the Hamiltonian explicitly as, \begin{align*} H(k)=\vec{h}(k)\cdot\vec{\sigma}=\left(\begin{array}{cc} h_z(k) & h_x(k)ih_y(k)\\ h_x(k)+ih_y(k) &h_z(k) \end{array}\right) \end{align*} The eigenvalues for $H(k)$ are $\pm\vec{h}(k)$. We assume a gap condition, namely, $\vec{h}(k)\neq 0$ for all $k$. In this case, for each $k$, there are two different eigenvalues, associated to onedimensional orthogonal eigenspaces. Since they are orthogonal to each other, we will focus on, \begin{align*} H(k)\psi=\vec{h}(k)\psi, \end{align*} explicitly, if we write $\psi=(u\ v)^t$, for $u,v\in\mathbb{C}$, we have \begin{align*} \left(\begin{array}{cc} h_z(k)+\vec{h}(k) & h_x(k)ih_y(k)\\ h_x(k)+ih_y(k) &(h_z(k)\vec{h}(k)) \end{array}\right)\left(\begin{array}{cc} u\\ v \end{array} \right)=0 \end{align*} Giving, if $\vec{h}(k)\neq h_z(k)$, \begin{align*} v=\frac{h_x(k)+ih_y(k)}{\vec{h}(k)h_z(k)}u. \end{align*} If we introduce a normalized vector $(x(k),y(k),z(k))=\vec{h}(k)/\vec{h}(k)$, then, \begin{align*} v=\frac{x(k)+iy(k)}{1z(k)}u. \end{align*} This gives a nontrivial solution whenever $u\neq 0$. Notice that this solution is valid if $z(k)\neq 1$. If $z(k)=1$, then $v=1$ and $u=0$ provides the right solution. Since $\psi(k)$ generates a onedimensional subspace, we have the freedom of rescaling the solution. In particular, we can set $u\equiv 1$, giving, \begin{align*} v(k)=\frac{x(k)+iy(k)}{1z(k)}. \end{align*} Hence we get a local solution, valid if $z(k)\neq 1$, \begin{align*} \psi(k)=(1, v(k)). \end{align*} Actually, $v=xiy/(1z)$ is associated to a local complex coordinate on the sphere $S^2$ coming from stereographic projection from the North pole and $v(k)$ is the local form of the map $k\mapsto \vec{h}(k)/\vec{h}(k)\in S^2$ in terms of this local coordinate system. Now the Berry connection acts on the local ``frame'' provided by $\psi$ (any vector on the associated eigenspace at $k$ can be written as a multiple of $\psi(k)$, whenever $\psi$ is defined), as\begin{align*} \nabla \psi= \frac{\langle\psi d\psi\rangle}{\langle \psi\psi\rangle} \psi\equiv \mathcal{A}\cdot \psi, \end{align*} Hence, \begin{align*} \mathcal{A}=\frac{\bar{v}(k)dv(k)}{1+v(k)^2}. \end{align*} The Berry curvature is obtained by taking the exterior derivative of $\mathcal{A}$, \begin{align*} \mathcal{F}&=d\mathcal{A}=\frac{d\bar{v}(k)\wedge dv(k)}{(1+v(k)^2)}\frac{v(k)^2d\bar{v}(k)\wedge dv(k)}{(1+v(k)^2)^2}\\ &=\frac{d\bar{v}(k)\wedge dv(k)}{(1+v(k)^2)^2}. \end{align*} In order to relate this to the expression wanted, we will use spherical coordinates on the sphere $S^2$, namely $x=\sin\theta\cos\phi$, $y=\sin\theta\sin\phi$, $z=\cos\theta$, then, \begin{align*} v=\frac{x+iy}{1z}=\frac{\sin\theta}{1\cos\theta}e^{i\phi}=\frac{\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}{\sin^2\left(\frac{\theta}{2}\right)}e^{i\phi}=\cot\left(\frac{\theta}{2}\right)e^{i\phi} \end{align*} and, \begin{align*} dv&=\frac{1}{2}\left(1+\cot\left(\frac{\theta}{2}\right)^2\right)d\theta e^{i\phi}+ivd\phi=\frac{1}{2}(1+v^2)d\theta \frac{v}{v} +ivd\phi,\\ d\bar{v}&=\frac{1}{2}(1+v^2)d\theta \frac{\bar{v}}{v} i\bar{v}d\phi \end{align*} so that, \begin{align*} \frac{d\bar{v}\wedge dv}{(1+v^2)^2}=i\frac{vd\theta\wedge d\phi}{(1+v^2)}, \end{align*} now \begin{align*} \frac{v}{1+v^2}=\frac{\cot\left(\frac{\theta}{2}\right)}{1+\cot^2\left(\frac{\theta}{2}\right)}=\cot\left(\frac{\theta}{2}\right)\sin^2\left(\frac{\theta}{2}\right)=\cos\left(\frac{\theta}{2}\right)\sin\left(\frac{\theta}{2}\right)=\frac{1}{2}\sin\theta, \end{align*} since $\sin(\theta)>0$ for $\theta\in(0,\pi)$. Therefore, \begin{align*} \mathcal{F}=\frac{i}{2}\sin(\theta(k)) d\theta(k)\wedge d\phi(k). \end{align*} Notice that $\sin\theta d\theta\wedge d\phi$ is nothing but the Riemannian volume form on the sphere $S^2$, which can be written as \begin{align*} \sin\theta d\theta\wedge d\phi&=\left(\vec{x}\cdot \frac{\partial \vec{x}}{\partial \theta}\times \frac{\partial\vec{x}}{\partial \theta}\right)d\theta\wedge d\phi=\varepsilon_{ijk}x^i\frac{\partial x^j}{\partial\theta}\frac{\partial x^j}{\partial\phi}d\theta\wedge d\phi\\ &=\frac{1}{2}\varepsilon_{ijk}x^i dx^j\wedge dx^k. \end{align*} If we take in the previous expression $x^i\rightarrow x^i(k)$ (mathematically, it means that we are taking the pullback by the map $k\mapsto \vec{x}(k)= \vec{h}(k)/\vec{h}(k)\in S^2$), we get \begin{align*} \varepsilon_{ijk}x^i\frac{\partial x^j}{\partial k_1}\frac{\partial x^j}{\partial k_2}dk_1\wedge dk_2=\frac{1}{\vec{h}^3}\varepsilon_{ijk}h^i\frac{\partial h^j}{\partial k_1}\frac{\partial h^k}{\partial k_2}dk_1\wedge dk_2, \end{align*} where we used skewsymmmetry of $\varepsilon_{ijk}$. Therefore, \begin{align*} \mathcal{F}=\frac{i}{2}\varepsilon_{ijk}x^i(k)\frac{\partial x^j}{\partial k_1}\frac{\partial x^k}{\partial k_2}dk_1\wedge dk_2=\frac{i}{2\vec{h}(k)^3}\varepsilon_{ijk}h^i(k)\frac{\partial h^j}{\partial k_1}\frac{\partial h^k}{\partial k_2}dk_1\wedge dk_2. \end{align*} It is now clear that \begin{align*} \mathcal{F}_{ab}(k)=\frac{i}{2\vec{h}(k)^3}\varepsilon_{ijk}h^i(k)\frac{\partial h^j}{\partial k_a}\frac{\partial h^k}{\partial k_b}. \end{align*} The factor of $i$ is just a matter of convention: usually, in the physics literature, the Berry connection is taken to be $i\mathcal{A}$. The expression found for the curvature also shows, explicitly, that the first Chern number is the degree/winding number of the map $\Phi:k\mapsto \vec{x}(k)= \vec{h}(k)/\vec{h}(k)\in S^2$, \begin{align*} \int_{\mathbb{T}^2}\frac{i\mathcal{F}}{2\pi}=\frac{1}{4\pi}\int_{\mathbb{T}^2}\Phi^{*}(\sin\theta d\theta\wedge d\phi)=\text{deg}(\Phi), \end{align*} where $\mathbb{T}^2$ is the first Brillouin zone (a twodimensional torus).
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dgwp
Updated on July 17, 2020Comments

dgwp about 2 years
Many texts quote the expression for the Berry curvature for a twolevel system, with Hamiltonian $\mathbf{h}(\mathbf{k})=(h_x,h_y,h_z)$ in terms of $\mathbf{k}=(k_x,k_y)$, as something like \begin{equation} F_{ij}=\frac{1}{2h^3}\epsilon_{abc}h_a\frac{\partial h_b}{\partial k_i}\frac{\partial h_c}{\partial k_j} \end{equation} However, I have been unable to find any explanation as to how exactly this is derived from the expression for the Berry connection \begin{equation} A_x=\frac{1}{2h(hh_z)}\left(h_y\frac{\partial h_x}{\partial k_x}h_x\frac{\partial h_y}{\partial k_x}\right)\quad (\text{and similar for }A_y) \end{equation} using \begin{equation} F_{ij}=\frac{\partial A_j}{\partial k_i}\frac{\partial A_i}{\partial k_j} \end{equation} I have tried inserting the expressions for $A_x$ and $A_y$ in the above, but just end up with a huge number of terms that won't reduce to anything like what I am aiming for. I have also tried doing it by manipulating components and using LeviCivita symbols, but I can't get rid of the prefactors $1/2h(hh_z)$. I'm sure it is possible to parameterise in terms of spherical polars in order to make things easier, but I would really like to do it in terms of cartesian coordinates to improve my knowledge of index notation and manipulation. (I apologise in advance if this question is more mathematics than physics, but I think it would be helpful to anyone entering the field of topological insulators etc. to see this worked out.)

Mark Mitchison almost 6 yearsThis question is missing some context, in particular the Hamiltonian for the twolevel system. Do you mean $H = \mathbf{h}(\mathbf{k}) \cdot \sigma$, where $\sigma = (\sigma^x,\sigma^y,\sigma^z)$ are Pauli operators?

Qmechanic almost 6 years$\uparrow$ Which texts? Which pages?

dgwp almost 6 yearsSorry Mark  yes, that is exactly what I meant.

dgwp almost 6 yearsQmechanic  standard books on topological insulators such as Bernevig & Hughes, Franz & Molencamp, plus numerous review and other articles.

dgwp almost 6 yearsJust noticed that Nakahara's geometry/topology book makes a similar leap from the Berry connection to curvature without showing any intermediate steps, albeit expressed in terms of differential forms.
