# Berry phase of generic two-dimensional gapless Dirac Hamiltonian

Reference: Topological Insulators and Superconductors, B. Andrei Bernevig, Taylor L. Hughes: Chapter 8, problem 1

The generic Bloch Hamiltonian $$H(k)=k_i\mathcal{A}_{ij}\sigma_j$$, with $$i\in\{1,2\}$$ and $$j\in\{1,2,3\}$$, is considered. Using the form $$H=d_i\sigma_i$$ with $$d_j=k_i\mathcal{A}_{ij}$$, the eigenvectors are, \begin{align} u_\pm&=\frac{1}{\sqrt{2d(d+d_3)}}\begin{bmatrix} d_3\pm d\\d_1-id_2 \end{bmatrix}, \end{align} where, \begin{align} d&=\sqrt{(k_iA_{i1})^2+(k_iA_{i2})^2+(k_iA_{i3})^2}=\sqrt{(k_xA_{11}+k_yA_{21})^2+(k_xA_{12}+k_yA_{22})^2+(k_xA_{13}+k_yA_{23})^2}\nonumber\\ &=k\sqrt{ (A_{11}^2+A_{12}^2+A_{13}^2)\cos^2(\theta)+(A_{22}^2+A_{12}^2+A_{23}^2)\sin^2(\theta)+2(A_{11}A_{12}+A_{22}A_{12}+A_{13}A_{23})\cos(\theta)\sin(\theta) } \end{align} The Berry connection is given by, \begin{align} A_i&=\frac{-1}{2d(d+d_3)}(d_2\partial_id_1-d_1\partial_id_2)=\frac{-1}{2d(d+k_iA_{i3})}(k_j\mathcal{A}_{j2}\mathcal{A}_{1i}-k_j\mathcal{A}_{j1}\mathcal{A}_{2i}). \end{align} Considering an anisotropic Fermi surface in lines with the generic anisotropy present in the Hamiltonian, the Berry phase is found as, \begin{align} \phi_b&=\oint dk_i A_i=\int d\phi k(\phi) A_\phi(k(\phi))=\int d\phi k(\phi) \bigg(A_y\frac{k_x}{k}-A_x\frac{k_y}{k}\bigg)\nonumber\\&=-\int d\phi \frac{k_x(k_j\mathcal{A}_{j2}\mathcal{A}_{12}-k_j\mathcal{A}_{j1}\mathcal{A}_{22})-k_y(k_j\mathcal{A}_{j2}\mathcal{A}_{11}-k_j\mathcal{A}_{j1}\mathcal{A}_{21})}{2d(d+d_3)}\nonumber\\ &=-\int d\phi \frac{-k_x^2(\mathcal{A}_{11}\mathcal{A}_{22}-\mathcal{A}_{12}^2)-k_y^2(\mathcal{A}_{11}\mathcal{A}_{22}-\mathcal{A}_{21}^2)+k_xk_y(\mathcal{A}_{22}\mathcal{A}_{12}-\mathcal{A}_{21}\mathcal{A}_{22}-\mathcal{A}_{12}\mathcal{A}_{11}+\mathcal{A}_{11}\mathcal{A}_{21})}{2d(d+d_3)}\nonumber\\ &\xrightarrow{\mathcal{A}\text{ is symmetric?}}-\int d\phi k^2\frac{\det(\mathcal{A}_{1:2,1:2})}{2d(d+d_3)}=-\int d\phi \frac{\det(\mathcal{A}_{1:2,1:2})}{2\Bigg(\stackrel{\displaystyle(a\cos^2(\theta)+b\sin^2(\theta)+c\cos(\theta)\sin(\theta))}{+\sqrt{a\cos^2(\theta)+b\sin^2(\theta)+c\cos(\theta)\sin(\theta)}(A_{13}\cos(\theta)+A_{23}\sin(\theta))}\Bigg)}. \end{align} This is where I get stuck, unlike the Eq. 8.12-8.13 given in the text. Is this the best way to proceed? Also, the Eq. 8.12 in the text seems to rely upon the symmetry of the matrix $$\mathcal{A}$$, which is not obvious to me. In the most generic case, if the the rotational symmetry of the space is not assumed (argument could go the other way, having asymmetric $$\mathcal{A}_{1:2,1:2}$$ implies asymmetric Fermi-surface and hence broken rotational symmetry), then clearly $$\mathcal{A}_{12}\neq \mathcal{A}_{21}$$.