Chern number for electronic energy bands with orthogonal states

Question

In the "lecture notes on Topological insulators by Asboth et all" the Chern number is defined on the basis of phase change of non-orthogonal states on a closed torus. Nevertheless, in calculation of Chern number for electronic energy bands, the different k-states in the Brillouin zone are orthogonal. These two arguments are incompatible. Any help would be appreciated.

Answer 1

The definition you refer to is relevant when you have a finite set of states which constitute a (finite) lattice on a torus; in that case, the relative phase between $|\psi_i\rangle$ and $|\psi_j\rangle$ is defined to be $$\gamma_{ij} := -\mathrm{arg}\left(\langle \psi_i|\psi_j\rangle\right)$$

The Berry flux through placquette $(i,j)$ is then defined to be $$F_{ij}:= -\mathrm{arg}\left(\exp\left[-i(\gamma_{ij}+\gamma_{(i+1),j}+\gamma_{(i+1)(j+1)}+\gamma_{i,(j+1)})\right]\right)$$ and the Chern number is $Q:= \frac{1}{2\pi}\sum_{ij}F_{ij}$.

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For a continuous space of states, these definitions require modification. As you say, $|\psi(\mathbf R)\rangle$ and $|\psi(\mathbf R + d\mathbf R)\rangle$ are generically orthogonal. What we then do is consider the inner product $\langle \psi(\mathbf R)|\psi(\mathbf R + d\mathbf R)\rangle$ to linear order in $d\mathbf R$, i.e. $$\langle \psi(\mathbf R)|\psi(\mathbf R+d\mathbf R)\rangle \sim \langle \psi(\mathbf R)|\psi(\mathbf R)\rangle - i\mathbf A(\mathbf R) \cdot d\mathbf R, \qquad \mathbf A(\mathbf R) := i \langle\psi(\mathbf R) | \nabla_\mathbf R \psi(\mathbf R)\rangle$$

where we choose the states $|\psi(\mathbf R)\rangle$ to be normalized for all $\mathbf R$. Proceeding from there, we find the relative phase as defined above to simply be $\mathbf A \cdot d\mathbf R$. Integrating this around an infinitesimal loop $\partial S$ yields the Berry phase $$\gamma_{\partial S}:= \oint _{\partial S} \mathbf A \cdot d\mathbf R$$ which can be expressed as the integral of the Berry curvature over the area enclosed by $\partial S$: $$\gamma_{\partial S} := \oint _{\partial S} \mathbf A \cdot d\mathbf R = \iint _S \mathbf F \cdot d\mathbf S, \qquad \mathbf F := \nabla_\mathbf R \times \mathbf A(\mathbf R)$$

Finally, the Chern number is defined as the integral of the Berry curvature over the entire parameter space $\scr P$ via $Q:= \frac{1}{2\pi} \iint _\scr P \mathbf F \cdot d\mathbf S$.

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As another question could you please explain how is $\mathbf A(\mathbf R)= -\mathrm{Im}\langle \psi(\mathbf R)|\nabla_\mathbf R \psi(\mathbf R)\rangle$ obtained?

In the above, we have tacitly assumed that $|\psi(\mathbf R)\rangle$ are normalized. That being the case, $\langle \psi(\mathbf R)|\psi(\mathbf R)\rangle = 1$, and so trivially

$$\nabla_\mathbf R \langle\psi(\mathbf R)|\psi(\mathbf R)\rangle = 0 = \langle \nabla_\mathbf R \psi(\mathbf R)|\psi(\mathbf R)\rangle + \langle\psi(\mathbf R)|\nabla_\mathbf R\psi(\mathbf R)\rangle$$ $$\implies \langle\psi(\mathbf R)|\nabla_\mathbf R\psi(\mathbf R)\rangle=-\overline{\langle\psi(\mathbf R)|\nabla_\mathbf R\psi(\mathbf R)\rangle}$$ which implies that $\langle\psi(\mathbf R)|\nabla_\mathbf R\psi(\mathbf R)\rangle$ is purely imaginary. Writing $\langle\psi(\mathbf R)|\nabla_\mathbf R\psi(\mathbf R)\rangle= i \mathrm{Im}\langle\psi(\mathbf R)|\nabla_\mathbf R\psi(\mathbf R)\rangle$ yields the alternate definition of $\mathbf A$.