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.
Chern Number – Calculating Chern Number for Electronic Energy Bands with Orthogonal States
topological-insulators
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The answer of David Aasen is correct, but let me add some comments which connect to your question of the relation of between the $\mathbb Z_2$ invariant $\nu$ and the first Chern-Number $C_1$.
Such a relation does not exist unless you require some extra symmetry than the generic symmetries usually required in the classification of topological insulators (such as time-reversal invariance in this case). Say the Hamiltonian is invariant under spin rotations along the $z$-axis (so a $U(1)$ subgroup of $SU(2)$ in left invariant), then the Hamiltonian can be block-diagonalized as
$H = \begin{pmatrix} H_\uparrow & \\ & H_\downarrow \end{pmatrix}, $
where the indices refer to spin-up and down degrees of freedom. Due to time reversal symmetry we have that $H_\downarrow(k) = H^*_\uparrow(-k)$. The system now consist of two copies of Quantum Hall effects with counter propagating edge states of opposite spin. As Davis Aasen says, the chern number is zero $C_1 = C_1^\uparrow + C_1^\downarrow = 0$. The difference however, the "spin Chern number", $C_1^\uparrow - C_1^\downarrow = 2C_{spin}$ can be non-zero and can be calculated by the Chern-numbers of the spin up/down sectors. As long as $S_z$ is preserved the spin Chern-number can be any integer $C_{spin}\in\mathbb Z$.
But if we add off-diagonal elements, and thus break the rotation symmetry along $z$, the invariant breaks down to $\nu = C_{spin}\,\text{mod}\,2\in\mathbb Z_2$ (as was shown by Kane and Mele). So topological trivial/non-trivial phases are characterized by even and odd spin-Chern numbers $C_{spin}$, not the original Chern number $C_1$. This however only makes sense when you have this extra symmetry.
Yes, there is a proof of that. The first one appeared by a beautiful paper of Hatsugai in 1993 https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.71.3697 for a particular special classes of models of the IQHE. More general proofs ensued, but it does turn out that the proofs rely on some non-trivial math, the most simple way to present it seems to be rooted in the context of some basic facts of complex analysis, as for example is presented in this paper: https://journals.aps.org/prb/abstract/10.1103/PhysRevB.83.125109
There you see for example that essentially the Cauchy integral formula lies at the heart of the bulk-boundary correspondence. More general proofs rely on more sophisticated math, e.g., K-theory or Fredholm theory.
Best Answer
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 $Q:= \frac{1}{2\pi}\sum_{ij}F_{ij}$.
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 $Q:= \frac{1}{2\pi} \iint _\scr P \mathbf F \cdot d\mathbf S$.
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 $\mathbf A$.