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.
1 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 $Q:= \frac{1}{2\pi} \iint _\scr P \mathbf F \cdot d\mathbf S$.
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 $\mathbf A$.
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$\begingroup$ Thank you very much. As another question could you please explain how is $A(R)=-Im <\psi (R)|\nabla_R \psi(R)>$ obtained? $\endgroup$– H. KhaniAug 6, 2021 at 9:53
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$\begingroup$ @H.Khani If I have answered your question, you may accept it by ticking the box next to the answer; if not, feel free to ask for additional clarification. $\endgroup$ Aug 8, 2021 at 13:58
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1$\begingroup$ @skachko Because that expansion is truncated to first order. If we didn’t stop at first order but continued to sum the entire series, then for any nonzero $d\mathbf R$ it would evaluate to zero. If this seems ad-hoc then it can be understood more naturally by thinking of the space of Bloch states as a $\mathbb C$ line bundle over the parameter space and then understand $\mathbf A$ as a connection on that bundle, but that is more sophisticated a viewpoint than called for in the OP. $\endgroup$ Aug 15, 2021 at 23:58