# Chern number for electronic energy bands with orthogonal states

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.

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 $$\mathbf A$$.

• Thank you very much. As another question could you please explain how is $A(R)=-Im <\psi (R)|\nabla_R \psi(R)>$ obtained? Aug 6, 2021 at 9:53
• @H.Khani I've updated my answer. Aug 6, 2021 at 14:45
• Dear J. Murray, Thanks very much again. Aug 8, 2021 at 8:42
• @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. Aug 8, 2021 at 13:58
• @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. Aug 15, 2021 at 23:58