In Appendix B of the paper (1), the authors compute the second Chern number $C_2$ of a band structure by manipulating the ground- and excited-state projection operators $P_{\text{G}}(\mathbf{k})$ and $P_{\text{E}}(\mathbf{k})$, where $\mathbf{k}$ is in the (four-dimensional) Brillouin zone. One step in the computation is to carry out a particular momentum integral: they write \begin{align} C_2 =&{} -\frac{\pi^2}{3}\epsilon^{ijkl}\int\frac{\mathrm{d}^4 k \,\mathrm{d}\omega}{(2\pi)^5} \left\{\frac{\mathrm{tr}\left[P_{\text{G}}\frac{\partial \text{P}_G}{\partial k_i}P_{\text{E}}\frac{\partial \text{P}_G}{\partial k_j}P_{\text{G}}\frac{\partial \text{P}_G}{\partial k_k}P_{\text{E}}\frac{\partial \text{P}_G}{\partial k_l}\right]}{(\omega+\mathrm{i}\delta-\epsilon_{\text{G}})^3(\omega+\mathrm{i}\delta-\epsilon_{\text{E}})^2}\\ +\frac{\mathrm{tr}\left[P_{\text{E}}\frac{\partial \text{P}_G}{\partial k_i}P_{\text{G}}\frac{\partial \text{P}_G}{\partial k_j}P_{\text{E}}\frac{\partial \text{P}_G}{\partial k_k}P_{\text{G}}\frac{\partial \text{P}_G}{\partial k_l}\right]}{(\omega+\mathrm{i}\delta-\epsilon_{\text{G}})^2(\omega+\mathrm{i}\delta-\epsilon_{\text{E}})^3}\right\}(\epsilon_{\text{E}}-\epsilon_{\text{G}})^4\\ ={}&\frac{1}{8\pi^2}\epsilon^{ijkl}\int\mathrm{d}^4 k\,\mathrm{tr}\left[P_{\text{E}}\frac{\partial \text{P}_G}{\partial k_l}\frac{\partial \text{P}_G}{\partial k_j}P_{\text{E}}\frac{\partial \text{P}_G}{\partial k_k}\frac{\partial \text{P}_G}{\partial k_l}\right],\tag{B10} \end{align} where $\delta\to 0$, and $\epsilon_{\text{G}}$, $\epsilon_{\text{E}}$ are constants (representing the energies of the flattened ground and excited bands respectively). Physically, we require $\epsilon_{\text{G}}<0<\epsilon_{\text{E}}$.

The projection operators are independent of $\omega$ and to go from one line to the next the authors carry out the frequency integral $$ I=\lim_{\delta\to 0}\int_{-\infty}^{\infty} \mathrm{d}\omega\phantom{,}\frac{1}{(\omega+\mathrm{i}\delta+a)^2(\omega+\mathrm{i}\delta-b)^3}, $$ where $a, b>0$. This integral is easily computed using standard contour integral techniques: the integration along the real axis can be completed with a large semicircle in the upper half plane; the integral over this semi circle vanishes for large radius, and since there are no poles within the integration domain (see diagram below), it follows that $I=0$.

However, this is clearly not what the authors have done, as this would imply $C_2=0$. In fact, it appears that they have carried out a Wick rotation $\omega\to -\mathrm{i}\omega$, so that they compute the integral $$ J=\int_{-\infty}^{\infty} \mathrm{d}\omega\phantom{,}\frac{1}{(-\mathrm{i}\omega+a)^2(-\mathrm{i}\omega-b)^3}. $$ In the process of Wick rotation, one of the poles actually crosses the integration contour, and the integral $J$ is actually non-zero, $$ J = -\frac{6\pi}{(a+b)^4}. $$

Before and after Wick rotation

How is it possible to justify this Wick rotation when it clearly changes the value of the integral?


(1) Topological Field Theory of Time-Reversal Invariant Insulators, arXiv:0802.3537


1 Answer 1


As explained in equation (3) of the QHZ paper, the relevent Green function for the DC response involves $$ G(i\omega, H)= \sum_n \frac{|n\rangle\langle n|}{i\omega-\epsilon_n}, $$ rather than the one with the $\omega +i\delta$ in the appendix. The sign in front of the $i\delta$ should really change as you cross the Fermi energy, so that the contour can be closed so as to enclose just the occupied states. I assume that QHZ were just a bit careless with the expression in the appendix.


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