Contractions in current-current correlator for Kubo formula

To calculate the conductance $$\sigma_{ij}(\mathbf{q}, \omega)$$ of e.g. a disordered electron gas using the Kubo formula, one must compute the (imaginary-time-ordered) current-current correlation function $$\left\langle J_i(\mathbf{q}, \tau) J_j(-\mathbf{q}, 0)\right\rangle$$, where the current operator is $$J_i(\mathbf{q}, \tau)= Q\sum_{\mathbf{k}} \frac{k_i}{m} \psi^\dagger_{\mathbf{k}-\frac{\mathbf{q}}{2}}(\tau)\psi_{\mathbf{k}+\frac{\mathbf{q}}{2}}(\tau)$$ with $$Q=-|e|$$ the electronic charge, $$\langle\cdots\rangle=\frac{1}{Z}\mathrm{Tr}[\mathrm{e}^{-\beta H}(\cdots)]$$ the thermal average, and I have omitted spin. This gives $$\left\langle J_i(\mathbf{q}, \tau) J_j(-\mathbf{q}, 0)\right\rangle= \frac{Q^2}{m^2}\sum_{\mathbf{k}\mathbf{k}'}k_i k_j' \left\langle \psi^\dagger_{\mathbf{k}-\frac{\mathbf{q}}{2}}(\tau)\psi_{\mathbf{k}+\frac{\mathbf{q}}{2}}(\tau) \psi^\dagger_{\mathbf{k}'+\frac{\mathbf{q}}{2}}(0)\psi_{\mathbf{k}'-\frac{\mathbf{q}}{2}}(0)\right\rangle.$$ The computation of the thermal expectation value of the four operators should be straightforward: I would expect that \begin{align} \left\langle \psi^\dagger_{\mathbf{k}-\frac{\mathbf{q}}{2}}(\tau)\psi_{\mathbf{k}+\frac{\mathbf{q}}{2}}(\tau) \psi^\dagger_{\mathbf{k}'+\frac{\mathbf{q}}{2}}(0)\psi_{\mathbf{k}'-\frac{\mathbf{q}}{2}}(0)\right\rangle &= \left\langle \psi^\dagger_{\mathbf{k}-\frac{\mathbf{q}}{2}}(\tau)\psi_{\mathbf{k}+\frac{\mathbf{q}}{2}}(\tau)\right\rangle\left\langle\psi^\dagger_{\mathbf{k}'+\frac{\mathbf{q}}{2}}(0)\psi_{\mathbf{k}'-\frac{\mathbf{q}}{2}}(0)\right\rangle \\ &\phantom{=}- \left\langle \psi^\dagger_{\mathbf{k}-\frac{\mathbf{q}}{2}}(\tau) \psi_{\mathbf{k}'-\frac{\mathbf{q}}{2}}(0) \right\rangle\left\langle\psi_{\mathbf{k}+\frac{\mathbf{q}}{2}}(\tau) \psi^\dagger_{\mathbf{k}'+\frac{\mathbf{q}}{2}}(0)\right\rangle \\ &= \delta_{\mathbf{q}, \mathbf{0}}G(\mathbf{k}, 0)G(\mathbf{k}', 0)\\ &\phantom{=}-\delta_{\mathbf{k}, \mathbf{k}'} G(\mathbf{k}-\frac{\mathbf{q}}{2}, \tau)G(\mathbf{k}+\frac{\mathbf{q}}{2}, -\tau). \end{align} However, in every book I have seen (e.g. [1], [2], [3]) only the second term is present and the first is not included. Why is this? I am probably missing something very obvious, but I can't see why it is not necessary to include this contraction.

References:

[1] Introduction to Many-Body Physics, P. Coleman

[2] Many-body quantum theory in condensed matter physics, H. Bruus and K. Flensburg

[3] Many-particle physics, G. Mahan, Third edition

• Perhaps $q\neq 0$? Oct 7, 2023 at 20:40
• In most cases you are actually interested in the long-wavelength limit $\mathbf{q}\to\mathbf{0}$, so I'm pretty sure that's not the reason. Oct 7, 2023 at 20:55
• Okay, I see. Just wanted to make sure you don't miss this possibility. Oct 7, 2023 at 20:55
• Can you please provide the full references (page and e.g. eq. number)? Oct 8, 2023 at 6:03

$$G({\bf k}, 0)$$ is even function of $${\bf k}$$: $$G({\bf k}, 0) = G(-{\bf k}, 0)$$. Such functions have the following property $$\sum_{{\bf k}} k_i G({\bf k}, 0) = 0.$$ This may be the reason why the first term is not included in the correlator. Its contribution is simply zero.
• Thanks for this. Generically, the $k_i/m$ should be replaced with the group velocity $\partial \varepsilon(\mathbf{k})/\partial k_i$, which need not be an odd function of k. Is there any reason for this sum to vanish in this case? Oct 7, 2023 at 21:22
• @xzd209 In the case when $\varepsilon({\bf k})$ is an even function of ${\bf k}$, its derivative with respect to $k_i$ is odd. I cannot immediately remember a system for which, provided that there are no external fields, the property $\varepsilon({\bf k}) = \varepsilon(-{\bf k})$ would not hold.