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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

xzd209
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1 Answers1

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$G({\bf k}, 0)$ 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.

Gec
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