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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

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  • $\begingroup$ Perhaps $q\neq 0$? $\endgroup$ Oct 7, 2023 at 20:40
  • $\begingroup$ 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. $\endgroup$
    – xzd209
    Oct 7, 2023 at 20:55
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    $\begingroup$ Okay, I see. Just wanted to make sure you don't miss this possibility. $\endgroup$ Oct 7, 2023 at 20:55
  • $\begingroup$ Can you please provide the full references (page and e.g. eq. number)? $\endgroup$ Oct 8, 2023 at 6:03

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – xzd209
    Oct 7, 2023 at 21:22
  • $\begingroup$ @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. $\endgroup$
    – Gec
    Oct 7, 2023 at 21:33
  • $\begingroup$ I'm pretty sure that property is broken in systems which lack inversion symmetry $\endgroup$
    – xzd209
    Oct 7, 2023 at 21:58
  • $\begingroup$ @xzd209 If a system of charged fermions does not have inversion symmetry, would such a system not have a non-zero current in the absence of a field? $\endgroup$
    – Gec
    Oct 7, 2023 at 22:21
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    $\begingroup$ You've made me realise that physically the reason these terms have been dropped is that the expectation value of the current must vanish if there is no applied field. However, I'm not sure how to show this mathematically - the question of inversion symmetry is also discussed here physics.stackexchange.com/questions/413712/… $\endgroup$
    – xzd209
    Oct 8, 2023 at 3:13

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