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Let us define

$$\tag{1} Q=i\nabla_\mathbf{k}\delta(\mathbf{k-k'})\left[\rho_{nm}(\mathbf{k'})-\rho_{nm}(\mathbf{k})\right], $$

where $\delta(\mathbf{k})$ is a Dirac delta, and $\rho_{nm}(\mathbf{k})$ is a reduced density matrix. I wish to show that $$\tag{2} Q =i\delta(\mathbf{k-k'})\nabla_\mathbf{k}\rho_{nm}(\mathbf{k}). $$ My strategy relies on the idea that $$\tag{3} Q=\nabla_\mathbf{k}\int d\mathbf{k}\,Q. $$ Let us integrate and differentiate $Q$ with respect to $\mathbf{k},$ $$ Q = \nabla_\mathbf{k}\int d\mathbf{k} \,Q \nonumber \\ = \nabla_\mathbf{k}\int d\mathbf{k} \left(i\nabla_\mathbf{k}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k'})-i\nabla_\mathbf{k}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k}) \right), \nonumber \\ = i\rho_{nm}(\mathbf{k'})\nabla_\mathbf{k}\int d\mathbf{k}\,\nabla_\mathbf{k}\delta(\mathbf{k'-k}) -i\nabla_\mathbf{k}\int d\mathbf{k}\,\nabla_\mathbf{k}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k}). \tag{4} $$ The first term vanishes, since the derivative of a delta function is odd. To evaluate the second term, we use the identity $$\tag{5} \int\frac{d}{dx}\delta(x-x')f(x)dx=-\int\delta(x-x')\frac{d}{dx}f(x)dx, $$ to obtain $$\tag{6} Q = i\nabla_\mathbf{k}\int d\mathbf{k}\,\delta(\mathbf{k-k'})\nabla_\mathbf{k}\rho_{nm}(\mathbf{k}). $$ Comparison with the first line of Eq. (4) yields $$\tag{7} Q = i\delta(\mathbf{k-k'})\nabla_\mathbf{k}\rho_{nm}(\mathbf{k}). $$ What if we had chosen to integrate and differentiate $Q$ with respect to $\mathbf{k'}$ instead? We will need the identities $$\tag{8} \frac{d}{dx}\delta(x-x')=-\frac{d}{dx'}\delta(x-x'), \qquad \delta(x-x')=\delta(x'-x). $$ Applying Eq. (8) to Eq. (1), $$\tag{9} Q = i\nabla_\mathbf{k}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k'})-i\nabla_\mathbf{k}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k}) \\ = -i\nabla_\mathbf{k'}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k'})-i\nabla_\mathbf{k}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k}) \\ = -i\nabla_\mathbf{k'}\delta(\mathbf{k'-k})\rho_{nm}(\mathbf{k'})-i\nabla_\mathbf{k}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k}). $$ Integrating and differentiating $Q$ with respect to $\mathbf{k'}$ this time, $$ Q = \nabla_\mathbf{k'}\int d\mathbf{k'} Q \nonumber \\ = \nabla_\mathbf{k'}\int d\mathbf{k'} \left(-i\nabla_\mathbf{k'}\delta(\mathbf{k'-k})\rho_{nm}(\mathbf{k'})-i\nabla_\mathbf{k}\delta(\mathbf{k-k'})\rho_{nm}(\mathbf{k}) \right) \nonumber \\ = \nabla_\mathbf{k'}\int d\mathbf{k'} i\delta(\mathbf{k'-k})\nabla_\mathbf{k'}\rho_{nm}(\mathbf{k'}) -i\rho_{nm}(\mathbf{k})\nabla_\mathbf{k}\nabla_\mathbf{k'}\int d\mathbf{k'}\delta(\mathbf{k-k'}), \tag{10} $$ where to obtain the first term in the last line we have used Eq. (5), and where the second term in vanishes because the integral just gives $1.$ Equating the integrands of the first and last lines of Eq. (10), we obtain $$\tag{11} Q=i\delta(\mathbf{k-k'})\nabla_\mathbf{k'}\rho_{nm}(\mathbf{k'}), $$ which, if $\mathbf{k}=\mathbf{k'},$ agrees with the desired result. That is, if $\mathbf{k}=\mathbf{k'},$ Eqs. (2), (7), and (11) unambiguously give $$\tag{12} Q=i\nabla_\mathbf{k}\rho_{nm}(\mathbf{k}). $$ However, if $\mathbf{k}=\mathbf{k'},$ then Eq. (1) gives $0$ immediately...

Even if my proof is flawed, I believe the result Eq. (2) is correct (see Aversa & Sipe 1995). Can anyone comment on the use of Eq. (3), or provide a simple alternative proof?

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I think you are forgetting the distributional identity $$ f(x) \delta'(x) = f(0)\delta'(x)- f'(x)\delta(x), $$ which is easily proved with the use of a suitable test function.

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  • $\begingroup$ The idea being that the desired result follows immediately if $\rho_{cc}(0)=0$? $\endgroup$
    – aRockStr
    Jun 18 '20 at 13:32
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    $\begingroup$ Yes. I think so. Just introduce a test function to make it rigourous. $\endgroup$
    – mike stone
    Jun 18 '20 at 13:35
  • $\begingroup$ Cheers, I'll give it a go :) $\endgroup$
    – aRockStr
    Jun 18 '20 at 13:35

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