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I am trying calculate the functional derivative of the expectation value of the energy, $$ E=\frac{\sum_{p,q}C_{p}^{*}C_{q}H_{pq}}{\sum_{p,q}C_{p}^{*}C_{q}\delta_{pq}} .$$

With respect to $C_{p}$, my first step is, $$ \frac{\delta E}{\delta C_{p}}=\frac{\sum_{p,q}\left(\frac{\delta C_{p}^{*}}{\delta C_{p}}C_{q}H_{pq}+C_{p}^{*}\frac{\delta C_{q}}{\delta C_{p}}H_{pq}\right)C_{p}^{*}C_{q}\delta_{pq}-\left(\frac{\delta C_{p}^{*}}{\delta C_{p}}C_{q}\delta_{pq}+C_{p}^{*}\frac{\delta C_{q}}{\delta C_{p}}\delta_{pq}\right)C_{p}^{*}C_{q}H_{pq}}{\left(\sum_{p,q}C_{p}^{*}C_{q}\delta_{pq}\right)^{2}}$$

Is this correct?

My second question is, which one of the following identities is correct; $ \frac{\delta C_{p}^{*}}{\delta C_{p}}=\delta_{p^{*}q}$ or $ \frac{\delta C_{p}^{*}}{\delta C_{p}}=\delta_{C_{p}^{*}C_{q}} $ ? I feel that the former is not correct as there is no variable as $p^{*}$.

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    $\begingroup$ Comment to the post (v2): This seems to be just ordinary differentiation, not functional differentiation. $\endgroup$ – Qmechanic Apr 3 '18 at 14:39
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Your mistake is that you can't take a functional derivative with respect to the function evaluated at a dummy variable. The left hand side of the second equation, therefore, has to be something like $\frac{\delta E}{\delta C_{p'}}$. Then you use the the relations \begin{align} \frac{\delta C_p}{\delta C_{p'}} &= \frac{\delta C_p^*}{\delta C_{p'}^*} = \delta_{pp'}\ \mathrm{and} \\ \frac{\delta C_p^*}{\delta C_{p'}} &=\frac{\delta C_p}{\delta C_{p'}^*} = 0. \end{align}

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