Editing Functional derivative (section)

====Coulomb potential energy functional====
The '''electron-nucleus''' potential energy is
<math display="block">V[\rho] = \int \frac{\rho(\boldsymbol{r})}{|\boldsymbol{r}|} \ d\boldsymbol{r}.</math>

Applying the definition of functional derivative,
<math display="block">\begin{align}
\int \frac{\delta V}{\delta \rho(\boldsymbol{r})} \ \phi(\boldsymbol{r}) \ d\boldsymbol{r}
& {} = \left [ \frac{d}{d\varepsilon} \int \frac{\rho(\boldsymbol{r}) + \varepsilon \phi(\boldsymbol{r})}{|\boldsymbol{r}|} \ d\boldsymbol{r} \right ]_{\varepsilon=0} \\[1ex]
& {} = \int \frac {\phi(\boldsymbol{r})} {|\boldsymbol{r}|} \ d\boldsymbol{r} \, .
\end{align}</math>
So,
<math display="block"> \frac{\delta V}{\delta \rho(\boldsymbol{r})} = \frac{1}{|\boldsymbol{r}|} \ . </math>

The functional derivative of the classical part of the '''electron-electron interaction''' (often called Hartree energy) is 
<math display="block">J[\rho] = \frac{1}{2}\iint \frac{\rho(\mathbf{r}) \rho(\mathbf{r}')}{| \mathbf{r}-\mathbf{r}' |}\, d\mathbf{r} d\mathbf{r}' \, .</math>
From the [[#Functional derivative|definition of the functional derivative]],
<math display="block">\begin{align}
\int \frac{\delta J}{\delta\rho(\boldsymbol{r})} \phi(\boldsymbol{r})d\boldsymbol{r}
& {} = \left [ \frac {d \ }{d\varepsilon} \, J[\rho + \varepsilon\phi] \right ]_{\varepsilon = 0} \\
& {} = \left [ \frac {d \ }{d\varepsilon} \, \left ( \frac{1}{2}\iint \frac {[\rho(\boldsymbol{r}) + \varepsilon \phi(\boldsymbol{r})] \, [\rho(\boldsymbol{r}') + \varepsilon \phi(\boldsymbol{r}')] }{| \boldsymbol{r}-\boldsymbol{r}' |}\, d\boldsymbol{r} d\boldsymbol{r}' \right ) \right ]_{\varepsilon = 0} \\
& {} = \frac{1}{2}\iint \frac {\rho(\boldsymbol{r}') \phi(\boldsymbol{r}) }{| \boldsymbol{r}-\boldsymbol{r}' |}\, d\boldsymbol{r} d\boldsymbol{r}' + \frac{1}{2}\iint \frac {\rho(\boldsymbol{r}) \phi(\boldsymbol{r}') }{| \boldsymbol{r}-\boldsymbol{r}' |}\, d\boldsymbol{r} d\boldsymbol{r}' \\
\end{align}</math>
The first and second terms on the right hand side of the last equation are equal, since {{math|'''''r'''''}} and {{math|'''''r&prime;'''''}} in the second term can be interchanged without changing the value of the integral. Therefore,
<math display="block"> \int \frac{\delta J}{\delta\rho(\boldsymbol{r})} \phi(\boldsymbol{r})d\boldsymbol{r} = \int \left ( \int \frac {\rho(\boldsymbol{r}') }{| \boldsymbol{r}-\boldsymbol{r}' |} d\boldsymbol{r}' \right ) \phi(\boldsymbol{r}) d\boldsymbol{r} </math>
and the functional derivative of the electron-electron Coulomb potential energy functional {{math|''J''}}[''ρ''] is,<ref name=ParrYangP248A.11>{{harvp|Parr|Yang|1989|loc=p. 248, Eq. A.11}}.</ref>
<math display="block"> \frac{\delta J}{\delta\rho(\boldsymbol{r})} = \int \frac {\rho(\boldsymbol{r}') }{| \boldsymbol{r}-\boldsymbol{r}' |} d\boldsymbol{r}' \, . </math>

The second functional derivative is
<math display="block">\frac{\delta^2 J[\rho]}{\delta \rho(\mathbf{r}')\delta\rho(\mathbf{r})} = \frac{\partial}{\partial \rho(\mathbf{r}')} \left ( \frac{\rho(\mathbf{r}')}{| \mathbf{r}-\mathbf{r}' |} \right ) = \frac{1}{| \mathbf{r}-\mathbf{r}' |}.</math>