One way is to use the functional derivatives, another way is:
Actually, the 1/2 accounts for double counting in the electron-electron integral, not present in the electron-nuclear integral since it is not a self energy.
Potential energy of a collection of charges q_i in the field of another collection of charges q'_j is
\sum_{i,j} q_i q'_j / |r_i-r'_j|
Whereas the *self* energy of the charges q_i is
\sum_{i,j>i} q_i q_j / |r_i-r_j| = 1/2 \sum_{i,j/=i} q_i q_j / |r_i-r_j|
where the j>i in the summation is required to avoid double counting pairs: e.g., includes q_1*q_2 term, omits q_2*q_1, etc.
And so in the continuous case, for the electron-electron self energy (Hartree energy) we have the 1/2:
E_H = 1/2 \int { rho(x) rho(x') / |x - x'| d^3x d^3x' } = 1/2 \int { rho(x) v_e(x) d^3x }
where v_e(x) = \int { rho(x') / |x - x'| d^3x' }
while for the electron-nuclear energy we have simply
E_en = \int { rho(x) rho_n(x') / |x - x'| d^3x d^3x' } = \int { rho(x) v_n(x) d^3x }
where v_n(x) = \int { rho_n(x') / |x - x'| d^3x' } = \int { Z delta(x'-x_n)/ |x - x'| d^3x' } = Z/|x - x_n|; x_n = nuclear position.
Use both. Possibly put this into the functional derivatives section.
One way is to use the functional derivatives, another way is:
Use both. Possibly put this into the functional derivatives section.