Explain more the 1/2 in E_H[n]

[Ondřej Čertík]

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.