Consider an approximate wave function of the form

 

and a Hamiltonian of the form

 

so that

 

and making a functional derivative with respect to Φ_l(r) gives

 

where two integrations by parts have been used to eliminate the Φ² δΦ in favor of δΦ²Φ.  Note that the j in the last integral can be divided out since the integrals over r an r’ are separable which allows the final integral to factor completely out of the sum so that with the definitions

 

 

 

 

we find

 

 

so that the bracket which needs to be set to zero contains v(r), not the 1/2 v(r) that is ultimately added up. The differential equation for Φ_i is then

 

Note that both E_l and the V_eff contain the full v(r) and

 

          Consider the variance in this same approximation  

 

The term inside the parenthesis is

 

 

in which parts from various i’s may cancel with each other to form a sum which goes to zero.  Suppose that we assume that each term in i  independently sums to give the variance which then becomes

 

 

including the rest of the variance and utilizing the assumptions that

 

which contains a problem in the last term only which involves a sum over all locations other than r in v which come from the r’s selected for the rest of the Φ‘s.  If we average over the probability of these occurring, this term becomes 1/2 of the preceding term to yield

 

 

and we see that the error for the minimum energy term is equal to half the standard deviation in the potential energy.  Making the same sorts of approximations directly in the variance, we find

 

 

which implies that the differential equation satisfied by Φ should be

 

 

The above results followed from the assumption that

 

 

that is from the notion that the errors from one Φ_i are uncorrelated with those from another.  If I were to assume that each Φ represents the wave function of an individual particle in the field of the others, the 1/2 would not be present.  Look again at the term inside the parenthesis. The term inside the parenthesis is

 

 

Crudely the ψ⁴ can be broken into its constituent Φ‘s and these moved inside the bracket as squares to say that the last sum goes to V_i(r_i) which in turn goes to the average to make this term go to zero.

          Suppose that for some configuration, the error is dominated by the fact that particles 1 and 2 are close, making v(r₁₂) very large.  There will be a large σ₁ and a large σ₂ which will not be independent at all.  The total variance contribution from this configuartion will be

 

 

which is a factor of two larger than our assumed value.  Suppose that we had assumed

 

 

then in this case only one term will contain v(r₁₂) and the variance will be nearly correct.  Note that multiplying by the above deduced factor of 2 on the assumption that these relatively rare events dominate makes the final variance at the minimum energy equal to the variance in the potential.  Suppose the sum is <V_i>-V(r₁)+ε and thus also the same  with respect to V(r₂), then the contributions are (2ε)² rather than 2ε² as assumed above, but now there are also a large group of other contributers which add and subtract to change these numbers.

Look again at the variance

 

 

the statistically independent quantities are the r’s.  Suppose r₁ moves to a new location, then r_1j and r_j1 both change so that n terms enter into the changed value in the squared quantity.  This allows us to semi-justify the Hartree equations above.