To proof the Hartree equations, we begin by rearranging eq4:
Noting that is normalised and simplifying the 1st and 2nd terms on RHS of eq15, we have
where and
Substituting eq16 and eq17 back in eq15, noting that (see this article for explanation)
where E is a functional of 2n independent variables, with n variables of and n variables of .
To find the minimum value of E subject to n constraints of , we apply the Lagrange method of undetermined multipliers to form the new functional , where
Question
Show that .
Answer
Taking the complex conjugate of F throughout, noting that E and hence F is real (recall that the functional F is obtained by subtracting the function , where from E),
Comparing the above equation with the original F,
Substituting eq18 in and , and noting that the 2nd order terms are approximately zero,
The minimum energy corresponding to F is when a small change in the functional’s input (change in and ) yields no change in the functional’s output, i.e. when , or
Using the Hermitian property of the operator
It can be easily shown, by expanding the Coulomb terms in the above equation, that 1st and 2nd Coulomb terms are the same, and that the 3rd and 4th Coulomb terms are the same. Therefore,
and can now be chosen arbitrarily. If we vary only the k-th function , eq20a becomes
Since is chosen arbitrarily, the only way to satisfy the above equation is for
Similarly, if we vary only the m-th function , where , we have
Repeating this logic to all other functions, we have a set of n simultaneous equations:
and another set of n simultaneous equations in the complex conjugate form:
Eq21 is known as the Hartree equations, and eq22, the complex conjugate form of the Hartree equations. The Hartree equations are sometimes written by changing the dummy variables and to and respectively. For example, eq21 can be expressed as