To proof the Hartree equations, we begin by rearranging eq4:

Noting that is normalised and simplifying the 1^{st} and 2^{nd }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 2*n* 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 2^{nd} 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 1^{st} and 2^{nd} Coulomb terms are the same, and that the 3^{rd} and 4^{th} 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