To proof the Hartree equations, we begin by rearranging eq4:
\cdots\phi_n^*(\mathbf r_n)\left(-\frac{1}{2}\sum_{i=1}^n\nabla_i^2-\sum_{i=1}^n\frac{Z}{r_i}\right)\phi_1(\mathbf r_1)\cdots\phi_n(\mathbf r_n)d\tau_1\cdots d\tau_n)
\cdots\phi_n^*(\mathbf r_n)\sum_{i=1}^{n-1}\sum_{j=i+1}^n\int\frac{\vert\phi_j\vert^2}{r_{ij}}d\mathbf r_j\phi_1(\mathbf r_1)\cdots\phi_n(\mathbf r_n)d\tau_1\cdots d\tau_n\;\;\;\;\;\;\;\;(15))
Noting that 
is normalised and simplifying the 1st and 2nd terms on RHS of eq15, we have
\vert\psi\right>=\sum_{i=1}^nH_i\;\;\;\;\;\;\;\;(16))
)
respectively, where \cdots\phi_n(\mathbf r_n))
, \left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(\mathbf r_i)d\mathbf r_i)
, 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 \phi_i(\mathbf r_i)d\tau_i=1)
, 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,
\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\delta\phi_i(\mathbf r_i)d\tau_i)
\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\phi_i(\mathbf r_i)d\tau_i+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\delta\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j)
\delta\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\delta\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j)
\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\delta\phi_j(\mathbf r_j)d\tau_id\tau_j)
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
\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\delta\phi_i(\mathbf r_i)d\tau_i+\sum_{i=1}^n\int\delta\phi_i^*(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\phi_i(\mathbf r_i)d\tau_i)
\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\delta\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j)
\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\delta\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j)
\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\delta\phi_j(\mathbf r_j)d\tau_id\tau_j=0\;\;\;\;\;\;\;\;(20))
Question
Show that 
is Hermitian.
Answer
Since 
is Hermitian and 
is real,
\psi\right]^*d\tau=\int\psi\left[\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\psi\right]^*d\tau-\varepsilon_i^*\int\psi^*\psi d\tau)
\psi d\tau-\varepsilon_i\int\psi^*\psi d\tau=\int\psi^*\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\psi d\tau)
Using the Hermitian property of the operator
\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\phi_i^*(\mathbf r_i)d\tau_i+\sum_{i=1}^n\int\delta\phi_i^*(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\phi_i(\mathbf r_i)d\tau_i)
\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\delta\phi_j(\mathbf r_j)d\tau_id\tau_j=0)
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,
\left[-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i+ \sum_{j\neq i}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_i(\mathbf r_i)d\tau_i)
\left[-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i+ \sum_{j\neq i}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_i^*(\mathbf r_i)d\tau_i=0}\;\;\;\;\;\;\;\;(20a))
)
and )
can now be chosen arbitrarily. If we vary only the k-th function )
, eq20a becomes
\left[-\frac{1}{2}\nabla_k^2-\frac{Z}{r_k}-\varepsilon_k+ \sum_{j\neq k}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{kj}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_k(\mathbf r_k)d\tau_k=0})
Since is chosen arbitrarily, the only way to satisfy the above equation is for
\frac{1}{r_{kj}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_k(\mathbf r_k)=0)
Similarly, if we vary only the m-th function )
, where 
, we have
\frac{1}{r_{mj}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_m(\mathbf r_m)=0)
Repeating this logic to all other functions, we have a set of n simultaneous equations:
\frac{1}{r_{ij}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_i(\mathbf r_i)=\varepsilon_i\phi_i(\mathbf r_i)\;\;\;\;\;\;\;\;(21))
and another set of n simultaneous equations in the complex conjugate form:
\frac{1}{r_{ij}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_i^*(\mathbf r_i)=\varepsilon_i\phi_i^*(\mathbf r_i)\;\;\;\;\;\;\;\;(22))
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
\frac{1}{\vert\mathbf r-\mathbf r'\vert}\phi_j(\mathbf r')d\tau'\right]\phi_i(\mathbf r)=\varepsilon_i\phi_i(\mathbf r))
,
,
and
in eq161 either analytically or numerically by letting
and
, and using the initial guess values of
.
and retain the lower root. Substitute
in terms of
.
, to solve for
and
for the next iteration.
