Proof of the Hartree equations

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

$E=\int\cdots\int\phi_1^*(\mathbf r_1)\cdots\phi_n^*(\mathbf r_n)\left(-\frac{1}{2}\sum_{i=1}^n\nabla_i^2-\sum_{i=1}^n\frac{Z}{r_i}\right)\phi_1(\mathbf r_1)\cdots\phi_n(\mathbf r_n)d\tau_1\cdots d\tau_n$
$+\int\cdots\int\phi_1^*(\mathbf r_1)\cdots\phi_n^*(\mathbf r_n)\sum_{i=1}^{n-1}\sum_{j=i+1}^n\int\frac{\vert\phi_j\vert^2}{r_{ij}}d\mathbf r_j\phi_1(\mathbf r_1)\cdots\phi_n(\mathbf r_n)d\tau_1\cdots d\tau_n\;\;\;\;\;\;\;\;(15)$

Noting that $\phi_i$ is normalised and simplifying the 1^st and 2^ndterms on RHS of eq15, we have

$\left<\psi\vert\sum_{i=1}^n\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\vert\psi\right>=\sum_{i=1}^nH_i\;\;\;\;\;\;\;\;(16)$

$\left<\psi\vert\sum_{i=1}^{n-1}\sum_{j=i+1}^n\int\frac{\vert\phi_j\vert^2}{r_{ij}}d\mathbf r_j\vert\psi\right>=\sum_{i=1}^{n-1}\sum_{j=i+1}^nJ_{ij}\;\;\;\;\;\;\;\;(17)$

where $H_i=\int\phi_i^*(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(\mathbf r_i)d\mathbf r_i$ and $J_{ij}=\int\int\frac{\vert\phi_i\vert^2\vert\phi_j\vert^2}{r_{ij}}d\mathbf r_id\mathbf r_j$

Substituting eq16 and eq17 back in eq15, noting that $\sum_{i=1}^{n-1}\sum_{j=i+1}^n=\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n$ (see this article for explanation)

$E=\sum_{i=1}^nH_i+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^nJ_{ij}\;\;\;\;\;\;\;\;(18)$

where E is a functional of 2n independent variables, with n variables of $\phi_i^*$ and n variables of $\phi_i$ .

To find the minimum value of E subject to n constraints of $\int\phi_i^*(\mathbf r_i)\phi_i(\mathbf r_i)d\tau_i=1$ , we apply the Lagrange method of undetermined multipliers to form the new functional $F[\phi_i^*,\phi_i]$ , where

$F[\phi_i^*,\phi_i]=E-\sum_{i=1}^n\varepsilon_i\left(\int\phi_i^*,\phi_id\tau_i-1\right)$

Question

Show that $\varepsilon_i^*=\varepsilon_i$ .

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 $g=0$ , where $g=\sum_{i=1}^n\varepsilon_i\left(\int\phi_i^*,\phi_id\tau_i-1\right)$ from E),

$F[\phi_i^*,\phi_i]=E-\sum_{i=1}^n\varepsilon_i^*\left(\int\phi_i^*,\phi_id\tau_i-1\right)$

Comparing the above equation with the original F,

$\varepsilon_i^*=\varepsilon_i$

Substituting eq18 in $F[\phi_i^*,\phi_i]$ and $F[\phi_i^*+\delta\phi_i^*,\phi_i+\delta\phi_i]$ , and noting that the 2^nd order terms are approximately zero,

$F[\phi_i^*+\delta\phi_i^*,\phi_i+\delta\phi_i]=F[\phi_i^*,\phi_i]+\sum_{i=1}^n\int\phi_i^*(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\delta\phi_i(\mathbf r_i)d\tau_i$
$+\sum_{i=1}^n\int\delta\phi_i^*(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\phi_i(\mathbf r_i)d\tau_i+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\delta\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j$
$+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\delta\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\delta\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j$
$+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\delta\phi_j(\mathbf r_j)d\tau_id\tau_j$

The minimum energy corresponding to F is when a small change in the functional’s input (change in $\phi_i^*$ and $\phi_i$ ) yields no change in the functional’s output, i.e. when $dF=F[\phi_i^*+\delta\phi_i^*,\phi_i+\delta\phi_i]-F[\phi_i^*,\phi_i]=0$ , or

$dF=\sum_{i=1}^n\int\phi_i^*(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\delta\phi_i(\mathbf r_i)d\tau_i+\sum_{i=1}^n\int\delta\phi_i^*(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\phi_i(\mathbf r_i)d\tau_i$ $+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\delta\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\delta\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j$ $+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\delta\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j$ $+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\delta\phi_j(\mathbf r_j)d\tau_id\tau_j=0\;\;\;\;\;\;\;\;(20)$

Question

Show that $-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i$ is Hermitian.

Answer

Since $-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}$ is Hermitian and $\varepsilon_i$ is real,

$\int\psi\left[\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\psi\right]^*d\tau=\int\psi\left[\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\psi\right]^*d\tau-\varepsilon_i^*\int\psi^*\psi d\tau$

$=\int\psi^*\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\psi d\tau-\varepsilon_i\int\psi^*\psi d\tau=\int\psi^*\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\psi d\tau$

Using the Hermitian property of the operator $-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i$

$dF=\sum_{i=1}^n\int\delta\phi_i(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\phi_i^*(\mathbf r_i)d\tau_i+\sum_{i=1}^n\int\delta\phi_i^*(\mathbf r_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i\right)\phi_i(\mathbf r_i)d\tau_i$ $+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\delta\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\delta\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j$ $+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\delta\phi_i(\mathbf r_i)\phi_j(\mathbf r_j)d\tau_id\tau_j$ $+\frac{1}{2}\sum_{j\neq i}\sum_{i=1}^n\int\int\phi_i^*(\mathbf r_i)\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_i(\mathbf r_i)\delta\phi_j(\mathbf r_j)d\tau_id\tau_j=0$

It can be easily shown, by expanding the Coulomb terms in the above equation, that 1^st and 2^nd Coulomb terms are the same, and that the 3^rd and 4^th Coulomb terms are the same. Therefore,

$dF=\int\sum_{i=1}^n\delta\phi_i^*(\mathbf r_i)\left[-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i+ \sum_{j\neq i}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_i(\mathbf r_i)d\tau_i$ ${+\int\sum_{i=1}^n\delta\phi_i(\mathbf r_i)\left[-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}-\varepsilon_i+ \sum_{j\neq i}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_i^*(\mathbf r_i)d\tau_i=0}\;\;\;\;\;\;\;\;(20a)$

$\delta\phi_i^*(\mathbf r_i)$ and $\delta\phi_i(\mathbf r_i)$ can now be chosen arbitrarily. If we vary only the k-th function $\delta\phi_k^*(\mathbf r_k)$ , eq20a becomes

${dF=\int\delta\phi_k^*(\mathbf r_k)\left[-\frac{1}{2}\nabla_k^2-\frac{Z}{r_k}-\varepsilon_k+ \sum_{j\neq k}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{kj}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_k(\mathbf r_k)d\tau_k=0}$

Since $\delta\phi_k^*(\mathbf r_k)$ is chosen arbitrarily, the only way to satisfy the above equation is for

$\left[-\frac{1}{2}\nabla_k^2-\frac{Z}{r_k}-\varepsilon_k+ \sum_{j\neq k}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{kj}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_k(\mathbf r_k)=0$

Similarly, if we vary only the m-th function $\delta\phi_m^*(\mathbf r_m)$ , where $m\neq k$ , we have

$\left[-\frac{1}{2}\nabla_m^2-\frac{Z}{r_m}-\varepsilon_m+ \sum_{j\neq m}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{mj}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_m(\mathbf r_m)=0$

Repeating this logic to all other functions, we have a set of n simultaneous equations:

$\left[-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}+ \sum_{j\neq i}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_i(\mathbf r_i)=\varepsilon_i\phi_i(\mathbf r_i)\;\;\;\;\;\;\;\;(21)$

and another set of n simultaneous equations in the complex conjugate form:

$\left[-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}+ \sum_{j\neq i}\int\phi_j^*(\mathbf r_j)\frac{1}{r_{ij}}\phi_j(\mathbf r_j)d\tau_j\right]\phi_i^*(\mathbf r_i)=\varepsilon_i\phi_i^*(\mathbf r_i)\;\;\;\;\;\;\;\;(22)$

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 $\mathbf r_i$ and $\mathbf r_j$ to $\mathbf r$ and $\mathbf r'$ respectively. For example, eq21 can be expressed as

$\left[-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}+ \sum_{j\neq i}\int\phi_j^*(\mathbf r')\frac{1}{\vert\mathbf r-\mathbf r'\vert}\phi_j(\mathbf r')d\tau'\right]\phi_i(\mathbf r)=\varepsilon_i\phi_i(\mathbf r)$

Proof of the Hartree equations

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