# Central Field In Many Electron Atoms

I was reading Hartree Self Consistent Field and came across "Atomic Physics by P. Ewart" (PDF). In Central Field Approximation page 8 it gives the formula for the Hamiltonian in two terms, the Hamiltonian due to central part and the rest of the perturbation as.

We recall that the hydrogen problem was solved using the symmetry of the central Coulomb field – the $$1/r$$ potential. This allowed us to separate the radial and angular solutions. In the many electron case, for most of the time, a major part of the repulsion between one electron and the others actstowards the centre. So we replace the $$1/r$$, hydrogen-like, potential with an effective potential due to the nucleus and the centrally acting part of the $$1/r_{ij}$$ repulsion term. We call this the Central Field $$U(r)$$. Note it will not be a $$1/r$$ potential. We now write the Hamiltonian $$\hat{H}=\hat{H}_0+\hat{H}_1 \tag{24}$$ $$\text{where }\quad\hat{H}_0 =\sum_i\left\{-\frac{\hbar^2}{2m}\nabla_i^2+U(r_i)\right\} \tag{25}$$ $$\text{and }\quad\hat{H}_1 =\sum_{i>j}\frac{e^2}{4\pi\epsilon_0r_{ij}} -\sum_i\left\{\frac{Ze^2}{4\pi\epsilon_0r_i}+U(r_i)\right\}\tag{26}$$

So if you calculate $$\hat{H}$$ you get $$\displaystyle \hat {H}= -\sum_i {\frac{h^2}{2m} \nabla^2_i } + \sum_i U{(r_i)} + \displaystyle \sum_{i>j} \frac{e^2}{4 \pi \epsilon_0 r_{ij}} - \sum_i U{(r_i)} - \sum_i \frac{Z e^2}{4 \pi \epsilon_0 r_{i}}$$

$$\displaystyle \hat {H}= -\sum_i {\frac{h^2}{2m} \nabla^2_i }+ \displaystyle \sum_{i>j} \frac{e^2}{4 \pi \epsilon_0 r_{ij}} - \sum_i \frac{Z e^2}{4 \pi \epsilon_0 r_{i}}$$ Now when I tried to calculate it. I took potential due to nucleus ( central ) as $$\displaystyle - \sum_i \frac{Z e^2}{4 \pi \epsilon_0 r_{i}} = \sum V_i$$ The electrostatic repulsion as $$\sum_{i>j} \frac{e^2}{4 \pi \epsilon_0 r_{ij}} = \sum B_{ij}$$ Some part of $$\sum V_i$$ is cancelled by central part of $$\sum B_{ij}$$say that part is $$\sum C_i$$. So the non central part left in the system is $$\sum_{i>j} \frac{e^2}{4 \pi \epsilon_0 r_{ij}}- \sum C_i$$

And the central part is $$\sum V_i - \sum C_i= \sum_i U{(r_i)}$$(say)

So the equation for $$\displaystyle \hat{H}= \hat {H^0} + \hat {H^1}$$ where $$\hat {H^0}$$ is central part and $$\hat {H^1}$$ is non-central part a.k.a perturbation.

put all of above in the equation we get $$\displaystyle \hat {H}= -\sum_j {\frac{h^2}{2m} \nabla^2_i } + \sum_i U{(r_i)} +\sum_{i>j} \frac{e^2}{4 \pi \epsilon_0 r_{ij}} - \sum C_i$$

But $$\sum V_i - \sum C_i= \sum_i U{(r_i)}$$ so

above equation becomes

$$\displaystyle \hat {H}= -\sum_i {\frac{h^2}{2m} \nabla^2_i } + \sum_i U{(r_i)} + \displaystyle \sum_{i>j} \frac{e^2}{4 \pi \epsilon_0 r_{ij}} + \sum U{r_i} - \sum V_i$$

$$\displaystyle \hat {H}= -\sum_i {\frac{h^2}{2m} \nabla^2_i }+ \displaystyle \sum_{i>j} \frac{e^2}{4 \pi \epsilon_0 r_{ij}} - \sum_i \frac{Z e^2}{4 \pi \epsilon_0 r_{i}}+ 2\sum U{r_i}$$

Where am I calculating wrong ?

With that said, I therefore would conclude that the "true" potential $$U\left(\vec{r}\right)$$ as presented in equations (25) and (26) not just be one of electrostatic potentials, but also the other terms such as that from $$H_2$$, the spin-orbit coupling, and other effects (such as hyperfine-coupling).
Even disregarding all these, the following 2 equations you have written are not consistent \begin{align} \sum B_{ij} & = \sum C_i + \left( \sum B_{ij} - \sum C_i \right), \\ \sum U \left( r_i \right) & = \sum V_i - \sum C_i. \end{align}
Full working: \begin{align} H & = H_0 + H_1 \\ & = \sum_i \left\lbrace -\frac{\hbar^2}{2m}\grad_i^2 \right\rbrace + \sum_{i>j} \left\lbrace \frac{1}{4\pi\epsilon_0}\frac{e^2}{r_{ij}} \right\rbrace + \sum_i \left\lbrace -\frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r_i} \right\rbrace \\ & = \sum_i \left\lbrace -\frac{\hbar^2}{2m}\grad_i^2 \right\rbrace + \sum_{i>j} B_{ij} + \sum_i V_i \\ & = \sum_i \left\lbrace -\frac{\hbar^2}{2m}\grad_i^2 \right\rbrace + \sum C_i + \sum_{i>j} \phi_{ij} + \sum_i V_i \quad \text{where}\ \sum_{i>j} \phi_{ij} = \sum_{i>j} B_{ij} - \sum C_i \\ & = \sum_i \left\lbrace -\frac{\hbar^2}{2m}\grad_i^2 + C_i + V_i \right\rbrace + \sum_{i>j} \phi_{ij} \\ &= \sum_i \left\lbrace -\frac{\hbar^2}{2m}\grad_i^2 + U\left(r_i\right) \right\rbrace + \sum_{i>j} \phi_{ij} \\ & = H^0 + H^1. \end{align}