PX262 - G5 - multielectron atoms

for two electrons, the hamiltonian has a form:

\hat{H} = - \frac{ℏ^{2}}{2 m} \nabla_{1}^{2} - \frac{ℏ^{2}}{2 m} \nabla_{2}^{2} - \frac{2 e^{2}}{4 π ϵ_{0} r_{1}} - \frac{2 e^{2}}{4 π ϵ_{0} r_{2}} + \frac{e^{2}}{4 π ϵ_{0} r_{12}}

where,
$\nabla_{1}$ and $\nabla_{2}$ represent the derivatives with respect to the coordinates of the two electrons;
$r_{1}$ and $r_{2}$ are the distances of the electrons from the nucleus, $r_{12}$ is the distance between the two electrons

with more electrons, the hamiltonian will be similar with kinetic terms and potential from nucleus-electron and electron-electron each electron
now, using this hamiltonian to solve the schrödinger equation
an approximation can be made where the interaction between electrons is ignored
the equation splits into two, one for each electron, same as in the hydrogenic atom
the electrons cannot be distinguished, and are placed in a common set of states in the lowest energy
any particle with half-integer spin has to obey the pauli exclusion principle, which states that the wavefunctions of two identical particles have to be antisymmetric with respect to the exchange of particles

u (r_{1}, r_{2}) = - u (r_{2}, r_{1})

consequently, only one particle can be in a particular state
for just two electrons, they can both be placed in $1 s$ state $(n = 1, l - 0, m_{l} = 0)$ as each can have a different z-component of spin
for more electrons, eg: $^{3}$ Li, with $3$ electrons, two can be in the ground state: $1 s^{2}$ (two electrons with $s = \pm \frac{1}{2} ℏ$ ), and another will be in $2 s^{1}$