How can we take a term which 'represents the repulsive coulomb interaction between the $i$th and $j$th electron'? Even in classical physics when a charge is moving it's electric field is not given by coulomb's law. So how can we use potential energy term derived from coulomb's law in Schrodinger equation of many-eletron atom. When the position of an electron is not defined classically how are we even using $|r_i−r_j|$? When we are talking about electric field created by an electron in an atom it's obviously not an classical electrostatic situation.

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    $\begingroup$ This is just what physics does - using simple toy models to get insight into real phenomena. Here, we have non-relativistic model, ignoring magnetic forces and radiation. More detailed model taking into account these things is much more difficult to formulate and analyze. $\endgroup$ Mar 18, 2023 at 17:23

1 Answer 1


It is important to remember that in QM the concept of movement, specially of trajectory is not present, that is entirely a classical and relativistic concept, that said, Coulomb's law is very useful and somewhat accurate in these situations. We can see its usefulness when we study the hydrogen atom where the Hamiltonian takes the form: $$ -\frac{\hbar^2}{2\mu}\nabla^2 -\frac{q^2}{4\pi\epsilon_0r} $$ For: $$\mu = \frac{m_pm_e}{m_p+m_e} $$ This works great because we take advantage of things like the center of mass and symmetries for the radial component, however, it is important to know that we are not by any means assuming or studying the proton and electron as two different entities, but rather we are studying the center of mass of the system like somewhat of an approximation which yields excellent results, however, in many-body quantum phenomena (much like in all phenomena) the Coulombic approach is rather useless for anything getting close to reality, that's where the Kohn-Sham equations and Density Functional Theory come in! Assuming that all electrons are the same and that the electron-eletron interaction is negligible due to the shielding effect then we define the equations as: $$(-\frac{\hbar^2}{2m}\nabla^2 + v_{eff}(r))\phi(r)_i = \epsilon _i \phi(r)_i $$ Here $\epsilon_i$ is the orbital energy associated to the Kohn-Sham orbital $\phi_i$ and the density of the N-particle system is given by: $$\rho(r) = \sum^{N}_{i}|\phi(r)|^2 $$ And the Kohm-Sham potential is given by: $$E(\rho) = T_s(\rho) + \int dr v_{ext}(r)\rho(r) + E_H (\rho) + E_{xc}(\rho)$$ Where $T_s$ is the Kohm-SHam kinetic energy given by: $$T_s(\rho) = \sum^N_i\int dr\phi^*_i(r)(-\frac{\hbar^2}{2m}\nabla^2)\phi_i(r) $$ As you can see, we need the energy operators to find the electronic density but they are defined in terms of the electronic density! So what is done to work around this is to work an iterative method where a certain set of potentials is proposed and we improve them as we go. Hope I was able to answer your question!


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