# How does electrostatic repulsion between electrons in "many electron atom" lead to coupling of individual orbital angular momentum vectors?

I just started studying $$LS$$ coupling scheme, book has described $$LS$$ coupling in following order

1. Firstly it mentions due to "spin-spin" correlation individual spin angular momentum vectors couples to from resultant spin angular momentum vector. i.e $$\vec{S}$$.

And quantum number $$S$$ takes values from $$|\vec{s_1}+\vec{s_2}+\vec{s_3}.....|_{min}$$ to $$(\vec{s_1}+\vec{s_2}+\vec{s_3}......)$$

2. Then it says As a result of Residual Electrostatic Interaction individual orbital angular momentum vectors of "optical" electrons are strongly coupled with one another to form a resultant orbital angular momentum vector $$\vec{L}$$ of magnitude $$\sqrt{L(L+1)} \hbar$$ which is constant of motion. My question arises here is how does residual electrostatics interaction which is repulsive electric potential between electrons in an atom, leads to the coupling of individual orbital angular momentum vectors ?

Reference :- Page 144 of the PDF or 140 of the book.

• I think its in the sense that while orbiting around the nucleus the swarm of electrons move in lock step--they can't get too close to each other or too further away without getting close to some others still-so their $\vec{L}$ couple strongly. Apr 28, 2020 at 13:40
• how does "can't get too far or too close" results in coupling? actually what is coupling I have seen various definition of it but can you give physical interpretation of it ? Apr 28, 2020 at 13:47
• Its the presence of correlation between individual $\vec{L_i}$-that is there is an interaction. Because of the localization of electrons in the nuclear potential, electrons affect each other's motion and thereby the angular momenta. Apr 28, 2020 at 13:52
• Okay, that makes sense. Thank you. But can you please write it down in answer section in a more elaborated way or send a link to any article/pdf that describes it the way you said? I can't find any reason online. Apr 28, 2020 at 13:55
• I deliberately didn't..I am not on a sure footing here..did it a long time ago..try googling LL,LS coupling schemes Apr 28, 2020 at 13:57

The Coulomb interaction doesn't result in coupling of $$\vec{l_1},\vec{l_2}....$$ to form $$\vec{L}$$. Instead it makes the coupling to happen in such a way that the $$\vec{L}$$ remains constant.