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.

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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.

  • $\begingroup$ 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. $\endgroup$
    – lineage
    Commented Apr 28, 2020 at 13:40
  • $\begingroup$ 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 ? $\endgroup$ Commented Apr 28, 2020 at 13:47
  • $\begingroup$ 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. $\endgroup$
    – lineage
    Commented Apr 28, 2020 at 13:52
  • $\begingroup$ 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. $\endgroup$ Commented Apr 28, 2020 at 13:55
  • 1
    $\begingroup$ I deliberately didn't..I am not on a sure footing here..did it a long time ago..try googling LL,LS coupling schemes $\endgroup$
    – lineage
    Commented Apr 28, 2020 at 13:57

1 Answer 1


Thanks to @lineage 's comment about seeing section 10.3 of book Quantum Physics of Atoms,Molecules Solids. for insights.

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.

This happens simply because in most quantum states the charge distributions of the electrons are not spherically symmetrical, and so they exert torques on each other. Since the space orientation of the charge distribution of an electron is related to the space orientation of its orbital angular momentum vector, there are torques acting between the angular momentum vectors. The torques do not tend to change the magnitude of the individual orbital angular momentum vectors, but only tend to make them precess about the total orbital angular momentum vector in such a way that its magnitude L' remains constant.

The question then arises: Which of the possible values of L' corresponds to the state of lowest energy?

There are opposing tendencies, but the basis of the one which usually dominates can be understood even from classical physics by considering two electrons in a Bohr atom. Two optically active electrons mov- ing in the same Bohr orbit tend to remain at op- posite ends of a diameter so as to minimize their Coulomb repułsion. As a result, their orbital angu- lar momenta tend to couple in such a way as to yield a maximum total orbital angular momentum. repulsion between the electrons, the most stable arrangement is obtained when the electrons stay at the opposite ends of a diameter. In this state of lowest energy, the electrons rotate together with individual orbital angular momentum vcctors parallel, and therefore with the magnitude L' of the total angular momentum vector a max- imum. This conclusion is confirmed by an analysis of the spectra produced by atoms with several optically active electrons. That is, for such atoms the residual Coulomb interaction produces a tendency for the orbital angular momenta of the optically active electrons to couple in such a way that the magnitude of the total orbital angular momen- tum L' is constant, and the energy is usually lowest for the state in which L' is largest.


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