j-j Coupling -Atomic Physics

J-J Coupling is obtained in two ways :

a) As a result of the stronger spin-orbit interaction, the orbital and the spin angular
momentum vectors of each individual electron are strongly coupled together to form a resultant angular monnentuin vector of magnitude :
                      √ j ( j + 1).h/2π

  where, j l – 1/2 and l + 1/2 that is, j takes half-integral values only.

This means that due to spin-orbit interaction, the unperturbed energy level is splitted into a number of well-spaced levels, each corresponding to a different combination of the possible j-values for the individual optical electrons; the level corresponding to all the electrons having their smaller j-value (j = l – 1/2)being lowest.
b) As a result of the residual electrostatic interaction and spin-spin corelation, the
resultant angular momentum  vectors of the individual electrons are less strongly
coupled with one another to form the total angular momentum vector J of the atom, of
magnitude = √ J(J + 1).h/2π

The  total angular momentum quantum number takes the values:

Let us know explain electronic configuration of 4p 4d under J – J coupling.
This gives four possible combinations :
{ 1/2 , 3/2 }, {1/2 ,5/2 } ,{3/2 ,3/2 } ,{3/2 ,5/2}
The four combinations of J values gives :

The complete splitting is shown in fig. below as :

Selection Rule in j -j Coupling :

(i) The parity of the configuration must change in an electric-dipole transition
(Laporte rule). This means that if only one electron jumps in the transition (as is usually
the case) then for this election we must have Δ= ± 1. If two electrons jump then,
Δl1 = ±1 and Δl2 = 0,± 2. This rule is exactly same as in L-S coupling.
(ii) Δj = 0, ±1 for the jumping electron, and Δj = 0 for all the other electrons.
(iii)  For the atom as a whole,
ΔJ=0, ± 1 but J = 0 <—/—>  J = 0.
The selection rules ΔS = 0 and Δ L = 0, ±1 no 
longer good quantum numbers.
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