Why filled shell of electrons has L=0 and S=0?

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The discussion centers on the filled shell of electrons exhibiting total orbital angular momentum (L) of 0 and total spin angular momentum (S) of 0. It is established that the spin momenta of the electrons cancel each other out, resulting in S=0, while the orbital momenta also cancel, leading to L=0. The S-L coupling rules indicate that for electrons with quantum numbers (l1,s1) and (l2,s2), the filled shell configuration results in a singlet state, which is energetically favorable due to the Hamiltonian's characteristics.

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runnerwei
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The text says that the spin momenta of those electrons cancel each other so S=0.
The text also says that the orbital momenta of those electrons cancel each other so L=0.
But, if there are electrons with quantum numbers (l1,s1) and (l2,s2), using S-L coupling, the L=l1+l2,l1+l2-1,...\l1-l2\,
S=s1+s2,s1-s2
How to reach the conclusion that L=0 and S=0 if the shell is filled?
Thanks a lot!
 
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Because it is rotationally invariant and, hence, must correspond to a singlet state.
 
As according to the coupling rule, the allowed values for S is S=s1+s2,s1-s2. As s1=s2=1/2, S can have two values, 0 and 1. So how would they say that S=0, rather than S=1?
 
runnerwei said:
As according to the coupling rule, the allowed values for S is S=s1+s2,s1-s2. As s1=s2=1/2, S can have two values, 0 and 1. So how would they say that S=0, rather than S=1?

Probably because the S = 0 state usually has lower energy than S = 1, but this depends on the details of the Hamiltonian.
 

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