Hund's rule and angular momentum coupling

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
daudaudaudau
Messages
297
Reaction score
0
Hi.

In Hund's second rule, it seems that we calculate the value of L simply by summing the [itex]L_z[/itex] components of the individual electrons. But L has to do with the eigenvalue of the L^2 operator, i.e. the eigenvalue is L(L+1). So how can this be correct?
 
Physics news on Phys.org
Meir Achuz said:
L is the maximum eigenvalue of L_z.
L(L+1) is the eigenvalue of a different operator, L^2.

Yeah that is exactly my point. We know that that L_z has some particular value. Now why is L=L_z ?