# Hund's rule and angular momentum coupling

1. May 25, 2010

### daudaudaudau

Hi.

In Hund's second rule, it seems that we calculate the value of L simply by summing the $L_z$ 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?

2. May 25, 2010

### Meir Achuz

L is the maximum eigenvalue of L_z.
L(L+1) is the eigenvalue of a different operator, L^2.

3. May 25, 2010

### daudaudaudau

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