- #1
Haorong Wu
- 413
- 89
At Griffiths QM 3rd edition, page 215 Problem 5.18(b), it reads:
(b) Hund's second rule says that, for a given spin, the state with the highest total orbital angular momentum (L), consistent with overall antisymmetrization, will have the lowest energy. Why doesn't carbon have L=2? Hint: Note that the "top of the ladder" (##M_L = L##) is symmetric.
However, where the top of the ladder is mentioned in the book? I can locate the ladder operator part, but I cannot find something relating to the symmetry of the top of the ladder.
In the solution, it says :
... by going to the top of the ladder: ##|2 2>=|1 1>_1 |1 1>_2##
OK, I'm totally confused now.
(b) Hund's second rule says that, for a given spin, the state with the highest total orbital angular momentum (L), consistent with overall antisymmetrization, will have the lowest energy. Why doesn't carbon have L=2? Hint: Note that the "top of the ladder" (##M_L = L##) is symmetric.
However, where the top of the ladder is mentioned in the book? I can locate the ladder operator part, but I cannot find something relating to the symmetry of the top of the ladder.
In the solution, it says :
... by going to the top of the ladder: ##|2 2>=|1 1>_1 |1 1>_2##
OK, I'm totally confused now.