Why is L=2 not acceptable for carbon according to Hund's second rule?

[Haorong Wu]

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.

 

[DrClaude]

The "top of the ladder" is simply the final state obtained by successive application of the raising ladder operator.

The symmetry comes from the fact that the only way to build the $| L = 2, M_L=2 \rangle$ state from single-electron states is
$$
| L = 2, M_L=2 \rangle = | l = 1, m_l=1 \rangle \otimes | l = 1, m_l=1 \rangle
$$
That state is symmetric with respect to particle interchange. What constraint does that put on the spin part of the state?

 

[Haorong Wu]

The "top of the ladder" is simply the final state obtained by successive application of the raising ladder operator.

The symmetry comes from the fact that the only way to build the $| L = 2, M_L=2 \rangle$ state from single-electron states is
$$
| L = 2, M_L=2 \rangle = | l = 1, m_l=1 \rangle \otimes | l = 1, m_l=1 \rangle
$$
That state is symmetric with respect to particle interchange. What constraint does that put on the spin part of the state?

Thanks, Drclaude.

I can see why it says $M_L = L$, now.
I understand since $S=1$ is symmetrical, then the spatial state should be antisymmetrical, and then $L=2$ is unacceptable.

 

[DrClaude]

I understand since $S=1$ is symmetrical, then the spatial state should be antisymmetrical, and then $L=2$ is unacceptable.

Correct.