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- Summary
- Question on Hund's rule to determine ground sate of electron configuration.

So as I can see from the literature there are two "methods" on how to apply Hund's rules to determine the ground state of an electron configuration.

Method 1: One determines all possible states due to Pauli's principle (wave function must be totally antisymmetric) using angular momentum addition rules. Then one can select the sate with highest total S and L and can then also select J according to the 3rd rule.

So far, so good. This is quite lengthy, especially when one has to add multiple spins together, but it works and is plausible.

But one can also find this approach, for example on wikipedia:

Method 2: One fills up the possible [itex] m_l [/itex] states beginning with the highest [itex] m_l [/itex]. First, spins with [itex] m_s = \frac{1}{2} [/itex] are assigned; when all [itex] m_l [/itex] are occupied once, [itex] m_s = -\frac{1}{2} [/itex] are assigned again beginning with the highest [itex] m_l [/itex]. Now one can calculate [itex] m_L = \sum{m_l} [/itex] and [itex] m_S = \sum{m_s} [/itex]. Now comes the confusing part:

The ground state simply has [itex] L=m_L [/itex], [itex] S=m_S [/itex].

I don't really get why the two methods are equal. Why is [itex] L=m_L [/itex]? According to angular momentum addition rules, L could also take higher values than that.

I may have found a hint on the solution to this, but I'm not sure if I'm on the right track and also I don't know how to generalize this and kind of proof the equivalence of the two methods: When for example one looks at the Clebsch Gordan coefficients for 1/2x1/2 and 1x1 one can see that [itex] L=m_L=1+1=2 [/itex] is antisymmetric and [itex] S=m_S=\frac{1}{2} + \frac{1}{2}=1 [/itex] is symmetric. So maybe this is due to a general property and the method works because it satisfies Pauli's principle?

Method 1: One determines all possible states due to Pauli's principle (wave function must be totally antisymmetric) using angular momentum addition rules. Then one can select the sate with highest total S and L and can then also select J according to the 3rd rule.

So far, so good. This is quite lengthy, especially when one has to add multiple spins together, but it works and is plausible.

But one can also find this approach, for example on wikipedia:

Method 2: One fills up the possible [itex] m_l [/itex] states beginning with the highest [itex] m_l [/itex]. First, spins with [itex] m_s = \frac{1}{2} [/itex] are assigned; when all [itex] m_l [/itex] are occupied once, [itex] m_s = -\frac{1}{2} [/itex] are assigned again beginning with the highest [itex] m_l [/itex]. Now one can calculate [itex] m_L = \sum{m_l} [/itex] and [itex] m_S = \sum{m_s} [/itex]. Now comes the confusing part:

The ground state simply has [itex] L=m_L [/itex], [itex] S=m_S [/itex].

I don't really get why the two methods are equal. Why is [itex] L=m_L [/itex]? According to angular momentum addition rules, L could also take higher values than that.

I may have found a hint on the solution to this, but I'm not sure if I'm on the right track and also I don't know how to generalize this and kind of proof the equivalence of the two methods: When for example one looks at the Clebsch Gordan coefficients for 1/2x1/2 and 1x1 one can see that [itex] L=m_L=1+1=2 [/itex] is antisymmetric and [itex] S=m_S=\frac{1}{2} + \frac{1}{2}=1 [/itex] is symmetric. So maybe this is due to a general property and the method works because it satisfies Pauli's principle?