This problem is 1.1b out of "Atomic Physics" by Budker, Kimball, and Demille. There are solutions in the book, but I am confused:(adsbygoogle = window.adsbygoogle || []).push({});

I'm asked to find the ground state configuration of Phosphorus, which is has 3 P-state valence electrons. Following Hund's rule, we want to find a state with largest total spin (S) and largest total angular momentum (L) (I use little l and s to refer to single particle states). So for 3 electrons, picking the largest S state is easy: we get S=3/2 (i.e. [tex]|m_s=1/2 \rangle|m_s=1/2 \rangle|m_s=1/2 \rangle [/tex]).

Since the spin part is chosen to be symmetric, we must construct an antisymmetric spatial wavefunction. We know all electrons are in the P manifold, so our choices of states for each particle are

[tex] l_i=1, m_{l_i}=1,0,-1 [/tex]

The authors go on to use a Slater determinant to find a totally anti-symmetric combination of these states for 3 particles, which coincides with the total angular momentum state [tex] |L=0, m_L=0\rangle [/tex]. Great! So then the ground state will be

[tex] |S=3/2, m_S= \text{4 possible values}\rangle|L=0, m_L=0\rangle [/tex].

But what the authors don't address is the total angular momentum L=2 state, which should also be anti-symmetric (since symmetry alternates between L=3,2,1,0). And also, since L=2>L=0, it should have a lower energy according to Hund's rules, no? That is my confusion.

Thanks.

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# Ground state of Phosphorus Problem

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