Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ground state of Phosphorus Problem

  1. Dec 27, 2012 #1
    This problem is 1.1b out of "Atomic Physics" by Budker, Kimball, and Demille. There are solutions in the book, but I am confused:

    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.

  2. jcsd
  3. Dec 27, 2012 #2
    Actually, I think I solved my own problem. It is true that for two identical spins the total angular momentum states follow the symmetry pattern:
    [tex] L_\text{total,max} \rightarrow \text {Symmetric} [/tex]
    [tex] L_\text{total,max}-1 \rightarrow \text {Antisymmetric} [/tex]
    [tex] L_\text{total,max}-2 \rightarrow \text {Symmetric} [/tex]
    [tex] \text{etc.} [/tex]

    However, for three particles, I'm pretty sure this statement is not true. For example, in my problem I had the addition of three l=1 particles. I think the states actually look like...
    [tex] L_\text{tot}=3 \rightarrow \text {Symmetric} [/tex]
    [tex] L_\text{tot}=2 \rightarrow \text {No def symmetry} [/tex]
    [tex] L_\text{tot}=1 \rightarrow \text {No def symmetry} [/tex]
    [tex] L_\text{tot}=0 \rightarrow \text {Antisymmetric} [/tex]

    Hence, only the Ltot=0 case would work for my problem.

    Does anyone know any more about this many-particle addition of angular momentum? E.g., maybe the alternating symmetry works for addition of even numbers of identical particles?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Ground state Phosphorus Date
A Several ground state calculations at once Aug 26, 2017
B What is the ground state? Aug 4, 2017
I How can an atom change from ground to excited state? Mar 12, 2017
A Overlap between ground states Dec 10, 2016