Talk:Multiplicity (chemistry)

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Oppose. The proposal is presumably based on the idea that multiplicity is just a special case of (near) degeneracy, corresponding to the number of spin orientations. However this is not always the case, as explained in the third paragraph of the intro to this article. For example, the ground state of the N atom is a ⁴S state, with a multiplicity of 4 but only one energy level. Therefore multiplicity does not always correspond to spin degeneracy, and multiplicity requires a separate article. Dirac66 (talk) 20:18, 10 September 2016 (UTC)[reply]

Hi Dirac66, As you know, degeneracy is not about the number of different energy levels but about the number of [occurences] of the same energy level. Maybe you want to give another example but the Term symbol ⁴S only says that L=0, 2S+1=4. Therefore there are (2S+1)*(2L+1)=4 degenerate "microstates" (See Term symbol). So in this case multiplicity gives the number of degenerate energy levels. For L>0 this would of course not be the case anymore. However multiplicity 2S+1 still gives you the degeneracy that arises from the spin S. Therefore without external magnetic field: For a state in an electronic system in an atom with a certain quantum number m_l the number of degenerate states is, due to the degeneracy of the spin values 2S+1.--Biggerj1 (talk) 05:05, 12 September 2016 (UTC)[reply]

It is true that I oversimplified by considering energy levels rather than states. The ground state of the N atom is ⁴S which is one energy level with 4 microstates (M_J = +3/2, +1/2, -1/2, -3/2) with only 1 level in the absence of a magnetic field, which can be separated into 4 levels (microstates) with a magnetic field. So in this case the multiplicity equals the degeneracy with a magnetic field.

However the ground state of the C atom is ³P, with L > 0 as you suggest. Here there are 3 levels (J = 0, 1, 2) without a magnetic field, and 1+3+5 = 9 microstates which form 9 levels in a magnetic field. So in this case the multiplicity equals the degeneracy without a magnetic field, contrary to the case for the N atom.

To return to the merge proposal under discussion, I think the above anslysis shows that the relation between multiplicity and degeneracy requires detailed analysis and more than one example, so it is really better to have a separate article on multiplicity to explain, and I still oppose the merge proposal. Dirac66 (talk) 15:18, 12 September 2016 (UTC)[reply]

Degeneracy counts the number of microstates with that have the same energy. For the state ⁴S of the electron shell you have four degenerate microstates (M_J = +3/2, +1/2, -1/2, -3/2) in the absence of a magnetic field. In the presence of a magnetic field those microstates have different energies E1, E2, E3 and E4 all with degeneracy one then...--Biggerj1 (talk) 11:07, 9 October 2016 (UTC)[reply]

Oppose. For practical reasons. The terms are used pedagogically, in quite different areas, e.g., degeneracy in the discussion of Jahn-Teller distortions. For the sake of utility in instruction, having two articles, accessed by different audiences for different purposes—but which will perhaps, both in future, better explain the relationships between the two concepts based on sourced presentations—this is my vote. Note, your arguments here, however well-informed, need to be based in sources. There is no common knowledge here. For instance, I looked at Peter Atkins presentation [Atkins, J. de Paula, R. Friedman (2009). Quanta, Matter, and Change: A Molecular Approach to Physical Chemistry, p. 80 and passim], and there is no apparent mention of mutiplicity in his subsection on degeneracy. Point is, however which way we as editors might desire to present things, we should present them as do the recognised experts. To persuade one way or another, present sources. Thanks and cheers. Le Prof 73.210.155.96 (talk) 00:27, 17 January 2017 (UTC)[reply]