Vanadium 50 said:
You're right, on that.
The fully antisymmetric state is unique, by inspection. Every other bit of my argument is invalid. (But I still suspect a theorem)
Here's a symmetry argument for the three spin-1 case.
First, we analyse by ##j## values (as before):
##j = 3## has ##7## levels
##j = 2## has ##5## levels
##j = 1## has ##3## levels
##j = 0## has ##1## level
There are numerous ways to meet the condition of degeneracies, assuming a common degeneracy across each level.
Then, we analyse by ##m## value:
##m = 3## is ##1## dimensional
##m = 2## is ##3## dimensional (cycles of ##110##)
##m =1## is ##6## dimensional (cycles of ##100## and ##11-1##
##m = 0## is ##7## dimensional (cycles of ##10-1## and ##000##)
Now, ##J^2## is symmetric and preserves the ##m##-value, hence preserves the above dimensionality, in terms of ##|j m \rangle ## eigenstates. That might be your theorem:
There is only one ##m = 3## state, hence ##j = 3## has degeneracy ##1## - i.e. no degeneracy (as previously established).
##m = 2## is ##3## dimensional. One dimension is for ##j = 3##, leaving two for ##j = 2##. Hence ##j = 2## has degeneracy ##2##.
Looking at ##m = 1## implies ##j = 1## has degeneracy ##3##.
And ##m = 0## implies ##j =0## has degeneracy ##1##.
Looking at the negative values gives the same answer. So, we have, as expected:
##j = 3## has ##7## levels with degeneracy ##1##
##j = 2## has ##5## levels with degeneracy ##2##
##j = 1## has ##3## levels with degeneracy ##3##
##j = 0## has ##1## level degeneracy ##1##
The same argument also works for the established result in the four-electron case.