Addition of angular momenta of fermions

Question

In my lecture notes, the example is given of finding the maximum total angular momentum $J$ for four identical fermions each with angular momentum $j = 5/2$. It explains that since $M_J$ is given by the sum of $m_j$ for each particle, then for the particles to occupy different states the maximum $M_J$ is $5/2+3/2+1/2-1/2=4$, this then gives to $J_{max}=4$. I am wondering how can know that there is not another state with $M_J=5$ with a larger value of $J$ or perhaps a state with a lower value of $M_J$ and larger value of $J$.

A similar problem is the addition of two identical spin-$1/2$ fermions. In this case there are two states with $M_s=0$ corresponding to $S=0$ and $S=1$. The $S=1$ state is ruled out by symmetry considerations.

Is there a similar principle for this case and in general for the addition of fermion momenta.

Answer 1

Your teacher wants you to work out the combinatorics as you breathe. The principle, evident from your second paragraph/example, is that all states involving symmetrized pieces of your four components die by Fermi statistics/antisymmetrization. A lowering operator does not change the symmetry, but the independent state orthogonal to what you find by lowering must have different symmetry.

So, lower the (vanishing!) state of four 5/2s; the orthogonal state with one 3/2 and three 5/2s must have symmetric pieces, so it too is vanishing. So lower to three 5/2s and a 1/2, likewise vanishing, or orthogonal to it, (haplessly better) to two 5/2, a 3/2, and a 1/2. Your next lowering will net you the orthogonal state with 5/2,3/2,1/2,-1/2 which, finally, can be antisymmetric, without vanishing. So the method consists in ensuring absence of symmetric combinations of $M_s$ s, and fishing surviving S s out of them.

Combining 4 sextets (2 × 5/2 +1) will give you $6^4=1296$ states, but only ${6\cdot 5 \cdot 4\cdot 3\over 4!}=15=9+5+1$ of those are fully antisymmetric, a tiny minority. The S=4 he already worked out for you by ordering his component $M_s$ s from larger to smallest. The trailing S s, all even spin, are evidently 2 and 0, respectively. The quintet has 3/2,1/2,-1/2,-3/2. The singlet has equal numbers of matched $M_s$ s. I'm not sure what general principle you seek.