# Symmetry of the total spin eigenstate for systems of three and more identical particles

I have just found very nice question and answer:

How to determine whether an eigenstate of total spin is symmetric or antisymmetric?

concering the relation of the value of the total spin and the symmetry of the total spin wave-function for a system of two identical particles.

Is there a general statement about the symmetry for systems of $$n$$ identical particles with spin $$I$$ and total spin $$J$$ $$(0\le J\le nI)$$?

This is in general a complicated problem. It involves the permutation group of $$n$$ objects and its irreducible representations can be constructed systematically but no so elegantly using methods based on Young diagrams. (Young tableaux are Young diagrams with the boxes filled.)

If you have $$n$$ spin-1/2 particles, the situation is somewhat simple. Using Schur-Weyl duality, the possible total spins $$S$$ and the number of times each appear is obtained from the Young diagrams with $$n$$ boxes on at most two rows, or alternatively the partitions of $$n$$ in at most two parts. Since Young diagrams are difficult to write here, I will use partitions.

Suppose $$n=5$$. The partitions of $$5$$ with at most $$2$$ parts are $$(5,0),(4,1)$$ and $$(3,2)$$. The possible values of $$S$$ are given by the differences $$\frac{1}{2}(5-0)=\frac{5}{2}$$, $$\frac{1}{2}(4-1)=\frac{3}{2}$$, and $$\frac{1}{2}(3-2)=\frac{1}{2}$$. These partitions are associated with dimension of irreps of $$S_5$$ as follows: $$(5,0): \hbox{dim}=1\, \qquad \hbox{fully symmetric}\\ (4,1):\hbox{dim}=4\,\qquad \hbox{mixed symmetry}\\ (3,2):\hbox{dim}= 5\, ,\qquad\hbox{mixed symmetry}\, .$$
(there are rules to compute the dimension of an irrep based on its tableaux, which amount to counting the number of semi-standard Young tableaux of that shape.)

From this we can deduce that $$S=5/2$$ occurs once, $$S=3/2$$ occurs $$4$$ times, and $$S=1/2$$ occurs twice. We can check the dimension: $$2^5=32$$ total states, with $$6$$ with $$S=5$$, $$4$$ sets of $$4$$ states with $$S=3/2$$, and $$5$$ sets of $$2$$ states with $$S=1/2$$ so $$6+16+10=32$$.

The states with $$S=5/2$$ are fully symmetric, but for the other values of $$S$$ the states are of mixed symmetries, i.e. permuting the spins of two particles in a state does not necessarily return a multiple of this state but in general a linear combination of other states.

If $$s\ne 1/2$$, then one needs to use plethysms, which are not so elementary. The idea is to think of states of a given $$s$$ as living inside the definining irrep of $$su(2s+1)$$, take symmetrized products of this, and decompose each piece back into $$S$$ pieces.

For instance, if $$s=1$$ and $$n=4$$, one embeds the 3 $$s=1$$ states inside the defining irrep of $$su(3)$$, and take $$4$$ products of this irrep. The associated partitions are now partitions of $$4$$ with at most $$2s+1=3$$ parts: $$(4,0,0),(3,1,0),(2,2,0)$$ and $$(2,1,1)$$.

They can be shown to correspond to the $$su(3)$$ irreps (in Dynkin notation) $$(4,0), (2,1), (0,2)$$ and $$(1,0)$$. In turn: $$(4,0)$$ occurs once, contains $$S=4,2,0$$ and these states are symmetric; $$(2,1)$$ occurs $$3$$ times and contains $$S=1,2,3$$ (dimension 15), and these states have mixed symmetry; $$(0,2)$$ occurs twice, contains $$S=0,2$$ (dimension 6), and these states have mixed symmetry; finally, $$(1,0)$$ occurs occurs $$3$$ times, contains $$S=1$$ (dimension 3) and states are also of mixed symmetry. Thus, there are $$3^4=81$$ states, with: $$\hbox{1 set of 15 symmetric states with} S=4,2,0 \, (15 \, \hbox{total})\\ \hbox{3 sets of 15 states of mixes symmetry with} S=3,2,1 \, (45 \, \hbox{total}) \\ \hbox{2 sets of 6 states of mixed symmetry with} S=0,2 \, (12\, \hbox{total}) \\ \hbox{3 sets of 3 states of mixed symmetry with} S=1 \, (9 \, \hbox{total})\, .$$ All the states of mixed symmetries are actually different mixed symmetries under permutation.

Sooooo... yes it's possible but requires advanced methods and quite a bit of technical work (to infer the S contents of each $$su(2s+1)$$ subpiece.)

This only deals with the spin parts. The spatial states must be constructed separately, will have mixed symmetries, and are to be combined with spin states of matching (for bosons) or antimatching (for fermions) mixed symmetries to produce properly symmetrized spatial+spin states.

As an example, consider the coupling of 3 spin-1/2 particles, which yields $$S=3/2,1/2,1/2$$. This can be done in multiple ways, which are equivalent but do not give the same sets of $$S=1/2$$ states. One solution is to first couple particles $$1$$ and $$2$$ to get $$S_{12}=1$$ or $$0$$, then couple $$S_{12}$$ to $$s_3=1/2$$. In this scheme, states are labelled by the intermediate value $$S_{12}$$ so we have, for instance: \begin{align} \vert 3/2,3/2\rangle &= \vert \textstyle\frac{1}{2}\textstyle\frac{1}{2}\rangle _1 \vert \textstyle\frac{1}{2}\textstyle\frac{1}{2}\rangle_2 \vert \textstyle\frac{1}{2}\textstyle\frac{1}{2}\rangle _3 \end{align} To get the first $$S=1/2$$ set, using $$S_{12}=0$$: \begin{align} \vert \textstyle\frac{1}{2}\textstyle\frac{1}{2}\rangle _{S_{12}=0}&= \vert 00\rangle_{12}\vert \textstyle\frac{1}{2}\textstyle\frac{1}{2}\rangle _3\\ &= \frac{1}{\sqrt{2}}\left[\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle_1 \vert \frac{1}{2}-\frac{1}{2}\rangle_2 -\vert \textstyle\frac{1}{2}-\frac{1}{2}\rangle_1 \vert \frac{1}{2}\frac{1}{2}\rangle_2 \right]\vert \textstyle\frac{1}{2}\textstyle\frac{1}{2}\rangle _3\\ &=\frac{1}{\sqrt{2}}\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle_1\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle_2\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle_3- \vert \textstyle\frac{1}{2}-\frac{1}{2}\rangle_1\vert \textstyle\frac{1}{2}\frac{1}{2}\rangle_2 \vert \textstyle\frac{1}{2}\frac{1}{2}\rangle_3\, . \end{align} To get the second $$S=1/2$$, use $$S_{12}=1$$: \begin{align} \vert \textstyle\frac{1}{2}\textstyle\frac{1}{2}\rangle _{S_{12}=1} &= C_{11;\frac{1}{2}-\frac{1}{2}}^{\frac{1}{2},\frac{1}{2}}\vert 11\rangle_{12}\vert \textstyle\frac{1}{2}-\frac{1}{2}\rangle _3+ C_{10;\frac{1}{2}\frac{1}{2}}^{\frac{1}{2}\frac{1}{2}} \vert 10\rangle_{12}\vert \frac{1}{2}\frac{1}{2}\rangle_3\\ &=\textstyle\sqrt{\frac{2}{3}}\vert \frac{1}{2}\frac{1}{2}\rangle_1\vert \frac{1}{2}\frac{1}{2}\rangle_2 \vert \frac{1}{2}-\frac{1}{2}\rangle_3- \frac{1}{\sqrt{3}} \left(\frac{1}{\sqrt{2}}\vert \frac{1}{2}\frac{1}{2}\rangle_1\vert \frac{1}{2}-\frac{1}{2}\rangle_2 +\frac{1}{\sqrt{2}}\vert \frac{1}{2}-\frac{1}{2}\rangle_1\vert \frac{1}{2}\frac{1}{2}\rangle_2\right) \vert \frac{1}{2}-\frac{1}{2}\rangle_3\\ &=\textstyle\sqrt{\frac{2}{3}}\vert \frac{1}{2}\frac{1}{2}\rangle_1\vert \frac{1}{2}\frac{1}{2}\rangle_2 \vert \frac{1}{2}-\frac{1}{2}\rangle_3 -\frac{1}{\sqrt{6}}\vert \frac{1}{2}-\frac{1}{2}\rangle_1\vert \frac{1}{2}\frac{1}{2}\rangle_2\vert\frac{1}{2}\frac{1}{2}\rangle_3 -\frac{1}{\sqrt{6}}\vert \frac{1}{2}\frac{1}{2}\rangle_1\vert \frac{1}{2}-\frac{1}{2}\rangle_2\vert\frac{1}{2}\frac{1}{2}\rangle_3\, . \end{align} You would get a different set of two $$S=1/2$$ states if you choose instead to couple first the second and third particles, to get $$S_{23}$$ as an intermediate value.

See also this table for states with up to 4 spin-1/2. Note that the 3-particles $$S=1/2$$ states are actually linear combinations of those I presented here, highlighting the lack of uniqueness in the construction.

• Thank you for the detailed response! If I understand correctly in general case the spin eigenstates are neither symmetric nor antisymmetric. Could you please give an example writing down the eigenstates for three identical particles with spin $\frac12$?
– drer
Commented Oct 27, 2022 at 6:43
• @drer added the example as required. Commented Oct 27, 2022 at 13:55
• Thank you! Do I understand correctly that the physical reason is that the spin operator in a general spin state does not commute with all pairwise particle exchange operators in the case $n>2$, so that the corresponding spin state is to be degenerate? It seems that the highest spin state $J=nI$ is an exception: it is non-degenerate and symmetric. Is it true?
– drer
Commented Oct 27, 2022 at 16:19
• “degenerate” is not the right word but it’s the correct idea. Once would say that the full Hilbert space decomposes into a sum of subspaces containing multiple copies of states with a given permutation symmetry. (There are two groups here: permutation and SU(2), and states transforming by a given representation of the permutation group also the same S.). You are right that, if there is more than one set of states with a given S and the permutation symmetry, permutations will mix states between those sets. Commented Oct 27, 2022 at 17:09