Consider a system of e.g. $N=3$ spin-1/2 particles. The state of the system $\vert\psi\rangle$ lives in a Hilbert space of dimension $2^N=8$.
Now, consider the collective spin operator $$\mathbf{J} = \sum\limits_{i=1}^{N}\mathbf{j}^{(i)}.$$ From this operator one can define the Dicke states $|J,M\rangle$ as the simultaneous eigenstates of $$\mathbf{J}^{2}|J,M\rangle = J(J+1)|J,M\rangle,$$ $$\mathbf{J}_{z}|J,M\rangle = M|J,M\rangle.$$ For three spins we have the six possible states $|3/2,3/2\rangle$, $|3/2,1/2\rangle$, $|3/2,-1/2\rangle$, $|3/2,-3/2\rangle$, $|1/2,1/2\rangle$, $|1/2,-1/2\rangle$. Here, note that the operator $\mathbf{J}$ lives in a 8-dimentional space, and the two apparently missing eigenstates come from the fact that $|1/2,1/2\rangle$ and $|1/2,-1/2\rangle$ come with multiplicity two. Therefore, the six Dicke states listed before form a basis for the 8-dimensional Hilbert space of $N=3$ spins. (further references: Collective angular momentum , Dicke states and indistinguishable particles)
As Dicke states form a basis, I would expect the possibility to write uniquely any spin state in this basis. However, this is not possible as the basis has degenerate vectors. It seems that initial information on the symmetry under particle exchange is lost in this basis change.
For example, three spin can combine as listed here: Adding 3 electron spins . As you can see from the rows dealing with $N=3$ spins, there are two states compatible with $|1/2,1/2\rangle$ and two compatible with $|1/2,-1/2\rangle$. Which is because they have different permutational symmetry.
Is this reasoning correct? Therefore, what can one say about the basis formed by Dicke states? Is there a typical operator to use, compatible with $\mathbf{J}^{2}$ and $\mathbf{J}_{z}$, whose eigenstates brake the above degeneracy?
