Simultaneous eigenstates of non commuting operators

Question

In the lecture notes of Quantum Physics 2,MIT, Prof B Zwiebach shows that the for the multi-particle state $$|\psi\rangle = \alpha\bigg[|+\rangle\otimes|-\rangle-|-\rangle\otimes|+\rangle\bigg]$$ the total spin operators $S_z^T$ , $S_x^T$, $S_y^T$ are all zero(T stands for total), now what is the reason of that? why does that happen? because the spin operators in this case have a common eigenstate with zero eigenvalue, and the spin operators do not commute for single-particle states.

Answer 1

If you put two spin-1/2 systems together, the result could have total angular momentum $s=0$ or $s=1$. In the former (called the spin singlet, since there's only one state with $s=0$), the composite system has no (spin) angular momentum at all, and so that state is annihilated by all of the angular momentum operators. This is the state you've written down.

the spin operators do not commute for single-particle states.

That's true, but that doesn't mean they don't have any common eigenstates; it means that there is not a basis of common eigenstates. Indeed this is a very simple example of non-commuting operators which nevertheless share at least one eigenstate in common.