What is significance of $\vec{S}^2$ of spin in quantum mechanic? The main question is as the title. I have read about the spin in quantum mechanics.  I know that the $[\vec{S}^2,\vec{S}_i]=0$, they commute, they can share the same Eigenvector, and so what?  I just wonder that just $\vec{S}_z$ is not enough? The other thing is what is the meaning of this $\vec{S}^2$, what is it measures?
 A: It measures the (square of the) magnitude of the spin angular momentum, ignoring the spin’s direction.
A: If we have a rotating body, it's classical rotational energy is 
$$
E_{rot} = \frac{L^2}{2\theta}
$$
Therefore, if we include a rotational particle in quantum mechanics it seams "natural" to use the same term and replace $L^2$ by it's operator -- the mathematical argument involves some group theory and so called generators [see e.g. the book of Sakurai for a simple derivation]. Since we consider to spin as a rotation, the same is true for a spinning particle. 
Furthermore, there are different coupling themes like $J = L + S$ where the spin and the orbital angular momentum form the total angular momentum $J$. In this case the Hamiltonian concludes a $J^2$ term and a $LS$ term as well. Hence, the operators $L$, and $S$ gain importance as you progress with your book.
A: As an addendum to the other answers: When we say a particle is "spin 0" (e.g. Higgs boson), "spin 1/2" (e.g. electron), "spin 1" (e.g. photon), etc., we are talking about their different values for this operator. In general a "spin $s$" particle is described by eigenstates of $\hat{S}^2$ with eigenvalue $\hbar^2 s (s + 1)$.
