Pauli-Lubanski operator and Spin angular momentum operator It is not so clear to me why I interpret Pauli-Lubanski $W^{\mu}$ operator as Spin operator. I mean, I know that $W^{2}$ is the Casimir operator and so together with $P^2$ define the irreducible unitary representation.
The thing that confuses me is that while the momentum is also an invariant operator, why do I need two spin operators, i.e. $W^{\mu}$ and $S^{\mu \nu}$ (which would be the spin angular momentum density)? If $W^{\mu}$ already gives me the spin values, why do I keep using $S^{\mu \nu}$?
 A: One way to see the difference is to look at the operators in a 3-d notation:
The boost components:
$$ K^i = S^{0i}, \quad i=1,2,3$$
The angular momentum operators
$$J^i =  \epsilon_{jk}^i S^{jk}, \quad i, j, k=1,2,3$$
The Pauli-Lubanski vector time component:
$$W^0 = \mathbf{J}\cdot \mathbf{P}$$
($\mathbf{P}$ is the momentum operator)
and the spatial components
$$\mathbf{W} = \mathbf{P}\times \mathbf{K} + E \mathbf{J}$$
($E = P^0$, the energy). From this representation, we can see that the spatial Pauli-Lubanski components coincide in the rest frame with the angular momentum at the rest frame, i.e., the spin. 
But the existence of a rest frame is exclusive to a massive particle; and this interpretaion is false in the massless case. In this case, there is no rest frame. Furthermore, the constraint $W^{\mu}P_{\mu} = 0$, together with the maslessness condition $P^{\mu}P_{\mu} = 0$, imply that the Pauli-Lubanski is necessarily proportional to the momentum vector:
$$W^{\mu} = \lambda P^{\mu}$$
Looking at the zero component, we get:
$$\lambda = \frac{\mathbf{J}\cdot \mathbf{P}}{E}$$ 
The helicity. 
Therefore for a massless particle the Pauli Lubansky vector does not coincide and is not proportional to the spin.
