Pauli-Lubanski operator and Spin angular momentum operator

Question

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 $S^{\mu \nu}$?

Answer 1

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.