Classical vs. Quantum use of the spin 4-vector I have a few basic questions about the Pauli-Lubanski spin 4-vector S.


*

*I've used it in quantum mechanical calculations as an operator, that is to say each of the components of S is a matrix operator that operates on an eigenvector or eigenspinor. But my question is about the utility of S in a classical sense, that is to say it represents the physical spin angular momentum. For example, in an electron's rest frame, is the spin 4-vector for the case spin-up along the z-axis given by S = (0, 0, 0, h/2) and for spin-down along x we have S = (0, -h/2, 0, 0) etc?

*I know that in the particle's rest frame S = (0, Sx, Sy, Sz) where the spatial components are the spin angular momentum 3-vector components. However, when we Lorentz boost S, the time component is no longer zero. In this boosted case, do the 3 spatial components still give the spin angular momentum 3-vector (analogous to the case for 4-momentum where the 3 spatial components always give the 3-momentum), or do the spatial components now mean something else? The reason I'm not sure is that some 4-vectors, e.g. 4-velocity, have spatial components that do not represent 3-velocity at all since they may be superluminal, etc.
 A: The time component of the Pauli-Lubanski vector is equal to the helicity
times the (three) momentum magnitude:
$w^0 = \lambda ||\mathbf{p}|| =\mathbf{j}.\mathbf{p}$
Where $\lambda$ is the helicity, $\mathbf{j}$  is the (total) angular momentum and
$\mathbf{p}$ is the three momentum. Please see the following article  by Carineña, Garcia-Bondía, Lizzi, Marmo and Vitale (the second formula of section 2). Please, see also, the next formula where the transformation of the spatial and time components of
the Pauli-Lubanski vector under a general boost is written:  
$ w^0 \rightarrow cosh(\xi)w^0 + sinh(\xi) \mathbf{n}.\mathbf{w}$.
$ \mathbf{w } \rightarrow \mathbf{w} - sinh(\xi)w^0 \mathbf{n} + (cosh(\xi)-1) (\mathbf{n}.\mathbf{w}) \mathbf{w}$.
Where  $\mathbf{w}$  are the spatial components of the Pauli-Lubanski vector. $\xi$ is the rapidity, $\mathbf{n}$  is the boost direction
Now it is easy to deduce the properties of the time component of the
Pauli-Lubanski by inspection:
1) For a spinless particle, this component
is identically zero in all reference frames:
2) For a massless particle, and a Lorentz transformation which preserves the momentum. The angular momentum rotates around the momentum vector (Wigner rotation) such that the helicity is conserved. This is because for a lightlike 4-momentum, the
Pauli-Lubanski vector must be proportional to the momentum vector,
therefore its time component does not change under a momentum preserving
Lorentz transformation.
Update
The reason is as follows: For a massless particle, the Pauli-Lubanskii
4-vector is  light-like. Taking into accout that it is always orthogonal
to the momentum 4-vector (which is also light-like in this case), the two
vectors must be proportional (two orthogonal light-like vectors must be proportional). 
The proportionality factor is just the
ratio between the helicity (time component of the of the Pauli-Lubanski
vector) and the energy (time component of the 4-momentum). This suggets
that when the kinetic energy of a particle is much larger than its
rest mass, the Pauli-Lubanski and the momentum vectors tend to be
aligned. In order to see that more explicitely, one can use the
expression of the Pauli-Lubanski spatial components in terms of the spin
and momentum vectors for a massive particle:
$\mathbf{w} = m \mathbf{s} + \frac{ \mathbf{p}.\mathbf{s}}{p_0+m}\mathbf{p}$.
From this formula it is clear that when the particle speed becomes large,
the second term dominates and the pauli-Lubanski spatial components  3-vector becomes almost
aligned with the  momentum spatial components  3-vector.
