Classical vs. Quantum use of the spin 4-vector

Question

I have a few basic questions about the Pauli-Lubanski spin 4-vector S.

1. 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?

2. 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.

Answer 1

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.