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.


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.


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.

  • $\begingroup$ Thanks for the reply! I'm still not sure about my initial question: in an electron's rest frame, is the spin 4-vector for the case spin-up along the z-axis given by W = mc(0, 0, 0, h/2) and for spin-down along x we have W = mc(0, -h/2, 0, 0) etc? $\endgroup$ – Zoot Dec 19 '12 at 21:44
  • $\begingroup$ Basically, yes, but please notice that the spatial spin components in the rest frame satisfy the angular momentum commutation relations, and the x and z components cannot be measured simultinuously. Thus the numerical values of the Pauli- Lubanski vector should be understood as expectations. $\endgroup$ – David Bar Moshe Dec 20 '12 at 4:27
  • $\begingroup$ Thanks again David. The reason for my questions is that I'm trying to visualize the "direction" of an electron's spin before and after boosts. By "direction" I mean the direction in which a measurement of spin (e.g. using Stern Gerlach device) would return "spin-up" 100% of the time. In the rest frame this "direction" is easily found using the 2 components of the spinor. But I am trying to determine the new direction in a boosted frame. So can I just boost the 4-vector W and then use the 3 spatial components to describe the new (boosted) direction in 3-space? $\endgroup$ – Zoot Dec 20 '12 at 16:39
  • $\begingroup$ The boost transformation of the spatial components of the Pauli Lubanski vector is the third equation copied from the reference article. For a massive particle, the spin is the value of the spatial component vector at its rest frame. Knowing the momentum and the Pauli-Lubanski vector, one can perform a boost to a frame where the particle is at rest and get its spin. In a frame where the particle is not at rest the spatial components of the Pauli Lubanski vector still satisfy the spin commutation relations since they are composed from the spin and an angular momentum, thus should be quantized. $\endgroup$ – David Bar Moshe Dec 21 '12 at 5:23
  • $\begingroup$ OK, so you're saying that the spin really only makes sense in the rest frame where it's the spatial components of the PL vector. But I hear things like "For fast moving electrons, the spin is aligned along the direction of motion". How can I show this? I was hoping to do a LARGE boost of the PL vector from the rest frame, and then use the 3 spatial components in the boosted frame to show that the component along the direction of motion dominates over the others, thus this shows that the spin has become aligned to the direction of motion. Is this correct? $\endgroup$ – Zoot Dec 21 '12 at 16:31

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