# A Physical interpretation of Pauli-Lubanski pseudo-vector

1. Nov 10, 2016

### spaghetti3451

$P^{\mu}$ generates translations and extracts the four-momentum of a particle when it acts on the momentum eigenstate of a particle.

$J^{\mu\nu}$ generates rotations and measures the spin angular momentum along the $i$-direction of a particle when it acts on the $i$-th direction spin angular momentum eigenstate of a particle.

The Pauli-Lubanski psuedo-vector is given by $W_{\mu}=\frac{1}{2}\epsilon_{\mu\sigma\rho\tau}J^{\sigma\rho}P^{\tau}$ such that
$W_{\mu}|P,j,j_{z}\rangle = -mJ_{i}|P,j,j_{z}\rangle,$ where $|P,j,j_{z}\rangle$ is a momentum space eigenstate representing a particle of spin $j$ at rest with $P^{\mu}=(m,0,0,0)$ and $m\neq 0$.

What is the physical interpretation of $W_{\mu}$?

To prove that $[J_{\mu\nu},W^{2}]=0$, an explicit form of $[J_{\mu\nu},W_{\rho}]$ is necessary. One way to obtain $[J_{\mu\nu},W_{\rho}]$ is to define $I=\frac{i} {8}\epsilon_{\alpha\beta\gamma\delta}J^{\alpha\beta}J^{\gamma\delta}$ and show that $W_{\rho}=[I,P_{\rho}]$ and $[J_{\mu\nu},I]=0$.

What is an easy way to show that $[J_{\mu\nu},I]=0$ using the epsilon symbol in $I$?

$I$ is a scalar as all the indices are $0$, so why can not say that $[J_{\mu\nu},W^{2}]=0$ trivially?

2. Nov 11, 2016

### vanhees71

The Pauli Lubanski vector is pretty intuitive, when you familiarize yourself with Wigner's analysis of the irreducible unitary representations of the orthochronous proper Poincare group, the space-time symmetry group of special relativity. It provides the infinitesimal generators of the socalled little group associated with the representation. For details, see Appendix B of my lecture notes on QFT,

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf