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A Physical interpretation of Pauli-Lubanski pseudo-vector

  1. Nov 10, 2016 #1
    ##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. jcsd
  3. Nov 11, 2016 #2

    vanhees71

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    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
     
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