The Pauli-Lubanski pseudovector coincides with intrinsic spin in the rest frame of the particle. In a more general frame, one defines a tetrad and projects the PL vector on it to define intrinsic spin components. I am trying to understand how this works for the massless case, but I cannot understand the choice of tetrad member s at equation 10.53 p.117 here: http://staff.science.uva.nl/~msnoek/GT/LectNotes_GrTh_2011_2.pdf: it does not verify the transversality condition of p.115 since $s^0>|s^3|$. I basically don't understand the whole massless discussion p.117 in comparison with the massive case where steps were clear and state label $\sigma$ was decomposed using the Casimir operator $\vec{S}^2$ and an additional operator $S^3$.

  • $\begingroup$ Your link is dead. $\endgroup$ – Trimok Oct 4 '14 at 15:31
  • $\begingroup$ Thanks Trimok for pointing that out, luckily I still had the page loaded and downloaded the pdf. Here's a link to the pdf: cjoint.com/data/0JfaHOJjKaZ.htm $\endgroup$ – Issam Ibnouhsein Oct 4 '14 at 22:31
  • $\begingroup$ Nobody? Any comment will be appreciated! :) $\endgroup$ – Issam Ibnouhsein Oct 6 '14 at 0:18
  • $\begingroup$ Your web link is dead. $\endgroup$ – Cosmas Zachos Apr 6 '16 at 23:01

For the massless case, one needs to show that $W^\mu = \lambda P^\mu$. Equation (10.53) provides a basis for an arbitrary four-vector and then expands $W^\mu$ in that basis. Imposing the two conditions $W\cdot P=0$ and $W \cdot W=0$ completes the proof by showing that all other "components" in that basis vanish.

  • $\begingroup$ Thanks for your help suresh. However, the basis they provide does not correspond to their definition of tetrad since $s_\mu p^\mu \neq 0$, i.e. the fourth four-vector fails to be transversal. I am suspecting that maybe the definition of tetrad should change in the massless case but this is not explicitly said in the document. $\endgroup$ – Issam Ibnouhsein Oct 6 '14 at 13:49
  • $\begingroup$ All that matters is that the 4 four-vectors are linear independent. $\endgroup$ – suresh Oct 6 '14 at 14:10
  • $\begingroup$ I see. I guess the 4 conditions in the massive case were here to ensure that the boosted tetrad coincides with the rest frame tetrad. However, I don't understand why we impose the two conditions $W\cdot P=0$ and $W\cdot W=0$. I don't get the explanation of the latter with finite dim reps. Furthermore, $\lambda$ is defined as the helicity operator, so $\lambda P^\mu$ makes no sense, indeed we are projecting $W^\mu$ on the tetrad so it should be $\lambda p^\mu$. $\endgroup$ – Issam Ibnouhsein Oct 6 '14 at 23:30
  • $\begingroup$ The condition $W\cdot P=0$ follows from the definition of the Pauli-Lubanski tensor. The condition $W\cdot W=0$ is something that needs extra input. This comes from representation theory as worked out by Wigner. If you require a finite-dimensional unitary irrep in the massless case, then $W \cdot W=0$. You either accept it or work through Wigner's paper (a worthy thing!). $\endgroup$ – suresh Oct 7 '14 at 2:15
  • $\begingroup$ Thanks a lot for all the details! One last question: don't you agree with my comment on the distinction between $p^\mu$ and $P^\mu$? I can understand they used notation $\lambda$ for two distinct things, the helicity operator p.117 when they equate it to $W^0$, and a scalar in the decomposition 10.54 of $W$. However, even if we assume that $\lambda$ is now a scalar, why do we choose four-vector $P^\mu$ to decompose $W$ whereas our tetrad was defined by the four-vector of eigenvalues $p=(1,0,0,1)$? $\endgroup$ – Issam Ibnouhsein Oct 7 '14 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.