# Spin as Poincaré invariant label

I was thinking about how we construct unitary representations for the Poincaré group in the case of massive particles. We move to a frame where the particle is at rest, and here the little group that leaves the momentum invariant is the group of spatial rotations. Thus, we conclude that we can label the states by the eigenvalue of $$\textbf{J}^2$$, $$s$$. However, when we move to an arbitrary frame by performing a Lorentz boost, we have to use the Pauli-Lubanski vector because $$\textbf{J}^2$$ doesn’t commute with the boost generators, so it isn’t a Casimir operator. The Casimir operator obtained from $$W^{\lambda}$$ has eigenvalues $$m^2s(s+1)$$, so we conclude that $$s$$ is actually Poincaré invariant (since $$m$$ also is). My question is, how is it possible that the eigenvalues of $$\textbf{J}^2$$ are invariant under arbitrary Poincaré transformations without $$\textbf{J}^2$$ being a Casimir operator (i.e. without $$\textbf{J}^2$$ commuting with every other Poincaré generator)? I’m probably missing something trivial, so I would appreciate it if someone could point it out.

Consider first a massless particle (we are going to generalize later). The generators of its little group are : $$\begin{equation} p^\mu = l^\mu_a J^a \qquad, \qquad l\mu = \frac{\lambda}{M} p\mu. \end{equation}$$ with $$[l^b\nu, l^c\rho ] = i (\eta^{bc} - \frac{\delta^{bc}}{M}) l{\nu \rho}$$. Since this algebra has vanishing commutator between generators associated with different mass eigenstates then you can choose as representation space a Hilbert space labeled by momentum eigenvector. Here obviously all these states will be orthogonal and belong to an irreducible representations spaces labelled by energy/mass eigenstate which form a complete basis in this representation space : $$|\vec p > < p|$$. Now if you choose as generator associated with angular momentum operator $$L^{\alpha}{\hspace{.7mm}\beta} = x^{\alpha}(\frac{\partial}{\partial x^{\beta}}- i M^{-1}[\hat J,\frac {\partial}{\partial x^\beta}])$$ then you will see since $$[\hat {M},L^{\alpha}{\hspace{.7mm}\beta}]$$ and $$[\hat J,\frac {\partial}{\partial x^\beta}]$$ vanish identically on any momentum eigenstate $$|\vec p > < p|$$ they also commute between themselves $$[x^\sigma(L{\sigma\tau}), L{\gamma\nu}] |\vec p > . So if you now boost or rotate $$|\vec p > using rotation or boost matrixes on this state their angular momenta have same eigenvalue up to overall factors depending on initial and final mass. In fact if $$(B^+)^\tau{\hspace{.5mm}\sigma}(R^{-1})^{\sigma}{\hspace{.8mm}\tau}|\vec 0> = |\vec 0>$$ where $$(B^+)^\tau{\hspace{.5mm}\sigma}= e^{+iP \cdot X}/\sqrt{(2E)}$$ and $$(R^{-1})^{\sigma}{\hspace{.8mm}\tau}= e^{iL \cdot Y}/\sqrt{(2E)}$$ Then writing $$[x^-(\tilde P),\tilde P_x]|\phi>_P=0=[x^-(\tilde P),[\tilde P_X + S,\tilde P]_X + [\tilde P_Y +S,\tilde P]Y ]|\phi>{PL}$$ leads to the following relation: $$[S+, S-]|\phi>= 2E(MB -BM)|\phi>$$ On an other hand if you take two operators of this type acting on $$|\phi>$$ but defined differently (say one corresponding to $$b$$) then taking their commutators $$[B^+ (\check R), B^-(\check R')]$$ and noting that $$\check R'=\check R^{-1}$$ should also be used gives: $$[S+, S-]|\phi>= 2E [M(RBR'-BR'R') - BM'(RR'-\color{magenta}{RR'})]\psi >= 0$$ This means again angular momentum should be conserved for this case even though in general both operators don't correspond to physical observables except for massless cases. This has some nontrivial consequences especially when dealing with scattering problems such as Wigner's little groups which explains why they produce spin conservations laws while neither energy nor angular momentum conservation holds generally unless particle masses are all identical (or gose into zero) or particles are at rest and interacting through contact interaction only so called Adler's zero condition ($$N=4$$). If neither one above conditions hold, then once again it turns out that there will still be some kind of conservation laws not just approximate symmetries.