A question about relativistic spin operator The question comes from Ryder's Quantum Field Theory, 2nd edition. The author was looking for relativistic spin operator. It was concluded that it cannot be $J^2:=\mathrm{J} \cdot \mathrm{J}$, where $J_i$ is the generator for rotation in Lorentz group.
p.56:

$\ldots \Sigma$ is, of course, the matrix representative of $\mathrm{J}$, so we conclude that the relativistic spin operator is not $\frac{1}{2} \mathrm{J}$. This is confirmed by the fact that $\mathrm{J} \cdot \mathrm{J}=J^2 $(c.f.(2.167)) does not commute with all the generators of the Lorentz group. For example, $\ldots$, $[J^2,K_1] \neq 0 $.

Later Ryder found the square of the Pauli-Lubanski operator is the Casimir, but still not the relativistic spin operator. (It was refered to literature)
I have a question about the logic here. Why the spin operator needs to commute with all the generators in the Lorentz group. If the spin operator commutes with the Hamiltonian, it is a conserved quantity. We can use the eigenvalue to label states (good quantum number). That is precisely $[H,J_i]=0$. Is that still insufficient, because it may be frame-dependent? 
 A: The answer is based on the following Ref (chapter $V$, pages $5,6$)
By relativistic spin operator, I think that this means one is searching an  expression for a $3$ dimensional operator $\vec S$, such as $S^2 = \frac{1}{m^2} W^\mu W_\mu$ (Here , $W_\mu$ is the Pauli-Lubanski (pseudo)four-vector
) with the usual $SU(2)$ commutation relation for the $S_i$. So $S^2$ is the ratio of two Casimirs and should commute with all generators.
With some additional assumptions, that the $\vec S$ transform as a (pseudo)vector (equation $(74)$ page $5$ of the Ref), and that the $\vec S$ commute with the momenta operator, one is able to find some expression for the operator $\vec S$ (see formulae $(75),(76),(77)$ page $5$ of the Ref).
Lorents Transformation is not trivial for $\vec S$, see formula $(92)$ page $6$ of the Ref.
A: The difficulty with labelling states using $ J ^2 $ is indeed because its not a good quantum number in different reference frames. 
By demanding that the operator, ${\cal O}$, you want to describe your states commutes with the Hamiltonian you are requiring that 
\begin{equation} 
\left[ P ^0 , {\cal O}  \right] = 0
\end{equation} 
However, ${\cal O}$ may not be the same at different values of momenta (i.e., not a good label if we want to consider different frames). In QFT we want states that can be described in all reference frames. Thus we require
\begin{equation} 
\left[ P ^\mu , {\cal O}  \right] = 0
\end{equation} 
In your case, the operator $ J ^2 $ does not commute with the spatial components of momenta,
\begin{equation} 
\left[ P ^i , J ^2 \right] \neq 0 
\end{equation} 
so it is not a useful label in a relativistic theory. Note that if the eigenvalues of $P _i $ are very small one can show that $J^2$ does indeed commute with the momenta, so it is a good label in Quantum Mechanics.
Instead you can extend the idea of $J^2 $ to $ W_\mu W ^\mu$ as mentioned by Trimok.
