How does a particle's spin z component changed under lorentz group I am reading weinberg's QFT, on the page 104, exercise 1, He said an observer $O $ see a W-boson (spin one and mass m) with momentum $p $ in the y direction and spin z-component $\sigma$ . a second observer $O^\prime$ moves relative to the first with velocity $v$ in the  z-direction. How does $O^\prime$  describe the W state?
I don't know how  the spin component change under the lorentz group.
 A: I'll give you the general solution, so you can work out the special case. 
We consider here a massive relativistic particle of mass $m$ and $4$-momentum $p$:
$$ p^2 = m^2$$
In the particle's rest frame, its momentum is $p_0 = (+m, 0, 0, 0)$ and its spin is $ s = (0, s_x,  s_y, s_z)$. 
We will denote the general boost that we want to apply on the particle by $\Lambda$, belonging to the fundamental representation of the Lorentz group ($4\times4$ matrix).
Let $\Lambda_0(p)$ be a boost which takes the particle from its rest frame to the laboratory  frame where it has a momentum $p$. This boost is not unique, we can choose any such boost for the solution, but we should remain with the same definition throughout the computation.
The spin of a massive relativistic particle is its angular momentum in its rest frame. Thus the method of finding how the spin rotates under a boost, is first to transfer it to the  laboratory frame then apply the general boost on it, then transfer it back to the rest frame,  therefore, we must apply the following set of boosts to the spin vector:
$$s' = \Lambda_0(\Lambda p)^{-1} \Lambda \Lambda_0(p)s$$
The most striking property of this triple combination of boosts is that it is actually a rotation. It is called the "Wigner rotation":
$$R(\Lambda , p) =  \Lambda_0(\Lambda p)^{-1} \Lambda \Lambda_0(p)$$
Thus the fourth component of the spin will stay $0$ as it should. In order to see that it is actually a rotation, we can apply it to the vector $p_0$, the first transformation will take it to $p$, the second to $\Lambda p$, therefore the third will take it back to $p_0$. A Lorentz transformation which keeps a time-like vector fixed has to be a rotation.
If one wants to consider the spin quantum mechanically, then the spin wave function of a particle of spin $j$ is a $2j+1$ vector $\Psi$. The spin wave function transforms under the representation $D_j$
$$ \Psi' = D_j(R(\Lambda , p)^{-1} )\Psi$$
This is how the Wigner rotation is realized quantum mechanically.
