How is it possible to change the direction of a spin by boosting? In Weinberg QFT section 2.5.5, he defines the states of momentum $p$ by
$$\Psi_{p,\sigma}=U\bigl(L(p)\bigr)\Psi_{k,\sigma}$$
up to some irrelevant normalisation, and $L(p)$ is the Lorentz transformation that takes us from $k^\mu=(m,0,0,0)$ to any $p^\mu$ with $p^2=-m^2$ and $p^0>0$. 
From hindsight I know that $\sigma$ labels the spin of the particle along the z-axis, and that $L(p)$ is a boost. So by the definition above this boost should not change our observed spin along the z-axis.
I have $J_3\Psi_{k,\sigma}=\sigma\Psi_{k,\sigma}$, so by definition I should have 
$$J_3\bigl[U\bigl(L(p)\bigr)\Psi_{k,\sigma}\bigr]=\sigma \bigl[U\bigl(L(p)\bigr)\Psi_{k,\sigma}\bigr]$$
But this is inconsistent with Poincare algebra.
Indeed, if I only boost along the $z$-axis then everything is okay, since $J_3$ and $K_3$, the generator of boosts along the $z$-axis, commute. 
But I cannot reach arbitrary $p^\mu$ by only boosting along the $z$-axis. I can only get to $p^\mu=(p^0,0,0,p^3)$.
I cannot hope to fix this by rotation around $z$-axis either, I must rotate around the $x$ or $y$ axis. But these also do not commute with $J_3$, so I still have a problem.
How can I reach states of arbitrary $p^\mu$?
As an example suppose I start with a definite spin state in the rest frame, so I have 
$$J_3\Psi_{k,\sigma}=\sigma\Psi_{k,\sigma}$$
Now I perform an infinitesimal boost in the $y$ direction to get to a new state
$$e^{iK_2p_2}\Psi_{k,\sigma}=(1+iK_2p_2)\Psi_{k,\sigma}$$
Now I measure my spin. I find 
$$J_3(1+iK_2p_2)\Psi_{k,\sigma}=\sigma (1+iK_2p_2)\Psi_{k,\sigma}-iK_1p_2\Psi_{k,\sigma}$$
The first term is what I want but the second term ruins everything. How can I resolve this issue?
 A: It seems that you're right, and I was wrong in the comments section. The answer on your question is simple: $\sigma$ doesn't mean the polarization for arbitrary momenta $p^{\mu}$, although it coincides with the polarization at rest. So, how to get the interpretation of the $\sigma$ label? 
By the definition,
$$
\tag 1 U(L(p))|k,\sigma\rangle = |p , \sigma \rangle,
$$
so $\sigma$ quantity of one-particle state, defined at rest, is unchanged under the standard Lorentz boost. The physical meaning of $\sigma$ for arbitrary momenta is thus not simple. In general, it doesn't coincide with the spin $\hat{\mathbf J}_{3}$ eigenvalue.
Realize first that the Little group of massive orbit of the Lorentz group is isomorphic to $SO(3)$ group, whose irreducible representations generators are $\hat{\mathbf J}_{i}^{\sigma \sigma{'}}$ with $\sigma = -s,...,s$). So that it seems that $\sigma$ is just the eigenvalue of $\hat{\mathbf J}_{3}$. But there is the problem.
We know that $\hat{\mathbf J}_{3}$ is changed under the general Lorentz boost as the component of antisymmetric tensor; in the result we come to the statement that $\sigma$ is defined as the eigenvalue of some operator, the action of which on one-particle state with $k_{\mu} = (m, \mathbf 0)$ coincides with the action of $\hat{\mathbf J}_{3}$. For the general momentum $$
\tag 2 P_{\mu} = (E,\mathbf P), \quad P^{2} = m^{2}, \quad P_{\mu} = L_{\mu}^{\ \nu}k_{\nu} = (Lk)_{\mu},
$$ 
however, this correspondence is not true, i.e., $\sigma$ isn't the eigenvalue of $\hat{\mathbf J}_{3}$ and thus for general momenta it isn't the spin. 
Maybe the operator which defines the $\sigma$ label is Pauli-Lubanski operator $\hat{\mathbf W}_{\mu}$ multiplied on $L^{-1}(p)$:
$$
\tag 3 \hat{\mathbf V}_{\mu} \equiv L_{\mu}^{\ \nu}(p)\hat{\mathbf W}_{\nu} = \frac{1}{2m}L_{\mu}^{\ \nu}(p)\epsilon_{\nu \alpha \beta \gamma}\hat{p}^{\alpha}\hat{J}^{ \beta \gamma}
$$
Really, note that due to invariance of $\epsilon$ tensor we have that
$$
L_{\mu}^{\ \nu}\epsilon_{\nu \alpha \beta \gamma} = \epsilon_{\mu \rho \delta \epsilon}L^{\rho}_{\ \alpha}L^{\delta}_{\ \beta}L^{\epsilon}_{\ \gamma},
$$
so that, by using relation
$$
(L^{-1}(p) p)_{\mu} = m\delta_{\mu 0}
$$
we have that
$$
\hat{\mathbf{V}}_{\mu} = \begin{cases} 0, \quad \mu = 0 \\ \frac{1}{2}\epsilon^{0\alpha\beta l}L_{\alpha}^{\ \delta}L_{\beta}^{\ \kappa}\hat{\mathbf J}_{\delta\kappa} = \hat{\mathbf J}_{l}, \quad \mu = l\end{cases}
$$
Thus you have that independently on the value of general 4-momentum $p$ operator $\hat{\mathbf V}_{3}$, by acting on the state $|p, \sigma\rangle$, will always give $\sigma$ value as eigenvalue of $\hat{\mathbf J}_{3}^{\sigma \sigma{'}} = \sigma \delta^{\sigma \sigma{'}}$, with $\sigma = -s,...,s$.
