Weinberg's approach to QFT starts with particles which are not necessarily more fundamental than fields but are known more for certain. A particle of a particular species can have different momenta $p$ in different frames but its internal degrees of freedom (like electron spin) is described by a discrete quantum number $\sigma$, and the one-particle state is denoted by $\Phi_{p,\sigma}$. A standard boost is defined to relate one-particle states with different momenta but the same quantum number of internal degrees of freedom $\Phi_{p,\sigma} = U(L(p))\Phi_{k,\sigma}$, where $k$ is a chosen standard momentum ($(m,0,0,0)$ for particles with mass $m$) and $L(p)$ is a standard boost that takes $k$ to $p$. With the non-compact boost components of the Lorentz group trivialized in this way, essentially particles are defined as finite-dimensional unitary representation of the little group, that is, the stabilizer of $k$.
My question is about the choice of the standard boost $L(p)$. Is there any physical consideration (it doesn't change the quantum number of internal degrees of freedom) that determines $L(p)$? Different choices of $L(p)$ differ by an element in the stabilizer of $p$ (which is $SO(3)$), the standard boost for one choice is not standard in another, hence mixed quantum numbers of internal degrees of freedom, does it matter for physics? Perhaps it's just redefinition of quantum number of internal degrees of freedom?
