One requires that a relativistic wave function should transform well under Lorentz transformation. Why we shouldn't rather have that it transforms well under Poincaré transformations? On Wu Ki Tung's book "Group theory in Physics" is written that, even if finite Lorentz group representations are not unitary with not-selfadjoint generators and hence they not correspond to any physical state $|\psi\rangle$, physical variables like position, momentum, or wave functions and fields should transform as finite dimensional representation of Lorentz group. I know that physical states arise naturally with the unitary irreducible representations of Poincaré group and are labeled by two indices (M,s) mass and spin. But physical states emerge also from the solution of relativistic wave equations involving $\psi(x)$. Then if I have a wave function $\psi(x)$ solution of these equation why i should require (if it is a condition that i impose) or, simply, it is straightforward that is a quantity that behaves well under elements of Lorentz group and not of the more general Poincarè's one? Does the translated (of a quantity $c$) wave function $\psi'(r)\equiv\psi(x-c)$ have some problems?
All the physical fields transform under the full Poincare group. Poincare group, as well as its Loretnz subgroup, is not compact, which means that it doesn't have finite-dimensional unitary representations. It is much more convenient to work with compact little subgroup of Poincare group, which is $SO(D)$ rotation group for the case of massive particles, and $SO(D-1)$ for the massless ones. See Weinberg I for the thorough discussion.