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?
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3$\begingroup$ I think physics.stackexchange.com/q/286078/50583 is related - you're letting yourself be misled by the imprecise diction: The physicist rarely considers it relevant to mention that on fields/wavefunctions there is not only a representation of the Lorentz group but also of the Poincaré group, but the translation part just acts trivially as the $\psi(x)\mapsto \psi(x-a)$ you wrote down. $\endgroup$– ACuriousMind ♦Commented Nov 17, 2016 at 23:59
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$\begingroup$ That was a big hint, thank you! I only can't menage to understand the fact that introduction of spatial homogeneity in representation of the groups gave a richer structure and led us to the classification of particles, but it simply act on $\psi$ in a trivial way without adding something new to our understanding. $\endgroup$– pier94Commented Nov 18, 2016 at 0:15
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$\begingroup$ I'm confused by the statement : "...even if finite Lorentz group representations are not unitary with not-selfadjoint generators and hence they not correspond to any physical state $|\psi\rangle$." Does it mean, for example, a finite dimensional representation $(0,1/2)$ or $(1/2,0)$ representing a Weyl spinor, do not correspond to a physical state? Why then is such a representation considered at all? $\endgroup$– SRSCommented Nov 18, 2016 at 8:38
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$\begingroup$ @SRS Maybe i'm a bit confused but let's give a try: because physical variables are not properly states, thus even if a finite Lorentz representation is not unitary and therefore should not be realizable as a physical state, it could be used for representing physical variables. See as reference the last part of chapter 10.3.2 of W.K.Tung's book. $\endgroup$– pier94Commented Nov 18, 2016 at 9:14
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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.
answered Nov 18, 2016 at 7:29
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$\begingroup$ I already searched something about it on volume I but without any success, do you remember chapters or section or even better pages? $\endgroup$– pier94Commented Nov 18, 2016 at 7:34
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$\begingroup$ @pier94 Read the 2nd chapter. Discussion of little groups and so on is in 2.5. $\endgroup$ Commented Nov 18, 2016 at 13:19