Weinberg Sec 5.1 - Why is the transformation of fields in QFT independent of position? In Weinberg's Quantum Theory of Fields Vol. 1, it is claimed that annihilation and creation fields should transform as
$$U_0(\Lambda, a)\psi_l^+(x)U_0^{-1}(\Lambda, a) = \sum_\bar{l}D_{l\bar l}(\Lambda^{-1})\psi_{\bar l}^+ (\Lambda x+a).\tag{5.1.6}$$
This is equation (5.1.6).
But I do not see why this is the only way the creation and annihilation operators can transform in order to build up a $H(x)$ that satisfies the cluster decomposition principle. In particular, why should $D(\Lambda^{-1})$ be position independent? Presumably we could write down a field with a transformation that is different at each point in position, and then contract the indices so that we have a local $H(x)$?
 A: Weinberg derives the Poincare transformation properties of the creation and annihilation operators in the previous chapter 4 (see Eq 5.1.11). These just follow from the group transformation properties of single particle states he discussed in chapter 2, there isn't a lot of leeway in choosing them.
For simplicity just consider a rotation $R$ at the reference momentum $k$ (the momentum chosen to fix the little group)
$$U(R) a(k,\sigma) U^{-1}(R)= \sum_{\bar{\sigma}}D^{(j)}_{\bar{\sigma\sigma}}(R)a(k,\bar{\sigma})$$
This $D^{(j)}$ matrix acts on the spin index in the quantum mechanical operators, and it is not position dependent. The $D_{l\bar{l}}$ matrix in your question acts on the field index, and it ultimately will come from the $D^{(j)}$ matrix given suitable chosen coefficient functions $u,v$. This is the content of eq (5.1.23) and figuring out the coefficient functions from this relation makes up the bulk of chapter 5.
In any case my answer is very specific to Weinberg's textbook, but the main point is that the representation on the field index must come from the representation on the quantum mechanical operators, and those are not compatible with having a spacetime dependence like you propose.
