Why do the components of fields transform trivially under translations? This is a follow-up from this question. Weinberg states in Quantum Theory of Fields Vol. 1 that creation and annihilation fields must transform as
$$U_0(\Lambda, a)\psi_l^+(x)U_0^{-1}(\Lambda, a) = \sum_\bar{l}D_{l\bar l}(\Lambda^{-1})\psi_{\bar l}^+ (\Lambda x+a).\tag{5.1.6}$$
Implicit in this statement is that under a pure translation $U_0(1, a)$, we have
$$\psi_l^+ \to \psi_l^+(x+a),$$
i.e. that the different components of $\psi_l$ do not mix. It is not obvious to me why we should assume this. Is there no non-trivial way in which the fields could transform under spacetime translations?
EDIT:
Following the discussion with @Prahar in the comments, I see that the fields cannot mix under a translation. However, it is still not clear to me that the fields must transform trivially under spacetime translations. That is, is there anything preventing the possibility that
$$\psi_l^+ \to e^{i\theta a} \psi_l^+(x+a)?$$
 A: In my world that is almost a definition. Fields are defined as sections of fiber bundles. Or more informally, assume that we have a representation of the Lorentz group $S$ and a representation of the Poincare group on "x". And then  consider
\begin{equation}
S \otimes x, 
\end{equation}
which obeys the transformation rule that you wrote.
The only part that can "transform" non-trivially under the translation algebra is an "x" part, which are just coordinates on our Minkowski space.  For example, a scalar field is just a map from $M \to \mathbb{R} $. When you do a Lorentz transformation $f$ transforms as $\Lambda^*f(x) =  f'(x) =   f(\Lambda x)$.
In some sense Weinberg is decomposing this tensor product, which is a valid representation, into a sum of the Poincare group representations.
We can consider something that has the dependence you described, but it would ruin the group composition rule and would not be a representation:
\begin{equation}
U(\Lambda_1,a_1) U(\Lambda_2,a_2) = U(\Lambda_1\Lambda_2,\Lambda_1 a_2+a_1)
\end{equation}
And consequently this tensor product would have no reason to decompose in terms of representations.
