Allowed Field Re-definitions in QFT I am trying to understand which field redefinitions are allowed in a QFT. The textbooks I have read appear to treat this topic flippantly. I assume that one cannot arbitrarily manipulate the expression for a field; rather, the redefinition should probably satisfy some criteria, such as leaving S-Matrix elements invariant and/or leaving the space of one particle states intact.
I came across a recent paper claiming that one could perform a field redefinition by acting a differential operator on the field. My confusion is as follows. Consider the following Lagrangian:
$$ \mathcal{L} = \frac{1}{2}\left ( \partial_{\mu}\phi\partial^{\mu}\phi - m^2 \phi^2\right ) - V(\phi). $$
Suppose one performed the following "field redefinition":
$$ \phi \rightarrow \phi + \frac{\partial_{\nu}\partial^{\nu}}{v^2}\phi. $$
The kinetic term for the original Lagrangian respects unitarity because it yields a propagator that does not fall off faster than $1/k^2$ in momentum space. Under this "redefinition", the propagator would fall off faster than $1/k^2$, which would manifestly violate unitarity due to spectral decomposition theorem.
My question: is this a valid field redefinition? If not, why?
My guess: it is not. I think it wouldn't be because it actually changes the location of the pole for physical single particle states in scattering amplitudes.
 A: After the field redefinition
$$
\phi \rightarrow \phi + \frac{\partial_{\nu}\partial^{\nu}}{v^2}\phi\tag{1}
$$
there is an apparent violation of unitarity because the propagator decays too fast in momentum space. But recall that field redefinitions don't affect the $S$ matrix (cf. this PSE post). Therefore, if you actually carry out the calculations with the redefined Lagrangian, you are bound to obtain several cancellations that restore unitarity. More precisely, unitarity was never lost, but it in the new variables it is not manifest.

The LSZ formula is valid as long as $\langle 0|\phi(x)|\boldsymbol p\rangle\neq 0$. For a normalised field, we have
$$
\langle 0|\phi(x)|\boldsymbol p\rangle=\mathrm e^{ipx}\tag{2}
$$
You can perform any redefition, as long as $\langle 0|\phi'(x)|\boldsymbol p\rangle\neq 0$. For example, $(1)$ is valid iff $v\neq m$, because
$$
\langle 0|\phi'(x)|\boldsymbol p\rangle=\left(1-\frac{m^2}{v^2}\right)\mathrm e^{ipx}\tag{3}
$$

Recall that in the Stückelberg formalism, we have a vector field with Lagrangian
$$
\mathcal L\sim F^2+(\partial\cdot A)^2\tag{4}
$$
After the field redefinition $A_\mu\to A_\mu+\partial_\mu \pi$, the longitudinal mode $\pi$ has a kinetic term
$$
\mathcal L_\pi\sim (\partial^2\pi)^2\tag{5}
$$
which apparently leads to a non-unitary theory (the scalar modes clearly have negative norm!). But, as you already know, the theory is unitary, because the scalar modes decouple: the $\pi$ fields don't contribute to $S$ matrix elements.
