Consider the field equation $(\partial^2-m_1^2)(\partial^2-m_2^2)\phi =0 $
Now let me make the field redefinition $\psi = (\partial^2-m_2^2) \phi$
The question is, will the S-matrix be invariant under this field redefinition?
The criterion for a valid field redefinition, as usually stated, is that it $\langle \Omega| \psi(x)|p\rangle$ is not zero, where $|p\rangle$ is the one particle state. This follows from the LSZ formula (see QMechanic's answer on equivalence theorem here): $$\left[ \prod_{i=1}^n \int \! d^4 x_i e^{ip_i\cdot x_i} \right] \left[ \prod_{j=1}^m \int \! d^4 y_j e^{-ik_j\cdot y_i} \right] \langle \Omega | T\left\{ \phi(x_1)\ldots \phi(x_n)\phi(y_1)\ldots \phi(y_m )\right\}|\Omega \rangle$$
$$~\sim~\left[ \prod_{i=1}^n \frac{i\langle \Omega |\phi(0)|\vec{\bf p}_i\rangle }{p_i^2-m_1^2+i\epsilon}\right] \left[ \prod_{j=1}^m \frac{i\langle \vec{\bf k}_j |\phi(0)|\Omega\rangle }{k_j^2-m_1^2+i\epsilon}\right] \langle \vec{\bf p}_1 \ldots \vec{\bf p}_n|S|\vec{\bf k}_1 \ldots \vec{\bf k}_m\rangle$$
Redefining the fields on both sides leaves the $S$-matrix $\langle \vec{\bf p}_1 \ldots \vec{\bf p}_n|S|\vec{\bf k}_1 \ldots \vec{\bf k}_m\rangle$ unchanged.
Now clearly the criterion $\langle \Omega| \psi(x)|p\rangle \neq 0$ holds.
On the other hand the original field theory has two particles of masses $m_1$ and $m_2$. This has a different particle spectrum and it would appear that the two theories are physically inequivalent.
Indeed, if one had started with the original equation one would have obtained a different set of one particle states.
So my questions are, 1. Are the $S$-matrices of the two theories still equivalent? 2. If not, what is the criterion for equivalence under field redefinition?
Is it that the Hilbert space of asymptotic states should remain unchanged?
