In classical mechanics the equation of motion of a free particle in polar coordinates is given by \begin{aligned} &m\left(\ddot{r}-r \dot{\theta}^{2}\right)=0 \\ &m(r \ddot{\theta}+2 \dot{r} \dot{\theta})=0 \end{aligned}
while in cartesian coordinates is given by \begin{aligned} &m\ddot{x}=0 \\ &m\ddot{y}=0. \end{aligned}
Now suppose we are given the equation \begin{aligned} &m\left(\ddot{x}-x \dot{y}^{2}\right)=0 \\ &m(x \ddot{y}+2 \dot{x} \dot{y})=0. \end{aligned}
If we know that $x$ represents distances and $y$ represents angle than we know that the last equation represents the motion of a free particle.
Suppose that we have the quantum field Lagrangian $$\mathscr{L}=\frac{1}{2}\left(\partial_{\mu} \phi\right)^{2}-\frac{1}{2} m^{2} \phi^{2} \tag 1$$
and suppose we do a field redefinition $\phi=F(\psi)$ such that the Lagrangian $(1)$ turns into
$$\mathscr{L}=\frac{1}{2}\left(\partial_{\mu} \psi\right)^{2}-\frac{1}{2} m^{2} \psi^{2} -\frac{g}{4 !} \psi^{4}. \tag 2$$
Now relabeling $\psi \rightarrow \phi$ the Lagrangian we arrive at
$$\mathscr{L}=\frac{1}{2}\left(\partial_{\mu} \phi\right)^{2}-\frac{1}{2} m^{2} \phi^{2} -\frac{g}{4 !} \phi^{4}. \tag 3$$
In quantum field theory the equivalence theorem says that field redefinitions does not change the theory. So in order for the Lagrangian $(1)$ and $(3)$ represents the same theory the fields $\phi$ in $(1)$ and $(3)$ have to have different meanings.
My question is, do fields have meaning in quantum field theory?
