with the same form as the original: $$(\Box + m^2) \phi^{\prime}(x) = j^{\prime}(x) \label{eomp}$$ It also satisfies the same renormalization conditions $$\begin{aligned}
\langle \phi, \vec{p} | \phi^{\prime}(x) | {\rm vac} \rangle & =
&
e^{i p\cdot x}
\\
\lim_{q^2 \rightarrow m^2} \, (q^2 - m^2) \,  \int {\rm d}^4 x \,
e^{i q \cdot x}
\, \langle {\rm vac}|{\rm T}[\phi^{\prime}(x)
\phi^{\prime}(0)]|{\rm vac} \rangle \,
& = &\, i  \label {renorm2}\end{aligned}$$

The field variable $\phi^{\prime}$ is as good a choice for the field variable as the original field $\phi$—its equation of motion is of the same form and it satisfies the correct renormalization conditions. It makes no difference to any [*physical*]{} amplitude whether one chooses to describe the physics in terms of the field $\phi$ or $\phi^{\prime}$. Thus, the masses of particles and possible bound states and S matrix elements for scattering states must be identical with either description. Going from one to the other amounts to nothing more than a change of variables.

While the physics clearly does not depend on which field is chosen, the propagator depends strongly on this choice: $$\begin{aligned}
\int {\rm d}^4 x \, e^{i q \cdot x}
\, \langle {\rm vac}| {\rm T}[\phi^{\prime}(x) \phi^{\prime}(0)]
|{\rm 