not connect the vacuum to a one particle state $$\langle \phi , \vec{p} | j(0) | {\rm vac} \rangle \, = \, 0 .
\label{cond}$$ This can be seen simply: $$\langle \phi , \vec{p} | j(x) | {\rm  vac}\rangle = \langle
\phi ,
\vec{p} | (\Box + m^2) \phi(x) | {\rm vac}\rangle  = (-p^2 + m^2)
\langle
\phi , \vec{p} | \phi(x) | {\rm vac} \rangle$$ where $p^2$ is the square of the four momentum of the state which is $m^2$.

Now we come to the crux of the issue. There is enormous freedom in the choice of field variables, and consequently the Green’s functions. In particular, we introduce a new renormalized field and a new source current according to: $$\begin{aligned}
\phi^{\prime}(x) & = & \phi(x) + a(x), \label{def}\\
j^{\prime}(x)& = & j(x) + (\Box + m^2)\,     a(x) \label{jpdef}\end{aligned}$$ where $a(x)$ is an operator such that $\langle {\rm vac} | a| \phi, \vec{p} \rangle = 0$. Thus, for example, $a(x)$ may be a multiple of the renormalized source $j(x)$ or $a(x)$ could have the form $a(x) = (\Box + m^2) b(x)$ where $b(x)$ is an arbitrary renormalized local composite operator. The new field and source satisfies an equation of motion 