vac}\rangle \, \nonumber\\
= \int {\rm d}^4 x \, e^{i q \cdot x}
\, \langle {\rm vac}|{\rm T}[\phi(x) \phi(0)] \rangle +\int {\rm
d}^4 x \,
e^{i q \cdot x}
\, \langle {\rm T}[\phi(x) a(0)] + {\rm T}[a(x) \phi(0)] + {\rm
T}[a(x) a(0)]|{\rm vac}\rangle\end{aligned}$$ Eq.(\[renorm2\]), which picks out the pole at $q^2 = m^2$, is obtained since by construction $a$ does not connect the vacuum to the one $\phi$ state. Clearly, this is necessary since the correlation functions for $\phi$ and $\phi^{\prime}$ satisfy the renormalization conditions in eqs. (\[renorm\]) and (\[renorm2\]). Off shell, however, there is no requirement that this term vanish and the two propagators will in general differ. Moreover, since the overall scale of $a$ is arbitrary it is clear that one can make the difference between the two descriptions arbitrarily large.

Let us make these ideas explicit by considering two examples from a theory in which the current $j$ is a static external source. In this case the energy of the system is given by $$\begin{aligned}
E=\int {\rm d}^3r {1\over 2} j(\vec r)\phi(\vec r)\end{aligned}$$ or $$\begin{aligned}
E=\int {\rm d}^3r {1\over 2} j(\vec r) G(\vec r,\vec
r^{\,\prime})
j(\vec r^{\,\prime}),\end{aligned}$$ where $G(\vec r,\vec r^{\,\prime})$ is the inverse of the operator 