$\nabla^2-m^2$. Let us first take $a(x)$ to be a simple function of $\vec x$, which is independent of $\phi$. Then $\langle \phi,\vec p|a(x)|{\rm
vac}\rangle=0$ and the renormalization conditions of eqns. (\[renorm\],\[renorm2\]) are satisfied. One may determine a new Lagrangean density ${\cal
L}^\prime$ and a new Hamiltonian ${\cal H}^\prime$ by starting with the original ${\cal L}$ and transforming the variables. Then the new energy $E^\prime$ is given by $$\begin{aligned}
E^\prime=\int {\rm d}^3r {1\over 2} [\vec \nabla
(\phi^\prime(\vec r)+a(\vec r)\cdot(\phi^\prime(\vec r)+a(\vec
r))
+ \nonumber\\
m^2(\phi^\prime(\vec r)+a(\vec r))^2 +2
j(\vec r)(\phi^\prime(\vec r) +a(\vec r))],\end{aligned}$$ and using the equation of motion (\[eomp\]) in the static limit leads to $$\begin{aligned}
E^\prime=\int {\rm d}^3r {1\over 2}j(\vec r)(\phi^\prime(\vec r)
+a(\vec r)).\end{aligned}$$ But Eq.(\[def\]) tells us that $E^\prime=E$. Even though the current $j^\prime$ of Eq.(\[jpdef\]) is different than $j$ the energy of the system does not depend on the choice of the function $a(\vec x)$.

A more interesting example is obtained by letting $\phi=(1+f(\vec x))\phi^\prime $ (or $a(\vec x)= -{f(\vec x)\over 1+f(\vec x)} \phi(x)).$ We place the static source j at the origin and choose $f(\vec x)$ to vanish at large values of $|\vec x|$ faster than $e^{-m|\vec x|}/|\vec x|$. This maintains the original value of the renormalized 