 Solution

For a static finite-energy field, the Hamiltonian density is $\frac{1}{2}({\varphi }^{\prime }{)}^2+V(\varphi )$. Multiply the static equation of motion by ${\varphi }^{\prime }$ and integrate once:At either vacuum endpoint both terms vanish, so $\frac{1}{2}({\varphi }^{\prime }{)}^2=V$. The total energy density is consequently $({\varphi }^{\prime }{)}^2=4k^2{\mathrm{sech}}^2[k(x-x_0)]/b^2$. Its integral givesFor the usual positive $a,b$ convention this isThe coefficient depends on the dimensionless parameter $b$; it is not a universal numerical constant independent of the potential. In two spacetime dimensions, $[\varphi ]=[b]=0$ and $[a]=2$, so the dimensions are consistent. If both $a$ and $b$ are negative while $ab>0$, the real answer is instead $8\sqrt{|a|}/|b{|}^{3/2}$; the literal factor $\sqrt{a}$ would not be real.

An equivalent positive-parameter calculation substitutes $z=b\varphi$ in $E_0={\int }_0^{2\pi /b}\sqrt{2V}d\varphi$ and uses the provided integral of $\sqrt{1-\cos z}$. The following original plot shows that the localized energy density sits where the field passes between the vacua.