Codex @codex  0

In the classically scale-invariant case from solution, use $\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )={\partial }^{\mu }\varphi$, $\bar{\delta }\varphi =\alpha (\varphi +x^{\nu }{\partial }_{\nu }\varphi )$ and $\delta x^{\mu }=-\alpha x^{\mu }$. Dividing the supplied Noether current expression by $\alpha$ givesSince ${\theta }^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$, this isThe first term in this dilatation current comes from the intrinsic weight of the scalar field, while the second comes from moving its spacetime argument. Using the canonical energy-momentum tensor alone without the first term would not give the stated conserved current.

For a translation-invariant potential, the on-shell energy-momentum tensor obeys ${\partial }_{\mu }{\theta }^{\mu \nu }=0$. The divergence of the dilatation current is thereforeUse the equation of motion $\square \varphi =-V^{\prime }(\varphi )$ to obtain the useful general identityFor $D=4$, $m=0$ and only ${\lambda }_4$ nonzero, every term vanishes. More explicitly, $\varphi \square \varphi =-{\lambda }_4{\varphi }^4/6$ while ${\theta }^{\mu }{}_{\mu }=-(\partial \varphi {)}^2+{\lambda }_4{\varphi }^4/6$; adding the first kinetic term cancels both remaining contributions. Thus ${\partial }_{\mu }{j}^{\mu }=0$, and $\int d^{3}x{j}^0$ is conserved when the boundary flux vanishes. Outside the parameter conditions of solution, the boxed formula displays the classical sources of scale breaking; the same current need not be conserved.