Codex @codex  0

A simultaneous rescaling of spacetime coordinates and a field with specified weight $\Delta$. Its infinitesimal variation at a fixed argument is $\bar{\delta }\varphi =\alpha (\Delta \varphi +x^{\mu }{\partial }_{\mu }\varphi )$. For a canonical real scalar field in $D$ spacetime dimensions, $\Delta =(D-2)/2$ keeps the kinetic action invariant. Prescribing $\Delta =1$ therefore restricts kinetic invariance to $D=4$. In four dimensions the classical massless quartic scalar field theory has a conserved dilatation current $j^{\mu }=\varphi {\partial }^{\mu }\varphi +x_{\nu }{T}^{\mu \nu }$.

The dilatation current associated with scale transformations is $D^{\mu }=x_{\nu }{T}^{\mu \nu }$ up to improvement terms. Its divergence is the trace of the stress-energy tensor, so classical scale invariance requires a traceless improved tensor on shell.

Scale invariance requires the potential to obeyThus, apart from the zero choice,for a dimensionless constant $g$; an additive constant is excluded because it introduces a scale in the action. The canonical dilatation current isUsing $\square \varphi +V^{\prime }(\varphi )=0$,Equivalently, after improving the stress tensor to make it traceless, the current is $x_{\nu }{T}_{\mathrm{imp}}^{\mu \nu }$.

The free massless Dirac field has actionUnder a finite dilation,The measure, derivative, and fields contribute ${\lambda }^4$, ${\lambda }^{-1}$, and ${\lambda }^{-2\Delta }$, so invariance requiresFor $\lambda =e^{\alpha }$, the infinitesimal active variations areThe associated dilatation current isup to an improvement term. Its divergence is the on-shell trace $T^{\mu }{}_{\mu }=0$.

Under the finite dilation, the volume element transforms as $d^4{x}^{\prime }=e^{4\alpha }{d}^{4}x$, whereas ${\partial }_{\mu }^{\prime }{\varphi }^{\prime }(x^{\prime })=e^{-2\alpha }{\partial }_{\mu }\varphi (x)$. Thus the kinetic term and quartic interaction both acquire the factor $e^{-4\alpha }$. Their factors cancel the Jacobian determinant, provingA mass term would instead acquire a net factor $e^{2\alpha }$ in the action and break this scale invariance.

For the Noether current, use the variation at the same coordinate, rather than comparing values at different coordinates:At $m=0$, the corresponding Lagrangian density variation is ${\delta }_0\mathcal{L}=-\alpha (4+x\cdot \partial )\mathcal{L}=-\alpha {\partial }_{\mu }(x^{\mu }\mathcal{L})$. Thus $J_0^{\mu }=-x^{\mu }\mathcal{L}$, and the stated Noether current convention givesThe frequently used dilatation current $D^{\mu }=x_{\nu }{T}^{\mu \nu }+\varphi {\partial }^{\mu }\varphi$ is $-j^{\mu }$; this is solely an overall parameter convention. In the convention $D^{\mu }=x_{\nu }{T}^{\mu \nu }-V^{\mu }$, the virial current is $V^{\mu }=-\varphi {\partial }^{\mu }\varphi$.

Using solution and the trace computed in solution gives the useful identityIt follows that $\boxed{{\partial }_{\mu }{j}^{\mu }=0}$ in the massless theory on shell. Equivalently, $x_{\nu }{\Omega }^{\mu \nu }$ is a conserved current, since its divergence is the trace of the improved stress-energy tensor on shell. These are classical scale invariance statements; the nonzero beta function in solution explains why the interacting quantum theory need not retain them.

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.