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The stress-energy tensor generates translations. One may improve it by adding a divergence without changing the translation charges under suitable boundary conditions, and choose it symmetric in a relativistic theory. If ${\Theta }^{\mu \nu }$ is symmetric and conserved, a spacetime vector field $\omega$ defines the current $j^{\mu }={\Theta }^{\mu \nu }{\omega }_{\nu }$. For a Conformal Killing vector field,Thus a traceless conserved stress-energy tensor produces all the conformal charges. The dilation current is $x_{\nu }{\Theta }^{\mu \nu }$, and the special conformal currents are $(2x_{\lambda }{x}_{\nu }-x^2{\eta }_{\lambda \nu }){\Theta }^{\mu \nu }$. A trace that is merely the divergence of a virial current gives scale invariance; conformal invariance requires the additional improvement that removes the trace. These statements distinguish an actual traceless tensor from a trace discarded without justification.

For a free massless scalar field, the canonical stress-energy tensor has trace $(1-d/2)(\partial \varphi {)}^2$. For $d>1$, the improved stress-energy tensor of a free massless scalar isIts improvement is identically conserved. On $\square \varphi =0$, the identity $\square {\varphi }^2=2(\partial \varphi {)}^2$ shows that its trace vanishes. In $d=2$, no improvement is needed: the canonical stress-energy tensor is already traceless. In Lorentzian null coordinates, conservation and tracelessness imply that its two surviving components depend separately on $x^{+}$ and $x^{-}$. After Euclidean continuation, these become the holomorphic stress-energy tensor $T(z)$ and its antiholomorphic counterpart $\bar{T}(\bar{z})$, withLocal conformal changes are arbitrary holomorphic and antiholomorphic coordinate changes, not just the three global Möbius transformation generators in either sector.

To make the free-field quantum normalization explicit, rescale the field to $\phi =\sqrt{4\pi }\varphi$, so that its plane contraction is $\phi (z,\bar{z})\phi (w,\bar{w})\sim -\log |z-w{|}^2$. Define $J=i\partial \phi$, so $J(z)J(w)\sim (z-w{)}^{-2}$. Normal ordering givesA single Wick contraction shows that $J$ is a weight-one primary operator. The two double contractions in $T(z)T(w)$ produce $\frac{1}{2}(z-w{)}^{-4}$; single contractions, followed by Taylor expansion at $w$, produce $2T(w)/(z-w{)}^2+\partial T(w)/(z-w)$. Thus the operator product expansion isThe same holds independently for $\bar{T}$, with $\bar{c}=1$. The undifferentiated noncompact scalar has a logarithmic contraction and a zero-mode subtlety; its derivative current has the ordinary primary transformation law.

The holomorphic stress-energy tensor implements infinitesimal conformal transformations through contour charges $Q_v=\frac{1}{2\pi i}\oint dzv(z)T(z)$. A primary field $\mathcal{O}$ of weight $h$ obeysFor example the free-field vertex operator $:e^{i\alpha \phi }:$ has holomorphic weight $h={\alpha }^2/2$, as follows by contracting $\partial \phi$ with the exponential twice. In correlators the contour may encircle each insertion, giving the conformal Ward identity $⟨T(z){\prod }_i{\mathcal{O}}_i(z_i)⟩={\sum }_i[h_i/(z-z_i{)}^2+{\partial }_{z_i}/(z-z_i)]⟨{\prod }_i{\mathcal{O}}_i(z_i)⟩$. This also yields the global conformal Ward identities for chiral correlators.

Writing $T(z)={\sum }_n{L}_n{z}^{-n-2}$ and integrating the stress-tensor operator product expansion gives the Virasoro algebraIts central charge is the quantum correction to the classical conformal algebra; the correction vanishes on $L_{-1},L_0,L_1$. Equivalently, in the pullback convention a finite holomorphic change $z\mapsto f(z)$ sends $T$ to $(f^{\prime }{)}^{2}T(f)+\frac{c}{12}\{f,z\}$, where $\{f,z\}=f^{\prime \prime \prime }/f^{\prime }-\frac{3}{2}(f^{\prime \prime }/f^{\prime }{)}^2$ is the Schwarzian derivative. This anomalous term vanishes for Möbius transformations. On a curved background the related trace anomaly is proportional to its curvature; it does not destroy the flat-space conformal symmetry of this massless theory.

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.

A virial current $V^{\mu }$ represents the trace of a conserved stress-energy tensor as a divergence. Then the dilation Noether current $D^{\mu }=x_{\nu }{T}^{\mu \nu }-V^{\mu }$ is conserved, since its divergence is $T^{\mu }{}_{\mu }-{\partial }_{\mu }{V}^{\mu }$. This establishes scale invariance; a traceless stress-energy tensor improvement is the extra condition needed for conformal currents.