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Let ${\varphi }_B=Z_{\varphi }^{1/2}\varphi$ define the renormalized field in Phi-fourth theory. For an $n$-point correlation function, with no composite operator insertions, this impliesThe bare correlation function is independent of the arbitrary renormalization scale $\mu$ when the bare parameters and external momenta are fixed. DefineThe definitions of the running mass and wave-function renormalization functions specify their signs and the factor of two. Differentiating $Z_{\varphi }^{n/2}{G}_n$ using the chain rule yields the Callan–Symanzik equationHere ${\hat{\beta }}_{\lambda }$ denotes the renormalization-group beta function itself, not the function multiplied by an additional $\lambda$. The Callan-Symanzik equation concerns dependence on the arbitrary reference scale, with the displayed external momenta kept fixed.

For the minimal subtraction scheme in dimensional regularization, writeso ${\lambda }_B={\mu }^{ϵ}F$ and $m_B^2=m^{2}H$. The ${\mu }^{ϵ}$ factor gives the engineering contribution $-ϵ\lambda$ to the renormalization-group beta function. Put ${\hat{\beta }}_{\lambda }=-ϵ\lambda +{\beta }_{\lambda }$. Scale independence of the bare coupling then givesThe coefficient of ${ϵ}^0$ is $f_1-\lambda f_1^{\prime }+{\beta }_{\lambda }$. Its vanishing proves the first of the simple-pole formulas for minimal-subtraction renormalization group functions:The coefficients of ${ϵ}^{-k}$, $k\ge 1$, also vanish and impose $(\lambda {\partial }_{\lambda }-1)f_{k+1}={\beta }_{\lambda }{f}_k^{\prime }$. These relations explain how the higher poles are consistent with a finite renormalization-group beta function.

For the mass, differentiate the second bare-parameter relation:Its finite coefficient is ${\gamma }_{m^2}-\lambda b_1^{\prime }$. ThereforeThe higher poles impose $\lambda b_{k+1}^{\prime }={\gamma }_{m^2}{b}_k+{\beta }_{\lambda }{b}_k^{\prime }$. The formulas for the running mass and renormalization-group beta function thus follow directly from bare-parameter independence rather than from an identification with diagram coefficients by name.

At the specified loop order, $f_1=3{\lambda }^2/(16{\pi }^2)$ and $b_1=\lambda /(16{\pi }^2)$. There is no wave-function renormalization at this order. HenceThe regulated renormalization-group beta function is ${\hat{\beta }}_{\lambda }=-ϵ\lambda +3{\lambda }^2/(16{\pi }^2)+O({\lambda }^3)$; after removing the regulator, ${\hat{\beta }}_{\lambda }={\beta }_{\lambda }$.

Now take $m^2=0$ and the four-dimensional limit. This massless surface is preserved since $\mu d{m}^2/d\mu =m^2{\gamma }_{m^2}$. Set $b=3/(16{\pi }^2)$, choose a reference scale ${\mu }_0$, and define the running coupling byDifferentiating $1/\lambda$ gives $d(1/\lambda )/d\log \mu =-b$, soThe zero-coupling solution is obtained by continuity. At one-loop accuracy ${\gamma }_{\varphi }=0$, and the chain rule turns the massless Callan-Symanzik equation intoThus the one-loop solution along a running-coupling trajectory isFor a prescribed endpoint coupling $\lambda$, substitute ${\lambda }_0=\lambda /[1+b\lambda \log (\mu /{\mu }_0)]$ on the right. The reference correlation function supplies arbitrary boundary data; the Callan-Symanzik equation determines their scale evolution, not their full momentum dependence. The running coupling formula is used on a branch on which its denominator is nonzero and the retained perturbative approximation is appropriate.

If one keeps $ϵ\ne 0$, the same characteristic calculation uses $d\lambda /d\log \mu =-ϵ\lambda +b{\lambda }^2$. Writing $r=\mu /{\mu }_0$ gives $\lambda (\mu )={\lambda }_0{r}^{-ϵ}/[1-(b{\lambda }_0/ϵ)(1-r^{-ϵ})]$, which tends to the displayed four-dimensional result as $ϵ\to 0$.

With a nonzero anomalous dimension, the characteristic equation instead reads $d{G}_n/d\log \mu =-n{\gamma }_{\varphi }(\lambda (\mu ))G_n$. Integration proves the characteristic solution of the massless Callan-Symanzik equation:For the stipulated ${\gamma }_{\varphi }=c{\lambda }^2$ and the same one-loop renormalization-group beta function, change variables using $d\log \nu =d\lambda /(b{\lambda }^2)$. The exponent's integral becomes $(c/b)(\lambda (\mu )-{\lambda }_0)$. Therefore the modified explicit solution isEquivalently its multiplicative factor is $\exp [-nc{\lambda }_0^2\log r/(1-b{\lambda }_0\log r)]$. This solves the equation with the stated truncated renormalization-group beta function and anomalous dimension; it does not introduce additional higher-order coefficients that were not specified.