Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 44 / 3 / Solution

A regulator and a renormalization condition introduce the reference mass scale $\mu$, even when the classical theory has no mass. Loop amplitudes contain dimensionless logarithms of momentum or distance ratios involving $\mu$. The resulting running coupling and field normalization compensate changes of this arbitrary reference scale.

Let ${\varphi }_0=Z_{\varphi }^{1/2}\varphi$ and $G_n=Z_{\varphi }^{-n/2}{G}_{n,0}$. Hold the bare parameters fixed and define $\beta (g)=\mu dg/d\mu$ and $\gamma (g)=\frac{1}{2}\mu d\log Z_{\varphi }/d\mu$. Assuming multiplicative field renormalization and no mixing or additive contact terms for the correlator, differentiating gives the Callan-Symanzik equationFor the dimensionless two-point factor put $r=p^2/{\mu }^2$. Its equation is $(-2r{\partial }_r+\beta {\partial }_g+2\gamma )C=0$. LetThe characteristic solution of the multiplicative Callan-Symanzik equation isTo check the sign, ${\partial }_{t}C(e^{2t}r,g)=2e^{2t}r{\partial }_{r}C$ equals $(\beta {\partial }_g+2\gamma )C$. The characteristic flow and its accumulated multiplier give precisely this evolution. Changing $\mu$ changes the dimensionless momentum and renormalized $g$; the same bare two-point correlation function is recovered after the compensating field normalization. An unnormalized renormalized correlator need not remain numerically identical under that change, but physical predictions do.

With a mass, write $v=m/\mu$ and define its running mass by $m^{\prime }(t)=\delta (g(t))m(t)$, $m(0)=m$. The dimensionless equation becomesIts flow is thereforeThe renormalization-group mass suppression criterion is ${\int }_0^t(\delta (g(u))-1)du\to -\infty$, for example an eventual bound $\delta \le 1-\eta$ with $\eta >0$. Then the mass argument on the right tends to zero. A regular massless limit, uniform along the limiting coupling trajectory, makes the mass negligible at high energies. Merely calling $\delta$ small without controlling this integrated exponent is insufficient.