Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 66 / 4

A loop integral in a theory with dimensionless classical couplings can contain a logarithmic ultraviolet divergence. After subtraction, an expression such as $\log (-p^2/{\mu }^2)$ remains; one must specify the momentum scale at which the measured or renormalized coupling is fixed. A logarithm of a dimensionful momentum alone is not meaningful. In dimensional regularization, the same necessity appears through ${\lambda }_{\mathrm{bare}}={\mu }^{d_{\lambda }}{Z}_{\lambda }\lambda (\mu )$ in the continued dimension, with $d_{\lambda }$ tending to zero in the physical dimension.

Thus the renormalization prescription introduces an arbitrary renormalization scale, even when no mass or length occurs in the classical action. Physical quantities cannot depend on its arbitrary choice: explicit logarithms cancel the implicit change of the running coupling, governed by $\mu d\lambda /d\mu =\beta (\lambda )$. A nonzero renormalization-group beta function is a quantum scale anomaly. An integration constant may encode the same data as a dimensionful scale, which is dimensional transmutation. The scale labels a renormalization condition, not a new arbitrary measurable parameter. At a true fixed point the coupling need not run; the use of a subtraction scale is then a convention rather than an intrinsic generated physical scale.

SetThe chain rule givesFor the ultraviolet expansion take ${\beta }_0>0$ and $L=\log ({\mu }^2/{\Lambda }^2)\gg 1$. The leading solution is $y\sim {\beta }_{0}L$. Substituting it in the next term gives $dy/dL={\beta }_0+{\beta }_1/({\beta }_{0}L)+O(\log L/L^2)$, whose integration yieldsChoosing $\Lambda$ removes the additive constant. This proves the two displayed leading terms of the two-loop asymptotic running coupling.

The printed remainder is too small in general. This can be checked even for an exactly two-loop renormalization-group beta function, so it does not depend on unknown higher-loop effects. Put $c={\beta }_1/{\beta }_0$. Separating variables and fixing the integration constant givesExpanding this identity one order further givesThe $\log L/L$ term cannot be absorbed into a fixed change of $\Lambda$. For example, with ${\beta }_0={\beta }_1=1$, the exact implicit relation is $y-\log (y+1)=L$, and the next term is $(\log L+1)/L$, which is not $O(1/L)$. The unspecified $O(y^{-2})$ contribution to the differential equation changes the constant coefficient of $1/L$, while the coefficient ${\beta }_1^2/{\beta }_0^3$ of $\log L/L$ is already fixed. Only special cases such as ${\beta }_1=0$ avoid this particular obstruction.

The asymptotic formula also presupposes an asymptotically free ultraviolet regime. If ${\beta }_0=0$, division by it is invalid and the leading scaling changes. If ${\beta }_0<0$, the weak-coupling asymptotic direction is generally toward the infrared and the logarithmic variable must be chosen accordingly. These qualifications distinguish the valid perturbative solution from an unconditional formula.

Use the canonically normalized Abelian coupling $g_1=-\sqrt{5/3}{g}^{\prime }$, with $g_2=g$ and $g_3=g_s$, and write ${\alpha }_i=g_i^2/(4\pi )$. The sign in $g_1$ is a generator convention; its square is what enters running and matching. Let ${\alpha }_5=g_5^2/(4\pi )$ and $L_X=\log (M_X^2/{\mu }^2)$. To leading order,Use exactly the common-flavor coefficients stipulated for this model:The same $n_f$ cancels from their differences. In particular $b_3-b_2=11/3$ and $b_2-b_1=22/3$, so eliminating the unified coupling givesAt the low reference scale, $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. With $w={\sin }^2{\theta }_W$ this meansThe last difference equation becomes $3(1-w)/(5\alpha )-w/\alpha =2(w/\alpha -1/{\alpha }_s)$. Solving givesSubstitute this into the first difference equation to obtainAll low-scale couplings on the right must be evaluated at the same specified $\mu$. A dimensionful $M_X$ cannot be fixed by dimensionless coupling values without such a reference scale. The result is independent of the common $n_f$ in these differences, although ${\alpha }_5$ is not. An inferred $M_X>\mu$ requires ${\alpha }^{-1}>8/(3{\alpha }_s)$, and perturbative matching requires a positive small ${\alpha }_5$.

This is the one-loop SU(5) prediction with common flavor coefficients. It is the prediction of the supplied simplified coefficient model, not the full matter-and-Higgs calculation for a realistic SU(5) grand unified theory. Actual representations, intermediate thresholds and higher-loop terms change the running and require the corresponding matching calculation.