Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 65 / 3 / Solution

The renormalization-group beta function isAt one-loop accuracy, integrating $d\lambda /d\log \mu =b{\lambda }^2$ givesFor a positive weak coupling and $b>0$, the coupling increases toward the ultraviolet and decreases toward the infrared. Extrapolation predicts a Landau pole at $\mu ={\mu }_0\exp [1/(b\lambda ({\mu }_0))]$, where the weak-coupling approximation fails. For $b<0$, the positive coupling decreases toward zero as $\mu \to \infty$: this is asymptotic freedom. Toward the infrared it instead grows, so perturbation theory eventually fails. The zero coupling is a fixed point. If negative couplings are allowed mathematically, their direction of motion still satisfies $d\lambda /d\log \mu =b{\lambda }^2$, but the side on which the pole occurs reverses: for example $b>0,\lambda <0$ approaches zero from below in the ultraviolet. Negative quartic coupling does not give a stable scalar potential.

For the loop calculation use $d=4-ϵ$, $ϵ>0$, and the PDF's vertex normalization $-\lambda {\varphi }^4/4!$. A Feynman parameter combines the denominators. Shift the Euclidean loop momentum to $\ell =k+xp$, and put $M_x^2=m^2+x(1-x)p^2$. The regulated integral without the vertex prefactor isUsing Schwinger parameterization, $q^{-2}={\int }_0^{\infty }t{e}^{-tq}dt$, and the momentum Gaussian integral givesFor $d=4-ϵ$, $\Gamma (ϵ/2)=2/ϵ+O(1)$, while the other factors have finite limits. Thus the massive scalar bubble pole in four dimensions isThis pole is independent of external momentum and mass and hence is canceled by a local quartic counterterm. There are three channels, corresponding to the three pairings of four external momenta. Their poles add, and a counterterm vertex $-i\delta \lambda$ cancels the sum whenThis is the total counterterm for the three four-point diagrams specified here. The minimal subtraction scheme removes their poles only. The one-loop tadpole is momentum independent and requires a mass counterterm, but does not produce one-loop wave-function renormalization; that is why it does not alter this one-loop coupling calculation.

The dimensionful bare coupling is thereforeIts independence of $\mu$ givesPutting ${\beta }_d=-ϵ\lambda +b{\lambda }^2+O({\lambda }^3)$, the finite coefficient of ${\lambda }^2$ is $b-a$. ThusThis is the one-loop quartic scalar beta function, positive for nonzero weak real coupling. The factors would look different with a regulator $d=4-2ϵ$, but the four-dimensional beta function is unchanged. For massless theory take nonexceptional external momentum, or an infrared regulator, so an infrared singularity is not confused with the ultraviolet pole.