= Solution
The <renormalization-group beta function> is
$$
\boxed{\beta(\lambda)=\left.\mu\frac{d\lambda}{d\mu}\right|_{\text{bare parameters fixed}}.}
$$
At one-loop accuracy, integrating $d\lambda/d\log\mu=b\lambda^2$ gives
$$
\boxed{\frac1{\lambda(\mu)}=\frac1{\lambda(\mu_0)}-b\log\frac\mu{\mu_0}.}
$$
For 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 $d=4-\epsilon$, $\epsilon>0$, and the PDF's vertex normalization $-\lambda\phi^4/4!$. 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 is
$$
I(p)=\mu^\epsilon\int_0^1dx\int\frac{d^d\ell}{(2\pi)^d}\frac1{(\ell^2+M_x^2)^2}.
$$
Using <Schwinger parameterization>, $q^{-2}=\int_0^\infty t e^{-tq}\,dt$, and the momentum <Gaussian integral> gives
$$
I(p)=\frac{\mu^\epsilon}{(4\pi)^{d/2}}\Gamma\left(2-\frac d2\right)
\int_0^1dx\,(M_x^2)^{d/2-2}.
$$
For $d=4-\epsilon$, $\Gamma(\epsilon/2)=2/\epsilon+O(1)$, while the other factors have finite limits. Thus the <massive scalar bubble pole in four dimensions> is
$$
\boxed{I(p)_{\rm div}=\frac1{8\pi^2\epsilon},\qquad
\left(\frac{i\lambda^2}{2}I(p)\right)_{\rm div}
=\frac{i\lambda^2}{16\pi^2\epsilon}.}
$$
This 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 when
$$
\boxed{\delta\lambda=\frac{3\lambda^2}{16\pi^2\epsilon},\qquad
\mathcal L_{\rm ct}^{(4)}=-\mu^\epsilon\frac{\delta\lambda}{4!}\phi^4.}
$$
This 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 therefore
$$
\lambda_B=\mu^\epsilon\left[\lambda+\frac{a\lambda^2}{\epsilon}+O(\lambda^3)\right],
\qquad a=\frac3{16\pi^2}.
$$
Its independence of $\mu$ gives
$$
0=\epsilon\left(\lambda+\frac{a\lambda^2}{\epsilon}\right)
+\beta_d\left(1+\frac{2a\lambda}{\epsilon}\right)+O(\lambda^3).
$$
Putting $\beta_d=-\epsilon\lambda+b\lambda^2+O(\lambda^3)$, the finite coefficient of $\lambda^2$ is $b-a$. Thus
$$
\boxed{\beta_d=-\epsilon\lambda+\frac{3\lambda^2}{16\pi^2}+O(\lambda^3),\qquad
\beta_{d=4}=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3).}
$$
This is the <one-loop quartic scalar beta function>, positive for nonzero weak real coupling. The factors would look different with a regulator $d=4-2\epsilon$, 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.
Back to article page
