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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.

Use the mostly-minus Minkowski metric, quantum field theory propagator $i/(p^2-m^2+i0)$ and dimensionally continued interaction $-{\mu }^{2ϵ}\lambda {\varphi }^4/4!$, with $d=4-2ϵ$. Take all four external momenta incoming, so ${\sum }_j{p}_j=0$, and set $s=(p_1+p_2{)}^2$, $t=(p_1+p_3{)}^2$, $u=(p_1+p_4{)}^2$.

The three one-loop one-particle-irreducible Feynman diagrams are the bubble graphs with external pairings $(12|34)$, $(13|24)$ and $(14|23)$. Their external quantum field theory propagators are omitted by amputation. Each Feynman diagram has symmetry factor $1/2$ from exchanging the two internal lines; the fixed external labels are already accounted for by the three distinct channels.

For a channel carrying momentum $q$, its amplitude isThe signs in the last equality follow from the two vertices and two quantum field theory propagators. Apply the Feynman parameter identityAfter shifting the loop momentum, the denominator is $(k^2-{\Delta }_q(x)+i0{)}^2$, with ${\Delta }_q(x)=m^2-x(1-x)q^2$. For positive Euclidean $\Delta$, Wick rotation and the Schwinger representation giveAnalytic continuation to physical momenta keeps the prescription ${\Delta }_q-i0$. Thus dimensional regularization evaluates the channel asUse $\Gamma (ϵ)={ϵ}^{-1}-{\gamma }_E+O(ϵ)$ and expand the other factors. The minimal-subtraction scalar four-point bubble channels result through the finite part isThe parameter integral is left unevaluated, as requested. The three channel contributions are $B(s)+B(t)+B(u)$. Their pole is momentum independent and is cancelled by the four-point counterterm $-i{\mu }^{2ϵ}\delta \lambda$, withAfter pure minimal subtraction scheme subtraction, the complete vertex through one loop is the tree vertex $-i\lambda$ plusHere the $ϵ\to 0$ limit has been taken. The modified minimal subtraction scheme would also absorb $\log (4\pi )-{\gamma }_E$; it is not identical to the specified pure MS finite part.

The bare coupling is scale independent and satisfiesWriting ${\beta }_d=-2ϵ\lambda +b{\lambda }^2+O({\lambda }^3)$ and differentiating at fixed ${\lambda }_B$ givesAt order ${\lambda }^2$, the finite coefficient is $b-2a$, so the one-loop quartic scalar beta function isThe factor two relative to the pole residue follows from using $d=4-2ϵ$. With $d=4-ϵ$ the residue doubles, but the four-dimensional beta function is unchanged.

For pure nonabelian gauge theory, the corresponding pure Yang-Mills one-loop beta function has the opposite sign:Gauge-boson self-interactions and the gauge-consistency ghost contributions give antiscreening. For a compact nonabelian simple factor $C_A>0$, the coupling decreases at high energy: asymptotic freedom. In contrast, positive scalar $\lambda$ increases toward the ultraviolet. Its one-loop extrapolation has a Landau pole, indicating breakdown of weak-coupling perturbation theory, not a proof by itself of nonperturbative triviality. Ultraviolet poles should be extracted at infrared-safe mass or nonexceptional kinematics; a scaleless massless zero-momentum integral cannot be used to read off this ultraviolet counterterm without separating its infrared behavior.

For $m^2>0$ and initially $\Re n>d/2$, Schwinger parameterization and the supplied Gaussian integral giveSubstitute $u=\alpha m^2$ to obtain the Euclidean massive loop integralFor a positive integer $n$, $\Gamma (n)=(n-1)!$. Outside the convergence region this expression defines its meromorphic continuation, used in dimensional regularization.

Combine the two denominators with a Feynman parameter and shift $k$ by $(1-x)p$. This givesThe gamma function has $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the remaining factors tend to $(4\pi {)}^{-2}$ and one. Hence the massive scalar bubble pole in four dimensions isThe pole is independent of the external momentum, as required for a local quartic counterterm.

Use the mostly-plus Minkowski metric and expand $e^{iS}$. The momentum-space Feynman rules are: an internal scalar propagator $i/(-p^2-m^2+i0)$; a four-scalar interaction vertex $-i{\mu }^{ϵ}\lambda$; a momentum conservation delta function at each vertex; and $\int d^{d}k/(2\pi {)}^d$ per independent loop momentum. Divide by the Feynman-diagram symmetry factor. All momenta can be taken incoming. External propagators are omitted when computing an amputated one-particle-irreducible vertex.

The connected one-loop one-particle-irreducible Feynman diagrams are the two-point tadpole diagram and the three four-point bubble diagrams, in the $s,t,u$ channels. Each has symmetry factor $1/2$. A tadpole attached to an external leg through a single bridge would be reducible, so is excluded here.

The Euclidean massive loop integral at $n=1$ has $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$, so $I_{d,1}=-m^2/(8{\pi }^2ϵ)+O(1)$. Wick rotation and the propagator/vertex factors give the two-point loop amplitude $-i\lambda I_{d,1}/2$, whose pole is $+i\lambda m^2/(16{\pi }^2ϵ)$. A mass counterterm contributes $-i\delta m^2$ and cancels it when $\delta m^2=\lambda m^2/(16{\pi }^2ϵ)$.

Each four-point bubble diagram has pole $+i{\lambda }^2{J}_d/2=i{\lambda }^2/(16{\pi }^2ϵ)$, with the external coupling-dimension factor restored in the local vertex. Summing its three channels gives the quartic counterterm $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. Therefore the one-loop massive phi-fourth counterterms are exactlyThere is no one-loop wave-function renormalization, since the tadpole is independent of external momentum. Vacuum graphs would additionally require a vacuum-energy subtraction.

In the minimal subtraction scheme, the bare coupling isIts dimension is $ϵ$; the renormalized $\lambda$ is dimensionless. Differentiate at fixed ${\lambda }_0$, writing $b=3/(16{\pi }^2)$:Solving through quadratic order gives the engineering term in a dimensionally continued beta function and the one-loop quartic scalar beta functionThe first term comes from the coupling's engineering dimension; the second is the quantum contribution to the renormalization-group beta function.

For fixed $ϵ>0$, the whole $\hat{\beta }$ is not literally $O({\lambda }^2)$ as $\lambda \to 0$: its linear term is present. Interpret the final instruction as keeping terms through quadratic order. Set ${\lambda }_{\ast }=ϵ/b=16{\pi }^2ϵ/3$. Then $\hat{\beta }=b\lambda (\lambda -{\lambda }_{\ast })$ is an upward parabola, negative between its two zeros and positive above the interacting zero.

For positive initial coupling ${\lambda }_r=\lambda ({\mu }_r)$, the quartic running coupling below four dimensions follows by solving the linear equation for $1/\lambda$:If $0<{\lambda }_r<{\lambda }_{\ast }$, then $\lambda (\mu )\to {\lambda }_{\ast }$ as $\mu \to 0$ and $\lambda (\mu )\to 0$ as $\mu \to \infty$. If ${\lambda }_r={\lambda }_{\ast }$ it stays fixed; the zero solution also stays fixed. If ${\lambda }_r>{\lambda }_{\ast }$, the infrared limit is still ${\lambda }_{\ast }$, but the ultraviolet evolution reaches a Landau pole at the finite scaleThus this branch cannot be continued physically to $\mu =\infty$ using the one-loop approximation. These are statements about the quartic coupling direction. The Wilson-Fisher fixed point of the full scalar theory also has a thermal relevant direction at the Wilson-Fisher fixed point; approaching the critical theory requires mass tuning. For negative initial coupling, the quartic potential is unstable: the same formula tends to zero from below in the ultraviolet and reaches a pole at finite scale toward the infrared.

Use $\hslash =c=1$ and normalize the interaction of quartic scalar field theory as $\lambda {\varphi }^4/4!$. For a renormalized $n$-point connected correlation function, adopt ${\varphi }_B=Z_{\varphi }^{1/2}{\varphi }_R$ and ${\gamma }_{\varphi }=\frac{1}{2}\mu d\log Z_{\varphi }/d\mu$. Independence of bare quantities from the renormalization scale gives the Callan-Symanzik equationwhere $\beta =\mu d\lambda /d\mu$ and ${\gamma }_{m^2}=\mu d\log m^2/d\mu$, with bare parameters fixed. A convention using proper vertices reverses the field-normalization term, so the choice of Green's function must be stated.

Solve it by characteristics: set $\mu (s)={\mu }_0{e}^s$, $d\lambda (s)/ds=\beta (\lambda (s))$, and $d\log m^2(s)/ds={\gamma }_{m^2}(\lambda (s))$. At fixed external momenta,This is the characteristic solution of the multiplicative Callan-Symanzik equation. If the field anomalous dimension vanishes, the function is constant along the running-parameter characteristics. At a massless fixed point with constant anomalous dimension, the normalization factor is a power of $\mu /{\mu }_0$. Choosing the scale close to a characteristic external momentum avoids large perturbative logarithms.

For the supplied minimal subtraction scheme relation in dimensional regularization with $d=4-ϵ$, let $a=3/(16{\pi }^2)$ and write ${\lambda }_B={\mu }^{ϵ}[\lambda +a{\lambda }^2/ϵ+O({\lambda }^3)]$. Differentiating at fixed bare coupling givesA consistent expansion yields $\beta =-ϵ\lambda +a{\lambda }^2+O({\lambda }^3)$. The $-ϵ\lambda$ term must be retained while it multiplies the pole; setting $ϵ=0$ prematurely loses the finite result. In four dimensions the one-loop quartic scalar beta function isThe one-loop two-point diagram is a momentum-independent tadpole diagram, so there is no field wavefunction renormalization at this order and ${\gamma }_{\varphi }=O({\lambda }^2)$. Differentiating the supplied bare-mass relation in the same way gives ${\gamma }_{m^2}=\lambda /(16{\pi }^2)+O({\lambda }^2)$.

Integrating $d\lambda /d\log \mu =a{\lambda }^2$ gives the running couplingFor positive coupling it grows toward the ultraviolet and decreases toward the infrared: the theory is not asymptotically free. Its one-loop Landau pole lies at ${\mu }_L={\mu }_0\exp [16{\pi }^2/(3\lambda ({\mu }_0))]$. Perturbation theory fails before reaching that pole, so the pole in this approximation alone does not prove the existence of an exact physical divergence. The corresponding one-loop running mass satisfiesIt follows from $d\log m^2/d\lambda =1/(3\lambda )$. The mass-independent subtraction scheme keeps these beta functions independent of the mass; physical low-energy matching can still require threshold treatment. Running parameters resum the leading logarithms within the regime of small coupling.

Use the Euclidean path integral with weight $e^{-S}$ and Fourier transform convention $\varphi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\varphi (p)$. The momentum-space Feynman rules are: an internal scalar propagator $1/(p^2+m^2)$; a quartic vertex $-\lambda$ together with $(2\pi {)}^d{\delta }^{(d)}(\sum p_i)$ for incoming momenta; and an integral $\int d^{d}k/(2\pi {)}^d$ over each independent loop momentum. Multiply by the Feynman-diagram symmetry factor; external propagators are retained for a full correlation function and removed for an amputated connected correlation function.

To enumerate the requested one-particle-irreducible Feynman diagrams, let $V$ be the number of quartic vertices and $I$ the number of internal edges. Four external legs imply $4V=2I+4$, while one loop order implies $I-V+1=1$. Thus $V=I=2$. Both internal edges must connect the two vertices: an alternative with a tadpole and one connecting edge would disconnect on cutting that edge. The only graphs are therefore the three pairings of labelled external legs, each with Feynman-diagram symmetry factor $1/2$.

The corresponding channel momenta are $q_s=p_1+p_2$, $q_t=p_1+p_3$, and $q_u=p_1+p_4$, with all external momenta incoming. This exhausts the connected one-loop four-point one-particle-irreducible Feynman diagrams.

For the Euclidean massive loop integral, first take $m^2>0$ and an integer $n>d/2$, where the integral converges. Schwinger parameterization givesThe supplied Gaussian integral then yieldsIntegrating over $s$, and using the Gamma function recurrence to obtain $\Gamma (n)=(n-1)!$, provesOutside its initial convergence range, the right-hand side defines the meromorphic continuation used in dimensional regularization; the divergent ordinary integral is not being assigned a convergent value.

For the scalar bubble integral, the Feynman parameter identity $1/(AB)={\int }_0^{1}dx[xA+(1-x)B{]}^{-2}$ followed by a translation of loop momentum givesWith $d=4-ϵ$, the Gamma function satisfies $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral tends to one. Hence the ultraviolet pole is independent of external momentum:One can also see why the same pole occurs without introducing a Feynman parameter: at large $k$, the difference between this integrand and $(k^2+m^2{)}^{-2}$ is ultraviolet integrable near four dimensions, so both have the same dimensional regularization pole. The positive mass avoids an infrared ambiguity in this argument.

In the quantum effective action convention, the tree-level four-point vertex is $+\lambda$, the negative of the amputated connected correlation function tree vertex. The three bubble corrections giveThus the pole counterterm is $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. The modified minimal subtraction scheme also subtracts the conventional finite $\log 4\pi -{\gamma }_E$ combination, or equivalently absorbs it into the subtraction-scale convention; that does not change the one-loop renormalization-group beta function. Using that scale convention, writeHere $\lambda$ is dimensionless, $[{\lambda }_B]=ϵ$, and the conventional constant scale factor is implicit. The one-loop tadpole is momentum independent, so wave-function renormalization does not contribute at this order. Differentiating at fixed ${\lambda }_B$, with $a=3/(16{\pi }^2)$, gives $0=ϵ(\lambda +a{\lambda }^2/ϵ)+\beta (1+2a\lambda /ϵ)+O({\lambda }^3)$. ThereforeFor the stable quartic scalar field theory, with $\lambda >0$, the one-loop quartic scalar beta function is positive: the running coupling increases toward the ultraviolet and decreases toward the infrared. The theory is not asymptotically free; extrapolating the one-loop flow gives a Landau pole.

Because the quartic coupling constant has mass dimension $ϵ=4-D$, introduce a renormalization scale $\mu$ and define the dimensionless renormalized coupling $g$ by minimal subtraction scheme:Here ${\lambda }_0$ is the fixed bare coupling constant. Equivalently the tree-level quartic term is $-{\mu }^{ϵ}g{\varphi }^4/4!$, and its one-loop counterterm is $-{\mu }^{ϵ}3g^2{\varphi }^4/(16{\pi }^2ϵ4!)$. This states explicitly the scale convention implicit in solution.

Put $a=3/(16{\pi }^2)$. Differentiating the bare relation at fixed ${\lambda }_0$ givesWrite $\hat{\beta }=-ϵg+b{g}^2+O(g^3)$. The coefficient of $g^2$ in the last identity is $a+b-2a$, so $b=a$. ThereforeThe first term is the engineering term in a dimensionally continued beta function. In four dimensions it disappears, leaving the positive one-loop quartic scalar beta function $3g^2/(16{\pi }^2)$. Thus the renormalized coupling increases with increasing renormalization scale near weak positive coupling. This quantum running is the obstruction to extending the classical scale invariance in solution to the interacting quantum theory.