How a mass scale appears without a classical mass parameter

Quantum renormalization replaces a dimensionless coupling specified at a reference scale with a dynamically generated scale. This explains how pure Yang–Mills can possess a characteristic mass scale. It does not determine whether the physical spectrum is gapped.

Prerequisite: running couplings from a first QFT course and physical states.

By the end:

• Integrate the one-loop running-coupling equation in the stated convention.
• Explain why a scale alone does not prove a mass gap.

Integrating the running coupling

For pure Yang–Mills, write the perturbative beta function as

$$\mu\frac{dg}{d\mu}=-b_0g^3+O(g^5), \qquad b_0=\frac{11C_A}{3(16\pi^2)}.$$

For $SU(N)$ in the usual normalization, $C_A=N$. The positive coefficient means the coupling decreases toward the ultraviolet. The original asymptotic-freedom calculations are those of Gross and Wilczek and Politzer.

At one loop,

$$\frac{d}{d\log\mu}\frac{1}{g^2}=2b_0.$$

Integrate between $\mu_0$ and $\mu$:

$$\frac{1}{g^2(\mu)} =\frac{1}{g^2(\mu_0)}+2b_0\log\frac{\mu}{\mu_0}.$$

Define an integration constant with units of energy,

$$\Lambda_{\text{1-loop}} =\mu_0\exp\!\left[-\frac{1}{2b_0g^2(\mu_0)}\right].$$

Then

$$g^2(\mu)=\frac{1}{2b_0\log(\mu/\Lambda_{\text{1-loop}})}.$$

The formula is useful where the coupling is small. Its apparent singularity near $\mu=\Lambda_{\text{1-loop}}$ indicates the failure of the approximation. It does not by itself establish a physical singularity.

What has been transmuted?

Specifying $g(\mu_0)$ and $\mu_0$ fixes the integration constant. The dimensionless coupling at a chosen scale can therefore be exchanged for a dimensionful parameter.

Check this directly. Differentiating the one-loop expression gives

$$\frac{d\log\Lambda_{\text{1-loop}}}{d\log\mu} =1+\frac{\beta(g)}{b_0g^3}=0$$

at that order. Different choices of reference scale describe the same running trajectory when the coupling is adjusted consistently.

Beyond one loop, the definition includes further coupling-dependent factors. The numerical value of $\Lambda$ depends on the renormalization scheme. A physical mass does not: the coefficient relating that mass to a scheme-dependent $\Lambda$ changes when the scheme changes.

This is why dimensionless physical ratios, such as ratios of two masses, are especially clean comparison quantities.

Where the missing physics lives

If a pure Yang–Mills theory has only the generated scale in its local massive spectrum, a mass can take the form

$$m=c\,\Lambda.$$

Dimensional analysis leaves the dimensionless coefficient $c$ undetermined. More fundamentally, it does not prove that a finite positive mass threshold exists. Establishing $c>0$ requires nonperturbative control of the relevant physical spectrum.

The possibilities “a scale exists” and “a positive gap exists” are therefore logically different. A scale can characterize a crossover or an observable without forcing every physical excitation to be massive.

The nonanalytic dependence

$$\Lambda\propto \exp\!\left[-\frac{1}{2b_0g^2}\right]$$

also helps explain why a finite power series in a weak coupling does not produce this scale. Every coefficient of its formal Taylor expansion at $g=0$ vanishes. This is an illustration of the difference between perturbative information and nonperturbative structure; it is not a general theorem that all nonperturbative quantities have this particular exponential form.

Connecting to a lattice

The ultraviolet cutoff of a lattice is of order $1/a$. To approach a continuum theory with a fixed physical scale, the bare parameters must be tuned as the spacing decreases. Perturbatively,

$$a\Lambda\sim \exp\!\left[-\frac{1}{2b_0g_0^2}\right]$$

up to higher-order factors and scheme-dependent normalization.

Thus the continuum trajectory approaches weak bare coupling. The long-distance theory can nevertheless remain strongly interacting. “Weak bare coupling” concerns fluctuations at the cutoff; it does not make every physical distance perturbative.

This observation is one reason a theorem proved only in a fixed strong-coupling lattice regime does not automatically establish the continuum result.

Try the running-coupling lab. Vary $N$ (the fundamental-representation dimension for $SU(N)$) and the scale ratio, then identify the region where the one-loop formula stops being a controlled approximation.

Exercises

1. Change the reference scale

Let $\mu_1=2\mu_0$. Find $g(\mu_1)$ at one loop and verify that both reference choices give the same $\Lambda_{\text{1-loop}}$.

Hint. Use the integrated equation before substituting in the exponential.

Solution. $g^{-2}(\mu_1)=g^{-2}(\mu_0)+2b_0\log2$. Therefore $\mu_1e^{-1/(2b_0g^2(\mu_1))}=2\mu_0e^{-\log2}e^{-1/(2b_0g^2(\mu_0))}$, equal to the original scale.

2. Why no perturbative mass term?

Does the essential singularity of $e^{-1/g^2}$ prove a mass gap?

Hint. Separate a possible functional dependence from the existence of a physical observable.

Solution. No. It shows that this function cannot be recovered by an ordinary power series around zero. One must still construct the theory, identify a spectral observable, and prove its positive threshold.

Continue at your own depth

Continuum scaling and reference scales develops this lesson without adding a prerequisite to the next lesson.

Read further

• Gross and Wilczek, ultraviolet behavior of non-Abelian gauge theories.
• Politzer, reliable perturbative results for strong interactions.
• David Tong, Gauge Theory, renormalization and Yang–Mills dynamics.

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