Renormalized gauge-invariant observables

Prerequisites: the continuum limit and distributions.

You will learn to identify the observable being constructed, expose a coincidence-limit problem in a Gaussian example, and keep smoothing separate from regulator removal.

Gauge invariance does not remove short-distance singularities

The trace of a curvature polynomial is gauge invariant, but that algebraic fact does not define a product of quantum fields at one point. A continuum field is generally a distribution. It acts on a smooth test function $f$:

$$\phi(f)=\int \phi(x)f(x)\,d^dx,$$

where the integral is distributional notation. Multiplying two distributions at the same point requires additional control.

For gauge theory, distinguish a smeared local composite such as a renormalized action density from a Wilson loop along an extended contour. The contour geometry, representation and regularization matter for the latter. A convergence statement about one kind of observable does not automatically give convergence of the other.

The Jaffe–Witten description, §4 and its footnote on printed p. 6 explicitly retains renormalization subtleties in the correspondence between classical gauge-invariant polynomials and quantum operators.

Worked example: subtracting the mean is only a first step

Let $X_\epsilon$ be a centered Gaussian variable with variance $v_\epsilon$, where $\epsilon$ denotes a regulator. Define the centered square

$$Y_\epsilon=X_\epsilon^2-v_\epsilon.$$

The Gaussian fourth moment is $\mathbb E X_\epsilon^4=3v_\epsilon^2$, so

$$\mathbb E Y_\epsilon=0,\qquad \mathbb E Y_\epsilon^2 =3v_\epsilon^2-2v_\epsilon^2+v_\epsilon^2 =2v_\epsilon^2.$$

If $v_\epsilon\to\infty$, centering removes the divergent expectation but not the divergent variance. Calling the subtraction “renormalization” does not establish convergence in a chosen space.

For two jointly Gaussian regulated fields at distinct points,

$$\mathbb E\!\left[ (\phi_\epsilon(x)^2-C_\epsilon(x,x)) (\phi_\epsilon(y)^2-C_\epsilon(y,y)) \right] =2C_\epsilon(x,y)^2.$$

This follows by the three pairings of the fourth moment: the disconnected pairing is subtracted, and two cross-pairings remain. If the right-hand side has a pointwise limit away from $x=y$, that still does not settle its extension to the diagonal as a distribution. In particular, this calculation is not an assertion that a centered pointwise square exists in every dimension.

State the renormalization prescription

A useful construction statement tells the reader what the regulated operator is, what subtractions or mixing are allowed, which normalization conditions fix finite terms, and in what topology its correlations converge. Operators with the same symmetries can mix. A finite change of normalization changes coefficients in a comparison, even when both prescriptions describe the same physics.

For a hypothetical sequence $O_{a,L}^{R}(f)$, an assertion might be convergence of every specified smeared $n$-point function as $a\to0$ and $L\to\infty$. It must also say whether test functions stay fixed in physical units, whether estimates are uniform in $L$, and whether positivity and covariance survive. Pointwise convergence at separated positions is a weaker statement.

Flow time is another scale

Gradient flow evolves a gauge field using an auxiliary smoothing parameter $t_{\mathrm f}$ with units of length squared. It is neither physical Euclidean time nor a sampling time. Even when flowed observables are useful and well behaved, a claim at fixed $t_{\mathrm f}>0$ is distinct from a claim about an unsmoothed local operator.

Lüscher, arXiv:1006.4518v3, §1 equations (1.1)–(1.4) and §3.4 equation (3.3) defines the flow and a reference scale through a flowed energy density. The positive-flow-time construction is a useful way to define and compare quantities; it does not by itself construct the unregulated four-dimensional target. Version v3 includes corrections recorded in the source history.

Exercise: identify the missing limit

An analysis fixes $\sqrt{8t_{\mathrm f}}=r$ in physical units, takes $a\to0$ and finds stable values of a flowed observable. Another analysis sets $t_{\mathrm f}=2a^2$ at every spacing. Are these the same limit?

Solution. In the first, the smoothing radius stays $r>0$ while the cutoff shrinks. In the second, the radius is $4a$ and tends to zero with the cutoff. Short-distance singularities can re-enter in the second limit; results proved uniformly only at fixed positive flow time cannot be transferred without a new estimate. The dimensionless ratio $t_{\mathrm f}/a^2$ distinguishes the two paths.

Continue with the limits visible

Read limits and uniformity to organize these parameters, then continuum scaling for reference-scale comparisons. When reading a research result, ask what observable is constructed before asking whether its formula resembles the desired theory.