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Yang–MillsA QFT.ORG PROJECT

What you need to know

This is a route finder, not an entrance examination. You can begin lesson 1 while using the explanations below. Familiarity with QED, a non-Abelian field strength, perturbation theory and renormalization is enough to start. Measure theory, spectral measures and constructive field theory are introduced here when needed.

After this page: identify a physical observable, subtract a vacuum contribution, distinguish a regulator from a physical parameter, and choose a mathematical bridge for one unfamiliar step.

Prerequisites: ordinary quantum mechanics, linear algebra and the QFT background described on Start. All examples below are teaching calculations, not calculations of a Yang-Mills mass.

1. Which quantity survives a gauge transformation?

Section titled “1. Which quantity survives a gauge transformation?”

Suppose a matrix-valued curvature transforms as Fμν(x)↦Ω(x)Fμν(x)Ω(x)−1F_{\mu\nu}(x)\mapsto \Omega(x)F_{\mu\nu}(x)\Omega(x)^{-1}. Is a matrix entry of FμνF_{\mu\nu} invariant? Is tr⁡FμνFμν\operatorname{tr}F_{\mu\nu}F_{\mu\nu}?

Work it out. A matrix entry usually changes under conjugation. But multiplication cancels the adjacent Ω−1Ω\Omega^{-1}\Omega, and cyclicity gives

tr⁡(ΩFμνFμνΩ−1)=tr⁡(FμνFμν).\operatorname{tr}\bigl(\Omega F_{\mu\nu}F_{\mu\nu}\Omega^{-1}\bigr) =\operatorname{tr}(F_{\mu\nu}F_{\mu\nu}).

This establishes a classical invariance statement. In a quantum theory the product at a point still needs a definition, usually involving smearing and composite-operator renormalization. A familiar symbol is not already a constructed operator.

Recovery route: physical states, then renormalized observables. The official problem asks for appropriate local gauge-invariant quantum fields; see Jaffe and Witten, section 4.

Consider the deliberately simple spectral model

⟨O(t)O(0)⟩=9+2e−t+e−3t,⟨O⟩=3,\langle O(t)O(0)\rangle=9+2e^{-t}+e^{-3t}, \qquad \langle O\rangle=3,

where time and energies use reciprocal units. The connected correlator is

C(t)=⟨O(t)O(0)⟩−⟨O⟩2=2e−t+e−3t.C(t)=\langle O(t)O(0)\rangle-\langle O\rangle^2 =2e^{-t}+e^{-3t}.

Without subtraction, the constant term dominates at large time and an effective decay rate tends to zero. After subtraction the asymptotic rate is 1. That rate is the lowest energy visible to this operator. Another state of energy 1/21/2 with zero overlap would be absent from this expression.

Recovery route: Euclidean correlators and operator overlap. If the replacement of a finite sum by an integral feels unfamiliar, use the spectral theorem bridge.

Use the site’s convention Dμ=∂μ−iAμD_\mu=\partial_\mu-iA_\mu, with the coupling outside the curvature. In dd dimensions, [Aμ]=1[A_\mu]=1 and [Fμν]=2[F_{\mu\nu}]=2 in mass units. Since

S=14g2∫ddx FμνaFμνaS=\frac{1}{4g^2}\int d^dx\,F_{\mu\nu}^aF_{\mu\nu}^a

is dimensionless, [g2]=4−d[g^2]=4-d. Thus g2g^2 has dimension one in three dimensions and zero in four. This dimensional statement alone determines neither a mass nor a nonperturbative spectrum. A source that places gg inside FF uses different engineering dimensions for its gauge potential.

Recovery route: conventions, notation crosswalk and dimensional transmutation. Tong’s Gauge Theory notes, section 2.1.1 provide broader physical background; they are optional recovery material.

4. How does Euclidean evolution select energies?

Section titled “4. How does Euclidean evolution select energies?”

Let H=diag⁡(0,2,5)H=\operatorname{diag}(0,2,5) and ψ=(0,1,1)/2\psi=(0,1,1)/\sqrt2. Matrix exponentiation gives

e−tH=diag⁡(1,e−2t,e−5t),⟨ψ,e−tHψ⟩=12e−2t+12e−5t.e^{-tH}=\operatorname{diag}(1,e^{-2t},e^{-5t}), \qquad \langle\psi,e^{-tH}\psi\rangle =\tfrac12 e^{-2t}+\tfrac12 e^{-5t}.

The lowest energy dominates at long positive time because its exponential decays most slowly. You have used the finite-dimensional version of the spectral theorem. Infinite-dimensional field theory adds continuous spectra and operator-domain issues; the idea survives but the hypotheses matter.

Recovery route: spectral theorem and semigroups. You need not know those theorems before beginning the course.

A spatial lattice with spacing aa and NsN_s sites has physical side length L=NsaL=N_sa. Suppose aa is halved and NsN_s doubled. The number of variables increases, the cutoff changes, and LL stays fixed. This is a step toward finer resolution, not toward unbounded physical space.

If a physical mass stays at m=1m=1, its lattice value amam is halved too. A vanishing lattice mass along that trajectory is compatible with a nonzero physical mass. The continuum lab makes the two controls independent.

Take H=diag⁡(0,1,4)H=\operatorname{diag}(0,1,4) and OΩ=(2,0,3)O\Omega=(2,0,3). What does vacuum subtraction leave? What mass can this operator see?

Hint: remove the first component before applying e−tHe^{-tH}.

Solution: (1−P0)OΩ=(0,0,3)(1-P_0)O\Omega=(0,0,3), so the connected spectral expression is 9e−4t9e^{-4t}. It sees energy 4 and misses the state at energy 1. A perfect fit does not establish that the full gap is 4.

If you can explain why that answer differs from the lowest eigenvalue, you have the central reading habit for this course: state both what the calculation establishes and what it cannot see. Continue to lesson 1 or organize a reading group.