For the convention in the question, a finite Yang-Mills gauge transformation acts covariantly on the field strength:or with and exchanged if the opposite convention is used for . Cyclicity of the matrix trace givesso the Yang-Mills theory Lagrangian is gauge invariant.
The quadratic gauge-field operator has zero directions along each gauge orbit. It therefore has no inverse on the full field space. Gauge fixing removes this degeneracy and produces a propagator, while the Faddeev-Popov determinant accounts for the corresponding Jacobian.
Requiring to transform as and using the gauge-field transformation law givesIndeed, differentiating produces two inhomogeneous derivative terms, and those cancel against the inhomogeneous part of the transformed connection.
At lowest derivative order, a general invariant effective Lagrangian through fourth order in the fields has the schematic formHere is any invariant symmetric rank-four tensor, the run over invariant contractions such as and , and the run over gauge- and Lorentz-invariant four-fermion contractions. The two sign symmetries forbid scalar cubic terms and Yukawa terms . The covariant kinetic terms automatically contain the allowed cubic and quartic interactions involving .
The canonical dimensions areand henceA coupling is relevant, marginal, or irrelevant when its dimension is positive, zero, or negative at the Gaussian fixed point. Thus gauge and scalar-quartic interactions are marginal in , the mixed two-scalar fermion bilinear is marginal in , and four-fermion interactions are marginal in . Quantum corrections replace this engineering classification near an interacting fixed point by the eigenvalues of its RG stability matrix.
The one-loop four-scalar one-particle-irreducible diagrams, together with permutations of the external scalar labels, are:
The first four are forced by the scalar covariant derivative and scalar potential. There is no ordinary Yukawa fermion box because the imposed symmetry forbids a one-scalar fermion vertex.
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