5

In this article they call weight $(h,\bar{h})=(1,1)$ fields marginal.

Why are these fields called marginal? Why are they to be distinguished.

Qmechanic
  • 220,844

2 Answers2

10

This terminology comes from renormalization group flow, where one has relevant, marginal, and irrelevant operators.

In CFT, operators with conformal weight $(1, 1)$ are known as marginal operators. More generally, operators of conformal weight $(h, \bar{h})$ are said to be relevant if $h + \bar{h} < 2$ and irrelevant if $h + \bar{h} > 2$. A (necessarily marginal) operator that preserves conformality is called truly marginal, or exactly marginal, etc, cf. Ref. 1.

References:

  1. P. Ginsparg, Applied Conformal Field Theory, arXiv:hep-th/9108028; Section 8.6.
Qmechanic
  • 220,844
8

They are called marginal because they correspond to "slight deformations" of a CFT, which do not break conformal invariance. Given a marginal field $\phi$, one can add to the action a term

$$ \delta S = \Delta \int_\Sigma \phi$$

which is just the operator integrated over the worldsheet, modulated by a deformation parameter $\Delta$. Since the (2D) integration measure transforms exactly oppositely to the marginal field, this term is conformally invariant, and hence produces a new theory that is still conformally invariant.

Therefore, it is possible to study new CFTs by producing them by disturbing those already known - the idea is to "probe" the CFT landscape starting from "nice" theories, e.g. rational CFTs.

ACuriousMind
  • 132,081