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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 ...


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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) ...


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I'm just learning this myself, but for the the first one, the thermal state I think just means that if you throw any field in the resulting space-time, it will immediately acquire the specified temperature. In Hawking's original calculation, he shows that this radiation will be dominated by the massless, lowest-spin particles available, which in our universe ...


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I would recommend the book Introduction to Conformal Field theory by Blumenhagen and Plauschinn. It is quite sort and can serve as a perfect introduction to CFT. It covers the basics of CFT in the first 3 chapters and then in the remaining 3 it goes on to introduce the CFT concepts that will appear most frequently in String theory. I believe the content of ...


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Both $h$ and $\tilde{h}$ are usually called weights. Their sum, $\Delta=h+\tilde{h}$ is the (scaling) dimension of the operator, while the difference, $s=h-\tilde{h}$ is called the spin. This is due to their association with scale transformations (dilatations) and rotations, respectively. To see this, note that the dilatation operator is given by ...


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One comes to this conclusion due to the fact that the contraction of a symmetric tensor with an antisymmetric one vanishes. Writing down the loop diagrams involves a contraction of both vertices. If you get expressions proportional to $\epsilon_{\mu\nu}\eta^{\mu\nu}$, this will be zero due to the fact that the metric is symmetric and the epsilon tensor is ...



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