# Tag Info

## Hot answers tagged conformal-field-theory

3

We can have CFTs with $c \neq {\tilde c}$ as long as $$c - {\tilde c} \in 24 {\mathbb Z}$$ This condition arises from modular invariance of the CFT when it is put on the torus. PS - In radially quantized CFTs, the adjoint condition is $L_m^\dagger = L_{-m}$ and ${\bar L}_m^\dagger = {\bar L}_{-m}$.

2

Well while it has similarities with the OPE, it is more than that. In fact, it satisfies the OPE limit when $z\to w_j$ for any $j$, since the OPE you are talking about tells you only the singular terms, while there are also infinitely many non-singular terms, i.e., schematically $$T(z)\phi(w,\bar w)=\frac{h_\phi}{(z-w)^2}\phi(w,\bar w)+\frac{1}{z-w}\... 2 A more "pedestrian" application is that the low-energy physics of a Heisenberg antiferromagnetic spin chain is described by a WZW theory. This is a very simple and concrete model which has been shown to accurately describe many real materials. See for example http://arxiv.org/abs/hep-th/9802014v1, http://arxiv.org/abs/1211.5421v1, or section 7.10 of ... 2 Comments to the post (v2): Ref. 1 is considering the d-dimensional real Euclidean space (\mathbb{R}^d,|\cdot|^2) with the standard norm$$|x|^2~:=~\sum_{\mu=1}^d (x^{\mu})^2~=~\sum_{\mu,\nu=1}^d x^{\mu}\eta_{\mu\nu}x^{\nu}, \qquad \eta_{\mu\nu} ~=~{\rm diag}(1,\ldots, 1),\tag{A}$$and inner product$$\langle x ,y\rangle~:=~\sum_{\mu,\nu=1}^d x^{\mu}\...

1

Recall that the (global) conformal group is given by $${\rm Conf}(p,q)~\cong~O(p\!+\!1,q\!+\!1)/\{\pm {\bf 1} \},\tag{1}$$ cf. e.g. this Phys.SE post. Using the embedding $\imath: \mathbb{R}^{p,q}\hookrightarrow \overline{\mathbb{R}^{p,q}}$ into the conformal compactifification $\overline{\mathbb{R}^{p,q}}$, one may show after a short calculation that the ...

1

There is a certain difference between the Euclidean and Lorenzian cases. In Euclidean QFT the correlation functions, the Schwinger functions, are all in a sense time-ordered (see Osterwalder–Schrader axioms). Intuitively one can say that this is because all separations are space-like. Say, if you consider the correlation function of one scalar, they are ...

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