When does the conformal dimension in 2d CFT have to be integer?

Question

I am quite confused. In all books on CFT it is stated that the conformal dimension $h$ of the chiral quasi-primary field has to be integer or half-integer. This seems to be crucial in certain proofs. See, for example, (6.142) in di Francesco, authors use $l+n+h_A=0$, where $l$ and $n$ are integers.

Also, if $h$ is not an integer, it's quite hard to imagine what $\phi(0)|0\rangle$ is; assuming that $$\phi(z)=\sum_{m\in \mathbb{Z}}z^{-m-h}\phi_m$$

However, as the discussion of Verma modules and minimal models starts, we begin considering the HWSs of $L_0$, the vectors $|h\rangle$ with arbitrary real values of $h$. I minimal models, $h$ takes certain rational values.

What am I missing? Are the fields corresponding to these states non-primary? When $h$ is integer and when is not?

Answer 1

Conformal dimensions do not have to be integer. Only the difference between the left and right dimensions, i.e. the conformal spin, should be integer for correlation functions to be single-valued.

In the Section 6.5 of di Francesco et al, all you need for defining normal ordering is that the OPE (6.124) of the two fields $A$ and $B$ involves integer powers of their separation. This assumption is independent from $A$ and $B$ having integer conformal dimensions. Rather, it means $$ h_{(AB)}\in h_A + h_B + \mathbb{Z} $$

Then there is an attempt to rewrite the normal ordering in terms of modes. In the mode decompositions (6.140), a priori $n+h_A$ and $p+h_B$ should be integer, but $n,p,h_A,h_B$ could be non-integer. Similarly, the mode numbers $m$ of $(AB)$ (as in $(AB)_m$) need not be integer, but $m+h_{(AB)}$ should be integer.

None of this relies on fields being primary or quasi-primary. (In general, conformal dimensions are defined as eigenvalues of $L_0$.)