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?


1 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 $L_0$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.