# Non-zero Euclidean commutator in 2D CFT?

In a Euclidean QFT, commutators of operators vanish for any spacetime separation. This can be argued very simply by using the path integral representation of the correlator, wherein operators become simple functions and hence can be easily moved around inside the integral.

Now, in a 2d CFT the two point correlator of a primary operator $$\mathcal{O}$$ with conformal weights $$h$$ and $$\bar{h}$$ looks like

$$\langle\mathcal{O}(z_1,\bar{z}_1)\mathcal{O}(z_2,\bar{z_2})\rangle=\frac{C}{(z_1-z_2)^{2h}(\bar{z}_1-\bar{z}_2)^{2\bar{h}}}$$ where $$C$$ is some normalizing constant.

We can exchange $$z_1$$ and $$z_2$$ in the above formula by rotating $$z_1$$ around $$z_2$$ by $$\pi$$: $$(z_1-z_2)\to (z_1-z_2) e^{i\pi},(\bar{z}_1-\bar{z}_2)\to (\bar{z}_1-\bar{z}_2) e^{-i\pi}$$

$$\langle\mathcal{O}(z_2,\bar{z}_2)\mathcal{O}(z_1,\bar{z_1})\rangle=e^{\pm 2\pi i s}\frac{C}{(z_1-z_2)^{2h}(\bar{z}_1-\bar{z}_2)^{2\bar{h}}}$$

where $$s=h-\bar{h}$$ is the spin of $$\mathcal{O}$$ and $$\pm$$ depends on the choice of the branch cut for the power functions.

Thus the commutator is

$$\langle[\mathcal{O}(z_1,\bar{z}_1),\mathcal{O}(z_2,\bar{z_2})]\rangle=\frac{C(1-e^{\pm 2\pi i s})}{(z_1-z_2)^{2h}(\bar{z}_1-\bar{z}_2)^{2\bar{h}}}$$

Clearly, the commutator is non-zero unless $$s \in \mathbb{Z}$$, which is inconsistent with our general expectation. What am I missing?

• The extra factor is $e^{\pm 2\pi i s}$ I believe, which then implies that the operators commute if $s \in {\mathbb Z}$ (and anti-commute if $s \in {\mathbb Z}+\frac{1}{2}$). Commented Feb 23, 2020 at 4:39
• @Prahar You are right! I missed the 2.
– user122637
Commented Feb 23, 2020 at 4:42
• You are missing nothing. Non-integer spins make commutators nonzero. Non-half-integer spin moreover make correlators multivalued. Commented Feb 23, 2020 at 23:06