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?

  • 1
    $\begingroup$ 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 $s \in {\mathbb Z}+\frac{1}{2}$). $\endgroup$
    – Prahar
    Feb 23, 2020 at 4:39
  • $\begingroup$ @Prahar You are right! I missed the 2. $\endgroup$
    – user122637
    Feb 23, 2020 at 4:42
  • 2
    $\begingroup$ You are missing nothing. Non-integer spins make commutators nonzero. Non-half-integer spin moreover make correlators multivalued. $\endgroup$ Feb 23, 2020 at 23:06

1 Answer 1


What you wrote is a correct explanation of why fields do not commute if their spins are not integer. A field with half-integer spin is called fermionic, and fermionic CFTs are the subject of a recent article by Runkel and Watts https://arxiv.org/abs/2001.05055. Fields with more general fractional spins are called parafermionic. Parafermionic fields not only do not commute, but also have multivalued correlation functions.

This illustrates the fact that in the axiomatic ("bootstrap") approach to CFT, most axioms can be relaxed, giving rise to generalizations. In this case, relaxing commutativity gives rise to fermionic and parafermionic CFTs.

  • $\begingroup$ I see that Euclidean correlators are multivalued for parafermionic operators. That means corresponding Lorentzian correlators (obtained by analytic continuation) are also multivalued. But real time correlators are physical observables and should be unambiguous, right? Or is it that the ambiguity is only in an overall phase which anyway cannot be discerned in experiments? $\endgroup$
    – user122637
    Feb 24, 2020 at 19:04
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    $\begingroup$ Good question. I am not sure that parafermions can be analytically continued to Lorentzian signature. Euclidean CFTs that are not guaranteed to make sense in the Lorentzian also include non-unitary CFTs. $\endgroup$ Feb 24, 2020 at 19:34

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