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  Spin-statistics theorem in axiomatic (Wightman) CFT

+ 2 like - 0 dislike

In Wightman QFT one can prove spin-statistics theorem rigorously.  I was wondering, whether a similar result holds in 2D CFT, namely, that $h,\bar{h}\in \mathbb{Z}$ for bosons and $h, \bar{h}\in\frac{1}{2}\mathbb{Z}$ for fermions?

Or put it differently, since we know that for spin $s$ it holds $s\in\mathbb{Z}$ or $s\in\frac{1}{2}\mathbb{Z}$ by spin-statistics theorem, do we also now that the scaling dimension $d$ is also either $d\in\mathbb{Z}$ or $d\in\frac{1}{2}\mathbb{Z}$? Then the result would follow from $d=h+\bar{h}$ and $s= h-\bar{h}$.

asked Mar 26, 2016 in Mathematics by Gytis (10 points) [ no revision ]

In 2D there is no spin-statistics theorem. This is because of the Boson-Fermion correspondence. In 2D one often has also nonstandard statistics described in terms of braid groups rather than permutation groups. scaling dimensions need not be half-integral.

@ArnoldNeumaier Thank you for reminding me of this. I have actually forgotten of this, and now I feel dumb. But still, if I take a book on CFT, e.g. Blumenhagen, Plauschinn "Intro to CFT", their example of a free boson on a cylinder on p. 46 shows that the associated fields have dimensions $(h,\bar{h}) = (1,0)$ and $(h,\bar{h})= (0,1)$. Moreover, on p. 58 they show that a free fermion has dimensions $(h,\bar{h}) = (1/2, 0)$ and $(h,\bar{h})= (0,1/2)$. The central charges of the theories are $c=1$ and $c=1/2$, respectively, which is also suggestive. Furthermore, a mathematical text like Kac "Vertex algebras for beginners" also introduces a supercommutator on p. 6 "to include fermions".

More precisely, I am mostly interested in such statement: if we use only commutators, does this imply that $h$ and $\bar{h}$ (or $d$ and $s$ as given above) are integers?

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