I have read The spin and weight of a primary field in CFT but it does not answer my question, short of a restatement of the question itself. So I hope this post does not risk being removed..

In Ginsparg's Applied Conformal Field Theory, as well as the book on Conformal Field Theory by di Francesco et al, the holomorphic and anti-holomorphic dimensions $h$ and $\bar{h}$ determine the ``spin'' $s = h - \bar{h}$. The justification given by Ginsparg is that the generator $\mathrm{i}(l_0 - \bar{l}_0)$ of $\mathrm{PSL}(2, \mathbb{C})$ is a generator of rotations in the complex plane, and working in the (simultaneous) eigenbasis of $l_0$ and $\bar{l}_0$, the corresponding eigenvalue is $h-\bar{h}$ which thus has the interpretation of spin.

When I first read this, I assumed that "spin" is just a term here. But when considering free bosonic or fermionic CFTs, one makes the association of the spin of the particle with this difference in holomorphic and anti-holomorphic conformal dimension.

Making an analogy with quantum mechanics, or field theory, I can see that orbital angular momentum could be connected to rotation, and by extension, so could spin. But it is not clear to me why this spin degree of freedom must necessarily correspond to the physical spin of the particle.

Further, what ensures that $h-\bar{h}$ will always be an integer or half integer? What ensures that it'll be non-negative?


1 Answer 1


So, as I now understand, it is (obviously) only in two dimensions that $h$ and $\bar{h}$ have the interpretation of the weights corresponding to left- and right- movers. And it is therefore clear that $h-\bar{h}$ is the spin.

However, the spin-statistics theorem says that spin is either integer or half-integer, so it isn't quite clear how $(h-\bar{h})$ is constrained to be an integer or half-integer, even for quasi-primary fields.

Edit: see comments.

  • 1
    $\begingroup$ I don't think it is. E.g. weights of the primary operator $:e^{i k X}:$ are both equal to $k^2 / 2$, where $k$ is basically arbitrary. In fact, I don't think that spin/statistics theorem is applicable here since we are dealing with infinite-dimensional representations (fields) of the infinite-dimensional conformal group. But I am not completely sure about this second part. $\endgroup$ Mar 17, 2015 at 11:43
  • $\begingroup$ Yes. So the spin statistics theorem is inapplicable here. The Lorentz group has all 1-dimensional representations. $\endgroup$ Mar 22, 2015 at 0:31

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.