# What do we mean by radial quantisation in CFT?

When we quantise QFT we do that in equal time slices. In CFT it is useful to use equal radius slices. Why is that the case? And what does it mean?

• – Qmechanic Feb 17 '20 at 17:13

The starting point is to consider a CFT on a sphere, $$S^{d-1}$$ so the manifold under consideration is the "cylinder" $$S^{d-1} \times {\mathbb R}$$ with metric $$ds^2 =- dt^2 + R^2 d\Omega_{d-1}^2$$ The next step is to perform a Wick rotation. Here, we rotate $$t \to - i \tau$$. This step is useful for many reasons, not least of which is that it brings all coordinates to an equal setting and the Euclidean path integral is convergent. The Wick rotated space is $$ds^2 = d\tau^2 + R^2 d\Omega_{d-1}^2.$$ Next, we change coordinates and define $$\tau = R \log(r/R)$$ so the metric takes the form $$ds^2 = R^2 \frac{dr^2}{r^2} + R^2 d\Omega_{d-1}^2 = \frac{R^2}{r^2} \left[ dr^2 + r^2 d\Omega_{d-1}^2 \right] .$$ The metric in the bracket is that of Euclidean flat spacetime, $${\mathbb R}^d$$.
So far, what I have done works for any quantum field theory. We now utilize the fact that our theory is conformal. This implies that the dependence of correlators on the Weyl factor is trivial (by which I mean, fixed by conformal invariance). We can therefore scale out the Weyl factor in the metric above, and consider the theory only on flat spacetime, $$d{\tilde s}^2 = dr^2 + r^2 d\Omega_{d-1}^2 = \delta_{\mu\nu} dx^\mu dx^\nu .$$ We are now ready to answer your question. In the flat metric described above, what is time? Well, the notion of time is determined from our starting point where time was described by the coordinate $$t$$ (to be precise, what I mean is that I am studying the theory from the perspective of an observer $$O$$ moving along the worldline with tangent $$k=\partial_t$$) which Wick rotates to $$\tau$$ which I coordinate transformed to $$r$$. Thus, on the final Euclidean plane, the observer $$O$$ experiences time to be $$r$$ (to be precise, $$O$$ travels along a worldline with tangent $$k = r \partial_r$$). We therefore quantize the theory w.r.t. constant $$r$$ slices. The final Euclidean plane described above is therefore called the "radial plane".
To summarize, A CFT on $$S^{d-1}\times {\mathbb R}$$ quantized on equal time slices can be described equivalently in terms of a CFT on $${\mathbb R}^d$$ quantized on equal radius slices.
You may also be wondering why we should be interested in CFTs on $$S^{d-1}$$ and not $${\mathbb R}^{d-1}$$ as one typically does in QFT. The problem is that CFTs are plagued with infrared divergences which arise from the infinite volume of $${\mathbb R}^{d-1}$$. To regulate these divergences, we start by considering CFTs on $$S^{d-1}$$ with radius $$R$$ and then take $$R \to \infty$$. In a QFT, the infrared divergences are regulated by giving fields a "mass", but this regulator breaks conformal invariance.
• Thanks! On your last comment, you say that CFT's are defined on $S^{d-1}$ with radius $R\rightarrow\infty$, else the regulation of infrared divergences break conformal invariance. Does that mean that CFT's can only be defined on AdS spaces (which nevertheless can be transformed into deSitter spaces?) – redhood Feb 17 '20 at 17:21
• I am not saying that CFTs can only be defined on $S^{d-1}$. I am only saying that it is - for some purposes - a good starting point. It depends a lot on what property of the CFT one would like to study. For instance, CFTs on $S^d$ (no time) are interesting if you want study free energies or entanglement entropies. – Prahar Mitra Feb 17 '20 at 17:25