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I am a little confused about the way commutators are derived in the radial quantization of a 2D CFT. I am trying to derive the relation $$\int_w \mathcal{R} \left(a(z)b(w)\right)dz = \left[A,b(w)\right]$$ where $A = \int a(z) dz.$

In order to clarify my question, I will follow the steps in reverse compared to what is done in the book "Conformal Field Theory" by Francesco and other authors.

Let's say we have two operators $a(z)$ and $b(w)$. Let us fix the point $w$ somewhere in the plane. Define two circular contours $C_1$ and $C_2$ around the origin such that the point $w$ sits in the annular region made by $C_1$ and $C_2$. Now define two integrals $\int_{C_1}a(z)b(w)dz$ and $\int_{C_2}b(w)a(z)dz$ and subtract them (contours shown in fig 1) as $$\int_{C_1}a(z)b(w)dz - \int_{C_2}b(w)a(z)dz.$$

enter image description here

Then morph the loops from fig 1 into those of fig 2 by moving the red and blue loops close to each other without changing the order of the operators (in order to maintain radial ordering). Finally, I break the integrals of fig 2 into two pieces, one piece is shown in fig 3 and the other one is in fig 4.

My question is this: Why do we neglect the integral corresponding to fig 4? In the limiting case where the blue arc $C_4$ and the red arc $C_5$ come very close, the integral corresponding to figure 4 should be $$\int_{C_5 \to C_4} [a(z),b(w)]dz.$$ Isn't the integral of fig 4 the major contributor to the commutator $\left[A,b(w)\right]$? If so, wouldn't the contribution of integral in fig 3 go to zero in the limit where $C_3$ shrinks to point $w$ because then fig 4 will become the entire commutator?

Yaman Sanghavi
  • 950

2 Answers2

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The OPE $${\cal R} a(z)b(w)~=~\ldots$$ is often a regular function for $z\neq w$, so then the contour in OP's Fig. 4 is zero. In fact the OPE is typically a truncated Laurent series in $z$ around $w$, cf. Ref. 1.

References:

  1. P. Di Francesco, P. Mathieu and D. Senechal, CFT, 1997; subsection 6.1.2 and eq. (5.72).
Qmechanic
  • 201,751
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Thanks to Qmechanic for the answer. I just want to clarify it a little more. I was having the impression that the OPEs are derived after going through this "radial ordered contour integral to commutator" procedure. But this is not correct. It is the other way around instead; we use a property of OPEs here, specifically the fact that the OPE of $\mathcal{R}(a(z)b(w))$ is regular at all points $\neq w$

OPEs are calculated independently of this "radial ordered contour integral to commutator" procedure. For example, we calculate the OPE $\partial_z \phi \partial_w \phi$ using the fact that the correlation function is $<\partial_z \phi \partial_w \phi> = \frac{1}{(z-w)^2}$. Since OPEs are radial ordered and since they are often regular at all points except $w$, we can firmly say that the integral corresponding to fig 4 i.e. $$\int_{C_5 \to C_4} [a(z),b(w)]dz = \oint_{C_4+C_5} \mathcal{R}(a(z)b(w))$$ has to vanish regardless of whether $[a(z),b(w)]$ is non-zero on the contour $C_5 \to C_4$.

But what about the limit when $C_3$ shrinks? Wouldn't the integral $$\int_{C_5 \to C_4} [a(z),b(w)]dz$$ become exactly equal to the commutator? In that case, there will be another contribution to fig 4 which was irrelevant up till now. That term is the green part in the figure below. Once $C_3$ starts to shrink, this green "almost" circular contour will start to get a significant contribution which would be exactly enough to cancel out the $\int_{C_5 \to C_4} [a(z),b(w)]dz$enter image description here

When $C_3$ was a finite-sized circle, although the green contour should have been there in principle, it was easily neglected since we could make $C_4$ and $C_5$ very close to each other and because of the finite size of $C_3$, the green contour would be significantly far from the singularity $w$. Once we shrink $C_3$, the green contour comes closer to $w$ and gets a significant value.

Yaman Sanghavi
  • 950