Radial ordered commutation relation

Question

In the book Conformal Field Theory of Francesco, Mathieu and Sénéchal, in Sec. 6.1.2, the authors state that the integral $$ \oint_w \mathrm{d}z~ a(z)b(w) ~=~ \oint_{C_1} \mathrm{d}z~ a(z)b(w) - \oint_{C_2} \mathrm{d} z ~b(w)a(z)\tag{6.15a} $$ can be seen as a commutator $$~=~[A,b(w)],\tag{6.15b} $$ where $$A~=~\oint a(z)\mathrm{d}z\tag{6.16}$$ is "the integral over space at fixed time (i.e., a contour integral of the field $a(z)$)". I feel confused about the meaning of $A$: If $A$ is a contour integral, what is the contour of the integral? The "fixed time" seems not specified in the definition of $A$.

Answer 1

1. Perhaps it would be more clear if Ref. 1 used the following notation: $$\begin{align}\oint_{|z-w|=\varepsilon}& \mathrm{d}z~{\cal R} a(z)b(w) \tag{6.15a}\cr ~=~& \oint_{|z|=|w|+\varepsilon} \mathrm{d}z~ a(z)b(w)\cr ~-~& (-1)^{|a||b|}\oint_{|z|=|w|-\varepsilon} \mathrm{d} z ~b(w)a(z)\tag{6.15b} \cr ~=~&A(|w|\!+\!\varepsilon) b(w)-(-1)^{|a||b|}b(w)A(|w|\!-\!\varepsilon) \tag{6.15c}\cr ~=:~&~[A(|w|),b(w)]_{SC},\tag{6.15d} \end{align}$$ where we have defined $$A(R)~:=~\oint_{|z|=R} \mathrm{d}z~a(z),\tag{6.16}$$ and $[\cdot,\cdot]_{SC}$ is a regularized super-commutator. Here $|a|$ and $|b|$ denote the Grassmann-parity of the operators $a$ and $b$, respectively.

2. In eq. (6.15a) the symbol ${\cal R}$ denotes radial ordering, $${\cal R} a(z)b(w)~:=~\left\{ \begin{array}{rcl} a(z)b(w)&{\rm for}&|z|>|w|, \cr (-1)^{|a||b|}b(w)a(z)&{\rm for}&|w|>|z|.\end{array}\right. $$
   The symbol ${\cal R}$ itself is often implicitly implied in CFT texts.

3. The non-radial-ordered OPE $a(z)b(w)$ is typically not well-defined/divergent for $|z|<|w|$. Therefore the formula $$\begin{align}{\cal R} a(z)b(w) ~=~&\theta(|z|\!-\!|w|)a(z)b(w)\cr ~+~&(-1)^{|a||b|}\theta(|w|\!-\!|z|)b(w)a(z)\end{align}$$ only makes sense if we define that "zero times ill-defined is zero".

4. The radial-ordered OPE ${\cal R} a(z)b(w)$ is typically a meromorphic function (possible with branch cuts). Integration contours can be deformed as long as they don't cross the position of other operator insertions, cf. Cauchy's integral theorem. The commutator (6.15d) is formally singular. It is regularized via point-splitting (6.15c).

5. Example: The holomorphic part of the bosonic string has non-radial-ordered OPE $$ X(z)X(w)~=~ -\frac{\alpha^{\prime}}{2}{\rm Ln} (z-w) \quad {\rm for} \quad |z|>|w|. $$ The radial-ordered OPE $${\cal R} X(z)X(w)~=~\left\{ \begin{array}{rcl} -\frac{\alpha^{\prime}}{2}{\rm Ln} (z-w)&{\rm for}&|z|>|w|, \cr -\frac{\alpha^{\prime}}{2}{\rm Ln} (w-z)&{\rm for}&|w|>|z|,\end{array}\right. $$ has a $\pm\pi i\alpha^{\prime}$ branch cut along $|z|=|w|$ because of the complex logarithm ${\rm Ln}$. This branch cut disappears when we consider derivatives of $X$.

References:

1. P. Di Francesco, P. Mathieu and D. Senechal, CFT, 1997; subsection 6.1.2.

2. D. Simmons-Duffin, TASI Lectures on the Conformal Bootstrap, arXiv:1602.07982; eq. (12).

Answer 2

It might be useful to recall that we work here under radial ordering, where the modulus of a complex arguments is related to the "time" coordinate. So one has the following definition: $$R(A(z)B(w)) := \left\{\begin{array}{ll} A(z)B(w) & \textrm{for}\;|z|>|w|\\ B(w)A(z) &\textrm{for}\;|w|>|z| \end{array}\right.$$ Then one can right $$\oint dz [A(z),B(w)] = \oint_{|z|>|w|} dz A(z)B(w) - \oint_{|z|<|w|} dz B(w)A(z)$$ So if you observe the contours of the right side expression: $$\oint dz [A(z),B(w)] = \oint_{C(w)}\; dz R(A(z)B(w))$$ The contour of the Integral needs to enclose $w$ but is otherwise arbitrary. It is not specified since its particular shape doesn't affect the result for holomorphic functions. You can check the details for this in the book by Blumenhagen and Plauschin "Introduction to CFT" which I believe has a nice explanation on all this.