For a $bc$ CFT (see e.g. Polchinski's string theory, section 2.5) given by \begin{equation} S=\frac 1{2\pi}\int dz^2 b \overline \partial c \end{equation} the energy-momentum tensor is: \begin{equation} T(z) = :(\partial b) c: - \lambda \partial (:bc:) \end{equation} where $\lambda$ is the weight of $b$.

I can't find a definition of $\partial:bc:$ that gives the correct result for the $Tb$ OPE: \begin{equation} T(z) b(0) \sim \frac \lambda {z^2} b(0) + \frac 1z \partial b(0) \end{equation}

My idea was that \begin{align} \partial :bc: &= \lim_{h\to0}\lim_{z_{12}\to0} \frac 1h \left[ b(z+z_{12}+h)c(z+h) - \frac 1{z_{12}} - b(z+z_{12})c(z) + \frac 1{z_{12}} \right] = \\ &=\lim_{z_{12}\to0} \lim_{h\to0} \frac 1h \left[ b(z+z_{12}+h)c(z+h)- b(z+z_{12})c(z) \right] = \\ &= \partial(bc) \end{align} Now, either the starting point is wrong, or exchanging the two limits is not allowed, because if so: \begin{align} T(z)b(0) &= :(\partial b)c: b(0) - \lambda \partial( :bc: ) b(0) \\ &= :(\partial b)c: b(0) - \lambda \partial( bc ) b(0) \\ &= :(\partial b)c: b(0) - \lambda [(\partial b(z)) c(z) b(0) + b(z) \partial (c(z) b(0)) ] \\ &\sim \frac 1z {\partial b(0)} - \frac \lambda z {\partial b(0)} + \frac \lambda {z^2} b(0) \\ &\sim \frac \lambda {z^2} b(0) + \frac 1z {\partial b(0)} (1-\lambda) \end{align}

A similar question was asked here, but I could not find how to apply the answer to the $Tb$ OPE.


1 Answer 1


The derivative of $: bc :$ is just that... the derivative of a normal ordered product. Perhaps there is a sense in which the derivative kills $bc(z) - :bc:(z)$ because it's an infinite constant thus allowing you to drop the colons. But if so, this would be a coincidence coming from the fact that $b$ and $c$ are low enough in dimension to only have one singular term in their OPE. More importantly, Wick's theorem is setup so that you can keep the normal ordering all the way through and not have to play these games.

Consider $$ :\partial b c :(z) b(0) \equiv \frac{1}{2\pi i} \oint \frac{dw}{w - z} \partial b(w) c(z) b(0) $$ which is one of the OPEs you have to compute. Fusing $c(z)$ and $b(0)$ in the integrand, the complicated terms which would yield nested normal ordered products all come with regular powers of $z$. Therefore for the calculation we're interested in, the above becomes \begin{align} :\partial b c :(z) b(0) &= \frac{1}{2\pi i} \oint \frac{dw}{w - z} \partial b(w) \left [ \frac{1}{z} + \dots \right ] = \frac{\partial b(z)}{z} + \dots \quad (1) \end{align} which is one of the terms you wrote down. For the exact same reason, $$ :bc:(z) b(0) = \frac{b(z)}{z} + \dots $$ which comes up in your second term. You just apply $-\lambda \partial$ to the above and $$ -\lambda \partial :bc:(z) b(0) = \lambda \frac{b(z)}{z^2} - \lambda \frac{\partial b(z)}{z} + \dots \quad (2) $$ is the result.

Now let's try adding (1) and (2) to get \begin{align} T(z) b(0) &= \lambda \frac{b(z)}{z^2} + (1 - \lambda) \frac{\partial b(z)}{z} + \dots \\ &= \lambda \frac{b(0)}{z^2} + \frac{\partial b(0)}{z} + \dots \end{align} where in the second line, we have Taylor expanded in order to have the operators in the numerator evaluated at the origin. In other words, your mistake had nothing to do with subtleties in the definition of $\partial :bc:$ and everything to do with accidentally writing $b(0)$ in a step that needed $b(z)$.

  • $\begingroup$ I see.. I didn't think of Taylor expanding before setting the argument to zero. Thank you for the detailed answer, it was really helpful! $\endgroup$
    – agc
    Dec 21, 2021 at 20:52

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