In theory of free scalar field $$ S = \int d^2x \;\partial_\mu \phi \partial^\mu \phi $$ $$ \langle \phi(z) \phi(w)\rangle \propto \ln(z-w) $$

Exist family of energy-momentum tensors (new term correspond to non-minimal coupling to gravity $\alpha R \phi^2$, example with linear coupling $R\phi$): $$ T_{\mu\nu} = \partial_\mu \phi \partial_\nu \phi + \alpha (\partial_\mu\partial_\nu - \eta_{\mu\nu}\Box) \phi^2 $$ Obviously $\partial^\mu T_{\mu\nu} = 0$, but $T_\mu^\mu \neq 0$ if $\alpha \neq 0$. And $$ T = T_{zz} =\partial_z \phi \partial_z \phi + \alpha (\partial_z\partial_z)\phi^2 = (1+2\alpha)\partial_z \phi \partial_z \phi + 2\alpha \phi \partial_z^2\phi $$

So this shift will affect standard OPE: $$ T(z)T(w) = \frac{1/2}{(z-w)^4} + \frac{2T(w)}{(z-w)^2} + \frac{\partial T(w)}{(z-w)} + \dots $$

To (I'm interested only in first term): $$ T(z)T(w) = \frac{1/2 (1+6\alpha+10\alpha^2)}{(z-w)^4} + \frac{-1/2 \ln(z-w)}{(z-w)^4} + \frac{2T(w)}{(z-w)^2} (1+??) + \frac{\partial T(w)}{(z-w)}(1+??) + \dots $$

And if one choose for example $\alpha = 1$, one obtain central charge $c = 17$ for free scalar field.. What about ln term in OPE?

Could someone to clarify and interpretate this confusion? Why we say that $c=1$ for free scalar filed?


1 Answer 1


The free scalar field can have any complex central charge, due to the linear modification of the energy-momentum tensor that you mention. The case $c=1$ is a bit special, in particular it allows compactification with an arbitrary radius, while for general $c$ the radius is quantized. See Section 4.1 of my review article for more details: https://arxiv.org/abs/1406.4290

In the context of string theory, the free scalar with $c\neq 1$ is sometimes called the linear dilaton theory.

  • $\begingroup$ Thank you! Could you refer to concrete equations? $\endgroup$
    – Nikita
    Feb 10, 2020 at 8:36
  • 1
    $\begingroup$ The tensor $T$ is given in (4.1.2). The extra term is linear. You use a quadratic extra term, which cannot be written in terms of the current $J=\partial \phi$ and its derivatives. Presumably this is why you get the $\ln$ term, which apparently breaks conformal symmetry, since your $TT$ OPE does not agree with the standard $TT$ OPE (2.2.10). Using the linear extra term, the central charge is given in terms of the coefficient $Q$ by (2.2.20). $\endgroup$ Feb 10, 2020 at 20:27

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