3
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thank you! Could you refer to concrete equations? $\endgroup$
    – Nikita
    Feb 10 '20 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 '20 at 20:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.