# Scales in Logarithmic CFTs

Logarithmic CFTs have OPEs (and operators) with logarithms. But to have logarithms one needs to have some scale to make the argument of the log a dimensionless quantity. But if the theory has a scale then how can that be a CFT? Standard CFT correlators don't have logs or exponentials precisely for that reason - they don't have any scale.

Standard CFT correlators do have logs since $z^\Delta=\exp (\Delta \log z)$. In logarithmic CFT the only difference is that $\Delta$ is a non-diagonalizable matrix rather than a number. If $\Delta$ is a matrix, the function $z^\Delta$ is still covariant under rescalings, in the sense that $(\lambda z)^\Delta=\lambda^\Delta z^\Delta$. Your logarithmic correlation functions are the matrix elements of $z^\Delta$, and they get mixed with one another by the multiplication with $\lambda^\Delta$.
Conclusion: the terms you get in logarithmic correlators when you do $z\to \lambda z$ are manifestations of the covariance under rescalings.
• One can always write $e^{\Delta}$ as $exp \, (\Delta \, log\, (\frac{z}{z_0}))$. But my question was what is the natural scale $z_0$ that goes into the argument of log and in that case why should it be called a conformal theory? – Physics Moron Apr 9 '18 at 14:49
• Any scale $z_0$ would do, and the choice of $z_0$ does not affect the physics. If you change $z_0$ in a correlation function, you only pick a simple overall factor that can be absorbed in field normalizations. In a CFT most fields are covariant under rescalings: only fields of dimension zero are invariant. – Sylvain Ribault Apr 10 '18 at 15:48