# Linear Dilaton CFT

I´m doing the exercises on the Tong lectures of String Theory, in particular Problem Sheet 2:

Consider the tensor: $T(z) = \frac{-1}{\alpha '} :\partial X(z) \partial X(z): - Q \partial^2 X$. By viewing the OPE of $TX$ show how $T$ transforms under an infinitesimal transformation $\delta z = \epsilon(z)$

So I'm trying to get the OPE:

$T(z)X(w) = [\frac{-1}{\alpha '} :\partial X(z) \partial X(z): - Q \partial^2 X] X(w)$

For the first part I have:

$\frac{-1}{\alpha '} :\partial X(z) \partial X(z): X(w) = \frac{-1}{\alpha '} 2\partial X(z) <\partial X(z) X(w) >$ Since I only have 2 possible simple contractions. Now I can use the propagator and tayloring $X(z) \sim X(w)$ i get: $\frac{-1}{\alpha '} :\partial X(z) \partial X(z): X(w) = \frac{\partial X(w)}{(z-w)}$

So far so good (I think), but for the other term I'm a little baffled:

$Q\partial^2 X(z) X(w) = Q [\partial (\partial X(z) X(w)) - \partial X(z) \partial X(w)]$ Where I seem to have a surface term, and the propagator. If I kill the surface term it looks like the OPE is:

$T(z) X(w) = \frac{\partial X(w)}{(z-w)} + \frac{Q \alpha'}{2(z-w)^2}$

Is this right? How do I know from here how does the $T(z)$ transforms?

• The question asks you to determine transformation of $X$, not $T$. – Prahar Oct 22 '15 at 14:03