# OPE in linear dilaton theory

I'm trying to do the following question from David Tong's problem sheets on string theory:

A theory of a free scalar field has OPE $$\partial X(z)\partial X(w) = \frac{\alpha'}{2}\frac{1}{(z-w)^2}+...$$ Consider the putative candidate for the stress energy tensor $$T(z) = \frac{1}{\alpha '}: \partial X (z) \partial X (z) : -Q \partial^2 X(z).$$ Use $$TX$$ OPE to determine the transformation of $$X$$ under conformal transformations $$\delta z = \epsilon(z)$$

Now to determine $$T(z)X(w)$$, I thought I would contract the normally ordered field with $$X(w)$$. So to get:

$$2\langle \partial X (z) \partial X (w) \rangle -Q^2 \partial^3 X(z).$$

Is that the correct way of proceeding? I'm not sure if I need to contract the not-normally ordered fields $$\partial^2 X(z)$$ with $$X(w)$$ as well?

Also, I don't quite understand how would I continue with this question after I worked out the OPE of $$TX$$.

So, from p.73 of Tong's notes you can see that an operator $$O(w)$$ should transform under $$\delta z =\epsilon(z)$$ by equation 4.12: $$\delta O(w) = -Res[\epsilon(z)T(z)O(w)]$$ meaning that knowing an operator's OPE with the stress tensor is knowing how it transforms under $$\delta z$$. So first we must calculate the OPE $$T(z)O(w,\bar{w})$$. The way we do OPEs is we sum over $$\textbf{all}$$ allowed contractions between the operators (but not between normal ordered operators). The OPE you're trying to do is $$T(z)X(w) = \left(\frac{1}{a'}:\partial X(z)\partial X(z): - Q\partial^2 X(z)\right)X(w)$$ so in the first term you can either contract $$X(w)$$ with the first $$\partial X(z)$$ or with the second, but not $$\partial X(z)$$ with $$\partial X(z)$$, so that will give you the same term (as I think you have already ghessed) therefore you can write the first term as $$(2/a'):\partial X(z):\langle \partial X(z)X(w)\rangle$$ where the normal ordering is now redundant. The second term only has one possible contraction and that is $$-Q\langle\partial^2 X(z)X(w)\rangle$$.
To evaluate these contractions you need $$\langle X(z) X(w)\rangle = \frac{a'}{2}\ln(z-w)$$ which can be differentiated to give $$\langle \partial X(z)X(w)\rangle = \partial_z\langle X(z) X(w)\rangle = \partial_z \frac{a'}{2}\ln(z-w) = \frac{a'}{2}\frac{1}{z-w}$$ $$\langle\partial^2 X(z)X(w)\rangle = \partial_z^2\frac{a'}{2}\ln(z-w) = -\frac{a'}{2}\frac{1}{(z-w)^2}.$$
The last ingredient you need to get only $$(w)$$ dependance on the RHS of $$T(z)O(w,\bar{w})$$ is to Laurent expand $$\partial X(z)$$ around $$w$$ by $$\partial X(z) = \partial X(w) + (z-w)\partial^2X(w)+O(z-w)^2$$. Sometimes this expansion will give you more singular terms but in this case we only need it to first order.
Putting everything together we see that $$T(z)X(w) = \frac{Qa'/2}{(z-w)^2}+\frac{\partial X(w)}{z-w}+n.s$$ meaning that $$X(w)$$ does not transform like a primary operator (because there is no $$X(w)$$ in the numerator of the first term, as you will probably see later there is a better candidate for a primary operator).
All that is left now is to calculate the residue of this OPE with $$\epsilon(z) = \epsilon(w) + (z-w)\partial \epsilon(w)+...$$ $$\delta X(w) = -Res\Big[(\epsilon(w) + (z-w)\partial \epsilon(w)+...)\left( \frac{Qa'/2}{(z-w)^2}+\frac{\partial X(w)}{z-w}+n.s \right)\Big]$$ $$=-Res\Big[ \frac{\partial\epsilon(w)Qa'/2+\epsilon(w)\partial X(w)}{z-w} \Big]$$ $$=-\frac{Qa'}{2}\partial\epsilon(w)-\epsilon(w)\partial X(w)$$