Derivation of $T(z)(TT)(w)$ in CFT I am trying to derive eq. (6.213) in Di Francesco's CFT book,
$$T(z)(TT)(w) \sim \frac{3c}{(z-w)^6}+\frac{(8+c)T(w)}{(z-w)^4}+\frac{3 \partial T(w)}{(z-w)^3}+\frac{4 (TT)(w)}{(z-w)^2}+\frac{\partial(TT)(w)}{z-w}.\tag{6.213}$$
I must have made an error, since my answer was
$$T(z)(TT)(w) \sim\frac{-3 c}{(w-z)^6}-\frac{8T(w)}{(w-z)^4}+\frac{5\partial T(w)}{(w-z)^3}-\frac{\partial^2 T(w)}{(w-z)}.$$
I have two confusions:

*

*In his OPE for $\partial T(x) T(w)$ (eq. 6.212) he has a term $\partial(TT)(w).$ Where does this come from? If you differentiate the $TT$ OPE's from earlier in the book you don't get this term because the expansion is truncated at the non-singular terms. Why this time do we keep going?

*He does not write the OPE for $T(x)\partial T(w)$ but just says you can get it from differentiation. However when I differentiate I get a $\partial^2 T(w)$ term that he must not have obtained. Why is this? What rule of differentiating OPEs am I breaking to get that term?

 A: Some introductory sources are too quick to say that only singular terms matter in chiral OPEs. Yes, meromorphic functions are determined by their singularities but it may well be the case that one needs a finite number of regular terms in order to use the OPEs during intermediate steps. It is convenient to start by finding out what this number is.
Looking at equation (6.159), singular terms in the final answer will arise in three ways.

*

*Going up to the first regular term in the $\partial T(x) T(w)$ OPE.


*Going up to the second regular term in the $T(x) T(w)$ OPE in the first line.


*Keeping only the first regular term in all OPEs of the second line.
The regular term in $A(x)B(w)$ is by definition $(AB)(w)$ so the second line yields
\begin{equation}
\frac{\frac{c}{2} T(w)}{(z - w)^4} + \frac{2 (TT)(w)}{(z - w)^2} + \frac{(T\partial T)(w)}{z - w}. \quad\quad (1)
\end{equation}
To tackle the first line, we plug in the known OPEs to get
\begin{align}
\frac{\frac{c}{2} T(w)}{(x - w) (z - x)^4} + \frac{c}{(x - w)^5 (z - x)^2} + \frac{4T(w)}{(x - w)^3 (z - x)^2} + \frac{2\partial T(w)}{(x - w)^2 (z - x)^2} + \frac{2(TT)(w) + (x - w) Y}{(x - w) (z - x)^2} - \frac{2c}{(x - w)^6 (z - x)} - \frac{4T(w)}{(x - w)^4 (z - x)} - \frac{\partial T(w)}{(x - w)^3 (z - x)} + \frac{(\partial TT)(w)}{(x - w) (z - x)}. \quad\quad (2)
\end{align}
I am being lazy with the $Y$ because I don't want to figure out what the second regular term in $T(x) T(w)$ is but we will come back to that.
Integrating (2) and adding it to (1), we get exactly the desired answer up to the $\frac{1}{z - w}$ term because we haven't solved for $Y$ yet. But this term needs to be the derivative of whatever $T(z)$ is acting on because the stress tensor generates translations. For more about how certain terms in stress tensor OPEs are universal (even when nothing is a Virasoro primary), see a previous question.
