Factor of $1/2$ in $TT$-OPE [closed]

I'm trying to calculate the TT OPE in a bosonic theory. I'm missing a factor of 1/2 in the least-singular term. We have (following Di Francesco)

$$\langle \partial \phi(z) \partial \phi(0) \rangle = \frac{-1}{4 \pi g}\frac{1}{z^2}$$

and

$$T(z) = -2 \pi g \, \colon \partial \phi \partial \phi \, \colon.$$

Performing Wick contractions, I'm able to obtain

$$T(z) T(0) = \frac{1/2}{z^4} - \frac{4 \pi g}{z^2}\, \colon \partial\phi(z) \partial \phi(0) \, \colon .$$

Now I want to expand the last normal-ordered term around $$0$$, but whereas I would write

$$\colon \partial\phi(z) \partial \phi(0) \, \colon = \, \colon \partial\phi(0) \partial \phi(0) \, \colon + z \, \partial \, \colon \partial\phi(0) \partial \phi(0) \, \colon$$

and obtain the incorrect OPE

$$T(z) T(0) = \frac{1/2}{z^4} + \frac{2 T(0)}{z^2} + \frac{2 \partial T(0)}{z},$$

$$T(z) T(0) = \frac{1/2}{z^4} + \frac{2 T(0)}{z^2} + \frac{ \partial T(0)}{z}.$$

I must be Taylor-expanding the normal-ordered product incorrectly. Can someone walk me through the steps of doing it the right way?

closed as off-topic by knzhou, Aaron Stevens, Cosmas Zachos, tpg2114♦Jul 19 at 12:29

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When you expand the normal ordered term, you have \begin{align} :\partial \phi(z) \partial \phi(0): &= :[ \partial \phi(0) + z \partial^2 \phi(0) ]\partial \phi(0): \\ &= : \partial \phi(0) \partial \phi(0): + z : \partial^2 \phi(0) \partial \phi(0): \\ &= : \partial \phi(0) \partial \phi(0): + \frac{z}{2} \partial \left( :\partial \phi(0) \partial \phi(0): \right) \\ &= T(0) + \frac{z}{2} \partial T(0). \end{align}
• Sorry to belabor this, but could you explain whence $\colon \partial^2 \phi(0) \partial \phi(0) \colon = \frac{1}{2} \partial \colon \partial \phi(0) \partial \phi(0) \colon$? Thanks so much! – Diffycue Jul 6 at 16:04
• For the same reason that $d(x^2) = 2x dx$ so that $x dx = \frac{1}{2} d (x^2)$. – Prahar Jul 6 at 16:05
• $(1/2 f^2)' = ff'$ ? – mike stone Jul 6 at 16:06