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Excuse me, I have calculated $a^g$ a lot of times, using the relation between $:\;:$ and ${}^{{}_\circ}_{{}^\circ} \; {}^{{}_\circ}_{{}^\circ}$. But I can't get the same result with the book. It is not too hard to get $$ :b(z)c(z'):-{}^{{}_\circ}_{{}^\circ} b(z)c(z'){}^{{}_\circ}_{{}^\circ} = \frac{\left(\frac{z'}{z}\right)^{\lambda-1}-1}{z-z'}. $$ Which can be found in Problem 2.13 on $P_{76}$. Take the limit $z\rightarrow z'$, we find $$ :b(z)c(z):-{}^{{}_\circ}_{{}^\circ} b(z)c(z){}^{{}_\circ}_{{}^\circ} = \lim_{z\rightarrow z'}\frac{\left(\frac{z'}{z}\right)^{\lambda-1}-1}{z-z'}=\lim_{z\rightarrow z'}\frac{\left(1-\frac{z-z'}{z}\right)^{\lambda-1}-1}{z-z'}=\lim_{z\rightarrow z'}\frac{\left(1-(\lambda-1)\frac{z-z'}{z}\right)-1}{z-z'} =\frac{1-\lambda}{z}. $$ $$ \partial (:b(z)c(z):)-\partial ({}^{{}_\circ}_{{}^\circ} b(z)c(z){}^{{}_\circ}_{{}^\circ}) = -\frac{1-\lambda}{z^2} $$ and $$ :\partial b(z)c(z):-{}^{{}_\circ}_{{}^\circ} \partial b(z)c(z){}^{{}_\circ}_{{}^\circ} = \lim_{z\rightarrow z'}\partial\frac{\left(\frac{z'}{z}\right)^{\lambda-1}-1}{z-z'} =\lim_{z\rightarrow z'}\left(-\frac{\left(\frac{z'}{z}\right)^{\lambda-1}-1}{(z-z')^2}+\frac{(1-\lambda)\left(\frac{z'}{z}\right)^{\lambda-1}}{z(z-z')}\right) =\frac{(\lambda-1)^2}{z^2}. $$ The energy-momentum tensor is $T(z)=:(\partial b)c:-\lambda\partial(:bc:)$.

Using the above results, we can express $T(z)$ in creation-annihilation normal ordering ${}^{{}_\circ}_{{}^\circ}\;{}^{{}_\circ}_{{}^\circ}$ $$ T(z)={}^{{}_\circ}_{{}^\circ} \partial b(z)c(z){}^{{}_\circ}_{{}^\circ}+\frac{(\lambda-1)^2}{z^2}-\lambda\left(\partial ({}^{{}_\circ}_{{}^\circ} b(z)c(z){}^{{}_\circ}_{{}^\circ})-\frac{1-\lambda}{z^2}\right) ={}^{{}_\circ}_{{}^\circ} \partial b(z)c(z){}^{{}_\circ}_{{}^\circ}-\lambda\partial ({}^{{}_\circ}_{{}^\circ} b(z)c(z){}^{{}_\circ}_{{}^\circ})+\frac{1-\lambda}{z^2} $$ $$ \Rightarrow a^g=1-\lambda $$ But in Polchinski's book,$a^g=\frac{1}{2}\lambda(1-\lambda)$. What wrong with my derivation?

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Check your algebra in the part where you need to compare $(\partial b) c$ between the two normal orderings, I got $\lambda(\lambda-1)/(2z^2)$ for the difference. The result follows immediately.

(By the way, thanks for the symbol ${}^{{}_\circ}_{{}^\circ}$, I've been looking for that forever.)

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  • $\begingroup$ Yeah, you are right. Thank you so much! I have made a mistake. $\left(\frac{z'}{z}\right)^{\lambda-1}$ should approximatively equal to $1-(\lambda-1)\frac{z-z'}{z}+(\lambda-1)(\lambda-2)\frac{(z-z')^2}{2z^2}$. But I have taken only to $\frac{z-z'}{z}$ order. You are welcome for the symbol ${}^{{}_\circ}_{{}^\circ}$. $\endgroup$
    – user28108
    Aug 11, 2013 at 11:32

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