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I'm trying to calculate the energy-momentum tensor of the $bc$ and $\beta\gamma$ CFT's, using a simple method of calculating the canonical expression and adding total derivatives (rather than the methods in this and this posts), I can reproduce the energy momentum tensor of $bc$ CFT, but not of the $\beta\gamma$ CFT. I follow the equations in Polchinski volume 1.

The action of the $bc$ CFT is given by Eq. (2.5.4): $$ S=\frac{1}{2\pi}\intop d^{2}zb\bar{\partial}c.\tag{2.5.4} $$ The canonical energy momentum tensor is given by: (Francesco (2.165)) $$ T_{\mu\nu}=\frac{\partial L}{\partial\left(\partial^{\mu}\Phi\right)}\partial_{\nu}\Phi-\eta_{\mu\nu}L\tag{2.165} $$ working in the $z,\bar{z}$ coordinate system, we have that $$T_{zz}=\frac{\partial L}{\partial\left(\bar{\partial}\Phi\right)}\partial\Phi$$ The variation of the Lagrangian with respect to $\bar{\partial}c$ is $$\frac{\partial L}{\partial\left(\bar{\partial}c\right)}=-b$$ (the minus sign is because $b$ and $c$ anti-commute), and we get that: $$T_{zz}=-b\partial c$$ This is also the result in Francesco Eq. (5.118). This is not the expression in Polchinski which is Eq. (2.5.11a) $$ T_{zz}=\partial bc-\lambda\partial\left(bc\right) $$ but we can add a total derivative $\left(1-\lambda\right)\partial\left(bc\right)$ to get: $$ T_{zz}=-b\partial c+\left(1-\lambda\right)\partial\left(bc\right)=-b\partial c+\partial bc+b\partial c-\lambda\partial\left(bc\right)=\partial bc-\lambda\partial\left(bc\right) $$ and this is the same as in Polchinski.

Trying to repeat the same steps for the $\beta\gamma$ CFT does not seem to work. The action is (Eq. (2.5.20)): $$ S=\frac{1}{2\pi}\intop d^{2}z\beta\bar{\partial}\gamma $$ The variation with respect to $\bar{\partial}\gamma$ is $\frac{\partial L}{\partial\left(\bar{\partial}c\right)}=\beta$ (without a minus sign, since $\beta$ and $\gamma$ commute), and so the energy momentum tensor becomes: $$ T_{zz}=\beta\partial\gamma. $$ (This is also the expression in arXiv:0708.4261 chapter 3, Eq. 6, page 6, though it is just from searching for searching the web, and I do not know if it is not a typo there). Adding a total derivative of $\left(-1-\lambda\right)\partial\left(\beta\gamma\right)$ will cause $$ T_{zz}=-\partial\beta\gamma-\lambda\partial\left(\beta\gamma\right) $$ which have a wrong minus sign compared to polcinski Eq. (2.5.23a): $$ T_{zz}=\partial\beta\gamma-\lambda\partial\left(\beta\gamma\right).\tag{2.5.23a} $$

Is there a mistake in my calculation?

If there is no mistake, why this simple approach of using the canonical expression of $T_{zz}$ and adding total derivatives works for the bc CFT but not for $\beta\gamma$? Was it pure luck in the first case?

Qmechanic
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Ronny
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1 Answers1

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Maybe you can calculate it by using the same trick in section 2.3, I will give my calculation process below. By a infinitesimal translation on the worldsheet $$ z \to z+\epsilon v(z,\bar{z}) $$ , $\beta$ and $\gamma$ transform as $$ \begin{aligned} \delta\beta=\epsilon(-\lambda (\partial v)\beta-v\partial \beta)\\ \delta\gamma=\epsilon(-(1-\lambda) (\partial v)\gamma-v\partial \gamma)\\ \end{aligned}$$ since they have conformal weight $h_\beta=(\lambda,0),h_\gamma=(1-\lambda,0)$. Now by using E.O.M of $\gamma$ we can throw out the contribution from the variation of $\beta$ and leaving only $$ \delta S= \frac{1}{2\pi}\int d^2z \beta \bar{\partial}(\delta\gamma) $$ With a careful calculation and again using the E.O.M of $\gamma$, one gets $$\begin{aligned} \delta S=\frac{-\epsilon}{2\pi}\int d^2z(1-\lambda)\beta\gamma(\partial\bar{\partial}v)+\beta\partial \gamma(\bar{\partial}v)\\ =\frac{\epsilon}{2\pi i}\int d^2z(-i)\left[\partial \beta\gamma-\lambda \partial(\beta\gamma)\right]\bar{\partial}v \end{aligned} $$ where I have used some IBP in this calculation, and note that I never exchange the position of $\beta$ and $\gamma$. Now you can easily see that by definition $j_a=i v^b T_{ab}$, we get the answer.

say
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