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Given the action $$ \mathcal{S_g}=\frac{1}{2\pi}\int{d^2x\left(b\bar{\partial}c+\bar{b}{\partial}\bar{c}\right)} $$ where $b$ and $c$ are ghosts. How can I calculate the energy-momentum tensor$$ T=2(\partial c)b+c\partial b? $$

I figure we should use the standard $$ T^\mu_\nu=\frac{\delta\mathcal{L}}{\delta(\partial_\mu\phi)} \partial_\nu\phi-\delta^\mu_\nu\phi$$ but in 2D this would reduce to the given form.

I'm not quite sure how to calculate that.

EDIT: The partials are defined as $$ \partial:=\partial_z\;\;\;\;\;\;\;\;\bar{\partial}:=\partial_{\bar{z}} $$ where $$ z:= \frac{1}{\sqrt{2}}(\sigma+\tau) \;\;\;\;\;\;\;\;\bar{z}:=\frac{1}{\sqrt{2}}(\sigma-\tau) $$ and $\sigma,\tau$ are the coordinates in our 2D CFT.

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If you wish to determine the stress tensor from the action, you will need to know how the $bc$ ghost system couples to gravity. To figure this out, we recall that the action arose from the BRST gauge fixing of the string path integral. If you follow through that procedure, we find the action $$ S_{gh} = \frac{1}{2\pi} \int d^2 \sigma \sqrt{g} b_{ab} \nabla^a c^b $$ where $b_{ab}$ is traceless and symmetric.

Let us verify that this reduces to the correct action if we set $g = e^{2\phi(z ,\bar z)} dz d{\bar z}$. Then, $d^2 \sigma \sqrt{g} = \frac{1}{2} d^2 z e^{2\phi}$ and $b = b_{zz}$ and ${\bar b} = b_{\bar z\bar z}$ are the only independent components. Similarly, we define $c = c^z$ and ${\bar c} = c^{\bar z}$. Note that $b$ has weight $(2,0)$ and $c$ has weight $(-1,0)$ as expected. Further $$ \nabla^z c^z = 2 e^{-2\phi} {\bar \partial} c . $$ Thus, $$ S_{gh} = \frac{1}{2\pi} \int d^2 z \left[ b {\bar \partial} c + {\bar b} \partial {\bar c}\right] $$ Good, so we have the correct action. The equations of motion are $$ \nabla^a b_{ab} = 0 , \qquad \nabla^a c^b + \nabla^b c^a - g^{ab} \nabla_c c^c = 0 . $$ We can now determine the stress tensor by varying the action w.r.t. the metric. Importantly, we need to preserve the tracelessness of $b_{ab}$ so we must have $$ \delta (g^{ab} b_{ab}) = 0 \quad \implies \quad g^{ab} \delta b_{ab} = - \delta g^{ab} b_{ab} . $$ Thus, we must vary $b_{ab}$ as well, while preserving the equation above. Then, we find $$ T_{ab} = - \frac{4\pi}{\sqrt{g}} \frac{ \delta S_{gh} }{ \delta g^{ab}} = - [ \nabla_c b_{ab} c^c + b_{ac} \nabla_b c^c + b_{bc} \nabla_a c^c ] . $$ It is easy to see that the stress tensor is traceless. We now work it out in a flat metric, $g = dz d{\bar z}$. Then, $$ T = T_{zz} = 2 \partial c b + c \partial b . $$


We may also determine the stress tensor without discussing the coupling of the $bc$ ghost system with gravity. The idea is to note that $T$ must be linear in $b$ and $c$ (since the action is linear in both $b$ and $c$). Further, $T$ is a $(2,0)$ operator so the most general form of the stress tensor is $$ T = \alpha \partial c b + \beta c \partial b $$ The constants $\alpha$ and $\beta$ are then fixed by computing the OPE of $T$ with $b$ and $c$ and requiring that $b$ and $c$ are primary operators with weights $(2,0)$ and $(-1,0)$ respectively.

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