# Expressions of action and energy momentum tensor in bc conformal field with central charge equals one

I have a question with conformal field theory in Polchinski's string theory vol 1 p. 51.

For $bc$ conformal field theory $$S=\frac{1}{2\pi} \int d^2 z b \bar{\partial} c$$ $$T(z)= :(\partial b) c: - \lambda \partial (: bc:)$$ with central charge $c=-3 (2 \lambda-1)^2 +1 =1$. Introducing $\psi$ and $\bar{\psi}$ to replace the anticommuting fields $b$ and $c$ as following $$b \rightarrow \psi =2^{-1/2} (\psi_1 + i \psi_2 )$$ and $$c \rightarrow \bar{\psi} =2^{-1/2} (\psi_1 - i \psi_2)$$ It is claimed that $$S=\frac{1}{4\pi} \int d^2 z \psi_1 \bar{\partial} \psi_1 + \psi_2\bar{\partial} \psi_2 (2.5.18b)$$ $$T=- \frac{1}{2} \psi_1 \partial \psi_1 -\frac{1}{2} \psi_2 \partial \psi_2 (2.5.18c)$$

I cannot obtain the above expressions of $S$ and $T$. Here is my derivations. First I try to recover the anti-commuting characters of fields $b$ and $c$ by $\psi_1$ and $\psi_2$. For $$bc+cb=0$$ I have $$\psi_1 \psi_1 + \psi_2 \psi_2 =0$$ Then for the action $$S=\frac{1}{4\pi} \int d^2 z \left( \psi_1 \bar{\partial} \psi_1 + i \psi_2 \bar{\partial} \psi_1 -i \psi_1 \bar{\partial} \psi_2 + \psi_2 \bar{\partial} \psi_2 \right)$$ (1) [Solved] Why the term $i \psi_2 \bar{\partial} \psi_1 -i \psi_1 \bar{\partial} \psi_2$ does not contribute to the action?

(2) How to derive (2.5.18c)?

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You know, it's completely fine for you to ask multiple questions about Polchinski in the course of your studying; there's really no need to start every question with a disclaimer like "another stupid question from..." :) –  joshphysics Jul 22 '13 at 23:02
Thank you! I will try! –  user26143 Jul 22 '13 at 23:19
For (1), you could try integration by parts and anti-commutativity of the fermions. –  Heidar Jul 22 '13 at 23:59
If $\psi_1$ and $\psi_2$ anti commutes, then I can let $i \psi_2 \bar{\partial} \psi_1 -i \psi_1 \bar{\partial} \psi_2$ disappears. But from $\{b,c\}=0$ I only get $\psi_1 \psi_1 + \psi_2 \psi_2=0$. –  user26143 Jul 23 '13 at 0:19
Because $b$ is the canonical momentum for $c$ - the derivative of the Lagrangian with respect to $\partial_0 c$ - in the same way as $p$ is the momentum for $x$ in mechanics. The canonical quantization says that $[p,x]$ is nonzero much like $\{b,c\}$. It's similar for real fermions $\psi$ except that $\psi$ plays the role of its own momentum. –  Luboš Motl Jan 31 '14 at 13:17