String theory $bc$ system CFT

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$bc$ CFT is an example of free-field CFTs. action is

$$S = \frac{1}{2\pi} \int d^2 z b \bar{\partial} c$$

How can we obtain equation of motion? Polchinski-volume 1-page 50 writes that

$$\bar{\partial}c(z) = \bar{\partial}b(z)=0$$, $$\bar{\partial} b(z) c(0) = 2\pi \delta^2 (z,\bar{z})$$

If this is true so the action will become zero.

This post imported from StackExchange Physics at 2015-04-28 15:20 (UTC), posted by SE-user farhad
retagged Apr 28, 2015

This post imported from StackExchange Physics at 2015-04-28 15:20 (UTC), posted by SE-user Danu

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First of all, the variation of the action produces the equations of motion. Inserting the solution of the equations of motion into the action simple provides you with the value of the action when evaluated on the solution. There's no problem with it being zero. However, when you do this, you should always worry about boundary terms, these are irrelevant for the equations of motion, but they do affect the value of the action when evaluated on a solution (i.e. on-shell).

Secondly, to derive the equations of motion, take the usual variation of the action:

$$2\pi \delta S = \int d^2z \left( \delta b \bar{\partial} c + b \bar{\partial} \delta c \right) = \int d^2 z \left( \delta b \bar{\partial} c - \left(\bar{\partial} b \right) \delta c \right),$$

where in the second line we have integrated by parts. Therefore the variation $\delta S/\delta b = 0$ leads to the equation $\bar{\partial} c = 0$, and the variation $\delta S /\delta c = 0$ leads to $\bar{\partial} b = 0.$

The last equation you write is not an equation of motion. The $\delta$ function is just the usual contact term.

This post imported from StackExchange Physics at 2015-04-28 15:20 (UTC), posted by SE-user Surgical Commander
answered Apr 28, 2015 by (155 points)