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I have three questions about the BRST symmetry in Polchinski's string theory vol I p. 126-127, which happen together

Given a path integral $$ \int [ d\phi_i dB_A db_A d c^{\alpha}] \exp(-S_1-S_2-S_3) \tag{4.2.3}$$ with $$ S_2 =-iB_A F^A (\phi) \tag{4.2.4}$$ $$ S_3 = b_A c^{\alpha} \delta_{\alpha} F^A(\phi) \tag{4.2.5} $$

the BRST transformation $$ \delta_B \phi_i = -i \epsilon c^{\alpha} \delta_{\alpha} \phi_i \tag{4.2.6a} $$ $$ \delta_B B_A=0 \tag{4.2.6b} $$ $$ \delta_B b_A = \epsilon B_A \tag{4.2.6c} $$ $$ \delta_B c^{\alpha} = \frac{i}{2} \epsilon f^{\alpha}_{\beta \gamma} c^{\beta} c^{\gamma} \tag{4.2.6d} $$

It is said

There is a conserved ghost number which is $+1$ for $c^{\alpha}$, $-1$ for $b_A$ and $\epsilon$, and 0 for all other fields.

How to see that?

The variation of $S_2$ cancels the variation of $b_A$ in $S_3$

I could get $i B_A \delta F^A$ in $\delta S_2$ and $(\delta_B b_A) c^{\alpha} \delta_{\alpha} F^A= \epsilon B_A c^{\alpha} \delta_{\alpha} F^A $ in $\delta S_3$. But there is a $c^{\alpha}$ in $\delta S_3$

the variations of $\delta_{\alpha} F^A$ and $c^{\alpha}$ in $S_3$ cancel.

How to see that?

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1 Answer 1

up vote 3 down vote accepted

answer for questions $1$,$2$, and $3a$

1) Looking at $2.7.22$ to $2.7.24$ (and also $2.7.18abc$), one define the ghost number $N^g = \frac{-1}{2\pi i} \int_0^{2 \pi}dz :b(z)c(z):$ , and that all the operators $c_n$ increase the ghost number by one. $[N^g,c_m] = c_m$, So the field $c(z)$, made of operators $c_m$, increase the ghost number by one unit.So $c^\alpha$ "has" ghost number +1.

Now, $B_A$ and $F^A$ do not change the nature of ghost states, so their ghost number is zero.

The total ghost number of the (Faddeev-Popov) action $S_3$ is zero, so $b_A$ and $c^\alpha$ must have opposite ghost number, so $b_A$ has ghost number $-1$

Finally, looking for instance, at equation $4.2.6c$, the ghost number must be the same for the two sides of the equation, so the ghost number of $\epsilon$ is $-1$ (you can check with the other equations $4.2.6x$ that it works)

2) We have $S_2 =-iB_A F^A (\phi) $. So

$\delta_BS_2 = -iB_A\delta_B F^A(\Phi)= -iB_A(\partial^i F^A) \delta_B \Phi_i = -iB_A(\partial^i F^A)(-i \epsilon c^{\alpha} \delta_{\alpha} \phi_i)$

$=-\epsilon B_Ac^\alpha (\partial^i F^A)\delta_{\alpha} \phi_i = -\epsilon B_Ac^\alpha \delta_\alpha F^A$

(Note that $\epsilon$ commute with the $B_A, F^A$)

The variation of $S_3$ due to $b_A$ is :

$\delta_B S_3 = \delta_B (b_A) c^{\alpha} \delta_{\alpha} F^A(\phi) = \epsilon B_Ac^\alpha \delta_\alpha F^A$

So The variation of $S_2$ cancels the variation of $b_A$ in $S_3$

[EDIT]

3)a) The variation of $S_3$ relatively to $c^\alpha$ is :

$b_A(\delta_B c^\alpha) (\delta_\alpha F^A) = b_A\frac{i}{2} \epsilon f^\alpha_{\beta\gamma}c^\beta c^\gamma(\delta_\alpha F^A) = b_A\frac{i}{2}\epsilon~ c^\beta c^\gamma [\partial_\beta, \partial_\gamma]( F^A) =b_A i\epsilon~c^\beta c^\gamma \partial_\beta \partial_\gamma( F^A)=0$

We have used $4.2.1$, and the fact that $c^\beta c^\gamma = - c^\gamma c^\beta $

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@user26143 : I have made an edit for the variation of $S_3$ relatively to $c^\alpha$ –  Trimok Aug 19 '13 at 15:11
    
In $\delta_BS_2 = -iB_A\delta_B F^A(\Phi)= -iB_A(\partial^i F^A) \delta_B \Phi_i $, is here a Taylor expansion? If yes, does $\phi_i$ stand for $\phi^i$? And does transformation (4.2.6a) apply to $i$-th power of $\phi$? or I mistaken? –  user26143 Aug 19 '13 at 17:18
1  
@user26143 : a) It is just the chain rule for partial derivatives, it is not a Taylor expansion, $i$ is not a power. b) Yes, for each function $F(\Phi)$, you can write $\delta_B F(\Phi) = - i\epsilon c^\alpha \delta_\alpha F(\Phi)$ –  Trimok Aug 19 '13 at 17:45
    
Thank you very much! –  user26143 Aug 19 '13 at 18:57

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