Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I met a problem of derving the Weyl transformation on the closed-string tachyon, Eq. (3.6.8) in Polchinski's string theory, vol 1, p 103.

Given the vertax operator of the closed-string tachyon

$$V_0 = 2g_c \int d^2 \sigma g^{1/2} e^{ik \cdot X} \tag{3.6.1}$$

It is said the Weyl transformation of Eq. (3.6.1) is

$$\delta_W V_0 = 2g_c \int d^2 \sigma g^{1/2} \left( 2 \delta \omega( \sigma) - \frac{k^2}{2} \delta_W \Delta (\sigma, \sigma) \right) [ e^{ik \cdot X(\sigma)} ]_r \tag{3.6.8}$$

here renormalized operator $[]_r$ and $\Delta (\sigma, \sigma)$ are defined as

$$ [ \mathcal{F}]_r = \exp \left( \frac{1}{2} \int d^2 \sigma d^2 \sigma' \Delta(\sigma, \sigma') \frac{ \delta}{\delta X^{\mu}(\sigma)} \frac{ \delta}{\delta X_{\mu}(\sigma')} \right) \mathcal{F} \tag{3.6.5} $$ $$ \Delta(\sigma,\sigma') = \frac{ \alpha'}{2} \ln d^2(\sigma,\sigma') \tag{3.6.6} $$ where $d(\sigma,\sigma')$ is the geodesic distance between points $\sigma$ and $\sigma'$.

I can get the first term in the big parathesis $()$ of Eq. (3.6.8), namely consider $$ \delta g^{1/2} = \frac{1}{2} \frac{1}{g^{1/2}} \delta g= \frac{1}{2} \frac{1}{g^{1/2}} g g^{ab} \delta g_{ab} = \frac{1}{2} \frac{1}{g^{1/2}} g g^{ab} 2 \delta \omega g_{ab} = 2 g^{1/2} \delta \omega $$ I try to get the second term in the big parathesis of Eq. (3.6.8) $$ \delta e^{i k \cdot X} = e^{i k \cdot X} i k^{ab} X^{ab} \delta g_{ab} = e^{i k \cdot X} i k^{ab} X^{ab} 2 \delta \omega g_{ab} = 2 i k \cdot X e^{ik \cdot X} \delta \omega,$$

but where do the $\Delta(\sigma,\sigma)$ and renormalized opeator $ [ e^{ik \cdot X(\sigma)} ]_r $ in (3.6.8) come from?

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

I think it works like this : Looking at the Weyl dependence of renormalized operators:

$$\delta_W[ \mathcal{F}]_r = [ \delta_W \mathcal{F}]_r + \frac{1}{2} \int d^2 \sigma ~~d^2 \sigma' \delta_W \Delta(\sigma, \sigma') \frac{ \delta}{\delta X^{\mu}(\sigma)} \frac{ \delta}{\delta X_{\mu}(\sigma')} [ \mathcal{F}]_r \tag{3.6.7}$$

We apply this formula to the operator $ \mathcal{F} = e^{i k.X(\sigma'')}$.

There is no explicit Weyl dependence of $e^{i k.X(\sigma'')}$, so the first term $ [ \delta_W e^{i k.X(\sigma'')}]_r$ is zero.

The second term is, using the fact that we keep only the $\sigma = \sigma' =\sigma''$ terms because $e^{i k.X(\sigma'')}$ depends only of $\sigma''$:

$$\frac{1}{2} \delta_W \Delta(\sigma'', \sigma'') (-ik^\mu)(-ik_\mu) [ e^{i k.X(\sigma'')}]_r$$

So, finally :

$$\delta_W[ e^{i k.X(\sigma'')}]_r = -\frac{ k^2}{2} \delta_W \Delta(\sigma'', \sigma'') [ e^{i k.X(\sigma'')}]_r \tag {1}$$

Now, we have :

$$V_0 = 2g_c \int d^2 \sigma g^{1/2} [e^{ik \cdot X(\sigma)}]_r \tag{2}$$

You already get the the Weyl dependence of $V_0$ relatively to $g^{\frac{1}{2}}$.

The Weyl dependence of $V_0$ relatively to $[e^{ik \cdot X(\sigma)}]_r$ is :

$$\delta_W V_0 = 2g_c \int d^2 \sigma g^{1/2} \delta_W[e^{ik \cdot X(\sigma)}]_r\tag{3}$$

So, finally, this dependence is, using $(1)$ :

$$\delta_W V_0 = 2g_c \int d^2 \sigma g^{1/2} (-\frac{ k^2}{2} \delta_W \Delta(\sigma, \sigma) [ e^{i k.X(\sigma)}]_r)\tag{4}$$

Remark : The formula $(3.6.7)$ comes directly from the formula $(3.6.5)$ and could be interpreted like this: you have 2 terms, the first corresponds to an explicit Weyl dependence of the operator, and the second term to a Weyl dependence via $\Delta(\sigma, \sigma')$

share|improve this answer
    
Thank you once more for your help! I have two questions (1) $V_0$ in the first line of (3.6.1) has no renormalization symbol $[]_r$, your Eq. (1) starts from $\delta_W [e^{i k \cdot X}]_r$, where does the $[]_r$ arise? (2) $e^{i k \cdot X}$ has no explicit Weyl dependence, is that because $k \cdot X = k^{\mu} X^{\nu} \eta_{\mu\nu}$, the metric is for the embeded space, not the 1+1 dimension one $\gamma_{ab} (1.2.21)$? –  user26143 Aug 5 '13 at 14:56
    
(1) The renormalization operation is a generalization of the normal ordering (see text p. 102 just before (3.6.5) The equation (3.6.1) is written with normal ordering$:e^{ik.X(\sigma)}:$, and we generalize it with renormalized operators $:[e^{ik.X(\sigma)}]_r$. Equation (1) is just equation $(3.6.7)$ with $\mathcal{F} = e^{i k.X(\sigma)}$, so you have a renormalized indice $_r$. –  Trimok Aug 5 '13 at 15:50
    
(2) You are perfectly right. There is no dependence on the worldsheet metrics for $ e^{ik.X(\sigma)}$ –  Trimok Aug 5 '13 at 15:51
    
Thank you very much! –  user26143 Aug 5 '13 at 15:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.