# A question about deriving Eq. (6.2.13) in Polchinski's string theory book volume 1

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I have a question about deriving Eq. (6.2.13) in Polchinski's string theory book volume I. It is claimed that

Now consider the path integral with a product of tachyon vertex operators, $$A_{S_{2}}^{n}(k,\sigma)=\left\langle [e^{ik_{1}\cdot X(\sigma_{1})}]_{r}[e^{ik_{2}\cdot X(\sigma_{2})}]_{r}\cdots[e^{ik_{n}\cdot X(\sigma_{n})}]_{r}\right\rangle _{S_{2}}\tag{6.2.11}$$ This corresponds to $$J(\sigma)=\sum_{i=1}^{n}k_{i}\delta^{2}(\sigma-\sigma_{i})\tag{6.2.12}$$ The amplitude (6.2.6) then becomes

$$A_{S_{2}}^{n}(k,\sigma)= iC_{S_{2}}^{X}(2\pi)^{d}\delta^{d}(\sum_{i}k_{i})\times ...$$

$...\exp(-\sum_{i<j} k_{i}\cdot k_{j}G'(\sigma_{i},\sigma_{j})-\frac{1}{2}\sum_{i=1}^{n}k_{i}^{2}G_{r}'(\sigma_{i},\sigma_{i}))\tag{6.2.13}$

where $C_{S_{2}}^{X}=X_{0}^{-d}(\det'\frac{-\nabla^{2}}{4\pi^{2}\alpha'})_{S_{2}}^{-d/2}$ and $G_{r}'(\sigma,\sigma')=G'(\sigma,\sigma')+\frac{\alpha'}{2}\ln d^{2}(\sigma,\sigma')$

Eq. (6.2.6) is

$$Z[J]=i(2\pi)^{d}\delta^{d}(J_{0})(\det'\frac{-\nabla^{2}}{4\pi^{2}\alpha'})^{-d/2}\times ...$$

$...\exp(-\frac{1}{2}\int d^{2}\sigma d^{2}\sigma'J(\sigma)\cdot J(\sigma')G'(\sigma,\sigma'))\tag{6.2.6}$

My question is: where do $X_0^{-d}$ and $G_r'$ come from in Eq. (6.2.13)? I could try to plug (6.2.12) into (6.2.6) to see all other term appears, but not $X_0^{-d}$ nor $G_r'$.

This post imported from StackExchange Physics at 2014-07-06 20:38 (UCT), posted by SE-user user26143

edited Jul 6, 2014

@user26143 there seems to be a problem with long equations, maybe an equation array could help? Not sure if it is a bug that could be fixed ...

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I fixed it by splitting the equations into separate lines manually, it's good enough to get by. The physics question is resolved by carefully dealing with the issue that the J-J propagator thing is singular when you go from a point to itself, which is presumably what the "r" subscript in this section does.

answered Jul 6, 2014 by (7,550 points)
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They come from different places. First, the $X_0^{-d}$ comes from the delta functions $\delta^d(J_0^{\mu})$.
$$J_0^{\mu}=\int_{M}d^{2}\sigma \,X_0 \left(\sum_{i=1}^{n}k_{i}^{\mu}\delta^{2}(\sigma-\sigma_i)\right)=X_0\sum_{i=1}^{n}k_i^{\mu}$$

Then, the deltas will be given by $\delta^d(X_0\sum_{i=1}^{n}k_i^{\mu})=X_0^{-d}\delta^d(\sum_{i=1}^{n}k_i^{\mu})$. This is how the $X_0^{-d}$ shows up.

Now, the renormalized Green's functions $G{´}_r(\sigma_i,\sigma_i)$ comes from the definition of the operator
$$\left[e^{ikX(\sigma_1)}\right]_{r}=\exp\left(\frac{1}{2}\int d^2\sigma 2d^2\sigma ' \Delta(\sigma,\sigma ')\frac{\delta}{\delta X^{\mu}(\sigma)}\frac{\delta}{\delta X_{\mu}(\sigma ')}\right)\,e^{ikX(\sigma_1)}=$$
$$=\exp\left(\frac{1}{2}\Delta(\sigma_1,\sigma_1)k_{1}^2\right)e^{ikX(\sigma_1)}$$

Then, for the self contraction terms, the $G(\sigma_i,\sigma_i)$ will be accompanied by the $\Delta(\sigma_i,\sigma_i)$, killing the divergence. The resultant will be the $G{´}_r(\sigma_i,\sigma_i)=G(\sigma_i,\sigma_i)+\Delta(\sigma_i,\sigma_i)$, the renormalized Green's function.

answered Feb 17, 2018 by (20 points)

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