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

+ 5 like - 0 dislike
257 views

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

asked Jul 6, 2014 in Theoretical Physics by user26143 (360 points) [ revision history ]
edited Jul 6, 2014 by Ron Maimon

@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 ...

1 Answer

+ 3 like - 0 dislike

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 Ron Maimon (7,535 points) [ no revision ]

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