Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

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

+ 6 like - 0 dislike
1136 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 (405 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 ...

2 Answers

+ 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 (7,720 points) [ no revision ]
+ 2 like - 0 dislike

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 Nogueira, Lucas (20 points) [ revision history ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverf$\varnothing$ow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...