# Question on Tachyon Correlator (Green Schwarz Witten)

1. Oct 16, 2013

### earth2

I'm reading through chapter 7 of Green-Schwarz-Witten and I have a problem with the derivation of the M-tachyon correlation function. Basically I'm trying to get 7.A.17 from 7.A.12 and eq 7.A.22 in the appendix of the first volume.

Basically I want to prove:

$$\langle\frac{V(k_1,y_1)...V(k_M,y_M)}{y_1...y_M}\rangle=\prod_{i < j}(y_i-y_j)^{k_ik_j}$$

with $$V(k_i,y_i)=e^{ik_iX(k_i)}$$

If I try to get this by just plugging everything into $$\langle:e^{A_1}:...:e^{A_M}:\rangle=e^{\sum_{i<j}<A_iA_j>}$$ using $$\langle X(y_i)^{\mu}X(y_j)^\nu\rangle=-\eta^{\mu\nu}log(y_i-y_j)$$ I somehow still have the $$\frac{1}{y_1...y_M}$$ factor in the formula above existing on the right hand side of the first equation. How is that cancelled in this approach? I dont see it.

More specifically I have

$$\langle\frac{V_1...V_M}{y_1...y_M}\rangle=\frac{1}{y_1...y_M}\langle V(k_1,y_1)...V(k_M,y_M)\rangle=\frac{1}{y_1...y_M}e^{(\sum_{i<j}k_i^\mu\langle X_iX_j\rangle k_j^\nu)}$$ $$=\frac{1}{y_1...y_M}e^{(\sum_{i<j}k_ik_jlog(y_i-y_j))}=\frac{1}{y_1...y_M}\prod_{i<j}(y_i-y_j)^{k_ik_j}$$
which is the result I want to have up to that nasty prefactor that I cant make sense of.

Edit: I also posted this on stackexchange but didnt get an answer :(
Btw, this is not a homework question. I'm reading it on my own.

Last edited: Oct 16, 2013
2. Oct 16, 2013

### fzero

You have to look back at eq (7.1.6), where the zero-mode part of the tachyon vertex operator was computed. The $\pm 1$ in the exponent of $z$ comes from normal-ordering the zero mode operators $\hat{x}^\mu, \hat{p}^\mu$. It is precisely this exponent that leads to the factors of $y_i$ in the n-pt function.

3. Oct 16, 2013

### earth2

Hi fzero,

Thanks for your answer but I'm not sure I understand it in the context of the equations above. Would you be so could to make it more precise w.r.t. the formulae I used above. That would be very nice since i'm banging my head against this for a while now. Moreover, this point is important for all the computations in this chapter after that...so i would really like to understand it.

4. Oct 16, 2013

### trimok

To fzero :

The question of earth2 is in fact, how to obtain $7.A.17$ from $7.A.12$ and $7.A.22$ (appendix of the first volume), because you have a curious overall term $y_1......y_M$

5. Oct 17, 2013

### earth2

Thanks Trimok for clearing that up :) As mentioned on phys stackexchange, I would like to understand this, since GSW say in the appendix that getting 7.A.17 is possible from 7.A.12 and 7.A.22... :)

6. Oct 17, 2013

### fzero

If we really want to understand this factor, then we should restore the factors of $\alpha'$ and note that it is

$$\prod_i y^{\alpha' k_i^2 }.$$

GSW refer to 7.A.22 as a mnemonic to recover 7.A.17, but don't actually tell you what value of $\lambda$ you need. It seems that the appropriate version of 7.A.22 is something like

$$\langle X^\mu (y_i) X^\nu (y_j) \rangle = - 2\alpha' \eta^{\mu\nu} \log (y_i-y_j) - 2\alpha' \eta^{\mu\nu} \delta_{ij} \log y_i.$$

The 2nd term computes the zero-mode contribution to a modified version of 7.A.12, which is something like

$$\begin{split} \langle :e^{A_1}: :e^{A_2}: \cdots :e^{A_M}: \rangle & = \exp \left[ \frac{1}{2} \sum_{i,j} \langle A_i A_j \rangle \right] \\ & = \exp \left[ \sum_{i<j} \langle A_i A_j \rangle \right] + \exp \left[ \frac{1}{2} \sum_i \langle A_i A_i \rangle \right] . \end{split}~~~(*)$$

The term in the correlator gives something like

$$\exp \left[ 2\alpha' \sum_{i<j} k_i\cdot k_j \log y_i + \alpha' \sum_i k_i^2 \log y_i \right] = \prod_i y_i ^{\alpha'k_i^2} \prod_{i<j} y_i^{2\alpha' k_i\cdot k_j },$$

which should be compared with 7.A.16.

This isn't a clean-cut derivation, since (*) introduces some divergences from the nonzero mode part that must be regularized away. For this reason, it is advisable to look at Polchinski, who is more careful about how to deal with the normal ordering regularization in the CFT correlators, as well as in the discussion of zero-modes and momentum conservation.

7. Oct 18, 2013

### earth2

Thanks fzero! Since both - Trimok and you - suggested to look at Polchinksi I will do so :) To the Library...!

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