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.

  • 1
    $\begingroup$ +1 : While the logic of the appendix is clear : the tachyon vertex $V_0$ is separated into a zero-mode $Z_0$ vertex, and a non-zero mode $W_0$ vertex, with $V_0=Z_0 W_0$, and so,to get the correlation function : $\langle V_0(k_1,y_1) ...V_0(k_M,y_M) \rangle$, one have to multiply the correlation function of the zero-modes ($7.A.16$) and the correlation function of the non-zero modes ($7.A.13$), it is not clear for me how to obtain $7.A.17$ from $7.A.12$ and $7.A.22$... $\endgroup$ – Trimok Oct 16 '13 at 11:14
  • $\begingroup$ @Trimok: I totally agree. I actually find it a bit weird, because they explicitely say that it is possible to get 7.A.17 from 7.A.12 and 7.A.22 on page 432 on the bottom. In fact this issue pops up in the later parts of chapter 7 for more complicated vertex operators as well..so I'm desperate to understand it :) $\endgroup$ – A friendly helper Oct 16 '13 at 11:41
  • $\begingroup$ A possible hint should be to consider $y_1....y_m$ as a "measure", which does not change the general "behaviour" ($(y_i-y_j)^{k_i.k_j}$). For instance, , with $V'(k,y) = V(k, \frac{1}{y})$, we have the same "behaviour", but with a different "measure". So, it maybe has to do with the different symmetries of string theory. $\endgroup$ – Trimok Oct 16 '13 at 16:33
  • $\begingroup$ Wouldn't that change $(y_i-y_j)^{k_ik_j}$ to $(y_i-y_j)^{-k_ik_j}$? Btw, thanks for intervening at the physicsforums :D $\endgroup$ – A friendly helper Oct 17 '13 at 5:59
  • $\begingroup$ In fact, this would give $(\frac{1}{y_i} - \frac{1}{y_j})^{k_i.k_j} = (\frac{y_j-y_i}{y_iy_j})^{k_i.k_j}$ (And in tree-level amplitudes, you have to take in account momentum conservation and mass shell conditions, so, I think that you will find that the measures are not so different ). In fact, in chapter $7.1$, one discusses open string, so tree amplitudes concerns vertex at the boundary (a circle) represented by quantities $0 \leq z_i \leq1$. In chapter $7.1.3$, one uses $7.1.1$ and $7.1.2$ in expression $7.1.15$ to get $7.1.26, 7.1.27$ . $\endgroup$ – Trimok Oct 17 '13 at 8:49

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