It is well known that operators in string theory are inserted at the origin. So, for example, the vertex operator for tachyon looks like $$:e^{ik.X(0, 0)}:$$ However, while computing the string amplitude we use an integrated version of the vertex operator i.e. $$\int d^{2}\sigma \sqrt{det(g)} e^{ik.X}$$ This seems like a contradiction to me. Please explain why are these two equivalent ?

  • $\begingroup$ I think you mean $\sqrt{g}$ and not $\sqrt{\sigma}$ and that should help you find the answer... $\endgroup$ Apr 3, 2020 at 12:07

1 Answer 1


Under the state-operator mapping we have the following identification $$ |k;0,0 \rangle \longleftrightarrow :e^{ik.X(0, 0)}:$$ where $:e^{ik.X(0, 0)}:$ is the normally ordered closed string tachyon vertex operator and $ |k;0,0 \rangle $ is the closed string tachyon ground state.

Now, why is $:e^{ik.X(0, 0)}:$ not used in computing string S-matrix elements?

The answer is that any S-matrix element must be invariant under Weyl and diffeormphism transformations, something like $$S(k)= \sum_{compact \\ topologies} \int \frac{[dXdg]}{V_{diff \times Weyl}} exp(-S_{x}-\lambda\chi):e^{ik.X(0, 0)}: $$ can't work for obvious reasons, namely that $S(k)$ is not a scalar under diff and Weyl symmetries because $:e^{ik.X(0, 0)}:$ is an operator of conformal dimensions $(\Delta=\frac{l_{S}^{2}k^{2}}{4} , \Delta=\frac{l_{S}^{2}k^{2}}{4})$.

The solution is to integrate $:e^{ik.X(0, 0)}:$ over the relevant worldsheet as

$$\int d^{2}{\sigma} \sqrt{g} :e^{ikX}:.$$ Now is easy to see why this is the correct guess.

In locally flat coordinates we have the replacement $$\int d^{2}{\sigma} \sqrt{g} e^{ikX} \rightarrow \int d^{2}{z} e^{ikX}$$ where we can turn $d^{2}{z}:e^{ikX}:$ into a tensor of type $(0,0)$ by noticing that $d^{2}{z}$ is of type $(1,1)$ and choosing $k^{2}=-m^{2}=\frac{4}{l_{S}}$.

Sleeping beauty: Please notice how beautiful is the fact that the precise value of the mass of the tachyon is enforced by consistency.

We conclude that the correct expression for a stringy S-matrix element should be $$S(k)= \sum_{compact \\ topologies} \int \frac{[dXdg]}{V_{diff \times Weyl}} exp(-S_{x}-\lambda\chi)\prod_{a=1}^{n} \int d^{2}{\sigma} \sqrt{g} :e^{ik_{a}X_{a}}: $$ As desired.

  • $\begingroup$ Actually you can get the integrated form naturally without having to guess if you use a ghost fields and ghost action as well. This is explained in witten's paper perturbative string theory revisited. $\endgroup$ Jun 4, 2020 at 3:56
  • $\begingroup$ Well, my humble opinion is the argument is not exactly a guess. The only way to "cancel" the conformal dimension of the tachyon operator is to contract it with a tensor of type (1,1) and locally there is no other choice than $d^{2}z$. Anyway, if you have found a more satisfactory explanation, is ok. $\endgroup$ Jun 4, 2020 at 4:09
  • $\begingroup$ Indeed, the argument interest me. Could you please point out the page in Witten's paper where the argument is elaborated? $\endgroup$ Jun 4, 2020 at 4:14
  • 1
    $\begingroup$ See section 2.5 and 2.5.2 for a complete understanding. He talks of two kinds of vertex operators integrated and non-integrated and shows their equivalence. $\endgroup$ Jun 4, 2020 at 12:13
  • 1
    $\begingroup$ Thanks :) I'm not too familiar with the theory of super Riemann surfaces, but now I see I should. Although the logic seems (at least to me) much more elaborated, I completely agree, the Witten's derivation is clear and much more powerful. $\endgroup$ Jun 5, 2020 at 4:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.