I understand that, from the state-operator correspondence, $$|0;p\rangle \;=\;:e^{ip.X(0,0)}: |0\rangle.$$ This is given in Polchinski, equation (2.8.9). I am now trying to understand the S-matrix for $2\times$tachyon $\rightarrow$ $2\times$tachyon scattering. My understanding is as follows.

We seek to calculate $\langle\psi_f|\psi_i\rangle$, where the initial and final wavefunctions are both two taychons. That is, we want to find $$\langle 0, q_1 ; 0, q_2 | 0,p_1;0,p_2\rangle \; =\; \langle0|:e^{-iq_1.X}: :e^{-iq_2.X}: :e^{ip_1.X}: :e^{ip_2.X}:|0\rangle$$ $$ = \int DXDg \; e^{-S_{Poly} [X, g]} \;V[-q_1, 0]V[-q_2, 0]V[p_1, 0]V[p_2, 0],$$

where $V[p, z] = \; :e^{ip.X(z, \bar{z})}:$.

However, this is not the result quoted in textbooks (e.g. Polchinski eqns 3.5.5, and 3.6.1). Instead the actual result is

$$\int DXDg \; e^{-S_{Poly} [X, g]} \;\int d^4z_i d^4\bar{z}_i V[-q_1, z_1]V[-q_2, z_2]V[p_1, z_3]V[p_2, z_4],$$

citing diffeomorphism invariance. I see that my expression is not invariant under diffeomorphisms, whilst the second one is. However, why is the second expression the correct result? I see it in textbooks referred to a guess, but a guess of what exactly? Surely what we have now is not the overlap of four tachyons $<\psi_f | \psi_i>$ which we sought to calculate, but rather the overlap of some strange superposition of infinitely many tachyons. How does this relate to the $2\times$ tachyon scattering amplitude at all?


1 Answer 1


The result without integration is a correlation function of $4$ operators in the auxiliary worldsheet CFT. It is not diffeo-invariant, as it should be, because the CFT is not diffeo-invariant.

The result with integration is the state of string theory, not CFT. String theory states are given by vertex operators integrated over the worldsheet.

If you're asking how we can see the last point, I suggest looking into canonical quantization of the string. Diffeomorphism invariance comes from the constraint that arises due to the use of Polyakov action, which is diffeomorphism invariant.

  • $\begingroup$ What exactly do you mean by a state/operator in string theory vs CFT? $\endgroup$
    – awsomeguy
    Aug 30, 2020 at 10:05
  • $\begingroup$ And so what string theory states are we calculating the overlap between here, in the last expression? Is $|0p_1;0p_2>$ not a state in the string theory? $\endgroup$
    – awsomeguy
    Aug 30, 2020 at 10:20
  • $\begingroup$ @awsomeguy hm... what you've written down looks like a legit state in string theory, yes. Note that it doesn't have a worldsheet point associated to it -- this state corresponds to an already integrated vertex operator in the CFT. To answer your first question -- string theory differs from the CFT by the presence of constraints. These can be derived from the Polyakov action, and the best way to quantize them is to use BRST (anystring theory textbook should contain details on BRST string quantization). In CFT, there are no such constraints. $\endgroup$ Aug 30, 2020 at 11:11
  • $\begingroup$ I see that the state $|0p_1; 0p_2>$ does not have a point associated to it - but can it not still be written as a local operator on the vacuum state as per the state-operator map? i.e. the state corresponds to a vertex that is not already integrated? $\endgroup$
    – awsomeguy
    Aug 30, 2020 at 11:34
  • $\begingroup$ @awsomeguy I’m not exactly sure what you mean. Integrated is the opposite of local, right? This stringy state corresponds to an integral of the CFT state over the worldsheet, but not to a local CFT operator insertion acting on the CFT vacuum $\endgroup$ Aug 30, 2020 at 11:36

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