# Why is the tachyon vertex operator $\int d^2z :e^{ik.X(z,\bar{z})}:$ integrated?

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?

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.
• 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? – awsomeguy Aug 30 '20 at 10:20
• 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? – awsomeguy Aug 30 '20 at 11:34