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The problem related to this post, but my question is even more elementary.

In p 101 of Polchinski's string theory vol I, it is stated

Using the state-operator mapping, the vertex operator for the closed-string tachyon is $$V_0 = 2g_c \int d^2 \sigma g^{1/2} e^{ik \cdot X} \rightarrow g_c \int d^2 z : e^{i k \cdot X}: \tag{3.6.1}$$

I miserably and completely don't get it. How to derive Eq. (3.6.1)?

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I think that it works like this :

In $(2.8.8)$, you have the state-operator correspondance :

$x^\mu|0\rangle \rightarrow~~X^\mu (0,0)$

Now, we can "exponentiate" it, like in $(2.8.9)$

$|0;k> \sim e^{ik.x}|0\rangle \rightarrow~~:e^{ik.X(0,0)}:$

In $(3.6.1)$, one speaks of the closest string tachyon, which is in fact precisely the state $|0;0>$(no excitation, level = $0$). Apply the vertex operator $e^{ik.x}$ for the ground state gives precisely $e^{ik.x}|0\rangle$, so it would explain the correspondance.

Criticism and precisions are welcome anyway, because I am not $100$% sure.

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  • $\begingroup$ Actually I don't get the point of state-operator correspondance (maybe I should open another thread). If I start from Eq. (2.8.3), the LHS is an operator, RHS is a path integral. We let the LHS of (2.8.3) acts on $|1 \rangle $, which is the LHS of (2.8.4), but why there is no $|1 \rangle $ on the RHS of (2.8.4)? $\endgroup$
    – user26143
    Commented Aug 3, 2013 at 20:36
  • $\begingroup$ It is shown in p111 that $|1 \rangle$ corresponds to a unit operator damtp.cam.ac.uk/user/tong/string/string.pdf $\endgroup$
    – user26143
    Commented Aug 3, 2013 at 23:58
  • $\begingroup$ In my answer, I have used the notation $|0\rangle$ as the ground state. Polchinkski notes it like $|1\rangle = |0;0\rangle$ (2.8.2). All this is the same thing. The correspondance is in fact, for instance 2.8.3 (skipping numerical terms) : $$\alpha^\mu_{-m}|0\rangle~\rightarrow~\alpha^\mu_{-m}~\rightarrow~\partial_m X^\mu(0)$$. The first correspondance is obvious, and the second corresponds to a residue calculus. The left term is a state, and the right term is an operator. $\endgroup$
    – Trimok
    Commented Aug 4, 2013 at 10:02
  • $\begingroup$ I don't see anything about a unit operator p.111 of David Tong ' course ! $\endgroup$
    – Trimok
    Commented Aug 4, 2013 at 10:05
  • $\begingroup$ "Our first task is to check whether the vacuum state is indeed equivalent to the insertion of the identity operator." (I took identity operator=unit operator, sorry for this inconvenience) $\endgroup$
    – user26143
    Commented Aug 4, 2013 at 10:11

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