Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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)? – user26143 Aug 3 '13 at 20:36
It is shown in p111 that $|1 \rangle$ corresponds to a unit operator – user26143 Aug 3 '13 at 23:58
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. – Trimok Aug 4 '13 at 10:02
I don't see anything about a unit operator p.111 of David Tong ' course ! – Trimok Aug 4 '13 at 10:05
"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) – user26143 Aug 4 '13 at 10:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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