I am reading Polchinski's String Theory Volume I. While I am learning the basics of string scatterings in Ch.3, I went back to chapter 2 and review the part of vertex operators. However, I found that I didn't really understand eq. 2.8.18 and the paragraph below when I tried to derive what he said.

In the book,

The state–operator isomorphism is an important but unfamiliar idea and so it is useful to give also a more explicit path integral derivation. Consider a unit disk in the $z$-plane, with local operator $A$ at the origin and with the path integral fields $φ$ fixed to some specific boundary values $φ_b$ on the unit circle. The path integral over the fields $φ_i$ on the interior of the disk with $φ_b$ held fixed produces some functional $Ψ_A[φ_b]$, $$Ψ_A[φ_b] = \int [dφ_i]_{φ_b} \exp(−S[φ_i])A(0) . $$ A functional of the fields is the Schrodinger representation of a state, so this is a mapping from operators to states. The Schrodinger representation, which assigns a complex amplitude to each field configuration, has many uses. It is often omitted from field theory texts, the Fock space representation being emphasized.

To go the other way, start with some state $Ψ[φ_b]$. Consider a path integral over the annular region $1 \geq |z| \geq r$, with the fields $φ_b$ on the outer circle fixed and the fields $φ_b$ integrated over the inner circle as follows:

$$\int[dφ'_b][dφi]_{φ_b,φ'_b} \exp(−S[φ_i])r^{−L_0−\tilde{L}_0}Ψ[φ'_b]. \quad(2.8.18)$$ That is, the integral is weighted by the state (functional) $r^{−L_0−\tilde{L}_0}Ψ[φ_b]$. Now, the path integral over the annulus just corresponds to propagating from $|z| = r$ to $|z| = 1$, which is equivalent to acting with the operator $r^{L_0+\tilde{L}_0}$. This undoes the operator acting on $Ψ$, so the path integral (2.8.18) is again $Ψ[φ_b]$. Now take the limit as $r → 0$. The annulus becomes a disk, and the limit of the path integral over the inner circle can be thought of as defining some local operator at the origin. By construction the path integral on the disk with this operator reproduces $Ψ[φ_b]$ on the boundary.

$$ \Psi[\phi_b] = \int d[\phi'_b]d[\phi_i]_{\phi_b,\phi'_b}\exp(-S[\phi_i])r^{-L_0-\tilde{L}_0}\Psi[\phi'_b] $$

  1. If I follow (A.1.15), I could not see how the factor $r^{-L_0-\tilde{L}_0}$ arises.

  2. As he mentioned, the integral means propagation from $|z|=r$ to $|z|=1$. My previous understanding is the propagation in $r$-direction is the time evolution so we act $\exp(-iHt)$ on it. The time direction is $\mbox{Im}(w)$ in the cylinder coordinate $(z=\exp(-iw))$, so we have $\exp(-iH \mbox{Im}(w)) = r^{-i(L_0+\tilde{L}_0)}$ ?

My sincere thank for your time and effort on explaining it. :)

Thank @Prahar for the comment. I guessed his is referring to euclidean world-sheet, thanks for confirming this.

  • 2
    $\begingroup$ Please include all relevant information into the post, in particular equations you are talking about - if I am a bit familiar with the field, I should not have to open a specific book to understand your question. $\endgroup$
    – ACuriousMind
    Dec 15 '15 at 13:58
  • $\begingroup$ We are working in Euclidean time. So the time evolution operator is $e^{-H \tau} = e^{- H \text{Im} w }$. Then, $r = |z| = e^{ \text{Im} w }$. Thus, $e^{- H \tau } = r^{-H} = r^{-L_0 - {\tilde L}_0}$ $\endgroup$
    – Prahar
    Dec 15 '15 at 15:42
  • $\begingroup$ Thanks for the comment. What do you think about my first question? $\endgroup$
    – user260822
    Dec 16 '15 at 4:40

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