[..quoting from Page 11 of Polchinski Vol2..]
Given $1+1$ conformal bosonic fields $H(z)$ one has their OPE as, $H(z)H(0) \sim -ln(z)$
Then from here how do the following identities come?
$e^{iH(z)}e^{-iH(z)} \sim \frac{1}{z}$
$e^{iH(z)}e^{iH(0)} = O(z) $
$e^{-iH(z)}e^{-iH(0)} = O(z) $
$\langle \prod_{i} e^{i\epsilon_i H(z_i)} \rangle_{S_2} = \prod_{i
I have pretty much no clue how these are derived!
(..I have done many OPE calculation earlier but this one beats me totally!..)
It would be great if someone can either show the derivation or give a reference where this is explained!
Also I am curious if there is a generalization of this to arbitrary even dimensional complex analytic complex manifolds...
Answer
I think there is a typo in the first formula. Let me propose this (partial) answer for the $3$ first formulae:
Because $H(z)H(0) \sim -ln(z)$, we may write the OPE for any pair of operators $F(H), G(H)$ functions of $H$ (in analogy with formula $2.2.10$ p.$39$ vol $1$)
$$:F::G: = e^{- \large \int dz_1 dz_2 ln z_{12} \frac{\partial}{\partial H(z_1)}\frac{\partial}{\partial H(z_2)}} :FG:\tag{1}$$
This gives, for $F = e^{i \epsilon_1 H(z_1)}, G = e^{i \epsilon_2 H(z_2)}$
$$:e^{i \epsilon_1 H(z_1)}::e^{i \epsilon_2 H(z_2)}: = (z_{12})^{\epsilon_1 \epsilon_2} :e^{i \epsilon_1 H(z_1)}e^{i \epsilon_2 H(z_2)}:\tag{2}$$
So, we have :
$$:e^{iH(z)}::e^{-iH(0)}:~ = \frac{1}{z}~:e^{i H(z)}e^{-i H(0)}: ~\sim \frac{1}{z}:e^{i H(0)}e^{-i H(0)}: \sim \frac{1}{z}\tag{3}$$
$$:e^{iH(z)}::e^{iH(0)}:~ = z~:e^{i H(z)}e^{i H(0)}: ~\sim z~:e^{2i H(0)}: \sim O(z)\tag{4}$$
$$:e^{-iH(z)}::e^{-iH(0)}:~ = z~:e^{-i H(z)}e^{-i H(0)}: ~\sim z~:e^{-2i H(0)}: \sim O(z)\tag{5}$$
[EDIT]
For the last equation, I think it is the same reasoning that the one done in Vol $1$, page $173,174$, formulae $6.2.24$ until $6.2.31$
[EDIT 2]
The formula $1$, and the formula $(2.2.10)$ are not formulae ad hoc. These are the consequence of a definition of the normal ordering, and the definitition of the contractions. These are the consequence of the general formulae $2.2.5$ to $2.2.9$, , for instance :
$$F = :F:+ ~contractions \tag{2.2.8}$$ $$:F::G: = :FG:+ ~cross-contractions \tag{2.2.9}$$
Now, we may specialize to holomorphic fields $Y(z)$, so that $Y(z)Y(0) \sim f(z)$, and write :
$$:F::G: = e^{ \large \int dz_1 dz_2 f(z_{12}) \frac{\partial}{\partial Y(z_1)}\frac{\partial}{\partial Y(z_2)}} :FG:\tag{6}$$ where $F$ and $G$ are functions of $Y$
The specialization to an holomorphic field does not change the logic and the calculus done in $2.2.5$ to $2.2.9$
No comments:
Post a Comment