Correlation functions - Polchinski equation 6.2.18

Question

At some point of Polchinski book, we are interested in calculate the following correlation function: $$\left\langle \prod_{j=1}^n[e^{ik_i\cdot X(z_i,\bar{z}_i)}]_r\prod_{j=1}^p\partial X^{\mu_j}(z_j'))\prod_{k=1}^q\bar{\partial}X^{\nu_k}(\bar{z}_k'') \right\rangle$$

In which the fields $X^\mu(z,\bar{z})$ are such that $\langle X^\mu(z,\bar{z})X^\nu(z',\bar{z}') \rangle= -\frac{\alpha'}{2}\eta^{\mu\nu}\ln|z-z'|^2$. Using usual methods of path integral it's possible to calculate: $$\left\langle \prod_{j=1}^n[e^{ik_i\cdot X(z_i,\bar{z}_i)}]_r \right\rangle = iC_{...}^X(2\pi)^{26}\delta^{d}(\sum_ik_i)\exp\left(-\frac{\alpha'}{2}\sum_ik_i^2\omega(\sigma_i)\right)\prod_{i<j}^n|z_{i}-z_{j}|^{\alpha'k_i\cdot k_j}$$ To calculate the first correlation function that I wrote we have to sum over all contractions, where $\partial X$ or $\bar{\partial}X$ must be contracted either with an exponential or with another $\partial X$ or $\bar{\partial}X$. But then Polshinski write the result of this contractinons: $$iC_{...}^X(2\pi)^{26}\delta^{d}(\sum_ik_i)\exp\left(-\frac{\alpha'}{2}\sum_ik_i^2\omega(\sigma_i)\right)\prod_{i<j}^n|z_{i}-z_{j}|^{\alpha'k_i\cdot k_j}\times \left\langle \prod_{j=1}^p[v^{\mu_j}(y_j) + q^{\mu_j}(y_j)]\prod_{k=1}^q[\tilde{v}^{\mu_k}(z''_k) + \tilde{q}^{\mu_k}(y_k'')] \right\rangle$$

Where $$ v^\mu(y) = -i\frac{\alpha'}{2}\sum_{i=1}^n\frac{k_i^\mu}{z-z_i}$$ and $q^\mu = \partial X - v^\mu$. But in this case he just wrote $$ iC_{...}^X(2\pi)^{26}\delta^{d}(\sum_ik_i)\exp\left(-\frac{\alpha'}{2}\sum_ik_i^2\omega(\sigma_i)\right)\prod_{i<j}^n|z_{i}-z_{j}|^{\alpha'k_i\cdot k_j}\times \left\langle \prod_{j=1}^p\partial X^{\mu_j}(z_j'))\prod_{k=1}^q\bar{\partial}X^{\nu_k}(\bar{z}_k'') \right\rangle $$

Well, Polchinski did not follow his own word, he just contracted the exponentials and then then contracted the $\partial X$'s. The expression $v^\mu$ INSIDE the expectation value to mo doesn't even makes sense, because the $v$'s are already the result of contractions of $\partial X$ with exponentials...

what on earth is happening?

Answer 1

Note that he drop the subscript $\langle\rangle_{S^{2}}\rightarrow\langle\rangle$ when he write $\langle (v+q)(\tilde v+\tilde q)\rangle$ so you should not interpret this as a true correlation function. It is just a notation tool that means: sum over all contractions of $q's$ using $-\eta^{\mu\nu}(z-z')^{-2}\alpha '/2$ so

$$ \langle \partial x^{\mu}(z) \partial x^{\nu}(z')\rangle_{S^{2}} \neq \langle (v^{\mu}(z)+q^{\mu}(z))(v^{\nu}(z')+q^{\nu}(z'))\rangle $$

The equation above becomes an equality only (up to a $\langle 1\rangle_{S^{2}}$ factor) if $v^{\mu}(z)=v^{\nu}(z')=0$, i.e. if there is no exponential.

Let us consider the case with two $\partial x$'s and various exponentials. The expectation value becomes

$$ iC_{...}^X(2\pi)^{26}\delta^{d}(\sum_ik_i)\exp\left(-\frac{\alpha'}{2}\sum_ik_i^2\omega(\sigma_i)\right)\prod_{i<j}^n|z_{i}-z_{j}|^{\alpha'k_i\cdot k_j}\times $$ $$ \times\langle (v^{\mu}(z)+q^{\mu}(z))(v^{\nu}(z')+q^{\nu}(z'))\rangle $$

where

$$ \langle (v^{\mu}(z)+q^{\mu}(z))(v^{\nu}(z')+q^{\nu}(z'))\rangle = v^{\mu}(z)v^{\nu}(z')+v^{\mu}(z)\langle q^{\nu}(z')\rangle+\langle q^{\mu}(z)\rangle v^{\nu}(z')+ $$ $$ +\langle q^{\mu}(z)q^{\nu}(z')\rangle $$

and $\langle q^{\nu}(z')\rangle=\langle q^{\mu}(z)\rangle=0$ while

$$ \langle q^{\mu}(z)q^{\nu}(z')\rangle = -\eta^{\mu\nu}\frac{\alpha '}{2}\frac{1}{(z-z')^{2}} $$

so the effect of replacing $\partial x$ by $(v+q)$, and treating $q$ as $\partial x$, i.e. contracting $q$'s as if they are $\partial x$'s, has the effect of taking account for the contractions of $\partial x$'s with the exponentials. The reason why this works is because the exponential is a kind of "eigenvector" under contractions with $\partial x$, where the "eigenvalues" are the $v$'s.

Another archetype is the correlation between one $\partial x$ and various exponentials. Using the formula proposed by Polchinski this gives

$$ iC_{...}^X(2\pi)^{26}\delta^{d}(\sum_ik_i)\exp\left(-\frac{\alpha'}{2}\sum_ik_i^2\omega(\sigma_i)\right)\prod_{i<j}^n|z_{i}-z_{j}|^{\alpha'k_i\cdot k_j}\times \langle (v^{\mu}(z)+q^{\mu}(z))\rangle $$

where now

$$ \langle (v^{\mu}(z)+q^{\mu}(z))\rangle = v^{\mu}(z)+\langle q^{\mu}(z)\rangle = v^{\mu}(z) $$

which confirms that $v^{\mu}(z)$ works as a "eigenvalue" and the exponentials as "eigenvector" of $\partial x^{\mu}(z)$. There is a physics behind this! The $\partial x^{\mu}(z)$ is a conserved current associated to momentum + winding, while the exponentials are, via state-operator correspondence, associated to states with well defined momentum and winding number.

The contraction of a current with a local operator is related, again by state-operator correspondence, to the conserved charge operator acting on the state.