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?


1 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

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


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

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.

  • 1
    $\begingroup$ Thanks! It sounds better. Now, these ideas brings me another point. Taking your example $\langle \prod_{i=1}^n e^{ik_iX} \partial X \rangle$, when we contract $\partial X$ with $\exp^{ikx}$, as you said we get, besides $v$, the $\exp^{ikX}$ again, and this will be used to compute the other part of the correlation function (that with only exponentials). I was resistant to accepting this, to me sounds strange, because in Wick contraction we contract 2 fields and dont get another one to be contracted again. So, couldn't I think in this String amplitudes in terms of wick contractions? $\endgroup$ Sep 6, 2020 at 8:10
  • 1
    $\begingroup$ I mean, does that means that we cannot think of Wick contractions in this context? Because it's tempting, since the OPE's of the exponetials matches exactly with the terms of result of the correlator - for instance $:e^{ik_1X(z)}: :e^{ik_2X(0)}: = |z|^{\alpha' k_1\cdot k_2}:e^{ik_1X(z)}e^{ik_2X(0)}:$ $\endgroup$ Sep 6, 2020 at 8:16
  • 1
    $\begingroup$ @BVquantization the Wick contraction comes as a theorem for free theories. Since we are dealing with free theories here Wick contraction is still valid. However the Wick contraction is about contracting fundamental*/free fields. The $ x$ is a free field but $e^{ikx} $ is not, so at best we can use the fact that $e^{ikx} $ is written as a functional of free fields $x $ and use Wick contractions for the free fields in $e^{ikx} $. $\endgroup$
    – Nogueira
    Sep 6, 2020 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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