I'm working with Polchinski and my question is related to chapter 2 of his book String Theory, Volume 1: Introduction to the Bosonic String.

What I believe to understand

An operator product expansion (OPE) yields something of the following form: $$ T(z)A(0,0)=\dotso+\frac{h}{z^2}A(0,0)=\frac{1}{z}\partial A(0,0)+\dotso,\tag{2.4.14} $$ with $h$ the weight. More precisely, $T(z)$ is defined as $$ T(z)=\frac{-1}{\alpha^\prime}\colon\partial X^\mu\partial X_\mu\colon \tag{2.4.4}$$ Now, a conformal normal ordering can be written as follows: $$ \colon\mathcal{F}\colon=\exp\left(\frac{\alpha^\prime}{4}\int\!\mathrm{d}^2z_1\mathrm{d}^2z_2\log|z_{12}|^2\frac{\delta}{\delta X^\mu\left(z_1,z^*_1\right)}\frac{\delta}{\delta X_\mu\left(z_2,z^*_2\right)}\right)\mathcal{F}\tag{2.2.7} $$

What I don't understand

How is the OPE expanded? Say, $F=X^\nu$, then $$ T(z)X^\nu(0,0)=\frac{-1}{\alpha^\prime}\colon\partial X^\mu\partial X_\mu\colon X^\nu(0,0), $$ which, to leading order, should be equal to $$ \frac{-1}{\alpha^\prime}\left(\partial X^\mu\partial X_\mu+\frac{\alpha^\prime}{4}\int\!\mathrm{d}^2z_1\mathrm{d}^2z_2\log|z_{12}|^2\frac{\delta}{\delta X^\mu\left(z_1,z^*_1\right)}\frac{\delta}{\delta X_\mu\left(z_2,z^*_2\right)}\partial X^\rho\partial X_\rho\right)X^\nu(0,0). $$ How is this evaluated? In particular: What am I supposed to do with the $\frac{\delta}{\delta X^\mu\left(z_1,z^*_1\right)}\frac{\delta}{\delta X_\mu\left(z_2,z^*_2\right)}\partial X^\rho\partial X_\rho X^\nu(0,0)$ part?

  • 4
    $\begingroup$ Well these functional derivatives in particular are easy to evaluate because they just give delta functions. Indeed $$\dfrac{\delta X^\mu(z,\bar z)}{\delta X^\nu(w,\bar w)} = \delta^\mu_\nu \delta^{(2)}(z,w)$$ $\endgroup$
    – Gold
    Commented Jan 27, 2023 at 13:23
  • $\begingroup$ Thank you for your comment. Why is $\frac{\delta X^\mu}{\delta X^\nu}=\frac{\delta \partial X^\mu}{\delta X^\nu}$? What happens with the partial derivative here? $\endgroup$
    – kalle
    Commented Jan 28, 2023 at 14:55
  • 2
    $\begingroup$ It simply isn't. The partials with respect to $z$ and $\bar z$ commute with the functional derivatives and they hit the delta function. $\endgroup$
    – Gold
    Commented Jan 28, 2023 at 15:09
  • 2
    $\begingroup$ You should use the Liebnitz rule (functional derivatives are derivatives, after all). In that case you have for example $$\frac{\delta}{\delta X^\mu\left(z_2,z^*_2\right)}\partial X^\rho(z,\bar z)\partial X_\rho(z,\bar z) = \frac{\delta[\partial X^\rho(z,\bar z)]}{\delta X^\mu\left(z_2,z^*_2\right)}\partial X_\rho(z,\bar z)+\partial X^\rho(z,\bar z)\frac{\delta[\partial X_\rho(z,\bar z)]}{\delta X^\mu\left(z_2,z^*_2\right)}=2\frac{\delta[\partial X^\rho(z,\bar z)]}{\delta X^\mu\left(z_2,z^*_2\right)}\partial X_\rho(z,\bar z),$$ after recognizing the two terms are really the same. $\endgroup$
    – Gold
    Commented Jan 28, 2023 at 18:39
  • 1
    $\begingroup$ Then you can commute the functional derivative with $\partial$ and evaluate this explicitly. After that you can take the other functional derivative. $\endgroup$
    – Gold
    Commented Jan 28, 2023 at 18:40


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