# Does differentiation and (conformal) normal ordering commute?

In Polchinski's String theory Volume I, chapter 2, he writes down the following OPE, $$X^\mu (z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2) = -\frac{\alpha'}{2}\eta^{\mu\nu}\ln|z_{12}|^2 + :X^\nu X^\mu(z_2,\bar{z}_2): + \sum_{k=1}^{\infty}\frac{1}{k!} \bigg[ (z_{12})^k :X^\nu\partial^kX^\mu(z_2,\bar{z}_2): + \ (\bar{z}_{12})^k :X^\nu\bar{\partial}^kX^\mu(z_2,\bar{z}_2): \bigg]$$

which he gets by taylor expanding $:X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2):$ since it can be written as a sum of holomorphic and antiholomorphic functions (since it is harmonic in $(z_1,\bar{z}_1)$).

But if we call the harmonic function $\ f(z_1,\bar{z}_1;z_2,\bar{z}_2) = :X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2):$, and taylor expand $\ f$ in $(z_1,\bar{z}_1)$ around $(z_1,\bar{z}_1) = (z_2,\bar{z}_2)$, then we get terms that go like ~ $\partial^k f$ ~ $\partial^k :X^\mu(z_1,\bar{z}_1)X^\nu(z_2,\bar{z}_2):$ evaluated at $(z_2,\bar{z}_2)$. Here the derivatives are outside the normal ordered expression while in Polchinski's expression, they are inside the normal ordered expression. Why are they equal? What am I misunderstanding?

Reference for the expression : Eq. 2.2.4 Pg 38, Polchinski Chapter 2 (hardcover) (2004 reprint I think).

Edit 1: $z_{12} = z_1 - z_2$ Edit 2: All partial derivatives are with respect to $z_1,\bar{z}_1$.

• the answer in general is 'no' except in very special circumstances. If you only consider primary operators then 'yes' they commute, but if you don't consider primaries then one needs to distinguish between two cases: whether the local Ricci curvature and/or it's covariant derivatives at the location of the operator insertion are zero or not. if either of these is not zero then normal ordering and differentiation do not commute. Sep 7, 2022 at 9:48

Yes, differentiation and normal ordering commute. You can prove this using the definition (2.2.7) of the normal ordering: $$:\mathcal{F}: = \exp \left( \frac{\alpha '}{4} \int d^2z_1 d^2z_2 \ln |z_{12}|^2 \frac{\delta}{\delta X^{\mu}(z_1 , \bar{z}_1)} \frac{\delta}{\delta X^{\nu}(z_2 , \bar{z}_2)} \right) \mathcal{F}$$
Using this expression, the commutativity of the normal order and differentiation boils down to commutativity of functional derivative and differentiation. Let us consider any functional $F$ of $X^{\mu}(z )$. Then $$\frac{\delta F[X^{\mu}(z )]}{\delta X^{\mu}(z_1)} = \lim\limits_{\epsilon \rightarrow 0} \frac{F[X^{\mu}(z) + \epsilon \delta(z-z_1) ]-F[X^{\mu}(z)]}{\epsilon}$$ (see for instance eq A.28 here). Then apply this for $F[X^{\mu}(z , \bar{z})] = \partial X^{\mu}(z)$: $$\frac{\delta \partial X^{\mu}(z)}{\delta X^{\mu}(z_1)} = \delta ' (z-z_1)$$ On the other hand, $$\partial \frac{\delta X^{\mu}(z)}{\delta X^{\mu}(z_1)} = \delta ' (z-z_1)$$ since $\frac{\delta X^{\mu}(z)}{\delta X^{\mu}(z_1)} = \delta (z-z_1)$. The two expressions are equal, which proves the claim.