Polchinski defines normal ordering in string theory as:

$$:X^\mu(z,\bar z)X^\nu(w,\bar w): = X^\mu(z,\bar z) X^\nu(w, \bar w) + \frac{\alpha'}{2} \eta^{\mu\nu} \log |z-w|^2$$

and for more complicated expressions one obtains the normal ordered expression via Wicks theorem (p. 39).

In the CFT-Context (e.g. compare with "Conformal Field Theory" by Di Francesco) normal ordering is defined as "regular part of the OPE".

How can we see that these definitions are equivalent?


2 Answers 2


This two definitions are not equal, and lead to different expressions for more complicated fields (composite ones) OPEs. This two definitions uses the same regularization (point-splitting regularization) but different subtraction schemes. The one adopted by Polchinski is given by subtracting the contractions between the fields, divergent and finite terms are subtracted in this procedure. Take as an example the following composite operators OPE:

$$ :\partial x^{\mu}(z)\partial x^{\nu}(z)::\partial x^{\rho}(w):=:\partial x^{\mu}(z)\partial x^{\nu}(z)\partial x^{\rho}(w):+:\partial x^{(\nu}(z):\eta^{\mu)\rho}\partial_z\partial_w\left(-\frac{\alpha'}{2}\log(|z-w|^2)\right) $$ $$ =:\partial x^{\mu}(z)\partial x^{\nu}(z)\partial x^{\rho}(w):-\frac{\alpha'}{2}\frac{\partial x^{\nu}(z):\eta^{\mu\rho}+\partial x^{\mu}(z):\eta^{\nu\rho}}{(z-w)^2} $$

expanding in $z\rightarrow w$ the numerator, we get

$$ =:\partial x^{\mu}(z)\partial x^{\nu}(z)\partial x^{\rho}(w):-\frac{\alpha'}{2}\frac{\partial x^{\nu}(w):\eta^{\mu\rho}+\partial x^{\mu}(w):\eta^{\nu\rho}}{(z-w)^2} $$ $$ -\frac{\alpha'}{2}\frac{\partial^2 x^{\nu}(w):\eta^{\mu\rho}+\partial^2 x^{\mu}(w):\eta^{\nu\rho}}{(z-w)}-\frac{\alpha'}{4}\partial^3 x^{\nu}(w):\eta^{\mu\rho}-\frac{\alpha'}{4}\partial^3 x^{\mu}(w):\eta^{\nu\rho}+\mathcal{O}(z-w) $$

subtracting the divergent part and then send $z\rightarrow w$, is the same thing as computing

$$ \left(:\partial x^{\mu}(w)\partial x^{\nu}(w):,:\partial x^{\rho}(w):\right)=\oint_{C(w)}\frac{dz}{2\pi i}\frac{:\partial x^{\mu}(z)\partial x^{\nu}(z)::\partial x^{\rho}(w):}{(z-w)} $$

and what we get is

$$ \left(:\partial x^{\mu}(w)\partial x^{\nu}(w):,:\partial x^{\rho}(w):\right)=:\partial x^{\mu}(w)\partial x^{\nu}(w)\partial x^{\rho}(w):-\frac{\alpha'}{4}\partial^3 x^{\nu}(w)\eta^{\mu\rho}-\frac{\alpha'}{4}\partial^3 x^{\mu}(w):\eta^{\nu\rho} $$

there is an extra term, usually called ordering terms, that appears in the right hand side. Note also that $:\partial x^{\mu}(z)\partial x^{\nu}(z):=(\partial x^{\mu}(z),\partial x^{\nu}(z))$. It is very important to notice that the $::$ ordering is associative and (anti-)commutative for bosons(fermions). The $(,)$ ordering is not associative, and not (anti-)comutative! This is why I prefer the Polchinski prescription, but the problem with the Polchinski prescription is that we need in advance to know what are the "fundamental" fields that are going to build any other local operator and their OPEs to define the contraction, while the $(,)$ ordering only require the knowledge of the OPEs, without picking a preferential "fundamental" field.

Usually the physics does not depend in how you define the ordering between various operators, but in the presence of interactions or non-linear constraints, the physics does depend in how you define the ordering of certain operators since composite operators are important for the dynamics, and they are sensitive under ordering.


The usual definition of normal ordered product is:

$$:X^\mu(z,\bar z)X^\nu(w,\bar w): = X^\mu(z,\bar z) X^\nu(w, \bar w) - \langle X^\mu(z,\bar z) X^\nu(w, \bar w) \rangle $$

As you said, this is the regular part of the OPE, since only the divergent part of two operators gives non vanishing contribution to the correlator. Of course

$$\langle X^\mu(z,\bar z) X^\nu(w, \bar w) \rangle=- \frac{\alpha'}{2} \eta^{\mu\nu} \log |z-w|^2$$

  • 3
    $\begingroup$ Just a note that this would only be true for free fields. $\endgroup$
    – Prahar
    Apr 27, 2017 at 5:07

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.