In the book Conformal Field Theory (authors: Philippe Di Francesco, Pierre Mathieu, David Senechal), a field $f(z)$ is primary if it transforms as $$f(z) \rightarrow g(\omega)=\left( \frac{d\omega}{dz}\right)^{-h}f(z)$$ under an infinitesimal conformal transformation $z \rightarrow \omega(z)$. The physical meaning of this definition is unclear to me. For instance, what's the difference between a primary field and a secondary field? And what's the significance of the fact that the energy-momentum tensor is not a primary field?

  • 1
    $\begingroup$ What do you mean by the "physical meaning"? Fields are not observables. However, being a primary field reflect in transformations property which in turn appear in the operator product expansions and in the expectation values. $\endgroup$
    – gented
    Aug 7, 2016 at 22:05
  • $\begingroup$ For example, why we call the term in the OPE of $TT$ which should not appear if $T$ is a primary field a anomalous term? anomalous for what, just for the transformation? $\endgroup$
    – Wein Eld
    Aug 7, 2016 at 22:13
  • $\begingroup$ The term anomaly is related to the fact that the Virasoro algebra can be related to extension of the de Witt algebra plus a central term (not entirely sure though, however that should be where the terminology comes from). $\endgroup$
    – gented
    Aug 7, 2016 at 22:29
  • $\begingroup$ Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Aug 8, 2016 at 4:34

1 Answer 1


In fact, primary fields(operators) sometimes are also called tensor field(operators). The name is justified as they transform in the same way that a tensor transforms under coordinate transformation.

To see that, we look at the transformation rule of primary fields, which by their definition, is $$\mathcal{O}'(z',\bar{z}')=(\partial_z z')^{-h}(\partial_{\bar{z}}\bar{z}')^{-\bar{h}}\mathcal{O}(z,\bar{z}).$$ Compare that to the transformation rules of a tensor with $n$ lower indices. $$T_{u_1u_2\dots u_n}'=\frac{\partial x^{v_1}}{\partial x'^{u_1} }\frac{\partial x^{v_2}}{\partial x'^{u_2}}\dots\frac{\partial x^{v_n}}{\partial x'^{u_n} }T_{v_1v_2\dots v_n}.$$ Note in the case of 2d conformal transformation, in terms of $z,\bar{z}$ coordinates, $z'$ is only a function of $z$. Thus $\frac{\partial x^{v_i}}{\partial x'^{u_i}}$ is non zero only when $v_i=u_i$. In other words, various $\frac{\partial x^{v_i}}{\partial x^{u_i}}$ are either $\frac{\partial z}{\partial z'}$ or $\frac{\partial \bar{z}}{\partial \bar{z}'}$. Therefore we can write $\frac{\partial x^{v_1}}{\partial x'^{u_1} }\frac{\partial x^{v_2}}{\partial x'^{u_2}}\dots\frac{\partial x^{v_n}}{\partial x'^{u_n} }$ as $(\frac{\partial z}{\partial z'})^h(\frac{\partial \bar{z}}{\partial \bar{z}'})^{\bar{h}}$, or with $(\frac{\partial z'}{\partial z})^{-h}(\frac{\partial \bar{z}'}{\partial \bar{z}})^{-\bar{h}}$ with $h+\bar{h}=n$.

Thus we just demonstrate that $\mathcal{O}(z,\bar{z})$ obeys the same transformation rule as a (component of a) tensor, and thus it is called a tensor field.

Finally, the direct consequence of $T_{zz}$ not being a primary field is that it does not transform as a tensor. Moreover because it does not transform as a tensor, its OPE can have a $z^{-4}$ term, which is related to the Casimir energy for a unitary CFT.

  • $\begingroup$ Well, you're just rephrasing the question as "the stress energy tensor not being a primary field means it doesn't transform as a primary field". $\endgroup$
    – gented
    Aug 9, 2016 at 8:48
  • $\begingroup$ Your answer is correct as long as we are looking at classical CFT's. When we are considering a quantum CFT, then the classical dimensions acquire anomalous dimensions, and hence tensor algebra cannot account for that,as it will only give is integer powers. Hence, your answer is true but only classically. $\endgroup$ Sep 16, 2018 at 12:29

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