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?
[Physics] the physical meaning of a primary field
conformal-field-theory
Best Answer
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.