# In a 2D CFT, is the free boson $X$ a primary field?

A primary field $$\mathcal{O}(w,\bar{w})$$ with weight $$(h,\bar{h})$$ is defined by having the following OPEs with the stress tensor:

$$T(z)\mathcal{O}(w,\bar{w})=\frac{h\mathcal{O}}{(z-w)^2}+\frac{\partial \mathcal{O}}{z-w},\qquad \bar{T}(\bar{z})\mathcal{O}(w,\bar{w})=\frac{\bar{h}\mathcal{O}}{(\bar{z}-\bar{w})^2}+\frac{\bar{\partial} \mathcal{O}}{\bar{z}-\bar{w}}.$$

Let us consider a theory of a free boson defined by:

$$S = \int d^2 z\ \partial X\bar{\partial}X.$$

Then $$\partial X$$ is a $$(1,0)$$ primary field with and $$\bar{\partial}X$$ is a $$(0,1)$$ primary field (see e.g. Tong's string theory notes, section 4.3.3).

The field $$X$$ itself has the OPE

$$T(z)X(w,\bar{w})=\frac{\partial X}{z-w},\qquad\bar{T}(\bar{z})X(w,\bar{w})=\frac{\bar{\partial} X}{\bar{z}-\bar{w}}.$$

(See e.g. Polchiski eq. 2.4.6). So, by the definition above, it should be a primary field with weight $$h=\bar{h}=0$$.

However, I've looked through many textbooks and lecture notes (e.g. Di Francesco, Blumenhagen, Polchinski, Tong...) and it is never explicitly said that $$X$$ is primary. So... Is it primary or not?

• If it is primary, then how come its descendants $$\partial X$$ and $$\bar{\partial} X$$ are also primary?

• If it is not primary, then how come its OPE is exactly the correct OPE for a primary field of weight $$(0,0)$$?

I do not agree that $X$ is a primary field. Primary field is defined by its transformation properties under the conformal group (see e.g. yellow book). In particular, under scaling transformation, a correlation function involving primary operators, transforms as $$\langle \mathcal{O}_1(\lambda x_1)\ldots\mathcal{O}_n(\lambda x_n)\rangle=\lambda^{-\Delta_1-\ldots-\Delta_n}\langle \mathcal{O}_1(x_1)\ldots\mathcal{O}_n(x_n)\rangle,$$ where $\Delta_i=h_i+\bar{h}_i$. It follows that for a primary field of dimension $(0,0)$ its two-point function must be constant. This is not true for $X$.

In fact, $X$ is not a local field in the sense of operator-state correspondence. The set of Virasoro primary operators in free boson CFT is $1,\partial X,\bar\partial X$ and the continuos family $e^{i\alpha X}$ for $\alpha\neq 0$.

• In light of your answer I have edited mine.... Is the definition given in the original question then not the correct definition? Feb 12, 2016 at 13:18
• Thank you for your answer! Could you please point me to a reference which explicitly states that $X$ is not a primary field? Polchinski (page 46-47, below eq. 2.4.17) states that the field $X$ "transforms as a tensor" and he defines a tensor operator and a primary field to be the same thing. I take it to mean that $X$ is a primary field, but it is stated in a confusing way. Is there an error in Polchinski?
– user90063
Feb 13, 2016 at 3:20
• Also, how does the 2-point function for $X$ transform exactly? It is my impression that $X(z)X(w)=\log(z-w)$ and this matches the general result that $A(z)B(w)=(z-w)^{-2h}$ if you take the limit $h \to 0$.
– user90063
Feb 13, 2016 at 3:20
• @Zoe, I'm not sure about a place where you can read that it is not a primary. From the two-point function of $X$ you can see two things. 1) In order to define it, one needs a scale, since $z$ and $w$ have dimension of length and so $\log|z-w|$ doesn't make sense without implicit assumption about a scale relative to which $|z-w|$ is measured. 2) The two-point function transforms inhomogeneously under rescalings, $\log(2|z-w|)=\log|z-w|+\log 2$. Feb 13, 2016 at 4:32
• @Zoe,@Akoben I don't have Polchinski on me, but I'll try to look up your reference later. I am used to think that $X$ is somewhat pathological by itself. For example, its two-point function does not satisfy cluster decomposition property. It is not a part of state-operator correspondence. In my opinion it is better to think of $X$ as very different from other operators, and thus not to call it a primary. Feb 13, 2016 at 4:41

To see why the descendants are primary, you can use $$\partial\left(T(z)X(w,\overline{w})\right) = T(z)\partial X(w,\overline{w}) = \frac{\partial^2 X}{z-w} + \frac{\partial X}{(z-w)^2}$$

And see that it is a primary field of weight $h = 1$, $\overline{h} = 0$, and similarly for the other field....