Is the free Bose field a conformal field? Let's consider the free Bose field 
$$S=\frac{1}{2}g\int d^2x\ \partial_\mu\varphi\partial^\mu\varphi.$$
The action apparently shows that the system has conformal invariance (at least at the classical level). And we can read off that the Bose field has conformal dimension $0$. But now, I read in a book (An introduction to Conformal Field Theory: With Application to String Theory, R. Blumenhagen & E. Plauschinn, pp 50, below Eq.(2.92)) that:

As we have mentioned earlier, the free boson $\varphi(z, \bar{z})$ is not a conformal field since its conformal dimensions vanish $(h, \bar{h}) = (0, 0)$.

My question is, does the vanishing conformal dimension indicate that the field itself is not a conformal field? How exactly do we define a conformal field?  
Furthermore, the two point correlator of a (quasi-)primary field with vanishing conformal dimension should have the form $\langle\varphi(x)\varphi(y)\rangle= \frac{C_{12}}{|x_1-x_2|}$, but for the free Bose field, on the other hand, we have $\langle\varphi(x)\varphi(y)\rangle= -\frac{1}{4\pi g}\ln (x-y)^2$. Why are they different? Maybe you will say, it is the difference for the correlator forms that indicates the free Bose field is not a conformal field. But I would like to know is there any deeper reason to explain why its correlator is different from that of a primary field and hence itself is not a conformal field?
 A: To be precise, the two-point function of (0,0) fields is really
$\langle \varphi(x) \varphi(y) \rangle \sim \log ( \mu^2 |x-y|^2 )  $
where $\mu$ is the IR cutoff of the theory. Note that this is the only dimensionally consistent equation one can write down. This immediately indicates why $\varphi(x)$ is not a well-defined operator in the quantum theory - its correlators are regulator dependent! It therefore cannot be an operator in the quantum theory since all physical operators are required to be regulator independent. 
That being said, it is convenient to continue to discuss the object $\varphi(x)$ as one may construct actual operators in the Hilbert space out of it, for instance $\partial_\mu \varphi(x)$ or $: \exp [ i k \varphi(x)] : $, etc. 
A: Indeed, the most obvious reason a 2D free boson field is not a proper field of a conformal field theory is that its two-point function is "wrong". This, however, should not itself be surprising - if we just start from a field invariant under conformal symmetry (which $\phi$ is), then we would from the "pure" CFT standpoint expect the theory to be trivial too. However, if we switch to a better description of the theory it becomes apparent how to read it as a proper CFT: The idea is that of usual bosonization, and we declare the "true" conformal field to be $V_\alpha = \, :\mathrm{e}^{\mathrm{i}\alpha\varphi}:$.
How do we know there's a "true" CFT hidden here? Well, we know that the space of states is not trivial! We have the currents $j(z) = \mathrm{i}\partial\varphi$ and its antiholomorphic counterpart. Expanding these in modes (since they fulfill $\partial j = \bar{\partial}\bar{j} = 0$) and integrating yields
$$ \varphi(z,\bar{z}) = \phi_0 - \mathrm{i}(a_0z 
 + \bar{a}_0z) - \mathrm{i}\sum_{n\neq 0}\frac{1}{n}\left(\mathrm{e}^{-nz}a_n + \mathrm{e}^{-n\bar{z}}\bar{a}_n\right)$$
Examining the commutation relations leads us to intepret $a_n$ as creation and $a_{-n}$ as annihilation operator for $n > 0$. The Hilbert space then must carry a representation of this operator algebra, and in particular it must decompose as the direct sum of Fock spaces built from an eigenvector of $a_0$.
Now, the exponentials in the mode expansion look rather "un-conformal". We usually want mode expansions like $a_nz^{-n-h}$. So we apply the cylinder-to-plane map $z\mapsto \mathrm{e}^{-z}$. This leads us to consider the theory of the conformal fields $V_\alpha$, and one could find that these obey the proper n-point functions for conformal fields. That is, the theory of 2D bosons on a cylinder is itself not naturally interpreted as a pure CFT, but its equivalent theory on a plane (with the usual conceit of CFT where we remove the origin and think of it as the infinite past) does possess such an interpretation.
A: This might be an old question but it appeared as a recommendation in a similar question. As my answer there states, it is possible to keep the bosonic field $\varphi$ around if you really want to but you must then resort to the language of logarithmic CFT.
