The neutrality condition and the (non)-vanishing of the one-point correlator for the bosonic vertex operator Consider the massless scalar field Hamiltonian,
\begin{align}
H = \frac{1}{2}\int \Pi^2- (\partial_x\phi)^2 dx
\end{align}
with $\Pi \sim \partial_t\phi$ the conjugate field of $\phi$. This Hamiltonian is treated in a number of texts, in particular Shankar's text on Bosonization on page 1841 and the Conformal Field Theory book by Di Francesco, Mathieu and Sénéchal (chapter 6 and 9). Furthermore, my question also relates to this text by Von Delft and Schoeller on bosonization.
Shankar defines the vertex operator as the normal ordered operator $:e^{i\alpha\phi}:$. He states (p1843) that the one-point correlator is unity, $\langle :e^{i\alpha\phi}:\rangle =1$, since the exponential can be expanded and only the first term does not vanish.
But just before that he also states that the two-point correlator
$$
\langle :e^{i\alpha\phi}: :e^{i\beta\phi}: \rangle 
$$
vanishes unless $\alpha + \beta = 0$. This is the neutrality condition and it follows from the fact that $\phi\rightarrow \phi + a$ is a symmetry of the Hamiltonian, which needs to be respected by the (two-point) correlators. That sounds convincing enough, but then why does the one-point correlator not vanish according to that same argument? 
To add to (my) confusion the normal ordered form is given by
$$
:e^{i\alpha\phi}:  = e^{i\alpha\phi_+}e^{i\alpha\phi_-}
$$
where $\phi_{\pm}$ contain all creation/annihilation operators. So using some rules of exponentiated operators (see appendix C of Von Delft) one can reorder the operators such that
$$
:e^{i\alpha\phi(z_1)}: :e^{i\beta\phi(w_1)}: = f(z-w) :e^{i(\alpha+\beta)\phi(w_1)}:  + \cdots
$$
with $f(z-w)$ a (singular) c-number. This is of course just the OPE. But that would imply that if the right hand side has a non-vanishing correlator, then the left hand side is non-vanishing as well. If the one-point correlator does not vanish, why does the two-point correlator need to obey the neutrality condition?
Di Francesco has some other, more elaborate proofs using Ward identities (Chapter 9) that show that indeed $\alpha+\beta = 0$ in order for the correlator to be non-zero. In particular that means that, according to their conventions, any $N$-point correlator vanishes unless the neutrality condition is satisfied. It could be that the different texts have different conventions that I'm missing and everything is fine. Still, a different question arises: If the exponentials in the normal ordered operators are expanded, and only the first (unit) term is kept, then surely the one-point correlator is just $1$? Why does this approach fail, according to their conventions? 
Let me also mention Von Delft and Schoeller. They also state that the one point correlator of the normal ordered operator $:e^{i\alpha\phi}:$ does not vanish. Instead they define the vertex operator (chapter 9) as
$$
V_\alpha = \left(\frac{L}{2\pi}\right)^{-\frac{\lambda^2}{2}}:e^{i\alpha\phi}: 
$$
with $L$ the system size (periodic boundary conditions). They state that "evidently $\langle V_\alpha\rangle =\delta_{alpha,0} $ in the limit of $L\rightarrow\infty$.". Does the neutrality condition then only hold in the $L\rightarrow\infty$ limit?. This would invalidate Di Fransesco's treatment, I presume.
So, in short, I'm quite confused on how to "merge" these different treatments. 
 A: Well, first of all it must be clearly stated the composite operators of which quantum field
theory you consider.  I can offer you one setup in which everything is perfectly clear: the
compactified free boson living on the   cylindrical world-sheet (= toroidal compactification of the closed string).  It then has to be said what is meant by normal ordering:  for me it will mean putting the annihilation operators of the oscillatory string modes right to the creation operators but not ordering the zero modes (i.e. the  position and the momentum of the string). Now the vanishing or non vanishing of   correlators like $\langle :e^{i\alpha\phi}: :e^{i\beta\phi}: \rangle$ or  $\langle :e^{i\alpha\phi}:  \rangle$ can be seen just by performing
calculations in the zero mode sector and the  result is compatible with the neutrality condition: the 2-point correlator vanishes unless $\alpha+\beta=0$ and 1-point correlator vanishes unless $\alpha=0$. How does it come about? Well, the Hilbert space of the theory
$H$ is the tensor product $L^2(S^1)\otimes F_L\otimes F_R$ (I am ignoring the winding zero mode sector for simplicity), where $S^1$ is the target circle on which the string is compactified and  $F_{L,R}$ are the Fock spaces for the left and right movers, respectively. In particular, the vacuum $\vert 0\rangle\in H$ can be written as the direct product $\vert 0\rangle_0\otimes \vert 0\rangle_L\otimes \vert 0\rangle_R$, where $\vert 0\rangle_0$ stands for  the normalized constant function in $L^2(S^1)$ and  $\vert 0\rangle_{L,R}$ are the Fock vacua. The vertex operator  $V(\alpha)=:e^{i\alpha\phi}:$ has now the  structure $V(\alpha)=e^{i\alpha\phi_0+i\alpha pt}: e^{i\alpha\hat\phi}:$ where the $\hat\phi$ is the oscillator part of the field $\phi$.  Let us now calculate
$$\langle :e^{i\alpha\phi}:  \rangle= \ _0\langle 0\vert e^{i\alpha\phi_0+i\alpha pt}\vert 0\rangle_0 \ _L\langle 0\vert\otimes\ _R\langle 0\vert  : e^{i\alpha\hat\phi}:\vert 0\rangle_R\otimes \vert 0\rangle_L=  \ _0\langle 0\vert e^{i\alpha\phi_0 +i\alpha pt }\vert 0\rangle_0 
=$$ $$=e^{\frac{i\alpha^2t}{2}}\int_{S^1}e^{i\alpha\phi_0}d\phi_0 = \delta_{\alpha,0}.$$
The calculation of $\langle :e^{i\alpha\phi}: :e^{i\beta\phi}: \rangle$ is slightly more involved but the result is indeed proportional to $\delta_{\alpha+\beta,0}$ because of the contribution from the zero mode sector.
