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.
