Klein factors and Conformal Field Theory

Consider the mode expansion of a (chiral) scalar field confined to a disc with circumference L: $$\phi(x) = \phi_{0} + p_{\phi} \frac{2\pi}{L} x + \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} e^{-(k_{n}a)/2} \left(e^{-ik_{n}x} \ b_{n}^{\dagger}+e^{ik_{n}x} \ b_{n}\right)$$ with $k_{n}=\frac{2\pi n}{L}$ , $\phi_{0}$ some "zero-mode", $p_{\phi}$ some "conjugate momentum" and $a$ some short-distance cut-off. The operators fulfill the following bosonic commutation relations $$\left[b_{n}^{\dagger} , b_{n'}\right]=\delta_{n,n'} \quad\text{and}\quad \left[\phi_{0},p_{\phi}\right]=i$$

(Fermionic) Vertex operators are defined by $$V_{\alpha}(x)=:e^{i\alpha\phi(x)}:$$ with $: \ ... \ :$ denoting normal ordering. Inserting the mode expansion of $\phi(x)$ into the definition of the vertex operator yields to lowest order in $\frac{a}{L}$:

$$:e^{i\alpha\phi(x)}: = \left(\frac{L}{2\pi a}\right)^{\Delta(\alpha)} e^{i\alpha\phi(x)}$$ with the "scaling dimension" $\Delta(\alpha)=\frac{\alpha^{2}}{2}$. The pre factor on right side in front of the exponential is sometimes called "Klein factor".

Now here are my questions (They may really be "Newbie"-CFT-questions;) ) :

1. Since the right hand side is only an approximation of of $:e^{i\alpha\phi(x)}:$ to lowest order I am wondering whether the left hand-side reproduces the correct (say) fermonic commutators in all cases and whether hand side only partially reproduces the correct fermonic commutators?

2. If the right-handside only indeed only partially reproduces the correct commutation relations how can we say that a certain product of fermionic operators (say a product of 3 fermionic operators) indeed obeys the correct sermonic commutators when written in the "bosonized language"?

3. What is the importance of the higher-order terms in $\frac{a}{L}$ in the "expansion" of the vertex operator?

4. Is all this a more general construction in CFT?

I am looking forward to your responses!

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The physical limit is $a\to 0$ so the terms in the operator that are subleading in $a/L$ go to zero and may be neglected.
But the $:\exp:$ operator is really "mostly" contained in the leading piece and the leading piece doesn't get subtracted in the definition of the operator, so the subleading terms in $a/L$ may be ignored just like the intuition suggests.
Dear Costa, the right prefactor (normalization) is clearly needed for the right normalization of the commutator, but that's it. The essence of the operator is the same, however. It's like if you used $X=x(a/L)^n$ instead of $x$, then the commutator $[X,p]$ will be $(a/L)^n \cdot i\hbar$ instead of $i\hbar$. But the special point about the exponential operators is that only the (divergent) overall normalization is affected by the normal-ordering effects. At least I hope so. – Luboš Motl Apr 4 '14 at 11:11