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;) ) :
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?
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"?
What is the importance of the higher-order terms in $\frac{a}{L}$ in the "expansion" of the vertex operator?
Is all this a more general construction in CFT?
I am looking forward to your responses!
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