Question on functional bosonization Let me be sort of specific. We consider a Weyl particles on $S^{1}$, with the following Hamiltonian
$$\mathcal{H}=v\int_{0}^{2\pi}\psi^{\dagger}(x)(-i\partial_{x})\psi(x)dx$$
Such particles have the property that 
$$\rho(x)=\psi^{\dagger}(x)\psi(x)$$
is ill defined, in the sense that the expectation value of $\rho(x)$ diverges. For this reson we change the definition of normal ordering to
$$\rho(x)=:\psi^{\dagger}(x)\psi(x):=\psi^{\dagger}(x)\psi(x)-\bar{\rho}(x)$$
Where the over bar is the expectation value. With this symbolism, we have that
$$[\rho(x), \rho(x')]=\frac{i}{2\pi}\delta'(x-x')$$
Istead of usual $[\rho(x), \rho(x')]=0$. We also know that for chiral bosons one has
$$[\varphi(x), \varphi(x')]=-i\pi~\mbox{sign(x-x')}$$
So by differentiating the bosonic commutator twice with respect to $x$ and then with respect to $x'$, we get precisely the commutator of densities, provided that
$$\rho(x)=\frac{1}{2\pi}\partial_{x}\varphi(x)$$
So far so good. Than one sees that
$$[\rho(x), \psi(x')]=-\psi(x)\delta(x-x')$$
In order to satisfy this commutation relation we have to make a clever guess. That is, if one asumes $\psi(x)=e^{i\varphi(x)}$, it is easy to show that indeed the algebra is satisfied. I'm pretty much satisfied with everithing by this point, it seems that we've found a canonical transformation and bla-bla-bla. What normally is then claimed in the literature that
$$\mathcal{H}=v\int_{0}^{2\pi}\psi^{\dagger}(x)(-i\partial_{x})\psi(x)dx=\frac{v}{4\pi}\int_{0}^{2\pi}(\partial_{x}\varphi(x))^{2}dx$$
And this is a place where I get stuck. So
$$\psi^{\dagger}(x)(-i\partial_{x})\psi(x)=e^{-i\varphi(x)}(-i\partial_{x})e^{i\varphi(x)}=e^{-i\varphi(x)}(\partial_{x}\varphi(x))e^{i\varphi(x)}=e^{-i[\varphi(x), \cdot]}\partial_{x}\varphi(x)=?$$
I.e. what is the commutator $[\partial_{x}\varphi(x), \varphi(x)]$? Does it even make sense? In this expression may also write
$$(\partial_{x}\varphi(x))e^{i\varphi(x)}=2\pi\rho(x)\psi(x)=2\pi\rho(x)\psi(x')\big{|}_{x'=x}=2\pi\psi(x')\rho(x)\Big{|}_{x'=x}-2\pi\psi(x)\delta(x-x')\Big{|}_{x'=x}=2\pi\psi(x)\rho(x)+?=e^{i\varphi(x)}\partial_{x}\varphi(x)+?$$
What am I doing wrong?
 A: I am basically following the first chapter of Bosonization  by Michael Stone. 
Starting from the bosonization formula:
$$\psi(x) = :e^{i\phi(x)}:$$
And the Hamiltonian
$$H = i \int dx  \psi^{\dagger}_R(x) \partial_x \psi_R(x) $$
In the bosonization of the composite terms, we need to point split and take care of the normal ordering according to:
$$ :e^{ia\phi(x_1)}::e^{ib\phi(x_2)}: =:e^{ia\phi(x_1)+ib\phi(x_2)}:e^{ab\log{(x_1-x_2)}} $$
Thus we have:
$$\begin{align*} 
\psi^{\dagger}_R(x) \partial_x \psi_R(x) 
& = \lim_{x' \to x} :e^{-i \phi(x')}: \partial_x :e^{i \phi(x)}:\\
&= \lim_{x' \to x} :e^{-i \phi(x')}:  :e^{i \phi(x)}:i  \partial_x \phi(x)\\
&=\lim_{x' \to x} (x'-x)^{-1}:e^{-i \phi(x') + i \phi(x)}: i  \partial_x \phi(x)\\
&= \lim_{x' \to x}  (x'-x)^{-1} \Big(1 - i (x'-x) \partial_x \phi(x) + O((x'-x)^2)\Big)i  \partial_x \phi(x)\\
& = \lim_{x' \to x}\frac{i}{x-x_1}+  \partial_x \phi(x) \partial_x \phi(x) + O(x'-x)
\end{align*} $$
The singular term is taken care of by the normal ordering of the right hand side:
Thus we get:
$$H = i \int dx (\psi^{\dagger}_R(x) \partial_x \psi_R(x) = i \int dx  :( \partial_x \phi(x))^2 :$$
Update: Proof of the normal ordering identity:
$$:e^{ia\phi(x_1)}::e^{ib\phi(x_2)}: = e^{ia\phi_{-}(x_1)+ia \phi_{+}(x_1)} e^{ib\phi_{-}(x_2)+ib \phi_{+}(x_2)}$$
where $\phi_{+}$ contains only creation operators and $\phi_{-}$ annihilation operators. In order to normal order the expression we need to commute between the second and third terms.
Using the Campbell–Baker–Hausdorff (CBH)
$$e^{A} e^{B} = e^{A+B} e^{[A, B]/2}$$
(for $[A, B] = const.$)
The relevant modes from the mode expansion
$$\phi_{-}(x)= \sum_{n>0} \sqrt{\frac{2}{n}} a_n e^{-\frac{nx}{L}}$$
$$\phi_{+}(x)= \sum_{n>0} \sqrt{\frac{2}{n}} a_n^{\dagger}e^{\frac{nx}{L}}$$
$L$ is the quantization box length which will be taken eventually to infinity
Using
$$[a_n, a_m^{\dagger}] = \delta_{mn}$$
We get:
$$[\phi_{-}(x_1), \phi_{+}(x_2)] = 2\sum_{n>0}\frac{1}{n}e^{-\frac{ n(x_1-x_2)} {L} } = 2 \log(1-e^{-\frac{ (x_1-x_2)} {L} })\rightarrow_{L\to \infty} 2\log(x_1-x_2) + \mathrm{const.}$$
