Edge theory of FQHE - Unable to produce Green's function from anticommutation relations and equation of motion? I'm studying the edge theory of the fractional quantum Hall effect (FQHE) and I've stumbled on a peculiar contradiction concerning the bosonization procedure which I am unable to resolve. Help!
In particular, consider the first few pages of X.G. Wen's paper "Theory of the edge states in fractional quantum Hall effects". Here, Wen defines a fermionic field $\Psi(x,t)$ in (1+1) dimensions in terms of a bosonic field $\phi(x,t)$ as
$$ \Psi(x,t) \propto e^{i\frac{1}{\nu}\phi(x,t)} .$$
The number $\nu$ is the filling fraction of the FQHE, which we shall set to $\nu=1/3$ for simplicity. The bosonic field fulfills the somewhat strange commutation relations
$$ [\phi(x,y),\phi(y,t)] = i\pi\nu\,\text{sgn}(x-y) $$
which are necessary to make $\Psi(x,t)$ anticommute like a proper fermion
$$ \lbrace \Psi(x,t), \Psi^\dagger(y,t) \rbrace = \delta(x-y) .$$
Moreover, the bosonic field satisfies the equation of motion
$$ (\partial_t-v\partial_x) \phi(x,t) = 0 .$$
Some operator algebra shows that $\Psi(x,t)$ is also a solution to this equation of motion. However, it appears to me that these two requirements, anticommutation and the equation of motion, already fix the Green's function of the fermion!
However, Wen goes on to note that these fermions have the Green's function (equation (2.12) in the paper)
$$ G(x,t) = \langle T(\Psi^\dagger(x,t) \Psi(0)) \rangle = \exp[\langle\frac1{\nu^2}\phi(x,t)\phi(0)\rangle] \propto \frac{1}{(x-vt)^{1/\nu}} .$$
I do not understand how this can be. After all, from the anticommutation relations and the equation of motion, we can calculate the Green's function to be
$$ G(x,t) \propto \frac{1}{x-vt} .$$
To do this, define Fourier modes $\Psi_k, \Psi^\dagger_k$, obtain the usual anticommutation relations for these, solve the equation of motion, and transform back into real space. The result will be as noted, and the exponent $1/\nu$ will be missing.

Where did the exponent $1/\nu$ go? What's wrong with calculating the Green's function from the anticommutation relations and the equation of motion?

Maybe there's something going on inside the $\delta(x-y)$ part of the anticommutation relations? If so, what exactly? Or maybe something about the ground state? Or something about the bosonization procedure as a whole?
 A: I understand that you want to compute the fermion propagator in the
operator formalism (in contrast to the path integral formalism where the
same result can be obtained). Then following José's remark, the
fermionization formula is correct, i.e., gives the canonical
anti-commutation relations iff it is normal ordered:
$\psi(z) = :\exp(i \frac{1}{\nu}\phi(z)): = \exp(i \frac{1}{\nu}\phi_+(z)) \exp(i \frac{1}{\nu}\phi_-(z))  $
where $\phi_+(z)$ ($\phi_-(z)$ ) contains only the  dependence the
creation (anihilation) field components.
The fermion propagator formula given in the question is a consequence of the product formula of two normally ordered exponentials:
$:\exp(ia\phi(z_1))::\exp(ib\phi(z_2)): =:\exp(ia\phi(z_1)+ib\phi(z_2)):exp(-ab\langle \phi(z_1) \phi(z_2)\rangle)$.
This formula can be easily verified independently for each mode using the Baker–Campbell–Hausdorff formula
Now, the computation is with respect to the boson vacuum and this is the
reason that the fermion propagator has the power dependence.
A: Here, I would like to make some additional remarks.
1) In eq (2.11) of the referred paper,  the correlation of the
boson field is given $\langle\phi(x,t)\phi(0)\rangle =-\nu \ln(x-vt)$.
This allows us to calculate
$G(x,t) = \langle T(\Psi^\dagger(x,t) \Psi(0)) \rangle = \exp[\langle\frac1{\nu^2}\phi(x,t)\phi(0)\rangle] \propto \frac{1}{(x-vt)^{1/\nu}}$.
2) It is not correct to write
$\lbrace \Psi(x,t), \Psi^\dagger(y,t) \rbrace = \delta(x-y)$, since here
$\Psi(x,t)$ is not the bare electron operator.
$\Psi(x,t) = exp(i\phi(x,t)/\nu)$ is only the projection of the 
bare electron operator into the low energy subspace. So we have
$ \Psi(x,t) \Psi^\dagger(y,t) = (-)^{1/\nu}\Psi^\dagger(y,t)   \Psi(x,t)
=-\Psi^\dagger(y,t)  \Psi(x,t)$ when $1/\nu =$ odd integer.
But $\lbrace \Psi(x,t), \Psi^\dagger(y,t) \rbrace = \delta(x-y)$ is not correct.
