Bosonization and Commutation Relation I'm playing a bit with bosonization $ψ→:e^{-φ}:$ and $ψ^*→:e^{φ}:$ in the sense that
$$
\Bigg\langle 0_\mathrm{F} \Bigg|∏_{i=1}^nψ(z_i)ψ^*(w_i)\Bigg|0_\mathrm{F}\Bigg\rangle = \Bigg\langle 0_\mathrm{B}\Bigg|∏_{i=1}^n:e^{-φ(z_i)}::e^{φ(w_i)}:\Bigg|0_\mathrm{B}\Bigg\rangle
$$
where the subscripts refer to fermionic/bosonic vacuum. I would like to know if it there is a way to recover
$$
\left\{ψ(z),ψ^*(w) \right\}= δ(z-w)
$$
in terms of bosons, something like
$$
\left\{:e^{-φ(z)}::e^{φ(w)}: \right\}= δ(z-w).
$$
 A: I have a recent discussion about this problem. 
Let me assume you already have the expectation value of the equal time anti-commutator 
\begin{equation}
\langle \{ \psi( x ) , \psi^{\dagger}( x' ) \} \rangle = \langle :e^{i \phi(x) }:, :e^{ -i \phi(x' )} :\rangle = i \delta( x - x' ) 
\end{equation}
This can be established for example through analytically continuing the Euclidean boson correlator $\ln | z_1 - z_2|$ with $t = i \epsilon$ trick, or operator formalism. 
For a single oscillator we have
\begin{equation}
:e^A: :e^B: = :e^{A+B}: e^{\langle A  B \rangle }
\end{equation}
generalizing this to the vertex operator
\begin{equation}
\begin{aligned}
\{:e^{i \phi(x) }:, :e^{ -i \phi(x' )} :\} &= (e^{\langle \phi(x) \phi(x')  \rangle } + e^{\langle  \phi(x') \phi(x)  \rangle }):e^{i (\phi(x) - \phi( x' )) }: \\
&= \langle :e^{i \phi(x) }:, :e^{ -i \phi(x' )} :\rangle : e^{i (\phi(x) - \phi( x' )) }:\\
&= i \delta( x - x' ) e^{i (\phi(x) - \phi( x' )) }\\
&= i \delta( x - x' )
\end{aligned}
\end{equation}
The last line is an unexpected way to reduce an operator to a c-number. 
