In 1+1 dimensions, the massless KG equation has the general solution $$\phi(x,t)=\int_{-\infty}^{\infty}dp/(4\pi E_p)[a_pe^{i(px-E_pt)}+a^{\dagger}_pe^{-i(px-E_pt)}]$$ where $E_p^2=p^2$. The operator coefficients $a_p$ and $a^{\dagger}_p$ satisfy the canonical commutation relation, namely $$[a_p,a^{\dagger}_q]=2\pi\delta(p-q).$$ Now it is said that $$\Psi(x,t)=:exp[i\phi]:$$ satisfies the anticommutation relation $$\{\Psi(x,t),\Psi(y,t)\}=0$$ for $x \neq y$. Here $:O:$ of an operator $O$ is the normal ordering. How is this anticommutation relation possible? I have no idea how to express $\Psi$ in terms of $\phi$ or deal with the normal ordering. Could anyone please help me how to prove the anticommutation relation $$\{\Psi(x,t),\Psi(y,t)\}=0~?$$ It is said to be related to the bosonization of fermions.



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