Assume we have 2 sets of non-interacting fermions which I show by $\psi^{\pm}$ and $\chi^{\pm}$ where we have $\left< \psi^{+}(z) \psi^{-}(0) \right>=\frac{1}{z}$ and similar for $\chi$. Now we bosonise the fermions by setting $\psi^{\pm}=e^{\pm i X}$ and $\chi^{\pm}=e^{\pm i Y}$ where $X$ and $Y$ are free bosons. I can see this procedure captures that $\psi^{\pm}$ anti commute with each other, however, I cannot see that it captures that $\psi$ and $\chi$ also anti commute. For example, if I set $j=\psi^{+} \chi^{+}=e^{iX} e^{iY}$, the OPE of $j$ with $\psi^{-}$ equals $- \frac{\chi^{+}}{z}$, however, using bosons $X$ and $Y$, I get $+ \frac{e^{iY}}{z}$. Or another issue that I have is that I do not see why $\psi^+$ and $\chi^+$ anti commute? Do you have any solution for what I am missing?
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You need to add extra Klein factors to the bosonized Fermi fields to make $\psi =\exp\{ iX\} $ and $\chi=\exp \{iY\} $ anticommute. You can, for example multiply $\chi$ by $\exp\{i\pi N(\psi)\}$ where $N(\psi)$ is the number operator for the $\psi$ fermions. The each time a $\psi$ is commuted through $\chi \exp\{i\pi N(\psi)\}$ you get a minus sign. These factors are ugly, and can often be ignored, as is explained in the lectures I linked to above
