# Bosonisation of two non-interacting Fermions

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?

## 1 Answer

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