# How Chern-Simons gauge field transform fermion to scalar?

In A.Zee's "QFT in a nutshell"book Page 324, after he wrote down the general Lagrangian, he said "in previous chapter, we learned that by introducing a Chern-Simons gauge field we can transform $\psi$ to a scalar field" However, I didn't found any clarification about this argument in previous chapter. Any comment or references are greatly appreciated.

• Please add more details to your question, not everyone has a Zee lying around: The general Lagrangian of what? $\psi$ is any fermion field? A Chern-Simons gauge field is the Chern 3-form here? How does Zee couple the Chern-Simons field to the fermion? – ACuriousMind Aug 6 '14 at 13:19
• The Lagrangian is just the usual form of describing electron in a magnetic field, since Zee want to argue Quantum Hall effect in that page. However, my question seems like a general argument which is independent of specific Lagrangian setup. The Chern-Simons term is just AdA, no AAA gauge self-interaction term .About the coupling thing, in previous chapter Zee argued the Chern-Simons term can be generated from gauged kinetic term by loop effects. I strongly recommend whoever interested reading Zee's book for explicit detail. – Curio Aug 7 '14 at 4:06

Yes, this is explained in the previous chapter (Fractional Quantum Hall effect)

Surprisingly, when we have a filling factor $\nu$ of the Landau levels of form $\frac{1}{3}$ or $\frac{1}{5}$, the quantum Hall fluid appears incompressible. This seems curious, because the integer quantum Hall effect appears only for integer values, corresponding to $\nu$ Landau levels completely filled (they have the degeneracy $\frac{BA}{2\pi}$, where $B$ is the magnetic field, and $A$ is the area occupied by the electrons).

So, something happens, calling $N_e$ the number of electrons ,when $\dfrac{N_e}{(\frac{BA}{2\pi})}= \nu$, with $\nu^{-1}$ odd.

Now, you have to notice that $\frac{BA}{2\pi}$ is simply the number of flux quanta $N_\phi$

So, finally, the number of flux quanta by electron, is : $\dfrac{N_\phi}{N_e} = \nu^{-1}$, a odd integer.

Now, if we look at the statistics, exchanging $2$ particles, we have the $(-1)$ term coming from the Fermi statistics, now multiplyed by an other $(-1)^{\nu^{-1}}$ term coming from a Aharanov-Bohm effect.

So, finally, one have a $(+1)$ term, which signals a Bose statistics.

A Dirac spinor in $(2+1)$ dimensions having $2$ degrees of freedom, we simply have to choose a bosonic field with $2$ degrees of freedom, that is a complex scalar field.

• Oh I see, that argument just sit above the Lagrangian..I should have read more carefully. Thank you very much for pointing that out. – Curio Aug 7 '14 at 20:18