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The main question is: why does nobody care about massive chiral fermions?

It is well-known that in QFT (in the axiomatic framework of Garding & Wightman) the quantum field transform according (possible reducible) finite representations of the universal covering of the (proper and orthochronous) Lorentz group $SL(2,\mathbb{C})$. The irreducible representations of $SL(2,\mathbb{C})$ are labeled with two half nonnegative integers $(A,B)\in \frac{1}{2}\mathbb{Z}_{\ge0} \times \frac{1}{2}\mathbb{Z}_{\ge0}$ (according to two copies of irreducible representations of $SU(2)$). In particular a left-handed chiral fermion transforms according to $(\frac{1}{2},0)$-representation and a right-handed chiral fermion transforms according to $(0,\frac{1}{2})$-representation. A Dirac fermion transforms according the reducible representation $(\frac{1}{2},0) \oplus(0,\frac{1}{2})$.

On the other hand, free quantum fields can be constructed from creation and annihilation operators acting on the Fock space made from the 1-particle Hilbert spaces. These 1-particle Hilbert spaces are the Hilbert spaces of irreducible unitary representations of the (universal covering) of the (proper and orthochronous) Poincare group $SL(2,\mathbb{C})\rtimes \mathbb{R}^4$. In particular, for massive particles, they are characterized by two numbers (mass and spin) $(m,s) \in \mathbb{R}_{>0} \times \frac{1}{2}\mathbb{Z}_{\ge0}$.

They are several examples in the literature of how to construct quantum fields transforming accoring the $(A,B)$-representation of $SL(2,\mathbb{C})$ which describes particles (in the Fock space) of some irreducible representation $(m,s)$ of $SL(2,\mathbb{C})\rtimes \mathbb{R}^4$. Moreover, there are constrains between the $(A,B)$ spin of the field and the spin $s$ of the particle described by such field. Also, it can be shown, that such fields (including the Fock Hilbert space and the transformations laws) satisfies all the Garding-Wightman axioms.

One canonical example is the Dirac field, which transform according the reducible representation $(\frac{1}{2},0) \oplus(0,\frac{1}{2})$ of $SL(2,\mathbb{C})$ and describes two different particles (fermion and antifermion) of spin $s=\frac{1}{2}$. BUT, we can reduce such $SL(2,\mathbb{C})$-representation and take only the left-handed chiral part (or the right-handed if you prefer) and we still have a Poincaré covariant field describing two different particles of spin $s=\frac{1}{2}$ but now the field transforms accoring to the irreducible representation $(\frac{1}{2},0)$ of $SL(2,\mathbb{C})$, and still satisfies all the Garding-Wightman axioms. In other words, the free chiral massive fermion field exists as a well-behaved QFT. (One way of doing that is taking the first two components fields of some free Dirac field in the chiral basis).

Hence, the question again is: Why nobody cares about it? In every QFT book, they explained with high detail the free Dirac field example, but nobody talks about this chiral massive free field. Why? The chiral massive free field is an example of a free field as good as the Dirac free field, or the free scalar field. In the literature, everybody claim that the chiral fields just can describe massless particles, which (at least for non-interacting fields) is false. Someone can argue that the Dirac field satisfies the Dirac equation and such equation only decouples (into two Weyl equations) for massless fields. But, who cares? Massive free chiral fermions still satisfies the relativistic Klein-Gordon equation (just any other free field) which encompasses the relativistic energy-momentum relation.

There is something like a theorem that interacting massive chiral fermion fields can not exists?

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  • $\begingroup$ Examples of massive chiral fermions are what people call Majorana fermions, they are of course well known and used. $\endgroup$ – TwoBs Apr 12 at 22:54
  • $\begingroup$ I think what @Diego is really noticing by its absence, is the charged massive chiral fermion... physics.stackexchange.com/questions/16221/… $\endgroup$ – Mitchell Porter Apr 12 at 23:52
  • $\begingroup$ My understanding is that Majorana fermions can have no charge. Otherwise their antiparticles are different. $\endgroup$ – Carl Brannen Apr 13 at 3:23

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