Antiparticles of spinors

Question

As far as I know, to couple scalar fields with photons, the fields must be complex, and have two degrees of freedom, which explains why the antiparticles exist. In the spinor cases, spinors themselves are already complex, and the right-handed spinors are antiparticles of left-handed spinors, but what about the complex conjugate of both right- and left-handed particles? Aren't they representing any particles?

From the perspective of degrees of freedom, complex scalar fields have two degrees of freedom corresponding to two types of spin $0$ particles, while spinor fields have 8 degrees of freedom, but they only correspond to two types of spin 1/2 particles, which needs only 2 degrees of freedom each. What about the rest of the 4 degrees of freedom?

Answer 1

As far as I know, to couple scalar fields with photons, the fields must be complex

Not necessarily. Schrödinger (Nature 1952, 169, 538) noticed that the scalar charged field interacting with the electromagnetic field can be made real by a gauge transform. See also my answer to Why can't a real scalar couple to the electromagnetic field? .

Let me also note that the number of degrees of freedom is not determined just by the number of components of the field, but also by the order of the differential equation of motion for this field. See, e.g., explanations in Feynman, Gell-Mann, Phys. Rev. 1958, 109, 193.

It is also relevant to your question that, generally, three out of four complex components of the spinor field can be algebraically eliminated from the Dirac equation in electromagnetic field, and the remaining component can be made real by a gauge transform (my article J. Math. Phys. 2011, 52, 082303).