particle physics - Why is a fermion field complex?

The Lagrangian of a fermion field is \begin{equation} \mathcal{L} = \overline{\psi} (i\gamma_{\mu} \partial^{\mu} - m)\psi \end{equation} It is said that the fermion field $\psi$ is necessarily complex because of the Dirac structure. I don't quite understand this. Why is the fermion field complex from a physical point of view? A complex field has two components, i.e., the real and imaginary components. Does this imply that all fermions are composite particles? For example, an electron is assumed to be a point particle that does not have structure. How can it have two components if it is structureless?

Answer

Any type of field can be complex, not only the fermions. The reason is the $U(1)_{EM}$ symmetry, i.e., the electromagnetic interactions.

The electric charge is the conserved quantity of the $U(1)_{EM}$ gauge symmetry in nature. A transformation of this symmetry is such that $$ \phi(x) \mapsto e^{iq\theta (x)} \phi (x) \\ \phi^{\dagger}(x) \mapsto e^{-iq\theta (x)} \phi^\dagger (x) $$ where $q$ is the electric charge of the field, and $\theta(x)$ is the gauge parameter. If $\phi(x)$ is a real-valued field, then the first and second equations should be identical, which implies $$ e^{iq\theta(x)} = e^{-iq\theta(x)} $$ This is only true if and only if $q=0$. For complex fields the charge would be opposite for their conjugates.

So, complex fields are charged and real fields are neutral. For example, after the electroweak breaking, the Higgs field is neutral therefore a real-valued boson, while W bosons, i.e., $W^\pm_\mu \equiv W^1_\mu \mp i W^2_\mu$, are charged, so complex-valued bosons. Neutrinos are neutral so they are real-valued fermions, but electrons are charged thus complex-valued fermions.