Klein-Gordon-Equation contains no Spin I have a question about an argument used in Schwabl's "Advanced Quantum Mechanics" concerning the properties of the Klein-Gordan-Equation (see page 120):

Since the eigenenergies of free solutions are $E= \pm \sqrt{p^2c^2+m^2c^4}$the energy states aren't bounded from below. But I don't understand why that  then K-G-equation provide a scalar theory that does not contain spin and then could only describe
particles with zero spin. 
Intuitively, I guess because that spin can't regard by a scalar equation but I find this "argument" too squishy and would like to know a more plausible argument. 
 A: The components of every field must satisfy the KG equation, regardless of its spin. This makes sense, since it is nothing more than just the Einstein energy momentum relation.
The idea is that the KG equation doesn't require that particles be spinors, so it can't be the full story since we know particles have spin (i.e. they transform under the 2 representation of the Lorentz group).Sure, it would hold for all components of a spinor, but any good theory should force us to conclude that an electron field is spin 1/2. 
However, maybe there is a more fundamental equation that encodes the spin information of a particle and yields the KG equation. This is the Dirac equation. The Dirac equation necessitates that (what we now call) spin 1/2 particles are represented by the direct sum of Weyl spinors. Specifically a direct sum of the $2$ and $\bar{2}$ representations of the Lorentz group. This precisely means that, say, the electron field necessitates the existence of an anti electron field, which together are described by a spin 1/2 field and an anti spin 1/2 field.  
