# 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.

• The spin of a field refers to the representation of the Lorentz group in the target space of the field. – Qmechanic Nov 3 '18 at 13:22

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.