The derivation of both Klein-Gordon equation and Dirac equation is due the need of quantum mechanics (or to say more correctly, quantum field theory) to adhere to special relativity. However, excpet that Klein-Gordon has negative probability issue, I do not see difference between these two. What makes Klein-Gordon describe scalar field while Dirac describe spin-1/2 field?
Edit: oops. Klein-Gordon does not have non-locality issue. Sorry for writing wrongly.
Edit: Can anyone tell me in detail why $\psi$ field is scalar in Klein-Gordon while $\psi$ in Dirac is spin-1/2? I mean, if solution to Dirac is solution to Klein-Gordon, how does this make sense?
Best Answer
Spin is a property of the representation of the rotation group $SO(3)$ that describes how a field transforms under a rotation. This can be worked out for each kind of field or field equation.
The Klein-Gordon field gives a spin 0 representation, while the Dirac equation gives two spin 1/2 representations (which merge to a single representation if one also accounts for discrete symmetries).
The components of every free field satistfy the Klein-Gordon equation, irrespective of their spin. In particular, every component of the Dirac equations solves the Klein-Gordon equation. Indeed, the Klein-Gordon equation only expresses the mass shell constraint and nothing else. Spin comes in when one looks at what happens to the components.
A rotation (and more generally a Lorentz transformation) mixes the components of the Dirac field (or any other field not composed of spin 0 fields only), while on a $k$-component spin 0 field, it will transform each component separately.
In general, a Lorentz transformation given as a $4\times 4$ matrix $\Lambda$ changes a $k$-component field $F(x)$ into $F_\Lambda(\Lambda x)$, where $F_\Lambda=D(\Lambda)F$ with a $k\times k$ matrix $D(\Lambda)$ that depends on the representation. The components are spin 0 fields if and only if $D(\Lambda)$ is always the identity.