# How can a Dirac field $\psi$ satisfy KG equation? What's wrong in my derivation?

The covariant form Dirac equation $$(i\gamma^\mu\partial_\mu-m)\psi(x)=0$$ can be multiplied from the left with the operator $$(i\gamma^\nu\partial_\nu+m)$$ and 4xpanding it out, to get $$(i\gamma^\nu\partial_\nu+m)(i\gamma^\mu\partial_\mu-m)\psi(x)=0,\\ \Rightarrow(\gamma^\nu\gamma^\mu\partial_\nu\partial_\mu+m^2)\psi(x)=0.$$ Changing the indices $$\mu\leftrightarrow\nu$$, we also have an equation $$(\gamma^\mu\gamma^\nu\partial_\mu\partial_\nu+m^2)\psi(x)=0$$ Now if we add them and use the anticommutaror formula, $$[\gamma^\mu,\gamma^\nu]_+=2\eta^{\mu\nu}$$, we find, $$\big([\gamma^\mu,\gamma^\nu]_+\partial_\mu\partial_\nu+2m^2\big)\psi(x)=(2\eta^{\mu\nu}\partial_\mu\partial_\nu+2m^2)\psi(x)=0$$ which is same as $$(\partial_\mu\partial^\mu+m^2)\psi(x)=0$$ the Klein-Gordon equation.

Now clearly this must be wrong. A Dirac field is spin-$$1/2$$ fermion field and it cannot satisfy KG equation. But which step is wrong is derivation? If this derivation is correct, how should I interpret this? Should I interpret that each of the four components of $$\psi(x)$$ behaves like a KG field? Please help!

But it's not quite the Klein-Gordon equation because $$\psi$$ is still a Dirac spinor, i.e has four components, not one like the scalar field in the Klein-Gordon equation.
• Should I say that each of the four components of $\psi$ behaves like a KG field? – mithusengupta123 Jan 19 '20 at 12:47