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!


Nothing in your derivation is wrong. This is what people mean when they say that the Dirac operator is the "square root" of the Klein-Gordon operator.

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.

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  • $\begingroup$ Should I say that each of the four components of $\psi$ behaves like a KG field? $\endgroup$ – mithusengupta123 Jan 19 at 12:47
  • $\begingroup$ @mithusengupta123 "Should" to what end? I mean, yes, that's how the equation works, each component obeys its own KG equation. But since the components are intermixed by Lorentz transformations and so talking about individual components is not an invariant way of talking about anything, I'm not sure what you're looking for here. $\endgroup$ – ACuriousMind Jan 19 at 12:53
  • $\begingroup$ Actually. I wondered if this piece of equation has any physical significance. But I think your comment above means there is nothing physically deep about it. It's not meaningful to talk about each component separately. $\endgroup$ – mithusengupta123 Jan 19 at 13:12

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