Kinetic energy operator in Dirac's relativistic quantum theory

In non-relativistic quantum theory $\hat{K}=\hat{p}^2/2m$, What is the Kinetic energy operator in Dirac's relativistic quantum theory?

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Dirac's equation is a relativistic equation so it is more natural to talk about the total relativistic kinetic energy, including the $E=m_0 c^2$ latent energy.
With this understanding, the question is equivalent to the question what is the Hamiltonian for a free Dirac particle. Because Dirac's equation says $$(p^\mu \gamma_\mu - m)\Psi = 0,$$ we may separate the temporal part $\mu=0$ and the spatial part, and we get $$(p^0 \gamma_0 - p^i \gamma^i - m) \Psi = 0$$ Multiply it by $\gamma_0$ from the left side to get $$p^0 \Psi = ( p^i \gamma_0\gamma^i +m\gamma_0)\Psi$$ and the whole parenthesis on the right hand side that acts on $\Psi$ is the operator of the total relativistic kinetic energy.
Thanks Luboš, and how one should subtract $m_0 c^2$ ? – richard Jul 29 '14 at 16:15
If you subtract $m_0 c^2$ from the operator i.e. if you add $(-m)$ – I used units with $c=\hbar=1$ everywhere - it just means that $m\gamma_0$ will change to $m(\gamma_0-1)$. Subtraction is subtract, just minus what you subtract. Note that when you do this subtraction, slow electrons will indeed have a very small "non-relativistic" kinetic energy because the corresponding $\Psi$ is "almost" annihilated by the $\gamma_0-1$ operator. – Luboš Motl Jul 29 '14 at 16:18