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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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1 Answer 1

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.

One must realize that relativity guarantees that such an equation has both positive-energy and negative-energy solution. One has to switch to quantum field theory i.e. "second quantize the Dirac field" to get a system where the full-fledged Hamiltonian is bounded from below.

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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

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