Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.