1
$\begingroup$

In Dirac field (Peskin and Schroeder), there is one equation in which it multiples the Dirac operator $$(-i\gamma^{\mu}\partial_{\mu}-m )$$ by $$(i\gamma^{\nu}\partial_{\nu}-m ),$$ obtaining $\partial^2+m^2 = 0$.

What does it mean?

$\endgroup$
4
$\begingroup$

It means that if a given field is a solution of the Dirac equation, then its components are automatically solutions of the Klein-Gordon equation.

This is basically how the Dirac equation is 'derived' (justified, really). The Klein-Gordon equation is nice and relativistically invariant, but the fact that it's second-order in time is awkward, inconsistent with the form we'd like to have for a Schrödinger equation, and ultimately undesirable. It's therefore desirable to have a "square root" of the Klein-Gordon operator, i.e. something which is first-order in time (and therefore in the form of a Schrödinger equation) and which naturally squares to the Klein-Gordon rendering of the relativistic energy-momentum dispersion relation. This is, of course, impossible using $c$-number coefficients, but having non-commutative operators makes the trick work ─ and of course, when you're finished, you're left with the Dirac equation.

$\endgroup$
  • $\begingroup$ then what about the spin of the particles? $\endgroup$ – Basant Acharya May 10 at 15:48
  • $\begingroup$ kg defines particles with integer spin (meson,photon, graviton) and dirac having half integer(electron).. Then where is the spin part in the solution if kg solution is dirac solution? $\endgroup$ – Basant Acharya May 10 at 15:52
  • $\begingroup$ The spin falls out naturally as a requirement of that generalization procedure. See your textbook for the details. $\endgroup$ – Emilio Pisanty May 10 at 15:53
  • $\begingroup$ ok thank you... $\endgroup$ – Basant Acharya May 10 at 15:55
1
$\begingroup$

All solutions of the free Dirac equation also solve the free Klein-Gordon equation. The reason is that the Klein-Gordon equation is the wave equivalent of Einstein's energy momentum equation, $E^2 = p^2 +m^2 $ obeyed by all free bodies in special relativity. Units such that c=1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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