It appears that zitterbewegung, a frequency associated with the total energy of a particle or system, is widely considered to be an unphysical quantity (e.g., Kobakhidze et.al.), @Lubos Motl, McMillan). However, a few physicists including Hestenes, Recami et.al.,consider it to be a fundamental, physical quantity.

It appears that, for example, the photon emitted in transitions between two states in an atom has precisely the frequency of the "beat frequency" (difference frequency) between the zitterbewegung frequencies of the two states.

Why is zitterbewegung considered unphysical by most physicists, and has there been a solid refutation of Hestenes' position that zitterbewegung is physical?

  • 2
    $\begingroup$ well, "beat frequencies" are differences, they do not depend on the absolute quantity $\endgroup$
    – lurscher
    May 18, 2018 at 16:08
  • $\begingroup$ You appear to be misreading several of your sources and slapping an ill-defined "unphysical" label on selected statements in them. Zitterbewegung is needless in QFT, but topical and useful in Dirac's equivalent relativistic fermion QM and results in specific terms in the correct hamiltonian, so, then "physical". At least one source indicates it is merely obsolete, as you may bypass it in QFT, not "unphysical". $\endgroup$ May 18, 2018 at 16:33
  • $\begingroup$ Some people claim that the Darwin term in fine structure of hydrogen is from the zitterbewegung. $\endgroup$
    – Mauricio
    May 18, 2018 at 16:37
  • 2
    $\begingroup$ @Mauricio. That's the point: "some people"="hole theory interpreters". It's a fine interpretation in that (holes in sea) picture. It is meaningless in QFT. Choose your language and use its tools. $\endgroup$ May 18, 2018 at 16:47
  • $\begingroup$ @CosmasZachos, maybe I misread, but here are some quotes: "The "Zitterbewegung" is completely unphysical."(Lubos Motl), and "One such unphysical effect, zitterbewegung, is discussed in Section 4.4" (McMillan). $\endgroup$
    – S. McGrew
    May 18, 2018 at 16:55


Your Answer

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

Browse other questions tagged or ask your own question.