The article " Type -II Dirac Fermions" in Physics 10, 74, July 5, 2017 suggests these particles have been "spotted" and are a low-energy condensed-matter particle that "break Lorentz invariance having no counterpart in the Standard Model. Are these true particles or maybe something else I am not understanding. The reason I ask is that the article is suggestive that there are many more of these "electron states" that may translate to more particles not accounted for by the Standard Model. Just curious.

  • 1
    $\begingroup$ Must be quasi-particles. Don't take them serious as particles. $\endgroup$ – JamieBondi Jul 13 '17 at 2:40

Every so often I see questions regarding exotic particles in condensed matter systems and a rush of particle physicists coming to claim that the particle is not "real". I want to tell the other side of the story.

To answer your question, yes, if you look at the field operators in the Standard Model, you won't find anything that looks like a Type-II Dirac Fermion, that much is true. If your definition of particle is something that has a field operator appearing in the Standard Model, then you stop here. However, if you want to study what happens when you take more than just a few of these particles (say $10^{23}$), like in a piece of copper or in a neutron star, you need to change your perspective.

Once you have many interacting particles, there is almost no hope of using the underlying Lagrangian (or Hamiltonian) to describe all the strange and new physics that comes from interaction. Consider that an analytic solution to the Helium atom is out of reach. Instead, you try to create an effective description, that approximates the interacting system. If you still want to do a Quantum treatment of the system, you need to write down a field theory, which inevitably uses field operators. By studying the physics of the interacting system (mostly by doing experiments), you can gain an idea of what field operators and interaction terms to use. These new field operators are the effective particles of your system (often called quasi-particles). Your underlying Lagrangian restricts your new field operators in some cases, but not all. That is why you can have electrons with fractional charge, Type-II Dirac Fermions, Anyons, Majorana Fermions, etc.

But wait a minute, let's take a step back. How do we know whether the underlying Lagrangian for the original system of particles (e.g. the Standard Model) is not an effective model as well? Turns out you don't know! In fact, just recognizing this fact is a core concept behind the Renormalization Group method.

So I advocate for you not to think of Type-II Dirac Fermions (or Skyrmions, Plasmons, Phonons, Polaritons, Excitons, Magnons, etc.) as new particles per se, but a new effective Lagrangian, with terms that have no analogue in the Standard Model.

| cite | improve this answer | |
  • $\begingroup$ That shed light on this mystery for me...many thank you's . But now you bring up a new mystery for me. These new effective Lagrangian's can and may in fact someday be used to enhance and create new properties and assemblages of semiconductors. The question now remains are these "effective Lagrangian's" fundamental like the electron or are they describing a "flow" or agitation like as in "current" "flowing" in an electrical wire? $\endgroup$ – Sedumjoy Jul 13 '17 at 13:32
  • $\begingroup$ by fundamental...I mean it would be hard to break up an "electron" into more "fundamental" parts right? Can we say the same thing about "effective Lagrangian's". $\endgroup$ – Sedumjoy Jul 13 '17 at 15:24
  • 1
    $\begingroup$ I'd say these properties have already been used in semiconductor device construction. All these theories have cutoffs, beyond which they break down and are not accurate. Even the standard model is expected to break down. Naturally the energy scale for the break down varies. For a semi conductor this may be on the scale of eV. For the standard model it is much much larger, at least near the Planck scale where gravity comes into play. $\endgroup$ – KF Gauss Jul 13 '17 at 15:29

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.