30
$\begingroup$

Quarks and leptons are considered elementary particles, while phonons, holes, and solitons are quasiparticles.

In light of emergent phenomena, such as fractionally charged particles in fractional quantum Hall effect and spinon and chargon in spin-charge separation, are elementary particles actually more elementary than quasiparticles?

Does the answer simply depend on whether one is adopting a reductionist or emergent point of view?

$\endgroup$

4 Answers 4

14
$\begingroup$

Elementary particles, like photons and electrons, are not more elementary in the sense that there are underlying theories, such as quantum spin model on lattice, from which they can be derived as an effective approximation (see for example arXiv:hep-th/0302201).

In particular, the string-net condensation provides a unified origin for gauge interactions and Fermi statistics: Both elementary gauge bosons (such as photons, gluons) and elementary fermions (such as electrons, quarks) can emerge as quasi-particles in a quantum spin model on lattice if the quantum spin model has a "string-net condensed state" as its ground state. An comparison between the string-net approach and the superstring approach can be found here.

There is a falsifiable prediction from the string-net theory: all fermions (elementary or composite) must carry gauge charges (see cond-mat/0302460). The standard model contain composite fermions that are neutral for $U(1)\times SU(2)\times SU(3)$ gauge theory. So according to the string-net theory, the standard model is incomplete. The correct model should contain extra gauge theory, such as a $Z_2$ gauge theory. So the string-net theory predicts extra discrete gauge theory and new cosmic strings associated with the new discrete gauge theory.

The emergence approach may also produce (linear) quantum gravity from quantum spin models (see arXiv:0907.1203). However, the emergence approach (such as the string-net theory), so far, fail to produce the chiral coupling between the $SU(2)$ weak interaction and the fermions.

$\endgroup$
9
  • 1
    $\begingroup$ Comments on the linked paper (of which I assume you are co-author): already in 1960, Skyrme showed how to get Fermions from bosonic fields, using a soliton in a scalar theory stabilized with quartic derivative terms (so not well defined), but where there is an extra topological term which makes the soliton (Baryon) a fermion. Balachandran Rajeev Nair, then Witten showed that this emerges in large N QCD with quarks, but Skyrme's model has no fundamental Fermions. Emergent fermions include bound-states of monopole-electron, and 1d examples a-plenty. $\endgroup$
    – Ron Maimon
    May 27, 2012 at 3:02
  • 1
    $\begingroup$ You can have Fermions emerge without Bosons and without Gauge fields, as in the 2d sine-Gordon model. This type of Fermionization of bosonic actions is typical in 2d. It is weird to say there should be an additional discrete gauge theory, and there is no argument I could see in the linked paper that supports this. Also Z_2 gauge symmetry does not give cosmic strings. This gauge theory does not have a real continuum limit. $\endgroup$
    – Ron Maimon
    May 27, 2012 at 3:06
  • 2
    $\begingroup$ @Ron: (a)When we give Skyrme-model a non-perturbative definition by putting it on a lattice and treating it as a bosonic model on lattice, then fermion can only emerge as a gauge charge. (The topo. term must come from underlying fermions.) In the string-net theory, we assume that the underlying degrees of freedom are bosonic (such as qubits). In this case fermion can only emerge as a gauge charge. (b)In 1D space the distinction between boson and fermion is not well defined. (c)All physical theories have a cutoff at the Plank scale. With finite cutoff, Z2 gauge theory does give cosmic strings. $\endgroup$ May 27, 2012 at 15:00
  • 1
    $\begingroup$ a) You can put the topological term in by hand in the lattice, so no gauge fields. b) The distinction is well defined in 1d--- a fermion obeys Fermi-Dirac statistics, a boson doesn't--- it only becomes ambiguous with infinite repulsive forces, like bosons with infinite delta-repulsion which are the same as Fermions, c) The "cutoff" exists, but it isn't a lattice, but some sort of infrared-ultraviolet mixing (as in string theory). You're right about the string--- it's the defect where there is the nontrivial Z2 bundle on the circle at large distances. $\endgroup$
    – Ron Maimon
    May 27, 2012 at 20:13
  • 1
    $\begingroup$ Hmm, I think when you say "emergent gauge field" you are thinking of the topological term like a nontrivial bundle structure on the particles, and so it's like a remnant discrete gauge field from some microscopic gauge field that's broken or something. This is a possible point of view, but discrete gauge fields don't have to have propagating modes. There is no emergent gauge theory in the skyrme model that I can see, all you get are fermionic solitons and massless pions. But this is still an interesting point of view. I'll read the paper again now that I get where you're coming from. $\endgroup$
    – Ron Maimon
    May 28, 2012 at 19:39
13
$\begingroup$

They are more elementary in the sense that there is no accepted underlying theory from which they can be derived as an effective approximation.

On the other hand, what is elementaty changes with time. At some time, protons and neutrons were considered to be elementary particles, whily they are now considered to be composed of quarks. There are various hypothetical theories in which the particles currently viewed as elementary are considered to be composed of even more elementary particles. The latter are called preons. See http://en.wikipedia.org/wiki/Preon

Thus if one of the preon theories would gain major acceptance, the currently accepted elementary partricles would get the status of quasiparticles of an effective theory (that would be the current standard model) deduced by coarse graining from the underlying preon theory.

$\endgroup$
2
  • $\begingroup$ In fractional quantum Hall effect, the collective behavior of electrons resulted in new particles. But at the same time, these particles having fractional charge seem to be constituents of electrons. So, it seems as if electrons and the quasiparticles are actually the same thing, and there is no distinction between elementary and quasiparticles. The same argument works for spin-charge separation, where an electron results in a spinon and a chargon, which can be viewed as constituents of an electron. $\endgroup$
    – leongz
    Mar 10, 2012 at 18:45
  • 2
    $\begingroup$ elementary = given by a field in the Lagrangian of a field theory from which the other particles can be derived. I haven't seen a Lagrangian for fractional Hall particles from which one can construct electrons as a bound state. So how can these particles be elementary? $\endgroup$ Mar 11, 2012 at 8:36
6
$\begingroup$

As you mention fractional quantum Hall effect, let me consider a system of $N$ electrons typical of a condensed matter system. Now think of your Hamiltonian as having two parts

$\hat{H} =\hat{H}_0+\hat{H}_{int}e^{-\zeta t}$ with $t >0$

so that you gradually switch-off the interacting part so that at large times you can map your complete Hamiltonian (with interactions) to your free Hamiltonian. If you can do that, the matrix elements of the interacting and non-interacting case will be identical.

In the Fermi liquid theory, the quasiparticles are understood as the excitations of an interacting many-body system. They correspond to the creation or annihilation of particles [electrons] and can be labeled by the same quantum numbers as the non-interacting states provided that:

  • The adiabatic procedure is valid (that is the energy of the state larger than the rate of change, $\varepsilon_{\mathbf{k}\sigma} \ll \zeta $, which is equivalent to assuming that $T\ll\zeta$ since typically $\varepsilon_{\mathbf{k}\sigma}\simeq T$.
  • The interactions do not induce transitions of the states in question, or in other words the life-time of the state must satisfy $\tau_{\text{life}} \gg \zeta^{-1}$.

Thus the quasiparticle concept only makes sense on time scales shorter than the quasiparticle life time and we must not thought of them as the exact eigenstates. On the other hand, electrons have infinite life time (understand infinite by very very large $\tau_{\text{life,e}}\simeq10^{26}$ years). The proper "elementary" particles are the electrons, not the quasiparticles. Again, quasiparticles refers just to the excitations of the system. Obviously, some properties of your system will be described only via the excitations of the whole system, these are the emergent properties you were talking about (fractional charge, fractional statistics...).

D.

$\endgroup$
3
  • 7
    $\begingroup$ This is not specific to quasiparticle. The concept of the top quark, say, also makes sense only on time scales shorter than its life-time. On longer scales, it is only a resonance. If only infinitely stable particles were considered elementary, we would just have up, down, the electron, and the neutrinos - not even photons. $\endgroup$ Mar 6, 2012 at 13:56
  • $\begingroup$ I agree with you, I had in mind more the theory of Fermi liquids and electronic quasiparticles in the answer. $\endgroup$
    – Dani
    Mar 6, 2012 at 18:28
  • $\begingroup$ @DaniH: I have to admit a prejudice up-front--- I hate "adiabatic switching", it's a useless and obsolete tool of the 1950s that never seems to go away. There is rarely a case where it is done properly, it is never physical, and it's only purpose is to regulate infinite time integrals. There is no physical effect which depends on it, in particular your "lifetime>tau" is a nonsense condition. No downvote, but please fix. $\endgroup$
    – Ron Maimon
    May 27, 2012 at 2:59
0
$\begingroup$

Quasiparticles exist only in condensed matter. Real particles can propagate also in vacuum. If you have an underlying medium in your modell for vacuum where you can describe the known real particles as emergent quasiparticles of some yet unknown real particles, this is another thing.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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