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  How do Dirac fermions arise in graphene, and, what significance (if any) does this have for high-energy physics?

+ 6 like - 0 dislike

Graphene has a honeycomb lattice (in the absence of defects and impurities). By considering the low-energy limit of the half-filled Hubbard model used to model the strongly interacting electron gas we find that the low-energy quasiparticles obey the dispersion relation for massless fermions. These details are all covered very nicely in a paper by Gonzalez, Guniea and Vozmediano (reference) among others.

It might seem like I'm answering the question. I'm following this line of exposition because I don't want to assume that this is a topic something commonly known or understood outside the condensed matter community. Any answers which elaborated on these basics would be very useful as they would help make the discussion more broadly accessible.

My primary question is more about the implications this fact has for high-energy physics, in particular the question of emergent-matter in theories of quantum gravity. The case of graphene is a canonical example in that regard where one obtains relativistic, massless excitations in the low-energy corner of an otherwise non-relativistic system - the 2D electron gas (2DEG).

Obviously I have my own beliefs in this regard and I will try to outline them in an answer. But I also want to solicit the communities views in this regard.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
asked Jan 19, 2011 in Theoretical Physics by Deepak Vaid (1,985 points) [ no revision ]
@space_cadet: Nice question! Massless fermions - does this imply that these guys behave like neutrino's in 2D?

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Robert Filter
@Robert - pretty much. There are all sorts of other nice things that happen - such as when you introduce defects in the lattice, the low-energy long-distance effect on the above-mentioned Dirac fermions is understood by introducing a non-abelian SU(2) gauge field - see for instance Gauge fields in graphene. So not only do you have "emergent" dirac fermions but also the "emergent" SU(2) gauge field which describes their interactions with lattice defects.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
@mbq the way I had originally stated by the question seemed slightly better it was after your edit (IMHO), so I have rolled it back. But yes, graphene should have a lowercase "g" :)

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346

2 Answers

+ 6 like - 0 dislike

The emergence of global symmetry at low energies is a familiar phenomena, for example Baryon number emerges in the context of the standard model as "accidental" symmetry. Meaning at low energies it is approximately valid, but at high energies it is not.

The reason this is the case is that it so happens that the lowest dimension operator you can write, with the matter content and symmetries of the standard model, is dimension 5. The effect is then suppressed by one power of some high energy scale - it is an irrelevant operator. This is a model independent way to characterize the possibility of the emergence of global symmetries at low energies.

We can then ask about Lorentz invariance - what are the possible violations of Lorentz invariance at low energies, and what is the dimensions of the corresponding operators. This depends on the matter content and symmetries - for the system describing graphene, there is such emergence. For anything containing the matter content of the standard model, there are lots and lots of relevant operators*, whose effect is enhanced at low energies - meaning that Lorentz violating effects, even small ones at high energies, get magnified as opposed to suppressed at observable energies.

Of course, once we include gravity Lorentz invariance is now a gauge symmetry, which makes its violation not just phenomenologically unpleasant, but also theoretically unsound. It will lead to all the inconsistencies which necessitates the introduction of gauge freedom to start with, negative norm states and violations of unitarity etc. etc.

  • At least 46, which were written down by Coleman and Glashow (Phys.Rev. D59, 116008). Relaxing their assumptions you can find even more. Each one of them would correspond to a new fine-tuning problem (like the cosmological constant problem, or the hierarchy problem).
This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user566
answered Jan 19, 2011 by anonymous [ no revision ]
+ 2 like - 0 dislike

The answer you'll get from most high-energy physicists is that there are no implications whatsoever. Lorentz invariance is extraordinarily well-tested: see, e.g., http://arxiv.org/abs/0801.0287. In particular, there are many relevant operators in the Standard Model that one would expect to be generated if physics at a high scale is not Lorentz-invariant. Even some irrelevant operators that one might naively expect to appear with order-one Planck-suppressed coefficients are constrained to have smaller coefficients. Adding in gravity only makes the problem worse. For instance, most attempts to generate emergent GR from nonrelativistic theories will have an extra scalar mode and run into massive phenomenological difficulties, because they aren't really gauging the full diffeomorphism group.

To be slightly more clear: there are cases (and the free relativistic fermion emerging in the long-distance limit of graphene is one of them) where lattice symmetries can forbid dangerous relevant operators. This shouldn't happen for the full Standard Model (I assume someone has written down a careful argument for this somewhere, but I don't know a reference offhand). Still, even for the graphene case there are irrelevant operators, and we have bounds on those too. Furthermore, once you start thinking about gravity you're more or less forced to give up the hope of an underlying highly symmetric lattice that forbids all the dangerous operators.

One more half-joking comment: this argument also tells you the correct answer to the FQXi essay contest "Is Reality Digital or Analog?," so if someone fleshes it out carefully they could possibly make up to $10k from it.

(It is a good question, by the way; there's an obvious conventional wisdom from effective field theory that explains why you don't see high-energy theorists pursuing this sort of thing much, but from the outside it might not be so clear why such ideas don't generate much interest.)

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Matt Reece
answered Jan 19, 2011 by Matt Reece (1,630 points) [ no revision ]
@Matt your points are well-taken (+1), but obviously there's a sizable number of folks who feel it does (Wen, Levin, Volovik ...) otherwise I wouldn't be posing this question in this general manner. Also, the issue of emergent GR adds another layer of complication and controversy. So if someone chooses to discuss it, that's wonderful, but I would be happy to leave GR out of the discussion for now in order to keep things simple. Your point about the irrelevant operators is important and I'll try to explain in an answer why and how I feel these operators can be suppressed by the dynamics.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
There is one example I can think of where Lorentz-violating operators play an important role in high-energy physics - the formation of LOFF condensates which purport to explain aspects of color superconductivity - work done by Rajagopal, Wilzcek, Bowers and Alford among others - reference

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
This is an example of spontaneous breaking of symmetry, where there is a symmetry at high energy that is broken at low energies. This is of course what happens to LI in the real world, which at low energies is not LI, but becomes so at high energies. What you are seeking is the opposite, a case where LI is valid at some energy range without being a true symmetry at the fundamental energy scale. This is unlikely to work because of all the relevant operators.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user566
@space_cadet: you say something about my "point about the irrelevant operators," but again, the real problem is relevant operators: those which, even if you set them to be small at short distances, become large at long distances. We don't see them. It's almost inconceivable that there could be a way out of this.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Matt Reece
@matt - thanks for that clarification. Again this relevant/irrelevant operator business is something I'm still hashing out in my head.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user user346
@space_cadet: One confusion might be that for the simple case of a free relativistic fermion from the honeycomb lattice, you don't have relevant operators. I've added a paragraph remarking on this.

This post imported from StackExchange Physics at 2014-04-01 16:44 (UCT), posted by SE-user Matt Reece

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