Talk:Spin–charge separation: Difference between revisions

‹See TfM›

Physics Start‑class Mid‑importance

This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics articles Start This article has been rated as Start-class on Wikipedia's content assessment scale. Mid This article has been rated as Mid-importance on the project's importance scale.

As far as I know, spin-charge separation was first observed directly in Cleaved-Edge overgrowth quantum wires, by Auslander et al., where the entire dispersion of the 1D modes was measured. This is reported in Science 295: Tunneling Spectroscopy of the Elementary Excitations in a One-Dimensional wire—Preceding unsigned comment added by 140.247.59.113 (talk • contribs) 07:48, October 1, 2007

Surely it can't be arbitrary which of the spinon or chargon is fermionic? After all, there's a huge difference between a Bose-Einstein and a Fermi-Dirac statistics!!! AnonyScientist (talk) 09:42, 23 August 2008 (UTC)[reply]

Note: IANAphysicist. I'm similarly confused. The article says that holons have zero spin, which implies that they are bosons, and thus the spinons have to be fermions (of spin 1/2, if the bound state is to look like an electron). Then it says it doesn't matter which is a fermion and which is a boson. Someone needs to collapse the wavefunction of this article, because it's in a superposition of inconsistent states. I don't feel right doing because I don't know enough about the subject, only that this inconsistency contradicts everything else I've read about quantum physics. Jpkotta (talk) 01:18, 21 September 2009 (UTC)[reply]

I thought those statistics only applied to situations with more than 2 dimensions, cf anyons. (Has its spin been measured?) Either way, if the electron has a spin of 1/2 and the holon has a spin of 0, it stands to logic that the spinon has a spin of 1/2. Is there a source for the statement marked as dubious? - MK (^t/_c) 14:34, 30 October 2009 (UTC)[reply]

Oh, and this article provides some readable insight as to how this works. - MK (^t/_c) 02:44, 31 October 2009 (UTC)[reply]

The article now has a certain amount of explanation of this. The main point is that you can construct your theory in either way, and you'll probably get different results in your calculations depending on how you assign the statistics, but the true physical properties obviously can't depend on your choice. If you could do perfect calculations making no approximations, then you'd get the same results either way. But in actual fact you always have to make approximations when dealing with strongly-interacting systems, and the techniques you can make and the approximations you're forced to use depend on which way you set up the problem in the first place. When you're done, you have to go and look at some experiments to decide which approximations give the best agreement with what happens in real life. [Note: I am a physicist.] Stevvers (talk) 14:45, 26 March 2010 (UTC)[reply]

It's been suggested that this article should be merged into the Cooper pair article. — V = I * R (talk) 22:46, 8 August 2009 (UTC)[reply]

I'm no physicist, but how are Cooper pairs (electrons that are bound together) related to the topic of this article, which is electrons behaving as a bound state of two independent quasiparticles? — Hex (❝?!❞) 21:54, 13 August 2009 (UTC)[reply]

They're not related. - MK (^t/_c) 02:44, 31 October 2009 (UTC)[reply]

I've made some substantial changes to the article, in the hope of making it a bit clearer. The main changes are:

1. I've rewritten the first couple of sentences to couch it in terms of deconfinement, which is the standard language that physicists use to talk about these things. The point is that you can always think of an electron as these two separate particles if you like, but in virtually every case the chargon and spinon are bound together and you can't separate them (like quarks in a proton). Only in certain situations do the two become deconfined, meaning that they are no longer in a bound state and can behave as separate particles (like quarks in the quark-gluon plasma).

2. I've removed the discussion of statistics from the part where the counterintuitiveness is pointed out. Statistics aren't so important, because this mostly applies in 1D. The main thing is that you can't build a chargon up out of electrons, unlike a Cooper pair, say. The discussion of statistics is in its own paragraph below.

I'm not sure that I'd really consider myself an expert on this, but I've removed the tag. Stevvers (talk) 14:53, 26 March 2010 (UTC)[reply]

Is the difference between the notation 'chargeon' and 'holon' that of charge? From my understanding, spinon and holon were the spin-charge separation theory particles, whereas a separate and as far as I know outdated or as of yet unobserved theory was particle-flux separation, which posited a chargeon and a fluxon. They were both apparently doubted shortly before one was observed so I assumed the latter precluded the former. - MK (^t/_c) 02:55, 30 March 2010 (UTC)[reply]

My understanding is just that the term 'holon' is only used for a particle that carries positive charge, whereas 'charg[e]on' (the two spellings are about equally common) can be used for either. It's true that in the 1D conductors where spin-charge separation has been seen, the carriers are usually holes and so the fractionalized charge carriers have positive charge, but since this article is supposed to be more general I think it makes more sense to use 'charg[e]on'. Stevvers (talk) 13:11, 30 March 2010 (UTC)[reply]

I didn't even notice I was spelling chargeon differently. I'm just used to g[a/u/o] being a hard g and g[e/i] being a soft g. So I read chargon as charg-on in my head and hypercorrected. As for the charge, perhaps an anti-holon will indeed be called a charg[e]on. We shall have to wait until someone undertakes an experiment using positrons instead :) - MK (^t/_c) 23:47, 30 March 2010 (UTC)[reply]