Conductance of an interacting quasi one dimensional wire

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Summary:

Why is it appropriate to discuss the conductance of an interacting quasi one dimensional wire using the method for a 1D fermi gas? We don't have a fermi gas anymore and more importantly we don't have a fermi liquid, as we are in 1D. So why do we still use 1D bands to explain the conductance?
Assuming the electrons are non interacting and spin degenerate, the conductance of a quasi one dimensional quantum wire is quantised in units of 2e^2/h. For small voltages, we simply count how many bands have their bottoms below the chemical potential and multiply this by 2e^2/h. This is due to the electron velocity and 1D density of states cancelling for all energies, when we do the integral over occupied energies for each occupied band.

Now we add electron-electron interactions. We don't have a fermi gas or even liquid now as we are in 1D. I naively thought that as the fermi gas/liquid no longer applies we couldn't rely on the above picture. But it appears that we roughly can. My question is this essentially; why?

I can sort of appreciate that it's a contact resistance so will be determined by the fermi liquid leads. Our wire itself is, as before, meant to be perfectly conducting, so the resistance should be determined by what goes on in the leads. (I'm still a little sketchy on this.) This aside, I still can't get my head around why we still talk about 1D bands and whether they have started to be filled. We don't have a fermi gas anymore and more importantly we don't have a fermi liquid, as we are in 1D. So why do we still use 1D bands to explain the conductance?

As a final remark, there are many cases in the literature, for example the 0.7 structure, where the fact the electrons interact does have some bearing. The possible explanations still talk about the 1D bands and mechanisms describing how they're filled. But we don't have a fermi gas or fermi liquid; why talk about 1D bands? Furthermore, the conductance is now not determined just by the fermi liquid leads, the interactions within the wire must now be taken into account.

Answers and Replies

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Twigg
Gold Member
Can you refer us to the text (if any) you are getting this from? A little context helps.

I'm not sure about the Fermi liquid part, but do you think the reason you can treat the interacting electrons as a fermi gas is simply that it's a mean-field picture? If it was a weak interaction, that would be appropriate.