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Fermi surface nesting and CDW/SDW/SC orders.

What is the definition of a nesting vector?

And why Fermi surface nesting gives rise to different orders at $T=0$?

(CDW: charge density wave; SDW: spin density wave; SC: superconductivity)

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2 Answers 2

up vote 8 down vote accepted

A nesting vector is a single vector in momentum space that connects large, parallel regions of a fermi surface. For example, a square lattice at half-filling has a fermi surface that looks like a diamond. If you have some scattering mechanism that takes an electron and adds a momentum $\vec{Q}$ to it, like a hamiltonian term $$g(\vec{k},\vec{Q}) \cdot c_{k+Q}^{\dagger}c_{k}^{\phantom{\dagger}}$$ then you can easily see that there's a single vector $\vec{Q}=\frac{1}{a}(\pi,\pi)$ that connects a whole lot of states in the Brillouin zone. In perturbation theory we get terms that go like $\frac{1}{E_k - E_{k+Q}}$, so if a single vector $Q$ connects lots of $k$ states (which have the same energy, by virtue of making up the fermi surface, and the states right around the fermi surface are the only ones that matter for low temperature stuff like CDW/SDW) then the static structure factor is going to be strongly peaked at that $Q$.

Fermi surface nesting gives rise to different orders because the scattering mechanisms of different materials are different. The term I wrote above didn't have any spin index on the operators. Spin preserving scattering like that would lead to a CDW. But if I had magnetic scattering that flipped the spin of the electron, I would have had an SDW state.

Lastly, superconductivity is a bit different. You don't need any nesting at all to get SC since it joins electron pairs of opposite spin and quasimomentum (or more generally time reversed pairs). This is why superconductivity is a much more robust phenomena than density waves.

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Thank you very much! –  ChenChao Dec 2 '12 at 6:09
    
As wsc said, you don't need nesting to get SC. However certain types of unconventional SC may result from partial Fermi surface nesting (although it is still unresolved in the literature). Broadly speaking in these cases (such as the pnictides) strong nesting leads to the SDW state and weak nesting results in a paramagnetic metal, but the region inbetween where there are fluctuations between the two states, the fluctuations are thought to pair electrons into a SC state - see for example papers by I. Mazin who is big into this idea –  Brendan Dec 2 '12 at 19:15
    
@Brendan: oops, I wrote an answer and had a system crash, and didn't post it until now. I was wondering if either you or wsc can explain why the diagram of the Fermi surface in the paper linked by OP doesn't match the ARPES Fermi surface from the textbooks I found by googling the metal compound described on page 7 of the linked paper. The incommensurate transition is also confusing me--- how do you identify a nesting vector which is incommensurate? I don't have a good intuition, I only know the primitive stuff. –  Ron Maimon Dec 3 '12 at 2:57
    
@RonMaimon: The OP only links to a google search for the phrase "fermi surface nesting," but I assume you mean the first arXiv paper which comes up, and their CeTe$_3$ fermi surface? Because this seems to match pretty well with the observed photoemission results of arxiv.org/abs/cond-mat/0403398. Which textbooks were you looking at? –  wsc Dec 3 '12 at 3:33
    
@wsc: I can't see the match (sorry, probably stupid) the Brouet paper is what I saw (and a textbook that reproduced the picture), and the nesting vector is supposedly linking the small diamond in the ARPES data with the big diamond to the right. It's an incommensurate transition, so I have no intuition for what the fermi surface, and I have not gotten the hang of these ARPES or numerical plots. In the simulation paper, the CeTe3 surface had a different nesting vector that overlayed what looked like a completely different part of the Fermi surface. I was mystified by the lack of matching. –  Ron Maimon Dec 3 '12 at 7:29

You should understand it in 1d first, because it's essentially a 1d phenomenon. Johannes and Mazin unfortunately have a fundamental misconception about this, and they say wrong things throughout their paper. Here are the false claims:

  • They claim that a line of Na atoms in 1d do not have a purely electronic instability if the ions are clamped in place. This is true, but it's implying falsely that someone expected otherwise. The Peierls instability is by definition an ion motion, and it obviously can't happen in a strictly periodic lattice (see below). They say the same thing about the realistic DFT simulations of real CDW materials, that the electrons 'even when prompted by being artificially placed in a periodic charge configuration relax to the symmetric state". Again, this shows a confusion about the origin of CDW configurations.
  • They falsely claim "In the Peierls picture, lattice distortion is a secondary effect that arises in response to an electronically driven charge redistribution that would occur regardless of whether or not the ions subsequently shift from their high symmetry positions." No.
  • The claim "In real materials, the electronic and ionic instabilities always occur simultaneously." This is true, but it is exactly what Peierls' argued.
  • They falsely claim that the wavevectors which produce experimental CDW's are not those which best line up the Fermi surface with itself (this is your question, see further below).

These confusions all come from one core misunderstanding--- they think the electrons organize themselves asymmetricially first, and then the ions follow along. This is incorrect. The ion positions determine the ground state of the electrons, and it is the reduced energy of the electrons in the deformed position that leads the CDW configuration to form.

Having said that, here is...

Peierls 1d argument

Imagine a line of atoms on integer positions sharing their electrons to make a conduction band. You have exactly one electron per atom, and the k-vectors range from $-\pi$ to $\pi$. The electrons are split in spin 50/50, so you get filling of the states in the first band up to exactly $\pm \pi/2$, and that's the Fermi surface (two points). This is called a half-filled band, since the states are filled duplicately until the exact half-way point.

In this situation there is an interesting phenomenon. Imagine that you deform the lattice by shifting all the atoms in the odd positions to the right by $\epsilon$. The new lattice is strictly only periodic with period 2 now, so the k vector range for the deformed lattice is reduced to $[-\pi/2,\pi/2]$. The band from before is now split into two bands of the longer period system, and most importantly, all the electrons are put in the lower band after the deformation, because the original band was half-filled.

The splitting of the old band is accompanied by lowering the energies of the electrons at the edge of the band, the dispersion relation E(k) for the original band, which was linear at $\pi/2$ before, is now split into two hyperbolas, one higher than the other. If the electrons filled up to $\pi/2 + .001$ instead, this wouldn't lower the energy, because the electrons in the lower band would lose, and the electrons in the upper band would gain a nearly equal amount.

But the Fermi surface is exactly at $\pi/2$, so the electronic energy is reduced by this deformation. The shape of the E(k) function is a hyperbola, and it's integral gives $\epsilon^2\log(\epsilon)$ reduction in energy. The $\log(\epsilon)$ is important, because it means that this will always beat the elastic energy cost of the deformation, which goes as $\epsilon^2$, so that this represents a true instability.

Note that the atomic motion is causing the electrons to reconfigure, so the critique in the paper you give is baseless. The 1d lattice is distorted, and this is what makes the electron energy go down. No distortion, no reduction in energy.

The 1d thing becomes an insulator after the band opens, because you opened a band gap, so you can't conduct electricity by giving electrons momentum. But you can still conduct electricity in a strange way, also described by Peierls. You can introduce a defect in the dimer-ordering of the atoms, since it breaks a symmetry you can do this, and then move this defect all the way from the right to the left of the 1-d system. This is a 1d version of a charge density wave.

The Peierls instability is a spontaneous condensation of coupled electron and phonon excitations. This is why it is called a "charge density wave", because the density (ion positions) and the charge (electron density) are together condensing into a strange pattern. It breaks the translational symmetry to a subgroup.

Nesting vector

Having seen the 1d stuff, the nesting vector is defined as the new periodicity of the deformed lattice, as indicated on the old lattice Brillouin zone. The vector in this case is a 1d vector going from $-\pi/2$ to $\pi/2$, which is the new period of the deformed lattice. You identify the two ends of this vector to get the new Brillouin zone.

For a 3d metal, the Fermi surface is a surface, and not a point. Introducing a modulated periodicity with spatial vector V (assumed commensurate) will lead to a band splitting of the old bands of the original metal which will cut the old Brillouin zone into parts (depending on the commensurate period). To the extent that the old Fermi surface happens to land right by the edge of a band, the electronic energy is lowered.

The displacement between the edges of these cuts is called the nesting vector. The Fermi surface nesting is asking how well the fermi surface meets itself under translation by V. This is not exactly what you should be asking (but it's close), you should ask, what is the electronic energy produced by a splitting of the Brillouin zone introduced by a periodic deformation? Since the main contribution is symmetric cancelling if the states go smoothly through the new division of bands (if the Fermi surface extends past $\pi/2$

This will reduce the electronic energy generically when the Fermi surface of the original metal has the property that it lies along the cut edges. This happens when the nesting vector links perpendicularly two nearly flat parts of the Fermi surface. Then the electronic/phononic deformation is exactly analogous to the 1d Peierls instability.

Johanes and Mazin's paper again

The main question Johannes and Mazin ask is: Why is the CDW ordering along a direction vector where the Fermi surface doesn't nest best? I am not sure if they got the right answer, because of he confusions in the rest of the paper. Looking at the ARPES results for the Fermi surface of TiCe compound didn't make things clearer for me, because I had a hard time interpreting the Fermi surface data, because the modulation of the CDW is incommensurate. According to other textbooks and papers, the modulation vector produces a nearly perfect nesting in the system (I take their word for it).

So this question is probably a non-issue, caused by a misidentification of the nesting vector in their paper. If you look at p7 top left, they show the calculated Fermi surface for the material. It doesn't resemble the Fermi surface I saw from the ARPES data, and it isn't labelled with k's so I am not sure where 0,0 k is (the lowest energy electronic state), or even which of their shapes interior is the Fermi sea. I think each of the four blue boxes is a Brillouin zone of the unperturbed lattice.

More general comments

The charge density wave ordering produces an insulator, because the gap opens exactly at the Fermi surface through the phonon condensation. The superconducting ordering is a complementary gap mechanism--- here it is electron pairing that leads to the gap, but the condensation is effectively of charged electron pairs rather than neutral phonons, and condensation gives a BCS superconducting state. The two situations are completely different effects.

There is a third way to produce a gap, by making a Mott insulator. This is due to electron electron interactions, and you can understand it as the reason that the nuclei in a solid don't make bands of their own, at least not in the ground state, so you don't have nuclear conduction opposite to the electronic conduction in solids, because the nuclei are in a Mott insulating phase. The electrons can have this happen too.

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