Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I am wondering what kind of two dimensional Fermi surface is called quasi one dimensional, what is its character? Also, when there are orbital hybridization taking place in lattice site, what are characters of hybridized Fermi surface?

share|cite|improve this question
up vote 3 down vote accepted

A quasi-one-dimensional Fermi surface is a Fermi surface whose topology is the same as the topology of a surface defined by an equation that only depends on one dimension.

enter image description here

In the picture above, taken from this paper, you see that the surface is effectively given by $$|k_x|\leq 1 $$ for the first picture. (Well, it should really be an equality if we talk about the surface itself but I wanted to discuss where the states are located, too.) It looks like $k_y$ doesn't really qualitatively matter. That's different from ordinary two- or three-dimensional Fermi surfaces given by inequalities such as $$\sqrt{k_x^2+k_y^2+k_z^2}\leq 1$$ Because $k_y$ and perhaps $k_z$ play no qualitative role in the quasi-one-dimensional Fermi surfaces, one may use various parts of intuition and formalism that are relevant for one-dimensional problems and just add the $y,z$ directions as dummy extra variables that don't affect anything qualitative.

I believe – but I may be wrong – that it's confused to talk about "characters of Fermi surfaces". In the context of orbital hybridization, "character" refers to the bonds themselves (the shape at the level of a single molecule), for example $sp^3$ has 25% s-character and 75% p-character. But when one talks about "character" of Fermi surfaces, the word "character" just means some more general properties – any properties, unspecified properties, not something particular.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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