Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Whereas I can calculate the Chern number of a quantum state (or band) from the integration of the Berry curvature in all space.

How can I infer the topology of the quantum state from this result? What is the physical meaning of a quantum state with non-zero Chern number?

share|improve this question
    
Sorry, I just don't see the long TeXed formula in TeX even though it looks OK in the preview - a strange bug that I am unable to fix. Can someone help? –  Luboš Motl Jan 24 '11 at 21:07
    
Not matter, the question could be undertood without any equation. Thank you for the help. –  Leandro Seixas Jan 24 '11 at 21:23
add comment

2 Answers

up vote 1 down vote accepted

I am in part trying to understand this myself. The Berry phase is computed from differential forms, such as the one-forms $\omega$ constructed from states $$ \omega~=~\langle\psi|d\psi\rangle $$ and with the covariant differential $D~=~d~+~\omega$ the two-forms $$ \Omega~=~D\omega~=~d\omega~+~\omega\wedge\omega $$ The tensor components of the 2-form $F$ are elements of a self-adjoint principal bundle $P$. The determinant of these elements $$ det\Big|1~+~\frac{ixF}{2\pi}\Big|~=~\sum_nc_jx^n $$ which is a characteristic polynomial which represents the Chern class. Each $c_n(P)$ is an element of $H^{2n}(M)$. So the curvature form for the Berry phase, or the Fubini-Study metric $\Omega=~dz\wedge d{\bar z}/(1~+~|z|^2)^2$ is evaluated $\int\Omega~=~2\pi i$ and gives $c_1~=~1$ So there is a nontrivial cocycle on the “2-level. For this projective geometry there are alternating Betti numbers $1,~0$ for even and odd.

If you had some product of states $\prod_n |\psi_n\rangle$, say in an entangled state etc, you could apply the differential $d$ up to $n$ times and form and $n$-form. For instance the product $|\psi_1,~\psi_2\rangle$ $=~|\psi_1\rangle|\psi_2\rangle$ defines the one-form $$ \omega~=~d|\psi_1,~\psi_2\rangle~=~d|\psi_1\rangle|\psi_2\rangle~+~|\psi_1\rangle d|\psi_2\rangle $$ and one could then build up a system of differential forms on various chains. The analogue of the projective geometry for this is a $G_2(V)$ Grassmannian and this continues up for n-product spaces.

share|improve this answer
    
Thank you for the answer. But, are there any relationship between Chern number and Genus (number of handles) or non-orientability? What's mean these numbers? –  Leandro Seixas Jan 29 '11 at 1:42
add comment

It means the system has nonzero Hall conductance.

share|improve this answer
    
Hi, welcome to Physics.SE! It would be nice if you could elaborate your answer because as it stands, there's very little information there. One line answers are discouraged here, for more information have a look at the faq. –  Kitchi Jul 9 '13 at 18:21
    
What is "the system" ? You need a parameter space in order to calculate a chern number. –  jjcale Jul 9 '13 at 18:38
add comment

Your Answer

 
discard

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.