1
$\begingroup$

I would like to understand the Berry curvature and the Chern number from mathematical geometry-topology.

I understand that in electronic QHE, there is a map from $k^2$ to a vector space where the eigenvectors "live". These eigenvectors are defined up to a phase (local gauge invariance). Why Berry curvature is defined like this?

$B=\epsilon_{ij}<\partial_{k_i}u|\partial_{k_j}u>$

Is it standard in mathematics? Where can I find it?

My motivation comes from topology appearing in non-electronic systems such as photonics, acoustics or mechanics, where Berry curvature plays a similar role than in electronics. I would like to explain it without electronics.

$\endgroup$

1 Answer 1

1
$\begingroup$

The Berry connection and Berry curvature only appear due to the wave nature of physical systems. That is why it also plays a role in photonics, acoustics and other classical wave equations.

The Berry connection and Berry curvature are a connection and a curvature in the mathematical sense on a vector bundle, commonly known as the Bloch bundle \begin{align*} \mathcal{E}_{\mathrm{Bloch}} = \bigsqcup_{k \in \mathrm{BZ}} \mathcal{H}_{\mathrm{rel}}(k) = \mathrm{span} \bigl \{ \varphi_1(k) , \ldots , \varphi_n(k) \bigr \} , \end{align*} which is constructed from gluing together the eigenspaces \begin{align*} \mathcal{H}_{\mathrm{rel}}(k) = \mathrm{span} \bigl \{ \varphi_1(k) , \ldots , \varphi_n(k) \bigr \} \end{align*} spanned by the eigenfunctions associated to the eigenvalues below the characteristic energy or frequency; in solid state physics, this is the Fermi energy. Here we have assumed that your characteristic energy or frequency lies in a bulk band gap, because then the dimensionality of $\mathcal{H}_{\mathrm{rel}}(k)$ is independent of $k$ and the relevant subspace $\mathcal{H}_{\mathrm{rel}}$ of your Hilbert space depends analytically on $k$. (In classical waves, you need to pay attention to the bands with linear dispersion around $k = 0$ and $\omega = 0$, though.)

Berry connection and Berry curvature are then, as mentioned before, just a connection associated to the curvature on this vector bundle.

$\endgroup$

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