Numerically calculating non-Abelian Berry curvature: Definition of multiplet in explicit $4\times 4$ system with 2-fold degeneracy?

Question

I am trying to use eq 16 of the following paper to calculate the Chern number of a 4x4 degenerate system: https://arxiv.org/abs/cond-mat/0503172 [1]. I believe this is the standard scheme used by many. In my sample system (I am using eq. 1 in https://arxiv.org/abs/1008.2026 [2]), I chose parameters such that the 2 bands above the gap are degenerate with each other, and the 2 bands below the gap are degenerate with each other.

I correctly used the method in the above paper [1] to make a MATLAB implementation of the Abelian case.

When I solve my 4-band system, I get four $4\times 1$ instantaneous energy eigenstates. When I plot the eigenvalues, I see '2' bands, because each is degenerate to another. So far, so good. My issue is that I am not sure what the authors mean when they say "multiplet".

However, I must now calculate the U(1) link variable in eq. 16 of [1] as: $$ U(k) = \frac{\det (\phi^* (k) \phi(k+\mu))}{|\det (\phi^* (k) \phi(k+\mu))|}, $$

where I chose $\phi$ to be the eigenstate of one (of four) levels, $^*$ is the complex conjugate transpose, and $\mu$ is a small parameter used to take the numerical derivative. However, the authors define $\phi$ as a "multiplet" $\{n_1, ...., n_M\}$, where $M$ is a dimension of the $M\times M$ non-Abelian Berry connection matrix $A=\psi^* d \psi$. I believe my problem arises here: What does "multiplet" mean in this context? I assumed it was just an equivalent term for "instantaneous eigenstate". So, I have four $4\times 1$ multiplets in this 2-fold degenerate system.

I assumed that $\phi^* (k) \phi(k+\mu)$ could be read as $|\phi(k)\rangle\langle\phi(k+\mu)|$, so that for a $4\times 1$ eigenvector $|\phi(k)\rangle$, I would get a $4\times 4$ matrix for $|\phi(k)\rangle\langle\phi(k+\mu)|$. Is this interpretation correct? I do not think it is, because the last two rows of all vectors are zero, making the determinant always zero, and therefore $U(k)$ ill-defined due to division by zero. I am guessing I got something wrong with my interpretation of the word "multiplet". It would make sense numerically if my multiplet was just the first 2 rows of each eigenvector (then, the denominator will be non-zero). However, I cannot justify this. Could someone clarify my confusion please? I have a feeling these "multiplets" should instead be $2\times 1$ vectors... but then, I don't know how to justify getting rid of 2 rows of numerical eigenvectors of a $4\times 4$ system. Thanks.

Answer 1

The multiplet refers to the set of $M$ states you want to calculate the non-Abelian Berry curvature for. Note that in its definition in the paper, $\psi=\left( |n_1\rangle, \dots, |n_M\rangle\right)$, each $|n_i\rangle=|n_i(k)\rangle$ is a Bloch state (i.e. what you call a $4\times 1$ vector). In this case, you most likely want $M=2$ for the two degenerate bands, such that $\psi$ is a $4\times 2$ matrix. This is the minimal set that will satisfy the gap-opening condition $E_n(k)\neq E_{n'}(k)$ for $n\in I$ and $n'\notin I$ where $I=\{n_1,\dots,n_M\}$ mentioned right before Eq. (16).