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

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.

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).
• Thank you for your answer. Just to make sure I understand correctly, could you verify the following for me? If I label my 4 energy eigenstates as A, B, C, D (eigenvalues of A = B, and C = D due to degeneracy), does this mean that my $\psi$ can only be either $(A,B)$ or $(C,D)$? Then, the $\phi^*(k)\phi(k+\mu)$ part of the numerator of $U(k)$ will be a $2\times 2$ matrix, right? Commented Mar 6, 2021 at 19:24
• @TribalChief Well, you could always consider the trivial case $\psi=(A,B,C,D)$. It won't be particularly revealing, but it is an allowed possibility. Generally, testing out choices for your multiplet becomes more interesting when there are more bands, or when you don't have exact degeneracies like this. Yes, the numerator of Eq. (16) is an $M\times M$ matrix. Commented Mar 6, 2021 at 19:36
• Thanks. I just wasn’t sure whether the numerical method took the commutator in the definition for non-Abelian Berry curvature into account. I also wasn’t sure how the $U(k)$ for A and B could be different if $\psi$ is defined using them both. Commented Mar 8, 2021 at 15:44