‘Proof’ that non-Abelian Berry phase vanishes identically

For a degenerate system with Hamiltonian $$H =H(\mathbf{R})$$ and eigenstates $$\left|n(\mathbf{R})\right\rangle$$ the non-Abelian Berry connection is $$A^{(mn)}_i=\mathrm{i}\left\langle m|\partial_in\right\rangle\tag{1}$$ and the non-Abelian Berry curvature is $$F_{ij} = \partial_i A_j-\partial_j A_i - \mathrm{i}\left[A_i, A_j\right]$$ in matrix notation or, including the state indices: $$F_{ij}^{(mn)} = \partial_i A_j^{(mn)}-\partial_j A_i^{(mn)}-\mathrm{i}\sum_k\left( A_i^{(mk)}A_j^{(kn)}-A_j^{(mk)}A_i^{(kn)}\right).\tag{2}$$ Substituting (1) into (2) gives \begin{align} F_{ij}^{(mn)} &= \mathrm{i}\partial_i\left\langle m|\partial_jn\right\rangle- \mathrm{i}\partial_j\left\langle m|\partial_in\right\rangle -\mathrm{i}\sum_k\left(\mathrm{i}^2\left\langle m|\partial_i k\right\rangle\left\langle k|\partial_j n\right\rangle-\mathrm{i}^2\left\langle m|\partial_j k\right\rangle\left\langle k|\partial_i n\right\rangle\right)\\ &=\mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle+\mathrm{i}\left\langle m|\partial_i\partial_jn\right\rangle -\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle-\mathrm{i}\left\langle m|\partial_j\partial_in\right\rangle +\mathrm{i}\langle m|\left(\sum_k|\partial_i k\rangle\langle k|\right)|\partial_j n\rangle-\mathrm{i}\langle m|\left(\sum_k|\partial_j k\rangle\langle k|\right)|\partial_i n\rangle\\ &=\mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle -\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle +\mathrm{i}\langle m|\left(-\sum_k| k\rangle\langle \partial_ik|\right)|\partial_j n\rangle-\mathrm{i}\langle m|\left(-\sum_k|k\rangle\langle \partial_jk|\right)|\partial_i n\rangle\\ &=\mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle -\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle -\mathrm{i}\sum_k \underbrace{\langle m| k\rangle}_{\delta_{mk}}(\langle \partial_ik|\partial_j n\rangle-\langle \partial_jk|\partial_i n\rangle)\\ &=\mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle -\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle- \mathrm{i}\left\langle \partial_i m|\partial_jn\right\rangle+\mathrm{i}\left\langle \partial_j m|\partial_in\right\rangle\\ &=0. \end{align} To go from the second to the third line I used $$0=\partial_i(1) = \partial_i\left(\sum_k |k\rangle\langle k| \right) = \sum_k |\partial_i k\rangle\langle k|+\sum_k |k\rangle\langle\partial_i k|.$$ Clearly I have done something wrong as the Berry curvature is not zero in all cases. Please could someone point out my error?

• I am not 100% sure about the resolution, but I am skeptical about the identity you use in the last line. Let's say the parameter of the curve you are following is $\lambda$. The state is in some degenerate subspace of the spectrum as it travels along a closed path. Now I agree you can choose an orthonormal basis $|k\rangle$ at some fixed $\lambda$ obeying the resolution of the identity. But are you sure that as you vary $\lambda$, these states remain an orthonormal basis? You need $\sum_k |k(\lambda) \rangle \langle k(\lambda) | =1$ to be true for all $\lambda$ for your identity to hold. Oct 24 at 4:13
• To be clear, it's not that the evolution will take you out of the subspace. You can always choose an orthonormal basis for every $\lambda$. My question is, if you take a set of states for a given $\lambda$, and then evolve those states adiabatically as you vary $\lambda$, will they remain an orthonormal basis? Oct 24 at 4:16

Your calculation is correct. I believe the problem is that the non-Abelian Berry curvature is useful only when it is defined in a sub-Hilbert space and that subspace is not the same at different parameter R. In this case, the identity you mentioned in the end no longer equals to zero.

• Do you have a source for any more details on this? Oct 26 at 9:34
• Sorry I'm not aware of any good references for this. But inspired by your question, I find it is more intriguing to write the non-Abelian Berry curvature in the following form $$\sum_{m,n}\Omega_{ij}^{(mn)}|m \rangle \langle n|=\hat{P}i(\partial_{i}\hat{P}\partial_{j}\hat{P}-\partial_{j}\hat{P}\partial_{i}\hat{P})\hat{P}$$, where $\hat{P}$ is the projection operator onto the sub-Hilbert space where the Berry curvature is defined.
– Leon
Nov 6 at 17:36

To expand on Leon's answer, it may be helpful to think in analogy with the usual $$U(1)$$ gauge theory. We can think of the non-abelian Berry connection as an $$SU(n)$$ gauge field, where $$n$$ is the number of bands. Our gauge transformations are a change of basis by unitary matrix $$U(k)$$ and the connection and curvature transform as $$A' = U^\dagger A U + i U^\dagger\partial_k U\\ F' = U^\dagger F U$$ respectively.

In matrix notation we may write the Berry connection $$A = i M^\dagger \partial_k M$$ where $$M$$ is a matrix of the eigenvectors, $$M_{ij} = |j\rangle_i.$$ If $$M$$ is unitary, corresponding to the condition $$\sum_k |k\rangle\langle k| = 1$$, we may perform a change of basis which takes $$A'\to0$$ (explicitly we change basis by $$M$$), and therefore $$F$$ must be zero since it is a gauge-covariant quantity. That is if we keep all bands, every Berry connection is "pure gauge". However, if we consider a sub-space of the bands, then $$M$$ is instead a projector onto those bands $$M = \sum_{k=1}^{n < N } |k\rangle\langle k|$$ where $$N$$ is the total number of bands, and in general there is not guaranteed to be a change of basis which nullifies $$A$$. Thus, $$F$$ can be non-zero in this subspace.

$$F_{ij}^{(mn)} = \partial_i A_j^{(mn)}-\partial_j A_i^{(mn)}-\mathrm{i}\sum_k\left( A_i^{(mk)}A_j^{(kn)} \mathbf+ A_j^{(mk)}A_i^{(kn)}\right).\tag{2}$$