# Calculating the Berry curvature in case of degenerate levels (Non abelian Berry curvature): issue

The Berry phase accumulated on a path can be described by a matrix when we look at adiabatic time evolution with a Hamiltonian with degenerate energy levels. The Berry phase matrix is given by $$\gamma_{mn}= \int_\mathcal{C} \left\langle m(R) \right | \nabla_R \left| n(R) \right \rangle . d R.$$

here $R$ parametrizes the said path and $A_{mn}= \left\langle m(R) \right | \nabla_R \left| n(R) \right \rangle$. Now what I want to do is calculate the Berry Curvature, something that, if I assume my path above is closed and has three determining coordinates$R_1$, $R_2$ , $R_3$ is $\vec{F}$ such that

$$\mathbf{\gamma}=\int_\mathcal{S} \vec{F}.d\vec{s}$$ note that the $\gamma$ and $F$ here are matrices and we're integrating over the surface $\mathcal{S}$ enclosed by curve $\mathcal{C}$

What is stopping me from applying the stokes theorem to $\gamma_{mn}$ and getting $\vec{F}_{mn}=\nabla_R\times A_{mn}$?

It is said that the answer contains a matrix commutator $[A_i,A_j]_{mn}$ c because this berry phase is non abelian. but I seem to be missing something fundamental.

Edit: Note: this also corresponds to problem 2 Chapter 2 of Topological insulators and superconductors by Bernevig and Hughes

The short answer is that only if the Berry curvature is defined by: (in matrix notation): $$F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} + [A_{\mu}, A_{\nu}]$$ it becomes gauge covariant, i.e., for a gauge transformation: $$A_{\mu} \rightarrow g^{-1}A_{\mu} g+g^{-1}\partial_{\mu}g$$ $g \in U(N)$ ($N$ is the degeneracy of the level), the curvature transforms as: $$F_{\mu \nu} \rightarrow g^{-1}F_{\mu \nu} g$$ The partial expression $F_{\mu \nu} = \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ will be covariant (actually invariant) only with respect to Abelian gauge transformations and not the whole $U(N)$ group.

The reason that the Berry curvature cannot be obtained as in the Abelian case by a direct application of the Stokes theorem is that the Stokes theorem does not exist(in the usual sense) in the non Abelian case becuse the holonomy needs to be path ordered:

$$\mathcal{Hol}(A) = \mathrm{P} e^{\int_C A_{\mu}dx^{\mu}}$$

However, it can be applied sequentially for an infinitesimal path. let us choose a square path in the $(xy)$ plane centered at $(x,y)$ with infinitesimal sides of length $\Delta x$ and $\Delta y$ as depicted in the picture. Therefore

$$\begin{matrix} \mathcal{Hol}(A) \approx e^{\int_a^b A_{\mu}dx^{\mu}} e^{\int_b^c A_{\mu}dx^{\mu}} e^{\int_c^d A_{\mu}dx^{\mu}} e^{\int_d^a A_{\mu}dx^{\mu}} \\\approx (1+A_x\Delta x-\frac{1}{2}\partial_yA_x\Delta x \Delta y) (1+A_y\Delta y+\frac{1}{2}\partial_x A_y\Delta x \Delta y) (1-A_x\Delta x-\frac{1}{2}\partial_yA_x\Delta x \Delta y)(1-A_y\Delta y+\frac{1}{2}\partial_x A_y\Delta x \Delta y) \\ \approx 1+(\partial_x A_y - \partial_y A_x + [A_x, A_y])\Delta x \Delta y\approx (1+ F_{xy}\Delta x \Delta y) \end{matrix}$$

Please notice that the commutator term is created due to the need to exchange the order of the second and the third expressions in the product in order to cancel $A_x \Delta_x$.

• This is a wonderful answer! However, when I try to follow the reductions of the holonomy for the infinitesimal square loop, I end up with lingering product terms $A_x^2\Delta_x^2 + A_y^2\Delta_y^2$ (i.e. when going to the second-to-last equality). What is the justification for throwing away these terms? To be specific, that means I get $\mathrm{Hol}(A) \approx 1 + (\partial_xA_y - \partial_yA_x + [A_x,A_y])\Delta_x\Delta_y - A_x^2\Delta_x^2 - A_y^2\Delta_y^2$. – daysofsnow Mar 15 '19 at 15:03
• Ah, as far as I can see, this is a small error: the proof actually only works if each of the exponentials are expanded to second order; presently, they are only expanded to first order. The "lingering" terms mentioned above cancel with these extra terms. As an example, the term $\mathrm{e}^{\int_a^b A_\mu \mathrm{d}x^\mu}$ should have been expanded to $1 + A_x\Delta_x - \frac{1}{2}\partial_yA_x\Delta_x\Delta_y + \frac{1}{2}A_x^2\Delta_x^2$. – daysofsnow Mar 15 '19 at 21:43
• @daysofsnow The integrals are expanded to the first order. The quadratic terms (proportional to $\Delta x \Delta y$ came from the fact that $A$ is the potential at the center of the square, while in the line integrals, we need the potential at the center of the edge, so these terms interpolate between these two points. The quadratic terms which should be taken from both contributions above should not cancel in the limit, because the field strength $F$ only approximates the holonomy for a small loop. The quadratic terms should add up to higher invariants. – David Bar Moshe Mar 17 '19 at 15:16
• I get that the $\Delta_x\Delta_y$ terms come from edge$\rightarrow$center interpolation, sure. However, in the your original answer, the product $(1 + A_x\Delta_x - \frac{1}{2}\partial_y A_x \Delta_x\Delta_y)(1 + A_y\Delta_y + \frac{1}{2}\partial_x A_y \Delta_x\Delta_y)(1 - A_x\Delta_x - \frac{1}{2}\partial_y A_x \Delta_x\Delta_y)\ldots$ equals $1 + (\partial_x A_y - \partial_y A_x + [A_x,A_y])\Delta_x\Delta_y - A_x^2\Delta_x^2 - A_y\Delta_y^2$. – daysofsnow Mar 18 '19 at 17:39
• @daysofsnow you are perfectly right thank you for the observation. – David Bar Moshe Mar 19 '19 at 7:44