# Adiabatic theorem and Berry phase

The adiabatic theorem is required to derive the Berry phase equation in quantum mechanics. Therefore the adiabatic theorem and the Berry phase must be compatible with one another. (Though geometric derivations are possible, they usually don't employ quantum mechanics. And while illuminating what is going on mathematically, they obscure what is going on physically.)

The question of degeneracy points is a little more subtle, but let me make one thing clear: **if one crosses a degeneracy point, the adiabatic theorem is no longer valid** and one cannot use the Berry phase equation that you have written in the question (the denominator will become zero at the degeneracy point).

Now, let us take the spin in magnetic field example as an illustration of the Berry phase. Suppose we have a spin-1/2 particle in a magnetic field. The spin will align itself to the magnetic field and be in the low energy state $E=E_-$. Now, we decide to adiabatically change the direction of the magnetic field, keeping the magnitude fixed. Adiabatically means that the probability of the spin-1/2 particle transitioning to the $E=E_+$ state is vanishingly small, i.e. $\hbar/\Delta t<<E_+-E_-$. Suppose now that the magnetic field traces out the loop below, starting and ending at the red point:

Berry Phase

In this case, one will pick up a Berry phase equal to:

\begin{equation} \textbf{Berry Phase}=\gamma = -\frac{1}{2}\Omega \end{equation}

where $\Omega$ is the solid angle subtended. This formula is proven in Griffiths QM section 10.2. However, it is not that important to understand the overall picture.

I chose this example because there a couple things to note that make it relevant to your question:

1) The adiabatic theorem is **critical** in this problem for defining the Berry phase. Since the Berry phase depends on the solid angle, any transition to the $E=E_+$ state would have destroyed the meaning of tracing out the solid angle.

2) The degeneracy point lies at the center of the sphere where $B=0$, where $B$ is the magnetic field. Though the spin may traverse any loop on the sphere, it cannot go through this degeneracy point for the Berry phase to have any meaning. This degeneracy point is ultimately responsible for the acquisition of the Berry phase, however. We must in some sense "go around the degeneracy point without going through it" for one to obtain a Berry phase.

The compatibility with the adiabatic theorem has been made clear by the answer from @Xcheckr. I just would like to add some insight to the degeneracy issue.

This degeneracy, which gives rise to the acquisition of a nontrivial Berry phase, lives in the abstract parameter space. It is in general not to be traversed by the interested closed path of adiabatic parameter evolution somehow because it is a singularity (topological defect) in terms of the adiabatic construction. And in regards to the spin-$\frac{1}{2}$ example of a $3D$ parameter space, the degeracy at $B=0$ is nothing but a magnetic monopole of the Berry curvature fictitious gauge field defined upon the parameter space. The presence of such a monopole renders the $U(1)$ bundle nontrivial, hence the $\pi_2(S^2)=\mathbb{Z}$ classification.

In simple words, once you have a monopole, no wonder you obtain nonzero Berry phase by calculating the fictitious magnetic flux covered by the subtended solid angle. If you know the Weyl semimetal, which has a $\vec{k}\cdot\vec\sigma$ Hamiltonian in $3D$ near the band crossing (Weyl point), you immediately know that the aforesaid $B=0$ degeneracy corresponds directly to the $k=0$ Weyl point.