How to calculate the ground states' Berry phases with doubly degeneracy, such as that due to the particle-hole symmetry or time reversal symmetry? Suppose the ground states of a system are doubly degenerate due to an anti-unitary symmetry $K$, which are $|\psi>$ and $|K\psi>$. If the system is an one-dimensional Fermion system and anti-unitary symmetry is a particle-hole symmetry, the degeneracy is often related to a pair of Majorana zero modes. 
It is known that exchanging two Majoranas would introduce a $\pm\pi/4$ phase into $|\psi>$ and $K|\psi>$. Do these phases are the Berry phase $i\int<\psi|d|\psi>=-i\int<K\psi|d|K\psi>$? Is this phase $i\int<\psi|d|\psi>$ related to the trace of the exchange? Or is it guaranteed to be $\pi/4$?
One more question: Does it make sense to calculate $i\int<\psi|d|\psi>$ when $\psi$ is one of the two degenerate ground states?
 A: Note that Berry's phase is only valid for non-degenerate states. Whenever you have a degenerate state, adiabaticity can no longer guarantee that an eigenstate's evolution will consist in staying in the same instantaneous eigenstate. An eigenstate will rather mix within its degenerate subspace. In this case, what one has to compute is a matrix that describes this mixing, i.e., given an eigenstate, under cyclic adiabatic evolution, the final state will in general be a superposition of the degenerate eigenstates. This is the Wilczek-Zee matrix $\mathcal{A}$ with entries $$\mathcal{A}_{ab} = i\langle a(\xi)|d|b(\xi)\rangle,$$ where $a$ and $b$ label degenerate eigenstates, $|a(\xi)\rangle$ is the instantaneous degenerate eigenstate corresponding to parameters $\xi$ and $d$ is the exterior derivative in the space of parameters. The evolution operator after a whole cycle $C$ of parameters is $$U_C = e^{\text{dynamical}}\mathcal{P}e^{i\oint_C A},$$ where $\mathcal{P}$ denotes path-ordered exponential. The classical reference on the Wilczek-Zee phase is http://dx.doi.org/10.1103/PhysRevLett.52.2111 and any book on geometric phases covers that topic.
