# Importance of the Pancharatnam–Berry Phase

As I understand it, the Pancharatnam–Berry phase first arises in the adiabatic approximation for the evolution of a quantum state. For the evolution of a quantum state parameterized by the set of parameters $\lambda_i, i = 1, \dots, N$, if the state evolves slowly enough (the details of when the adiabatic approximation holds I don't want to get into), then the projection of our state onto the eigenbasis of our Hamiltonian holds as the state evolves, plus a phase factor.

That is to say, given some initial state $$|\Psi(0)\rangle = \sum_k c_k|\phi_k(0)\rangle,$$ evolving according to the time-dependent Hamiltonian $H(\lambda(T))$,the state at a later time (when the approximation holds) will be $$|\Psi(t)\rangle = \sum_k c_k e^{i \gamma_k(t)}e^{i \theta_k(t)}|\phi(\lambda(t))\rangle.$$ Here $\theta_k(t) = i/\hbar\int_0^t E_n(\lambda(t'))dt'$ is the dynamic phase and $\gamma_k(t) =i \int_0^t \langle \phi_k(\lambda(t')) | \dot{\phi}_k(\lambda(t')) \rangle dt'$ is our geometric phase.

If the change of our system with time is periodic, usch that after some time $T$ we get $\lambda(T) = \lambda(0)$, then the geometric phase $\gamma_k(T) = \gamma_B$ becomes the Pancharatnam–Berry phase. It's roughly analogous to parallel transporting a vector along a sphere. If I'm not mistaken, one of the important features of the berry phase is its gauge invariance.

Here is where my question comes in: Why is this important? Why does this specific phase coming out of periodic evolutions hold such importance. How might this be more important than just the general feature of a geometric phase in adiabatic evolutions? Or what important applications and uses of the Berry phase have come about?

The gauge invariance of the Berry phase $\gamma$ is just the manifestation of its geometrical interpretation. Simply speaking, the phase $$\tag 1 \gamma = \oint\limits_{C} d\mathbf \lambda\cdot \mathbf{A}_{\lambda}, \quad \mathbf{A}_{\lambda} = \langle\psi(\mathbf \lambda) |\frac{d}{d\mathbf \lambda}|\psi (\mathbf \lambda)\rangle$$ depends only on the geometry of the space of the states $|\psi(\mathbf \lambda)\rangle$ (precisely, on the manifold $M$ of the parameters $\mathbf \lambda$ and on the $n$-dimensional space of the eigenstates of $n$-dimensional hamiltonian $H$). This means that any path-dependent variations of $\mathbf A_{s}$ contributes to zero. But this means that any variation $$\tag 2 \mathbf{A}_{\lambda}\to \mathbf{A}_{\lambda} + \frac{d}{d\mathbf \lambda}\theta(\mathbf \lambda)$$ (assuming it is not "large") leaves the phase $(1)$ unchanged.
From this in particular clearly follows that the only case when the phase $\phi$ is non-zero is when the mapping $M\to CP^{n-1}$, where $M$ is the manifold of the $\mathbf{\lambda}$ on the states $\psi$ and $CP^{n-1}$ is the space of defined modulo the phase eigenstates of $n$-dimensional complex hamiltonian, can't be continuously deformed to the mapping $M\to S^{2n-1}$, where $S^{2n-1}$ is the space of unit eigenstates with the phase being fixed. Indeed, if such lifting can be performed, then we can define multiply the state $|\psi\rangle$ on the continuous phase factor $e^{i\theta(\mathbf \lambda)}$, $$|\psi(\mathbf \lambda)\rangle \to e^{i\theta(\mathbf{\lambda})}|\psi(\mathbf \lambda)\rangle$$ so that the phase $(1)$ will vanish identically. But this corresponds to the transformation $(2)$. The remained non-trivial Berry connection $\mathbf A_{\lambda}$ is a pole-like field and it can't be eliminated by the smooth phase.