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?


1 Answer 1


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.

As for the applications, I'm not familiar with solid state physics so much (where, I'm sure, the Berry phase is frequently used in order to describe many phenomena), so the only application about which I can tell you is the semiclassical description of the chiral anomaly by the Berry phase. The description is based on the fact that for 2-dimensional Weyl fermions in presence of EM fields the anomaly breaks symplectic structure in the same way as the Berry phase does (because of pole in the expression for the Berry connection).


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