In a quantum system at the $n$-th eigenstate, an adiabatic evolution of the Hamiltonian sees the system remain in the $n$-th eigenstate of the Hamiltonian, while also obtaining a phase factor. The phase obtained has a contribution from the state's time evolution and another from the variation of the eigenstate with the changing Hamiltonian.

The second term corresponds to the Berry phase and for non-cyclical variations of the Hamiltonian it can be made to vanish by a different choice of the phase associated with the eigenstates of the Hamiltonian at each point in the evolution. However, if the variation is cyclical, the Berry phase cannot be cancelled, it is invariant and becomes an observable property of the system.

From the Schrödinger equation the Berry phase $\gamma$ can be calculated to be

$$\gamma[C] = i\oint_C \! \langle n,t| \left( \vec{\nabla}_R |n,t\rangle \right)\,\cdot{d\vec{R}} \,$$

where $R$ parametrizes the cyclic adiabatic process. It follows a closed path $C$ in the appropriate parameter space.

What is the Berry phase for a quantum system without any external fields? In other words, what is the Berry phase for a particle in a box that completes a cycle in real space without the influence of external fields?

  • $\begingroup$ I doubt it is possible to specify such phase in a general case, because any wave packet in a box will disperse and will not return to its original state up to a phase after some time. $\endgroup$ – Alexey Sokolik Nov 22 '16 at 10:11
  • $\begingroup$ Well, in this case, $R$ is the space of real numbers and so $d\vec{R}$ is simply the infinitesimal line element. Does it make sense to compute the integral for some generic integrand and see if it's zero? $\endgroup$ – nightmarish Nov 22 '16 at 10:38
  • $\begingroup$ Maybe it is possible in the path integral formulation, where $\vec{R}(t)$ is the particle trajectory. Berry phase should manifest itself somehow in the path integral, but I don't know the details. Maybe this can help: sciencedirect.com/science/article/pii/000349169290363Q $\endgroup$ – Alexey Sokolik Nov 22 '16 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.