# What is curved in Berry Curvature?

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Can anyone explain to me what is actually "curved" when we speak of a Berry Curvature?

This post imported from StackExchange Physics at 2014-03-22 17:09 (UCT), posted by SE-user ChickenGod
The surface or skin of the berry, to form usually a spherical like shape. The actual curvature depends on the type of berry and the individual berry picked.

This post imported from StackExchange Physics at 2014-03-22 17:09 (UCT), posted by SE-user Mew

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It's the curvature of a connection on a principal U(1) bundle over parameter space.

In describing the quantum Hall effect, we have a Hamiltonian, which depends on a number of parameters $H(R_1,R_2,..R_N)$. Suppose we have the system in its ground state. We now vary the parameters adiabatically (slowly!). As we vary the parameters, we can think of tracing out a curve $(R_1(\lambda), (R_2(\lambda)...(R_N(\lambda))$ in parameter space. As we twiddle the parameters we actually evolve the state using the Schroedinger equation. If we transport it round a closed curve in parameter space, i.e. we return to our starting parameters, we find that the state picks up a phase factor relative to the starting state. (Phase factors live in $U(1)$ the group of unit modulus complex numbers).

The mechanism that allows us to go from a state at one set of parameters to a state at another set is called a connection. In this case the connection is provided by the Schroedinger equation. Saying that the connection has curvature just means that transport of a state round a closed curve using this connection doesn't quite get you back to the state you started with (in fact mathematically the object which defines the curvature is obtained by transport round a little closed parallelogram).

This post imported from StackExchange Physics at 2014-03-22 17:09 (UCT), posted by SE-user twistor59
answered Mar 11, 2013 by (2,490 points)
Let me say it in more heuristic words. When we make a round trip, the wave function may return to itself up to a transformation, a phase in this case. This is similar to the case of parallel transport around curved loops that also rotates the original vector. That rotation is the result of the usual Riemann curvature, $\delta V^a \sim R^a_{bcd} V^b dS^{cd}$, and similarly curvature (field strength) of gauge fields and similar fields results in rotations in other spaces induced by the round trips around infinitesimal loops.

This post imported from StackExchange Physics at 2014-03-22 17:09 (UCT), posted by SE-user Luboš Motl

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