0
$\begingroup$

Parallel transporting a state along a geodesic doesn't introduce any anholonomy angle, that's what I learned in general relativity. In quantum mechanics, this anholonomy for states are related to the Berry phase. As one takes the state $$ \begin{pmatrix}\cos(\frac{\theta}{2}) \\ \sin(\frac{\theta}{2})e^{i \phi} \end{pmatrix} $$ And performs a parallel transport along the equator of the Bloch sphere (or along any other great circle for that matter) on obtains a $-\pi$ Berry phase. But I thought there was no anholonomy when you go along a geodesic, here along the great circles of the Bloch sphere.

The $-\pi$ Berry phase makes sense if we see it from the other viewpoint that an electronic wave function gets a $\exp(-i \pi)$ as it rotates $2\pi$. What I'm missing about the parallel transport in this case?

$\endgroup$

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.