# Classical - quantum analogy of Berry phase

I am happily convinced about the first "elementary geometry" example they provide. Now I am trying to understand how the rotation of these classical vectors on a sphere can be extended to the consideration of eigenspinors in massive Dirac systems ($H(\mathbf{k})=\mathbf{n}(\mathbf{k})\cdot\mathbf{\sigma}$). I have worked out the math for the latter and am reasonably happy about it. But I would like to establish a clear visual analogy between the rotating vector in the first case to the spinors in the second case.
On a first look, I don't see if the analogy holds true anymore. This is because in the quantum case, the spins will point up or down along the direction of the vector $\mathbf{n}(\mathbf{k})$. In the elementary geometry case however, the vectors are in the local tangent plane on each point on the sphere.
• If I write n(k) in terms of spherical coordinates, the ground state is for eg. $[\sin\theta e^{-\imath\phi}, -\cos\theta ]^T$. By global phase do you mean an additional phase multiplying this vector? In that case, why would the global phase define a unit vector in the 2D tangent space? (Sorry if this is too trivial. I am new to this field.) – arigato Oct 4 '15 at 20:38
• Yes, I mean a global phase multiplying the state as you wrote. This is of the form $\mathrm{e}^{\mathrm{i}\zeta}$ for some angle $\zeta \in [0,2\pi]$. Therefore $\zeta$ defines a point on the unit circle or equivalently a unit vector in the plane. The Berry phase arises because parallel transport of this unit vector along two different adiabatic paths through the state space does not commute, which arises due to the curvature of the Bloch sphere. – Mark Mitchison Oct 5 '15 at 11:35