A spin one-half sits in a field of fixed magnitude whose direction is carried slowly…

PHYSICS · Subjective · Fermi University Physics · Quantum Mechanics · Quantum dynamics and its formulations · Symmetry, angular momentum and spin

The Structure of Quantum Mechanics, Ch Slow change, sudden change and the Berry phase

Problem

A spin one-half sits in a field of fixed magnitude whose direction n^(β,ϕ)\hat n(\beta,\phi) is carried slowly round a closed loop made of three arcs on the sphere of directions: from the north pole (β=0\beta=0) down the meridian ϕ=0\phi=0 to the equator, along the equator from ϕ=0\phi=0 to ϕ=π/2\phi=\pi/2, and back up the meridian ϕ=π/2\phi=\pi/2 to the north pole. Use the text's eigenstate with spin along n^\hat n, ∣+⟩=cos⁡β2 ∣ ⁣↑⟩+eiϕsin⁡β2 ∣ ⁣↓⟩.|+\rangle=\cos\tfrac\beta2\,|\!\uparrow\rangle+e^{i\phi}\sin\tfrac\beta2\,|\!\downarrow\rangle. (a) Compute both components of the Berry connection, Aβ=i⟨+∣∂β+⟩A_\beta=i\langle+|\partial_\beta+\rangle and Aϕ=i⟨+∣∂ϕ+⟩A_\phi=i\langle+|\partial_\phi+\rangle. (b) Evaluate γ+=∮C(Aβ dβ+Aϕ dϕ)\gamma_+=\oint_C(A_\beta\,d\beta+A_\phi\,d\phi) arc by arc. (c) Find the solid angle Ω\Omega the loop encloses without doing an integral, and check your answer to (b) against γ+=−12Ω\gamma_+=-\tfrac12\Omega. (d) The loop runs through the north pole, where ϕ\phi has no value. Explain why that causes no trouble for the ∣+⟩|+\rangle written above.

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