 Solution

Set ${\vartheta }_{\alpha }(t)={\hslash }^{-1}{\int }^t{E}_{\alpha }(s)ds$. Differentiating the expansion produces ${\dot{a}}_{\beta }|{\phi }_{\beta }⟩+a_{\beta }|{\dot{\phi }}_{\beta }⟩-i{E}_{\beta }{a}_{\beta }|{\phi }_{\beta }⟩/\hslash$ for each term. The energy term cancels $H|\Psi ⟩$ in the Schrödinger equation. Projecting the remainder with $⟨{\phi }_{\alpha }|$ and using orthonormality yieldsThe diagonal term is purely imaginary and supplies the geometric phase. Off-diagonal instantaneous eigenstate couplings transfer population; differentiating the eigenvalue equation gives $⟨m|\dot{n}⟩=⟨m|\dot{H}|n⟩/(E_n-E_m)$ when the levels are distinct.