Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 / 1

An initially occupied, isolated instantaneous energy eigenstate follows its changing eigenspace, acquiring a dynamical phase and generally a Berry phase. For smooth nondegenerate evolution, a useful local smallness condition isA nonvanishing spectral gap and suitably smooth, slow driving throughout the evolution are needed; the local inequality alone is not a universal long-time error bound. Gap closure invalidates the simple nondegenerate approximation. A separated degenerate eigenspace may mix internally even when transitions out of it are suppressed.

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.

Use normalized instantaneous modes ${\phi }_p(x,t)=\sqrt{2/L}\sin (p\pi x/L)$ and expand $\Psi ={\sum }_p{a}_p{e}^{-i{\vartheta }_p}{\phi }_p$. Differentiating at fixed $x$ givesFor $n\ne p$, integration usesso the moving-wall derivative coupling isFor $n=p$ the normalization derivative and the remaining integral cancel exactly, giving zero. The moving integration endpoint contributes no boundary term because every mode vanishes there. ThereforeThe real coupling matrix is antisymmetric, as required by conservation of total probability.

For first-order excitation in a linearly expanding infinite well, retain only the ground-state amplitude on the right. For the second level, the coupling coefficient is $4\dot{L}/(3L)$. With $L=vt$,The relative dynamical phase is $-\alpha /t$ plus a constant. Up to an irrelevant phase,The lower endpoint is an oscillatory improper limit: substituting $u=\alpha /s$ gives ${\int }_{\alpha /t}^{\infty }{e}^{-iu}du/u$. A physical initial state cannot occupy a zero-width well. Prepare the ground state at $t_0>0$, integrate from $t_0$, and then take the formal $t_0\downarrow 0$ limit of the probability. For $t/\alpha \ll 1$, endpoint asymptotics give $P\sim 16t^2/(9{\alpha }^2)$; the first-order expression is reliable only while total excitation remains small.