Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 / 1

Let $\Delta ={\Omega }_0-\omega$ and $R(t)=e^{-i\omega t{S}_z/\hslash }$. Differentiating $|\Psi ⟩=R|{\Psi }_R⟩$ gives the rotating-frame Hamiltonian operator $R^{†}HR-i\hslash R^{†}\dot{R}$. The angular momentum commutator yields $R^{†}{S}_{\pm }R=e^{\pm i\omega t}{S}_{\pm }$, while $-i\hslash R^{†}\dot{R}=-\omega S_z$. Thus the rotating-frame Rabi Hamiltonian isThis transformation is exact for the circular drive; no rotating-wave approximation has been used.

Define $\Omega =\sqrt{{\Delta }^2+{\Omega }_R^2}$. Squaring the two-by-two matrix gives $H_R^2={\hslash }^2{\Omega }^{2}I/4$, soHere $|\pm {⟩}_R$ is the corresponding normalized eigenstate at $t=0$. In the rotating frame only its quantum phase changes. In the laboratory frame its spin direction also rotates about $z$ at angular velocity $\omega$: $⟨\mathbf{S}{⟩}_{\pm }=\pm \hslash ({\Omega }_R\cos \omega t,{\Omega }_R\sin \omega t,\Delta )/(2\Omega )$. An arbitrary initial state is the same fixed linear combination of these two solutions. If $\Omega =0$, the rotating-frame Hamiltonian operator vanishes and the entire doublet is degenerate; evolution is just $R(t)$.

Take $\omega >0$ and $T=2\pi /\omega$. A full spin one-half rotation gives $R(T)=-I$, an essential sign. The cyclic phase of a spin in a rotating field has factors and total quantum phasesTo separate the adiabatic dynamical phase, set $B=\sqrt{{\Omega }_0^2+{\Omega }_R^2}>0$ and $\cos \theta ={\Omega }_0/B$. Then $\Omega =B-\omega \cos \theta +O({\omega }^2/B)$. The instantaneous laboratory energies are $\pm \hslash B/2$, so the adiabatic dynamical phases are $\mp BT/2$. Subtracting them from ${\Phi }_{\pm }$ givesThe oriented solid angle $\mathcal{A}$ is swept out by the field direction. These are the Berry phases for the eigenstates parallel and antiparallel to that direction. Retaining the spinor sign makes the quantum phase vanish modulo $2\pi$ when the field does not trace a nontrivial loop. The assumption $\omega \ll B$ suffices for adiabatic following; near degeneracy it must not be silently dropped.^[ref]

The angular momentum commutators used in the frame transformation are representation-independent. Therefore $H_R=\Delta S_z+{\Omega }_R{S}_x$ still holds. Rotating its axis onto $z$ gives $2s+1$ eigenvalues and exact laboratory solutionsNow $R(T)=(-1{)}^{2s}I$, so $e^{i{\Phi }_{\mu }}=(-1{)}^{2s}{e}^{-i\mu \Omega T}$. In the adiabatic limit the dynamical phase is $-\mu BT$ andIndeed $2\pi s+2\pi \mu \cos \theta$ and $-2\pi \mu (1-\cos \theta )$ differ by $2\pi (s+\mu )$, an integer multiple of $2\pi$. Integer spins return with a positive rotation sign and half-integer spins with a negative one; the geometric phase need not vanish for either.