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For a normalized, smoothly chosen instantaneous energy eigenstate $|\psi (\mathbf{R})⟩$, the Berry connection has components $A_j=i⟨\psi |{\partial }_{R_j}\psi ⟩$. Its line integral around a closed parameter path $C$ gives the Berry phase in a periodic choice of state vector:Normalization makes the inner product $⟨\psi |{\partial }_{R_j}\psi ⟩$ purely imaginary, so the Berry connection is real. Changing the representative to $e^{i\chi (\mathbf{R})}|\psi ⟩$ changes the Berry connection to $A_j-{\partial }_{R_j}\chi$. A single-valued phase factor along the closed path changes the line integral by an integer multiple of $2\pi$, leaving $e^{i\gamma }$ unchanged. If the chosen vector is not periodic, include the endpoint phase:Here $\psi$ denotes the instantaneous energy eigenstate representative, with its dynamical phase removed. This is freedom to choose a representative of a state ray. ^[ref].

A Berry phase is the geometric phase acquired when a nondegenerate energy eigenstate follows a closed cycle of parameters of a Hamiltonian operator in the ideal limit of the adiabatic theorem. The energy eigenstate must stay separated from other energy eigenstates by a nonzero spectral gap, and the parameters must change sufficiently slowly. Returning the Hamiltonian operator to its starting value returns the state to the same ray, but its quantum phase can remember the journey. The total quantum phase contains both the dynamical phase and the Berry phase.

The beauty of the Berry phase is that, within the adiabatic theorem's limit, it depends on the oriented path of the parameters rather than how long the journey takes. For a spin one-half qubit, the Berry phase of a Bloch-sphere latitude turns that memory into half a solid angle. Berry-phase interference with a reference path converts this geometry into a detection probability.

Michael Berry's 1984 paper formulated the geometric phase for cycles governed by the adiabatic theorem and derived the solid angle result for spin. Its bibliographic details are Proceedings of the Royal Society of London A, volume 392, pages 45–57, DOI 10.1098/rspa.1984.0023. ^[ref]; ^[ref]. The wider concept of geometric phase does not require every cycle to obey the adiabatic theorem: Aharonov and Anandan formulated a geometric phase for general cyclic quantum evolution in 1987. ^[ref].

An isolated qubit that acquires only a global phase has unchanged measurement probabilities. To reveal the Berry phase, use two coherent paths of a Mach-Zehnder interferometer, represented by a path qubit. In one arm the internal spin one-half qubit follows the loop; in the reference arm it stays in its starting energy eigenstate. Assume equal arm amplitudes, identical final internal states, cancelled relative dynamical phases, and calibrated fixed apparatus phases. In the model above, hold the reference Hamiltonian operator at its initial value for the same duration. Both arms then have energy eigenvalue $-\Delta /2$ and acquire the same dynamical phase $\Delta T/(2\hslash )$, so their relative dynamical phase vanishes in the ideal limit of the adiabatic theorem.

The joint quantum state before recombination then factors asThe Berry phase is now a relative quantum phase between the two paths. The calibrated plus output projects onto $|+⟩=(|\mathrm{ref}⟩+|\mathrm{loop}⟩)/\sqrt{2}$. The Born rule givesFor the Berry phase of a Bloch-sphere latitude, the plus-port probabilities at $\theta =0,60^{\circ },90^{\circ },180^{\circ }$ are respectively $1,1/2,0,1$. This ideal Mach-Zehnder interferometer measurement cannot distinguish $\gamma$ from $-\gamma$ or $\gamma +2\pi$ by itself; scanning a known reference phase can resolve the sign. An uncancelled relative dynamical phase would shift the fringe as well.

An eigenstate of the rotating-frame Rabi Hamiltonian returns to the same laboratory ray after a full drive period. Its quantum phase includes both the rotating-frame energy and the sign of a full spin rotation. For spin $s$ and magnetic label $\mu$, the factor is $(-1{)}^{2s}{e}^{-i\mu \Omega T}$. Expanding the rotating-frame splitting to first order in the drive frequency and subtracting the adiabatic dynamical phase leaves $-\mu$ times the oriented solid angle of the field loop. Dropping the full-rotation sign gives a spurious half-integer-spin offset.

For an energy eigenstate with instantaneous energy eigenvalue $E(t)$, the dynamical phase isIt comes from the Schrödinger equation and depends on the time spent evolving. In an ideal cycle under the adiabatic theorem, a periodic choice of instantaneous energy eigenstate gives $|\Psi (T)⟩=e^{i(\delta +\gamma )}|\Psi (0)⟩$, with Berry phase $\gamma$.

A geometric phase is a quantum phase associated with the geometry of a path of state rays, after separating the dynamical phase. The Berry phase is its cyclic case governed by the adiabatic theorem; more general cyclic evolution can also have a geometric phase.

For the infinite square well of length $a$, the energy eigenvalues are $E_n=n^2{E}_1$, with $E_1={\pi }^2{\hslash }^2/(2m{a}^2)$. At $t_{\ast }=\pi \hslash /E_1$, the dynamical phase is $e^{-i{E}_n{t}_{\ast }/\hslash }=(-1{)}^n$. Since the midpoint parity operator has eigenvalues $(-1{)}^{n+1}$ in the same complete orthonormal basis, the time-evolution operator is exactly $U(t_{\ast })=-R_a$ on every state. At $2t_{\ast }$ every phase is one, giving a full revival.

Thus a state becomes its reflected copy, up to a global phase, at the half revival. Its position probability density is reflected, while its mean momentum changes sign because $P{R}_a=-R_{a}P$. For an initially reflection-symmetric probability density, the initial and half-revival position distributions agree even when their probability currents point in opposite directions.

Expanding a quantum state in instantaneous energy eigenstates separates dynamical phases from the coupling $A_{mn}$. orthonormality makes $A$ anti-Hermitian. For distinct eigenvalues, differentiating the eigenvalue equation gives $A_{mn}=⟨m|\dot{H}|n⟩/(E_n-E_m)$. Its off-diagonal entries drive nonadiabatic transitions; diagonal entries describe phase changes of the basis and generate the Berry phase.

A theoretical physicist whose work includes geometric phases in quantum mechanics.

Insert the expansion with amplitudes $c_1,c_2$ and the two unperturbed dynamical phases into $i\hslash \dot{\Psi }=(H_0+V)\Psi$. Differentiation of the explicit phases cancels $H_0$, leavingEquivalently, with $\Omega =(\omega -{\omega }_0)/2$,These interaction-picture equations conserve $|c_1{|}^2+|c_2{|}^2$. They are exact for the circularly rotating magnetic drive; a rotating-wave approximation has not been used.

Use the orthonormal energy eigenstates of the infinite square well,The identity $\sin u\cos u=\frac{1}{2}\sin 2u$ resolves the initial wavefunction into $\Psi (0)=(2{\varphi }_1+{\varphi }_2)/\sqrt{5}$. Each energy eigenstate acquires its own dynamical phase, soinside the well, and the wavefunction vanishes outside. The energy probabilities are $4/5$ and $1/5$ at every time. ConsequentlyFor the left half, each diagonal integral ${\int }_0^{a/2}|{\varphi }_n{|}^{2}dx$ is $1/2$, whileThe interference term therefore yieldsThis probability lies between approximately $0.1605$ and $0.8395$, with period $2\pi \hslash /(3E_1)$. A conserved energy distribution is compatible with time-dependent position probabilities because the relative quantum phase keeps evolving.

Put $\gamma =2m\hslash \omega$. If ${\psi }_n$ is a normalized energy eigenstate at $E_n$, the factorization from part (a) givesIt follows inductively that $v_n=(a_{+}{)}^n{\psi }_0$ has squared norm ${\gamma }^{n}n!$. The commutator with $H$ establishes its energy eigenvalue at each step. Henceis normalized and has $E_n=(n+\frac{1}{2})\hslash \omega$. Distinct levels are orthogonal because $H$ is a self-adjoint operator. In particular,

To decompose the specified initial wavefunction, retain this exact phase convention for the creation and annihilation operators. Their difference givesSince $a_{-}{\psi }_0=0$ and $a_{-}{a}_{+}{\psi }_0=\gamma {\psi }_0$,This quadratic excitation of an oscillator ground state, $\Psi (x,0)=C_0^{-1}{X}^2{\psi }_0$, contains only the $n=0$ and $n=2$ energy eigenstates. The printed initial function has an unspecified overall normalization, so the Born rule requires division by its squared norm. After wavefunction normalization the state isThe minus sign is consistent with the paper's convention: its $a_{+}$ is $i$ times the usual real-coordinate raising operator multiplied by $\sqrt{\gamma }$, so its ${\psi }_2$ has the opposite sign to the usual real second harmonic oscillator eigenfunction. The energy measurement probabilities are independent of this choice of phase.

Therefore the complete answer isEvery other energy has probability zero. At any later time the two coefficients only acquire their separate dynamical phases, so these probabilities remain the same. A direct check of the normalization isThis also confirms the factor of three in the squared norm of the two-component expansion.

The identity ${\cos }^{2}x=\frac{1}{2}+\frac{1}{4}{e}^{2ix}+\frac{1}{4}{e}^{-2ix}$ gives the momentum eigenstate expansionUsing the orthonormal basis from part (a), the wavefunction normalization isHence $|A|=2/\sqrt{3\pi }$. The wavefunction normalization leaves a global phase free; the positive real choice used in the requested expansion givesThe two nonzero eigenvalues of the momentum operator have the same energy eigenvalue $E_{\star }=2{\hslash }^2/m$. Under unitary time evolution, their coefficients acquire the same dynamical phase. With $\Omega =E_{\star }/\hslash =2\hslash /m$,At $t=0$, $1+\cos (2x)=2{\cos }^{2}x$ recovers the initial wavefunction. Its probability density, useful below, is

The ground state is the single energy eigenstate ${\psi }_0$. Its coefficient in the normalized wavefunction is $\sqrt{2/3}$, so the Born rule givesThis probability is independent of $t$ because unitary time evolution changes only the dynamical phases of the energy eigenstates.

Under unitary time evolution, each energy eigenstate acquires its own dynamical phase. Since $E_2=E_{-2}=E_{\star }$, the wavefunction isAt $t=0$, the identity $1-\cos (2x/R)=2{\sin }^2(x/R)$ recovers the normalized initial wavefunction.

By the Born rule, the probability for a measured energy eigenvalue is the squared norm of the orthogonal projection onto its whole energy eigenspace. ThusEvery other energy eigenvalue has probability zero. These probabilities are independent of $T$: unitary time evolution under the unchanged Hamiltonian operator changes the dynamical phases but not the squared expansion coefficients. The two equal-energy terms describe one measurement outcome.

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.

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.

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.