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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.

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

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 $\gamma$.

For $H(t)=\hslash \omega (a^{†}a+1/2)-Fx{e}^{t/\tau }$ on $t\le 0$, start in the unforced ground state at $t=-\infty$. Define $x_{01}=\sqrt{\hslash /(2m\omega )}$ and $\eta =F{x}_{01}/(\hslash \omega )$. Time-dependent perturbation theory givesThus the first excited-state probability in the original oscillator basis tends to zero for a rapid turn-on and to ${\eta }^2$ for slow turn-on, within the weak-force approximation $|\eta |\ll 1$.

This finite adiabatic limit is not a failure of the adiabatic theorem: the instantaneous ground state is displaced and has a nonzero projection onto the original excited states. In fact a linear force preserves a coherent state, whose exact original-basis displacement at $t=0$ is $\alpha =i\eta \omega \tau /(1+i\omega \tau )$. It follows either from $\dot{\alpha }=-i\omega \alpha +iF{x}_{01}{e}^{t/\tau }/\hslash$ or from the forced oscillator equations. Consequently the exact original-basis probability is $P_1^{(0)}=|\alpha {|}^2{e}^{-|\alpha {|}^2}$. The instantaneous oscillator basis has displacement $\eta$, so its mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/[1+(\omega \tau {)}^2]$, which vanishes adiabatically.

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.

A Landau-Zener transition occurs when a changing parameter sweeps two coupled quantum states through an avoided crossing. In a two-state description,the uncoupled energy levels cross at $t=0$, but the coupled energy eigenvalues have minimum separation $2|\Delta |$. The adiabatic theorem predicts following an instantaneous energy eigenstate during a sufficiently slow sweep; a fast sweep can retain the original uncoupled-state character and thus change instantaneous energy eigenstate.

For an ideal linear sweep extending from $t=-\infty$ to $+\infty$, the probability of remaining in the initial uncoupled state is $\exp [-2\pi |\Delta {|}^2/(\hslash |\alpha |)]$. The exponent compares the square of the coupling with the rate of change of the uncoupled energy difference. This model helps estimate when a molecular passage through an avoided crossing or a controlled qubit sweep ceases to obey the adiabatic theorem.

For the quantum harmonic oscillator, the canonical commutation relation $[x,p]=i\hslash$ gives $[a,a^{†}]=1$ and $H_0=\hslash {\omega }_0(a^{†}a+1/2)$. Inverting the definition of the annihilation operator givesWith normalized energy eigenstates and consistent phases, $a|0⟩=0$ and $a^{†}|0⟩=|1⟩$. Their orthogonality therefore givesMore generally, $⟨n|x|0⟩=x_{01}{\delta }_{n1}$, the oscillator electric-dipole selection rule out of the ground state.

To derive the transition amplitude, insert the expansion in stationary energy eigenstates into the Schrödinger equation with $H=H_0+V_1(t)$. The derivatives of the explicit phases cancel the $H_0$ terms. Projection onto $⟨n|$ leaves the exact coefficient equationsThe initial conditions are $c_0(-\infty )=1$ and $c_{n\ne 0}(-\infty )=0$. To first order in the perturbation, replace the amplitudes on the right by these initial values. Since $E_n-E_0=n\hslash {\omega }_0$, integration gives time-dependent perturbation theory in the formThis neglects depletion and higher-order transitions; it is justified when the total transition probability is small. Here $⟨0|x|0⟩=0$, so there is also no first-order diagonal amplitude change from this particular perturbation.

For the exponential electric field, the electric-dipole interaction is $V_1=-qEx{e}^{t/\tau }$, with $\tau >0$. Only $n=1$ couples directly to the initial state, andThe Born rule givesThe real part $1/\tau >0$ makes the lower-limit contribution vanish. ThusSet $u={\omega }_0\tau$ and $\eta =qE{x}_{01}/(\hslash {\omega }_0)$. The dimensionless form is $p_1/{\eta }^2=u^2/(1+u^2)$, increasing monotonically from zero to one, with half the asymptotic value at $u=1$. A sufficient weak-field condition for the entire curve is $|\eta |\ll 1$.

For ${\omega }_0\tau \ll 1$, the force has acted for too little time to appreciably change the state at the instant of measurement. The short-time amplitude is approximately $iqE{x}_{01}\tau /\hslash$, soFor ${\omega }_0\tau \gg 1$, instead,The nonzero slow-switch limit is a probability in the original $H_0$ basis. It does not contradict the adiabatic theorem. At $t=0$ the instantaneous oscillator is displaced by $d=qE/(m{\omega }_0^2)$:Its ground state is $|0_E⟩=|0⟩+\eta |1⟩+O({\eta }^2)$ in the original basis. Adiabatic following therefore predicts exactly the finite original-basis limit found above.

The exponential switch-on of a driven quantum oscillator also permits an exact check. A linear force preserves a coherent state, with original-basis displacement $\alpha (0)=i\eta u/(1+iu)$. Thus the exact original-basis probability is $|\alpha {|}^2{e}^{-|\alpha {|}^2}$, whose weak-field limit is the expression above. In the instantaneous displaced basis the mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/(1+u^2)$ and does tend to zero for slow switching. The two bases measure different notions of excitation.

For a fixed Hilbert space, a bounded self-adjoint Hamiltonian operator with a finite step $H_i\to H_f$ at $t=0$ does not deliver an instantaneous rotation of the quantum state. Integrating the Schrödinger equation across a short interval givesFor a normalized quantum state and $\Vert H(t)\Vert \le M$ in operator norm, the difference has norm at most $2ϵM/\hslash$, which tends to zero. The time derivative can jump even though the quantum state is continuous. Here $\hslash$ is the reduced Planck constant. An instantaneous replacement of the eigenbasis changes coordinates, not the vector being represented.

For a real switch occupying $0\le t\le \tau$, remove any real scalar contribution $a(t)I$ that produces only a global phase, and write $|\tilde{\psi }(t)⟩=e^{i{\int }_0^{t}a(s)ds/\hslash }|\psi (t)⟩$. The same integral argument gives the sufficient bounded-model estimateThus $\epsilon \ll 1$ controls the sudden approximation up to a global phase. For any normalized final energy eigenstate, its probability amplitude differs from the instantaneous-switch value by at most $\epsilon$, and its Born rule probability differs by at most $2\epsilon$. These are sufficient error bounds, not necessary criteria. For the fixed-gap two-level sudden quench, any switching path $H(t)=-\Delta \mathbf{n}(t)\cdot \mathbf{\sigma }/2$ with a unit vector $\mathbf{n}(t)$ has operator norm $\Delta /2$, giving $\epsilon \le \Delta \tau /(2\hslash )$.

This estimate is deliberately for bounded Hamiltonian operators. An unbounded energy spectrum does not supply a finite global $M$; relevant quantum states, operator domains and excited energy eigenvalues require separate control. The adiabatic theorem also requires hypotheses about the path and its spectral gaps; there is no universal switch-duration criterion covering all spectra. The cold-atom collapse and revival after a lattice quench illustrates how a single switch can be sudden for one degree of freedom and slow for another.

Let the atomic beam travel along $x$, with velocity $v>0$, and let $z$ measure displacement along the dominant magnetic field gradient. For a fixed effective magnetic moment component ${\mu }_z$, the magnetic dipole potential energy is $U\simeq -{\mu }_z{B}_z$. Taking its negative partial derivative givesA spatially uniform magnetic field can produce torque and Larmor precession, but cannot produce this translational separation. In a quantum state following a local internal energy eigenvalue $E_n(B)$, the more general expression is $F_z=-{\partial }_z{E}_n$, with effective magnetic moment $-\partial E_n/\partial B$. Treating that moment as constant is an approximation with a physical regime, not an extra measured fact. Assume the local magnetic field direction stays nearly fixed across the sampled beam, and that internal states follow it without transitions, as in the adiabatic theorem. The scalar gradient model describes the beam's neighborhood; it omits the full magnetic field's transverse components and fringe regions.

Assume a constant $G$ over magnet length $\ell$, followed by a field-free drift $D$. Neglect transverse motion at entry and changes in longitudinal velocity. Newton's second law gives transverse acceleration $a_z={\mu }_{z}G/M$ for whole-atom mass $M$. The magnet transit time is $t_m=\ell /v$, and the drift time is $t_d=D/v$. The magnet supplies both a displacement and an exit transverse velocity:The $\ell D$ contribution is motion with the acquired transverse velocity, not continued acceleration outside the magnet. For the plate-at-exit geometry, set $D=0$. For a nonzero entrance transverse velocity $w$, add $w(\ell +D)/v$. These terms show why initial collimation and flight geometry matter when inferring a magnetic moment from a deposit.

Choose hypothetical parameters $M=1.79\times 10^{-25}\mathrm{kg}$, $G=300\mathrm{T}{\mathrm{m}}^{-1}$, $\ell =0.035\mathrm{m}$, $D=0.100\mathrm{m}$, and $v=500\mathrm{m}{\mathrm{s}}^{-1}$. Approximate the two effective magnetic moments by ${\mu }_z=\pm {\mu }_B$, where the Bohr magneton is ${\mu }_B\simeq 9.274\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$. Then $|a_z|\simeq 1.55\times 10^4\mathrm{m}{\mathrm{s}}^{-2}$, $t_m=70\mu \mathrm{s}$ and $t_d=200\mu \mathrm{s}$. The positive displacement is about $0.0381\mathrm{mm}$ inside the magnet plus $0.2176\mathrm{mm}$ during drift: $d=0.2557\mathrm{mm}$, with full separation $2d=0.5114\mathrm{mm}$. These chosen values illustrate the model and are not reconstructed 1922 measurements.

The exit transverse velocity is only about $1.09\mathrm{m}{\mathrm{s}}^{-1}$, small compared with the chosen forward velocity; this checks the small-deflection assumption. Halving $G$ halves $d$, doubling $v$ divides $d$ by four, and setting $D=0$ removes most of this example's separation. A real pole has fringe magnetic fields and a position-dependent gradient; these require a more complete trajectory calculation.