University Quantum Mechanics II · The Adiabatic Approximation and Berry's Phase · 14.8

The Aharonov-Bohm effect as a geometric phase

This is where Berry's formula stops being abstract. A charge that never meets a magnetic field comes back from a trip around a solenoid carrying the enclosed flux as a phase. We derive that phase from one gauge transformation, show that no state and no loop shape can change it, and read it off shifted electron fringes.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Take a long solenoid of flux Φ, wall it off so no wavefunction can enter, and carry a charge q around it in a small box. On the path B = 0, so no Lorentz force acts, but A cannot vanish: Stokes' theorem demands ∮A ⋅ dr = Φ on every loop that encloses the solenoid, and outside it A = (Φ/2πr)φ̂. Around the box A is a pure gradient, so Unit 3's gauge transformation removes it: each eigenstate is the field-free one times exp[(iq/ħ)∫A ⋅ dr′], taken from the box's centre R, with its energy unchanged.

Fed to Berry's formula, those states give a connection 𝒜ₙ = (q/ħ)A(R) that is the same for every state, and a phase γ = (q/ħ)∮A ⋅ dR = qΦ/ħ that is the same for every loop winding once. The Berry curvature is (q/ħ)B, zero wherever the box can go; all of it sits inside the solenoid, a hole in parameter space, so only the winding number matters. Split a beam around the solenoid and the arms differ in phase by qΦ/ħ: fringes slide under a fixed envelope and repeat every h/q of flux, 4.14 × 10⁻¹⁵ Wb for an electron.

The model stops where its idealisation does. A real solenoid is finite and leaks return flux onto the path, and a wave that enters the field feels a force as well. Tonomura's 1986 test sealed a toroidal magnet in superconducting niobium, and flux quantisation in units of h/2e, quoted from superconductivity, made the measured shifts exactly 0 or π.

Simple definition: The Aharonov-Bohm effect is the phase qΦ/ħ that a charge q acquires on a loop enclosing a magnetic flux Φ its wavefunction never enters; it is the loop's Berry phase, and no force acts anywhere along the way.
Example: Electrons passing either side of a shielded flux Φ = 1.0 × 10⁻¹⁵ Wb acquire a relative phase eΦ/ħ = 1.52 rad, which moves their fringes by Φ/(h/e) = 0.242 of a period; a flux of h/e = 4.14 × 10⁻¹⁵ Wb moves them by exactly one.

Vector potential outside an ideal solenoidA = (Φ/2πr) φ̂, for r ≥ b

Its curl vanishes everywhere outside the solenoid, so B = 0 on every outside path; yet no single-valued gauge removes it, because the loop integral is Φ on every circuit.

A in T m, Φ in Wb = T m², r the distance from the axis in m, b the solenoid radius; derived from ∮A ⋅ dr = Φ and symmetry

Box eigenstates by a gauge transformationψₙ(r, R) = e^(iqχ/ħ) ψₙ⁰(r − R), with χ = ∫ A(r′) ⋅ dr′ from R to r

Derived with Unit 3's gauge factor: since ∇χ = A, it cancels A in the Hamiltonian, so Eₙ cannot depend on where the box is; only the phase of the state does.

ψₙ⁰ the field-free box state of energy Eₙ; R the box centre in m; χ in T m², taken inside the box where B = 0

Berry connection of every box state𝒜ₙ(R) = i⟨ψₙ|∇_R ψₙ⟩ = (q/ħ) A(R)

The label n has dropped out. Unlike a spin's −mΩ, every state of the box carries the same connection, set by the physical vector potential where the box sits.

𝒜ₙ in m⁻¹, q in C, A in T m, ħ in J s; derived here from ⟨p⟩ = 0 in a real bound state

The Aharonov-Bohm phaseγ = ∮ 𝒜ₙ ⋅ dR = (q/ħ) ∮ A ⋅ dR = wqΦ/ħ

Derived: the same for every state and every loop shape, zero when w = 0. No power of α appears, because the phase is exact in the model rather than a correction.

γ in rad; w the number of times the loop winds round the solenoid; C × Wb ÷ J s = C V s ÷ J s = 1

Two-path fringes on a screenI(y) = I₀(y)[1 + cos(2πy/s − qΦ/ħ)]

Derived for equal arm amplitudes: fringes slide by (qΦ/h)s under an envelope that stays put, and the pattern repeats whenever Φ changes by h/|q|, which is 4.136 × 10⁻¹⁵ Wb for an electron.

I₀(y) = |ψ₁|² + |ψ₂|², the slowly varying background; y the screen position and s the fringe spacing, in m

Flux trapped by a superconducting ring (quoted)Φ = n h/2e, n an integer

An electron then gains eΦ/ħ = nπ, so the fringes shift by 0 or π and by nothing in between: the signature Tonomura's shielded toroids showed.

h/2e = 2.0678 × 10⁻¹⁵ Wb; quoted from superconductivity, where the carriers are pairs of charge 2e

A path on which no force acts

Take a long solenoid of radius b and flux Φ, and surround it with a wall no wavefunction can cross. Outside, B = 0 exactly, so a classical charge there feels no Lorentz force and follows the same orbit whatever Φ is. The vector potential cannot follow B to zero. Stokes' theorem gives ∮A ⋅ dr = Φ on every loop around the solenoid, and symmetry then fixes A = (Φ/2πr)φ̂ for r ≥ b. For b = 1.0 μm and 10 mT inside, Φ = πb²B = π × 10⁻¹⁴ Wb, or 3.14 × 10⁻¹⁴ Wb, and at r = 5.0 μm the vector potential is 1.0 × 10⁻⁹ T m where B is zero. Locally A is a gradient, A = ∇(Φφ/2π), so on any patch that does not wrap around the solenoid it can be gauged away. Globally it cannot: φ grows by 2π on each circuit, so that gauge function is not single-valued. The field-free region has a hole, and the hole is where the effect lives. Even classically the coupling qṙ ⋅ A = (qΦ/2π)dφ/dt is a total time derivative: it leaves every orbit alone but adds qΦ/2π to the action per radian swept, which a quantum phase can register.

Gauging the vector potential out of the box

Now confine the charge, of mass M, in a small box centred at R, well clear of the solenoid: H(R) = (p − qA)²/2M + V(r − R). Around the box B = 0 and there is no hole, so χ(r) = ∫A(r′) ⋅ dr′, taken along any path from R to r inside that patch, is well defined and ∇χ = A. Unit 3's gauge transformation, quoted in one line, is the identity (p − qA)e^(iqχ/ħ)ψ = e^(iqχ/ħ)(p + q∇χ − qA)ψ. With ∇χ = A the vector potential cancels, so ψₙ(r, R) = e^(iqχ/ħ)ψₙ⁰(r − R) is an eigenstate of H(R) whenever ψₙ⁰ is a field-free box state, with the same energy Eₙ. Two consequences follow at once. The spectrum depends on neither R nor Φ, so the dynamic phase −EₙT/ħ of a circuit lasting T is the same whether or not the loop encloses the flux. And because χ is measured from the box's own centre, ψₙ returns to itself when R completes the loop: the states are single-valued on the path, which Berry's formula requires.

The connection is the vector potential, for every state

Berry's formula from this unit's geometric-phase topic, γₙ = i∮⟨ψₙ|∇_R ψₙ⟩ ⋅ dR, needs the gradient of the state with respect to R. R enters twice: as the lower limit of χ, giving ∇_R χ = −A(R), and as the shift in ψₙ⁰(r − R), giving −∇ψₙ⁰. So ⟨ψₙ|∇_R ψₙ⟩ = −(iq/ħ)A(R) − ∫ψₙ⁰∇ψₙ⁰ d³r, and for a real bound state the last integral is ½∫∇(ψₙ⁰)² d³r = 0, which is ⟨p⟩ = 0 in disguise. The connection is 𝒜ₙ = (q/ħ)A(R), and γₙ = (q/ħ)∮A ⋅ dR = qΦ/ħ. The label n has gone: a spin's −mΩ depends on m because the loop turns its states differently, but here one gauge factor multiplies every box state alike. The derivation needs slow transport, ħ|⟨m|∂H/∂t|n⟩| ≪ (Eₘ − Eₙ)², which for an electron in a 10 nm box means v ≪ 6 × 10⁴ m s⁻¹. At 100 m s⁻¹ a loop of radius 5 μm takes 0.31 μs, and the 3.76 meV lowest energy of motion along one side alone gives a dynamic phase of 1.8 × 10⁶ rad that changes with T, while γ does not.

Curvature that lives where the box cannot go

Stokes' theorem turns the loop integral into a flux: γₙ = ∫(∇_R × 𝒜ₙ) ⋅ da = (q/ħ)∫B ⋅ da over any surface spanning the loop. The Berry curvature is (q/ħ)B, and it is zero at every point the box can reach. All of it sits inside the solenoid, in a region the wall cuts out of parameter space. That is the counterpart of the previous topic's monopole: there a spin's curvature was sourced by the degeneracy at zero field, here by a line of flux the box is forbidden to enter. The consequence is topological. Deform the loop freely without crossing the solenoid and γ cannot change; wind it w times and γ = wqΦ/ħ; leave the solenoid outside and γ = 0; reverse the sense and γ changes sign. The size is set by the flux in units of h/q, not by any field on the path: a solenoid of radius 1.0 μm needs only 1.32 mT inside to give an electron γ = 2π, and at 1 T a disc of radius 36 nm already holds h/e. No small coupling is expanded, so no power of α sets the size.

Two arms and a screen

An electron interferometer does the same job without a box. Split a beam so the arms pass on opposite sides of the solenoid, and recombine them on a screen. Each arm lies in a region with no hole, so each wave carries its own gauge factor exp[(iq/ħ)∫A ⋅ dr] along its own path. Those single-arm phases depend on the gauge: add ∇Λ to A and both shift by (q/ħ)[Λ(end) − Λ(start)]. Their difference, (q/ħ)∮A ⋅ dr = qΦ/ħ, does not. With equal amplitudes the screen shows I(y) = I₀(y)[1 + cos(2πy/s − qΦ/ħ)]: the fringes slide by qΦ/h of a period under an envelope that stays put, because the flux bends no path. The pattern repeats whenever Φ changes by h/|q|, 4.136 × 10⁻¹⁵ Wb for an electron. The π × 10⁻¹⁴ Wb solenoid of the first section moves electron fringes by 7.60 periods, but a snapshot shows only the 0.60; the whole number comes from counting fringes past a fixed point while the current is ramped. As ħ → 0 the fringe spacing, proportional to the wavelength h/p, shrinks to nothing and the shift becomes invisible: the classical statement that no force acts.

Tonomura's sealed toroid, and the ideal solenoid

The model's solenoid is infinitely long and its wall perfectly impenetrable. A real solenoid has ends, and its return flux spreads over the region the electrons cross, so early tests such as Chambers' 1960 fringe shift behind a magnetised iron whisker drew the objection that stray field, not A, moved the fringes. Tonomura and colleagues closed that gap in 1986; their experiment is quoted here, not derived. They used a micrometre-scale toroidal ferromagnet, whose flux circulates round the ring and has no ends, sealed in niobium and then copper. Below niobium's critical temperature of 9.2 K, a table value, the superconductor confines the field, and the copper keeps the electron wave out of the magnet. A superconducting ring traps flux only in multiples of h/2e = 2.07 × 10⁻¹⁵ Wb, quoted from superconductivity, where the carriers are pairs of charge 2e, and an electron then gains eΦ/ħ = nπ. Electron holography compared the waves through the hole and outside it and found phase shifts of 0 or π and nothing in between: a pattern no leakage force could produce.

Change one variable at a time

Make the relationship visible.

Interactive model

Enclosed flux Φ, in units of h/e0.25 h/e

Probe charge |q|, in units of e1 e

Set Φ = 0.50 h/e and an electron's fringes move half a period, bright where they were dark: the π of a shielded toroid. Set the charge to 2 and the same flux moves a whole period, so the snapshot looks untouched. Ramp Φ to 1.00 and the dot walks a fringe while the envelope stays put.

FLUX Φ1.03 10⁻¹⁵ Wb

PHASE qΦ/ħ1.57 rad

FRINGE SHIFT qΦ/h0.25 periods

SNAPSHOT SHIFT, MOD 10.25 period

Live interpretationFLUX Φ: 1.03 10⁻¹⁵ Wb. PHASE qΦ/ħ: 1.57 rad. FRINGE SHIFT qΦ/h: 0.25 periods. SNAPSHOT SHIFT, MOD 1: 0.25 period

Catch the common trap

Explain before calculating.

A charge q in a small box sits in its third energy eigenstate. The box is carried slowly around a long, impenetrable solenoid of flux Φ on a loop that circles the solenoid twice before closing. What is the Berry phase?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA walled-off solenoid of radius 0.80 μm has a uniform field of 12 mT inside, and an electron biprism sends waves past either side of it. Find the flux, the Aharonov-Bohm phase and the fringe shift, then the field that would shift the pattern by exactly one fringe.

Flux: Φ = Bπb² = 12 × 10⁻³ T × π × (0.80 × 10⁻⁶ m)² = 2.41 × 10⁻¹⁴ Wb.
Phase: eΦ/ħ = (1.602 × 10⁻¹⁹ C)(2.413 × 10⁻¹⁴ Wb) ÷ (1.0546 × 10⁻³⁴ J s) = 36.7 rad. Units: C Wb ÷ J s = C V s ÷ J s = 1, as a phase must be.
Fringe shift: Φ/(h/e) = 2.413 × 10⁻¹⁴ ÷ 4.136 × 10⁻¹⁵ = 5.83 periods. Switching the field on, five whole fringes pass a fixed point and then 0.83 of the next; a single snapshot shows only the 0.83, a phase of 5.24 rad modulo 2π.
One fringe needs Φ = h/e, so B₁ = (h/e)/(πb²) = 4.136 × 10⁻¹⁵ Wb ÷ 2.011 × 10⁻¹² m² = 2.06 × 10⁻³ T. Check: 12 mT ÷ 2.06 mT = 5.83, the shift found above.

AnswerΦ = 2.41 × 10⁻¹⁴ Wb, phase 36.7 rad, a shift of 5.83 fringes (0.83 visible in one snapshot); 2.06 mT shifts the pattern by exactly one fringe.

MediumA thin impenetrable solenoid on the z axis carries Φ = 0.40 h/e. Electrons go from a source at (−4, 0) μm to a detector at (5, 1) μm along two straight-segment arms, the upper via (0, 3) μm and the lower via (0, −2) μm. Find each arm's phase (q/ħ)∫A ⋅ dr, their difference, and the effect of the gauge change A → A + C x̂ with C = 2.0 × 10⁻¹⁰ T m.

With A = (Φ/2πr)φ̂ and the φ-component of dr equal to r dφ, A ⋅ dr = (Φ/2π)dφ: each arm contributes Φ/2π times its change in polar angle, whatever its length.
Polar angles: source 180°, upper apex 90°, lower apex 270°, detector arctan(1/5) = 11.31°. Upper arm: 180° to 90° to 11.31°, Δφ = −168.69°. Lower arm: 180° to 270° to 371.31°, Δφ = +191.31°. They differ by 360°: together the arms enclose the solenoid once.
For q = −e, with eΦ/ħ = 0.40 × 2π = 2.513 rad: upper phase = −2.513 × (−168.69/360) = +1.178 rad; lower phase = −2.513 × (191.31/360) = −1.336 rad. Difference = 2.513 rad = 0.40 × 2π, a shift of 0.40 fringe, as for any layout that encloses the solenoid once.
The gauge change has zero curl, so B is untouched. Each arm gains (q/ħ)C(x_D − x_S) = −(1.602 × 10⁻¹⁹ C)(2.0 × 10⁻¹⁰ T m)(9.0 × 10⁻⁶ m) ÷ (1.0546 × 10⁻³⁴ J s) = −2.735 rad.
New arm phases: −1.557 rad and −4.070 rad. Both have changed; their difference is still 2.513 rad. As Φ → 0 the arm phases vanish before the gauge change, and after it they are equal, so the shift is zero.

AnswerUpper arm +1.18 rad, lower arm −1.34 rad, difference 2.51 rad (0.40 fringe). After the gauge change the arms read −1.56 and −4.07 rad, and the difference is still 2.51 rad.

HardA toroidal magnet has a rectangular core 2.0 μm wide and 30 nm thick, magnetised round the ring to 0.80 T. Electron holography compares the wave through its hole with the wave outside. Predict the phase difference for the bare magnet; then for the magnet sealed in niobium below 9.2 K, assuming the shield traps the whole number of quanta h/2e nearest the bare flux; then for a probe of charge 2e.

The two electron paths form a loop that threads the ring once, so it encloses the core flux once: Φ = 0.80 T × (2.0 × 10⁻⁶ m × 30 × 10⁻⁹ m) = 4.80 × 10⁻¹⁴ Wb.
Bare magnet: eΦ/ħ = 72.9 rad, or Φ/(h/e) = 11.61 fringes. A snapshot shows the fractional 0.61 of a fringe, a phase of 3.81 rad modulo 2π.
Sealed magnet: Φ/(h/2e) = 4.80 × 10⁻¹⁴ ÷ 2.0678 × 10⁻¹⁵ = 23.21, so n = 23 and Φ = 23 × 2.0678 × 10⁻¹⁵ Wb = 4.756 × 10⁻¹⁴ Wb. Screening currents in the niobium trim the flux by 4.4 × 10⁻¹⁶ Wb, only 0.92%.
Electron phase: eΦ/ħ = (e/ħ)(23h/2e) = 23π, which is π modulo 2π. The fringes through the hole sit half a period from those outside, bright where the others are dark. A sample trapping 24 quanta, 4.963 × 10⁻¹⁴ Wb, would show 0; the bare 3.81 rad is no longer possible.
Charge 2e: (2e)(nh/2e)/ħ = 2πn = 46π, no visible shift at all. The shield's own pairs see an Aharonov-Bohm phase that is a whole multiple of 2π, and that single-valuedness is where the quantisation comes from. Check: n = 0 gives no shift for any probe.

AnswerBare: 4.80 × 10⁻¹⁴ Wb, 72.9 rad (11.61 fringes, 3.81 rad visible). Sealed: 23 quanta, 4.756 × 10⁻¹⁴ Wb, phase 23π, a shift of exactly π. Charge 2e: 46π, no shift.