Aharonov-Bohm Effect

Equivalent forms

\Delta\varphi = \frac{2\pi\,\Phi_B}{\Phi_0},\quad \Phi_0 = \frac{h}{q}

\psi \to \psi\,e^{i\frac{q}{\hbar}\int \mathbf{A}\cdot d\boldsymbol{\ell}}

A field that is exactly zero along every path the particle takes still changes the outcome. Topology — how the path winds the flux — beats local force.

Symbol	Name	SI unit	Dimension	Range
$\Delta\varphi$	Interference phase shiftoutput Relative quantum phase between the two enclosing paths	rad	1	—
$q$	Particle charge	C	I*T	1.602176634e-19 – 1.602176634e-19
$\Phi_B$	Enclosed magnetic flux	Wb	ML^2T^-2*I^-1	0 – 2e-14
$\mathbf{A}$	Magnetic vector potential B = ∇ × A; nonzero even where B = 0	T*m	MLT^-2*I^-1	—

Where it holds

Requires phase coherence over the two paths (mesoscopic or vacuum-beam scale). The flux quantum for a single electron charge

is \Phi 0 = h/e \approx 4.14e-15\,\mathrm{Wb}

; for superconducting Cooper pairs (charge 2e) it is half that. Decoherence from inelastic scattering washes out the effect in large/warm systems.

Discovery

Yakir Aharonov · David Bohm · 1959

Aharonov and Bohm, working in Bristol, pointed out that the quantum phase depends on the vector potential A, not just the field B — so a region with B = 0 but A ≠ 0 should still produce observable effects. It was so counterintuitive that some physicists insisted the potentials must be unphysical. Robert Chambers saw the predicted fringe shift in 1960. The cleanest confirmation came in 1986 when Akira Tonomura's team at Hitachi used a superconductor-clad toroidal magnet to fully confine the flux, eliminating every loophole — the fringes still shifted.

Real-world applications

Mesoscopic ring conductors show AB conductance oscillations (basis of some qubits)
Foundation of the geometric/Berry phase concept
Flux quantization and SQUID magnetometers
Topological quantum computing braiding phases

Common misconceptions

The electrons are NOT passing through a region of nonzero B — that's the whole point
The vector potential A is gauge-dependent, but the loop integral $\oint A\cdot d$ ℓ $= \Phi is$ gauge-invariant and physical
There is no classical force and no momentum transfer — only a phase, visible solely through interference

Experimental verification

Chambers (1960) electron biprism; Tonomura et al. (1986) used toroidal ferromagnets fully clad in superconductor so the Meissner effect confined all flux, leaving zero leakage field on the electron paths — fringe shifts of exactly

\pi

between holes enclosing odd flux quanta were observed.

Derivation

In a magnetic field the canonical momentum is

p - qA

.

The wavefunction along a path acquires a phase factor

\exp [(iq/\hbar )\int A\cdot d

ℓ].

For two paths 1 and 2 meeting at a detector, the phase difference

is \Delta \varphi = (q/\hbar )[\int _{1}A\cdot d

ℓ

- \int _{2}A\cdot d

ℓ

] = (q/\hbar )\oint A\cdot d

ℓ around the closed loop.

By Stokes' theorem

\oint A\cdot d

ℓ

= \int (\nabla \times A)\cdot dS = \int B\cdot dS = \Phi _B

, the enclosed flux.

Hence

\Delta \varphi = q\Phi _B/\hbar = 2\pi \Phi _B/\Phi 0

with

\Phi 0 = h/q

— even though

B = 0

everywhere the electrons actually travel.

What if…

What if the solenoid carries zero current?

$\Phi _B = 0$ , $\Delta \varphi = 0$ — fringes sit in their field-free positions. Ramp the current and they slide sideways.

What if you used Cooper pairs (charge 2e)?

The flux quantum halves to h/2e, so a given flux produces twice the phase shift — exactly what's seen in superconducting rings.

What if there were no quantum coherence?

No interference pattern exists, so the phase is unobservable. The effect is purely quantum — classically the electrons feel nothing.

1

Flux for a π phase shift (electron)

Given ·

q:: $e = 1.602e-19\,\mathrm{C}$
target:: $\Delta \varphi = \pi$

Find · enclosed flux

\Phi _B

Steps

$\Delta \varphi = q\Phi _B/\hbar = \pi$ ⟹ $\Phi _B = \pi \hbar /q$
$= \pi (1.0546e-34)/(1.602e-19)$
$\approx 2.07e-15\,\mathrm{Wb}$ = half a flux quantum

Result ·

\Phi _B = \pi \hbar /e = \Phi 0/2 = h/2e \approx 2.07e-15\,\mathrm{Wb}

2

Phase from one flux quantum

Given ·

Phi B:: $\Phi 0 = h/e = 4.136e-15\,\mathrm{Wb}$

Find ·

\Delta \varphi

Steps

$\Delta \varphi = 2\pi \Phi _B/\Phi 0 = 2\pi \times (\Phi 0/\Phi 0)$
$= 2\pi$ (one whole fringe), so the pattern looks unshifted at integer flux quanta

Result ·

\Delta \varphi = 2\pi

— fringes shift by exactly one full period

de Broglie Wavelength→Time-Dependent Schrödinger Equation→Canonical Commutation Relation→