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The Aharonov-Bohm effect is the phase $q\Phi /\hslash$ acquired when a charge $q$ encircles magnetic flux $\Phi$, even when the magnetic field vanishes along the charge's path.

In the magnetic Aharonov-Bohm effect, a coherent electron wave takes two routes around inaccessible magnetic flux. Recombination converts their relative wave phase into displaced interference fringes. In 1959, Yakir Aharonov and David Bohm proposed a shielded solenoid geometry and identified the obstruction to removing the magnetic vector potential throughout the accessible region.^[ref]

Use International System of Units, electric charge $q$, and magnetic minimal coupling $H=(-i\hslash \nabla -q\mathbf{A}{)}^2/(2m)$. Let ${\gamma }_1,{\gamma }_2$ run from the same source $S$ to the same detector point $D$. The oriented closed curve $C$ follows ${\gamma }_1$ and then ${\gamma }_2$ backwards. For a static magnetic vector potential, the magnetic contribution to the relative wave phase isHere ${\Phi }_C={\int }_{\Sigma }\mathbf{B}\cdot d\mathbf{S}$ is the oriented magnetic flux through a spanning surface. Equal or calibrated nonmagnetic propagation phases are assumed. For an electron, $q=-e$, where $e>0$ is the elementary charge; reversing $C$ reverses the sign. The observable is the gauge-invariant holonomy $e^{i\delta }$, rather than a unique local value of the magnetic vector potential.

The accessible region is a multiply connected region: $C$ cannot contract to a point without entering the excluded flux region. Stokes theorem does not set the line integral to zero, because a spanning surface must cross that excluded region, where the magnetic field need not vanish. There is real magnetic flux in the apparatus; only the accessible paths are idealized as free of the magnetic field. The exterior of a finite straight solenoid also permits routes around its ends, so its exact topology and leakage differ from those of an infinite solenoid or an excluded solid torus.

An explicit model makes the obstruction visible. Outside an infinite flux tube of radius $a$, choose cylindrical coordinates $(r,\theta ,z)$ andIts curl vanishes there, but a loop with winding number $n$ has ${\oint }_C\mathbf{A}\cdot d\mathbf{l}=n\Phi$. Locally $\mathbf{A}=\nabla (\Phi \theta /(2\pi ))$; the angular coordinate changes by $2\pi$ on a circuit. For generic $q\Phi /\hslash$, that candidate gauge removal changes the wave function's single-valued boundary condition. On a simply connected region, a smooth static magnetic vector potential with zero curl instead has a globally defined scalar primitive and is a pure gauge potential. Thus local zero magnetic field and globally removable magnetic vector potential are different conditions.

For an original interference example, take mutually coherent scalar amplitudes with constant individual intensities $I_1,I_2$ and an imposed carrier phase $ky$, as in off-axis holography. ThenThe interference fringe spacing is $d=2\pi /k$. An electron enclosing $\Phi =h/(4e)$ has $\delta =-\pi /2$, shifting maxima by $d/4$ for this orientation. Increasing the magnetic flux by $h/e$ changes $\delta$ by $-2\pi$ and restores the same pattern: the electron Aharonov-Bohm effect has magnetic flux quantum $h/e$. This uses $h=2\pi \hslash$, the Planck constant. It does not require the source's magnetic flux to vary continuously during a single exposure.

For $I_1=1$, $I_2=1/4$, the extrema are $I_{\max }=9/4$, $I_{\min }=1/4$, and the interferometric visibility is $4/5$. A fixed Aharonov-Bohm phase translates these fringes without changing their contrast. Averaging unresolved fluctuating phases can reduce the contrast; a darkened pattern alone cannot identify enclosed magnetic flux. These formulas assume coherence, indistinguishable alternatives and stable beam envelopes.

The appeal of the Aharonov-Bohm effect is that a local Lorentz force calculation along either ideal path misses information carried by their global arrangement. It gives a physical example of holonomy and supplies the encircling phase used in Chern--Simons flux attachment. This alone does not establish intrinsic topological order. The static comparison also matters: changing magnetic flux can induce an electric field, so a time-dependent experiment requires the scalar-potential contribution as well as the displayed spatial line integral.

For the same mutually coherent, equally polarized monochromatic waves, the wave interference minimum occurs at phase difference $\delta =(2m+1)\pi$, with integer $m$. The wave amplitudes subtract, giving $I_{\min }=I_1+I_2-2\sqrt{I_1{I}_2}$. Complete cancellation requires equal individual wave intensities; $1$ and $0.25$ leave $0.25$ in their common unit. A dark band is a local minimum, and the average across an interference fringe spacing still obeys conservation of energy.

Use the schematic below to count the water traversals explicitly. Each stationary tube has laboratory length $\ell$. In the displayed setting, upper-tube water flows right and lower-tube water flows left, each at speed $u>0$. Beam A goes right through the upper tube, exchanges tubes at the far optics, and returns left through the lower tube: it follows the flow twice. Beam B goes right through the lower tube and returns left through the upper tube: it opposes the flow twice. Each complete beam path therefore contains total water length $L=2\ell$, not $\ell$ or $4\ell$. The connecting air paths and fixed optics are assumed to contribute no flow-dependent phase difference.

Let $v_{+}$ and $v_{-}$ be positive phase velocity magnitudes along and against the water flow:For a laboratory source of frequency $\nu =c/{\lambda }_0$, where ${\lambda }_0$ is its vacuum wavelength, define the signed interference fringe offset $N$ by the propagation phase difference $\Delta \varphi =2\pi \nu \Delta t=2\pi N$. This is an equivalent phase delay between steady beams; no stopwatch resolves their transit-time difference. Subtract the zero-flow baseline and expand to first order:Thus one shifted fringe corresponds to one cycle of phase difference. If the stationary pattern has spatial interference fringe spacing $d$, a simple readout model is $I(x)=I_0[1+\mathcal{V}\cos (2\pi x/d+2\pi N)]$, where $\mathcal{V}$ is the interference visibility. The maxima shift by $-dN$ with this choice of screen coordinate. A sign assigned to “rightward” motion depends on the image orientation; the magnitude in fringe units does not.

Reversing both flows gives $N(-u)=-N(u)$. The displacement between equal and opposite settings therefore has magnitudeThe factor $2$ from comparing opposite flows is additional to the factor $2$ already included in $L=2\ell$. Repeating a round trip in just one flowing tube would instead put a beam with the flow once and against it once; that geometry does not produce the two same-sign contributions drawn here.

Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

Let $I_0$ be the intensity from either fibre alone at the observation point, assuming equal output intensities, matching polarizations and negligible drift. The screen samples the relative phase $\Phi (X)$ between the two outputs. For an ideal monochromatic laser, first-order optical coherence is complete andThe interference fringes are approximately equally spaced in the small-angle far field, with a slowly varying diffraction envelope. If the output separation is $a$ and the screen distance is $D$, the interference fringe spacing is approximately ${\lambda }_{0}D/a$. Equal phases at the two input ends do not ensure a central bright fringe: the extra fibre contributes carrier phase $2\pi n_{\mathrm{eff}}(l_2-l_1)/{\lambda }_0$ in addition to the free-space optical path difference.

For a finite linewidth, the time-averaged interference signal becomesHere $g^{(1)}$ may be defined for the slowly varying field envelope, with the carrier phase already included in $\Phi$. A realistic laser therefore gives reduced interferometric visibility; for $V=1/2$, its extrema are $I_0$ and $3I_0$.

For a Lorentzian line shape of frequency full width at half maximum $\Delta \nu$, temporal optical coherence givesThe relevant index is the group index. Using the intended nondispersive approximation $n_g\simeq n_{\mathrm{eff}}=2$,This delayed-fibre linewidth measurement assumes the loss of contrast comes from linewidth. Unequal output intensities, polarization mismatch or mechanical phase fluctuations also lower interferometric visibility. If dispersion matters, the stated phase index alone does not specify the necessary group index.

A line with lifetime broadening has a Lorentzian line shape. Write its half-width at half-maximum in angular-frequency units as $\kappa >0$. The Wiener-Khinchin theorem, or the Fourier transform of the Lorentzian, givesTherefore in vacuumThe sketch is a cosine fringe pattern about baseline $I_0$, bounded by $I_0[1\pm e^{-\sqrt{2}\kappa |y|/c}]$. Its central value is $2I_0$, its spatial interference fringe spacing is $\lambda /\sqrt{2}$, and its visibility decays on length $c/(\sqrt{2}\kappa )$. For an isolated transition whose population lifetime is $T_1$ and which has no extra dephasing, $\kappa =1/(2T_1)$, so the field coherence time is $2T_1$, not $T_1$. Introducing $\kappa$ avoids confusing population, wave amplitude and spectral-width conventions when no numerical linewidth is given.

For each beam in vacuum, its magnetic field is ${\mathbf{B}}_j={\hat{\mathbf{k}}}_j\times {\mathbf{E}}_j/c$. Write the two complex scalar contributions, including the beam splitter factor, as $u_1=f(t-t_1)/\sqrt{2}$ and $u_2=f(t-t_2)/\sqrt{2}$. Since their electric fields are both along $z$,The optical-cycle and ensemble-averaged Poynting vector isThe mean individual light intensities are equal, soThus energy flows along the bisector of the two beam directions, crossing a detector plane normal to $x$. Such a sensor can in principle record the variation across its $y$ coordinate. Its actual normal energy flux is $I(y)/\sqrt{2}$ in the normalization used above; this fixed factor does not change interferometric visibility.

There is a practical resolution condition in addition to the direction of energy flow. The interference fringe spacing $\lambda /\sqrt{2}$ is very small, so direct pixels would need spacing no larger than $\lambda /(2\sqrt{2})$ as required by the Nyquist–Shannon sampling theorem, or optical magnification and sufficient angular aperture to preserve both incident directions. An ordinary optical camera with much larger effective pixels averages the interference fringes. The Poynting-vector argument establishes the allowed measurement geometry, not guaranteed resolution with any optical camera.

Each mutually incoherent patch of an extended star produces a displaced double-slit interference pattern. The detector adds their light intensities. For two apertures separated by a projected interferometric baseline $D$, a source direction offset by a small angle ${\vartheta }_x$ changes their phase difference by $2\pi D{\vartheta }_x/\lambda$. The normalized sum over the source's angular brightness distribution is its first-order optical coherence, whose magnitude is the interferometric visibility for equally bright channels.

For a uniform circular stellar disk of full angular diameter $\alpha$, the normalized angular Fourier transform gives the uniform-disk stellar visibilityHere $J_1$ is the order-one Bessel function of the first kind. This follows by integrating $e^{i(2\pi D/\lambda ){\vartheta }_x}$ over the disk and dividing by its area. In particular, its power series gives $2J_1(x)/x=1-x^2/8+O(x^4)$.

For an ordinary slit separation $d=1\mathrm{mm}$, using the paper's $\alpha =22.6\times 10^{-8}\mathrm{rad}$ and observing wavelength $\lambda =570\mathrm{nm}$,Even $d=10\mathrm{mm}$ gives a visibility loss of only $1.94\times 10^{-5}$. Thus millimetre separation leaves the interference fringes almost indistinguishable from those of a point source; it gives very little sensitivity to the diameter.

The movable mirrors in the Michelson stellar interferometer collect light at a large adjustable separation $D$ and redirect it into closely spaced output beams. They increase the sampling interferometric baseline independently of the fixed output slit spacing $d$: $D$ sets sensitivity to source structure, whereas the output optics set the screen's interference fringe spacing. The first Bessel function zero is $x_1=3.83170597$, henceThe interference fringes disappear at this baseline even though each beam separately remains bright.

The value $3.08\mathrm{m}$ assumes a uniform disk. The original question supplies a diameter without a surface brightness law; real limb darkening changes the precise visibility zero. A one-dimensional uniformly bright strip of the same full angular width instead has a sinc function visibility and first zero $D=\lambda /\alpha =2.522\mathrm{m}$. That is a different source model, not the circular-disk result. The robust conclusion without a surface brightness model is that a baseline of a few metres, rather than millimetres, is required.

For a stationary rotating apparatus driven by a source with optical coherence and proper frequency ${\nu }_0$, define ${\lambda }_0=c/{\nu }_0$. The source wave phase advances by $2\pi {\nu }_0\Delta \tau$ between the emission times of the two contributions that meet at one readout spacetime event. Define the signed phase lag $\Delta \varphi$ of the $+$ contribution and the corresponding displacement $N$ in units of one interference fringe byThe sign of a photographed spatial displacement also depends on the imposed fringe orientation; this definition fixes the theoretical sign through the delayed beam. In an equal-amplitude output, a representative light intensity is $I(x)=I_{\max }[1+\cos (Kx-\Delta \varphi )]/2$, where $K$ sets the chosen spatial interference fringe spacing $2\pi /K$. Moving $\Delta \varphi$ by $2\pi$ shifts the pattern by one fringe. A fixed output exactly at a bright or dark fringe has zero first-order intensity slope, so a known phase bias or a spatial fringe readout is useful for determining a small signed delay.

Choose a square of side $0.30\mathrm{m}$, a rectangle $0.40\mathrm{m}$ by $0.20\mathrm{m}$, ${\lambda }_0=633\mathrm{nm}$ and $\Omega =+10\mathrm{rad}{\mathrm{s}}^{-1}$, with angular velocity parallel to the positive normal vectors. Both perimeters are $1.20\mathrm{m}$, but the areas are $0.090{\mathrm{m}}^2$ and $0.080{\mathrm{m}}^2$. Using $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, the leading predictions areThese are original model choices and predictions, not measured points. The square has $9/8$ times the response despite the equal path lengths. Reversing from $+\Omega$ to $-\Omega$ gives $N(-\Omega )-N(+\Omega )=-2N(+\Omega )$: the displacement magnitude between opposite rotations is twice the displacement magnitude from rest. Confusing these two comparisons introduces a factor-of-two error.