The Aharonov–Bohm effect

Intuition

Split an electron beam round a long, thin solenoid and let the two halves meet on a screen. Outside the solenoid there is no magnetic field at all, so no force acts on the electrons. Yet switching the current on shifts the interference fringes. The vector potential outside is not zero: it circles the solenoid, and although it is the gradient of a function on each side, no single-valued function does for the whole way round. The two paths pick up phases $(q/\hbar)\int\mathbf{A}\cdot d\mathbf{l}$ that differ by $q/\hbar$ times the flux inside. Aharonov and Bohm predicted it in 1959, and Tonomura’s electron holography confirmed it beyond doubt in 1986. In quantum mechanics the potentials act, through the phases they give round closed loops, even where the fields vanish.

Walk round a spiral staircase back to where you started: you are above the same spot, but a floor higher. Only a whole loop reveals it. Round the solenoid the electron’s phase climbs by $q\Phi/\hbar$ in the same way, while every step looks like level ground.

Electron fringes on a screen, the intensity against the position in fringe widths, from two paths passing either side of a long solenoid: dashed with no current, solid with a quarter of a flux quantum $h/e$ inside. The whole pattern shifts by a quarter of a fringe, though no field reaches the paths.

Two paths from a source to a point on the screen pass on opposite sides of a region enclosing the magnetic flux $\Phi$ , with $\mathbf{B}=0$ on both paths. The interference between them gains the phase

\Delta\varphi=\frac{q}{\hbar}\oint\mathbf{A}\cdot d\mathbf{l}=\frac{q\Phi}{\hbar}

Properties

The phase depends only on the flux enclosed, not on the shape of the paths or on how the field is spread inside.
It is periodic: a flux of $h/|q|$ , for an electron $h/e=4.14\times10^{-15}$ Wb, gives and leaves the pattern as it was.

The Aharonov–Bohm phase

\mathbf{B}=0\ \text{on both paths}\ \Longrightarrow\ \varphi_{1}-\varphi_{2}=\frac{q\Phi}{\hbar}

Where the field vanishes the potential is, along each path, a gradient, so each path’s wave is the field-free one times the phase $(q/\hbar)\int\mathbf{A}\cdot d\mathbf{l}$ along it — the gauge transformation of the last lesson. The two phases differ by the integral round the loop the two paths make together, which Stokes’s theorem turns into the flux inside.

Proof steps

\psi_{k}=\exp\Bigl(\frac{iq}{\hbar}\int_{k}\mathbf{A}\cdot d\mathbf{l}\Bigr)\,\psi_{k}^{(0)}

Along path $k$ , $\mathbf{A}=\nabla\chi_{k}$ with $\chi_{k}=\int_{k}\mathbf{A}\cdot d\mathbf{l}$ ; is the wave with no potential.

\varphi_{1}-\varphi_{2}=\frac{q}{\hbar}\Bigl(\int_{1}-\int_{2}\Bigr)\mathbf{A}\cdot d\mathbf{l}=\frac{q}{\hbar}\oint\mathbf{A}\cdot d\mathbf{l}

Out along path 1 and back along path 2 is one closed loop.

\oint\mathbf{A}\cdot d\mathbf{l}=\int(\nabla\times\mathbf{A})\cdot d\mathbf{S}=\Phi

Stokes’s theorem; the curl is $\mathbf{B}$ , nonzero only inside the solenoid.

I\propto1+\cos\Bigl(\delta+\frac{q\Phi}{\hbar}\Bigr)

$\delta$ is the field-free phase difference: the fringes shift by $q\Phi/2\pi\hbar$ of a fringe.

Applications

Practice

No Field, Still an Effect

Outside a long solenoid $\mathbf{B}=0$ , yet the vector potential circles it; electrons passing on either side interfere with a phase set by the flux inside.

Try it

In the Aharonov–Bohm experiment, the electrons are deflected by a magnetic force outside the solenoid.