# Why isn't Dirac credited with the discovery of the Aharonov-Bohm effect?

Above equation (8) of Dirac's famous 1931 paper in which he proposes his quantization condition for magnetic monopoles, he says "the change in [an electron's] phase around [a] closed curve [is] $2 \pi n + e/\hbar c \int {\bf H} \cdot d{\bf S}$." This sure sounds like the Aharonov-Bohm effect to me. And he clearly understood that the effect would still work if the magnetic field itself vanishes along the electron's path, because his entire magnetic charge quantization argument boils down to "magnetic monopoles can be thought of as endpoints of infinitely thin magnetic flux tubes, and electrons can't pick up a nontrivial phase as they circle around these flux tubes, even from far away." So why does the Wikipedia article credit the discovery of the effect to Ehrenberg and Siday in 1949, and then to Aharonov and Bohm in 1959?

• Perhaps because the AB effect is a (proposed) experiment? – Keep these mind Sep 11 '16 at 6:48
• a question for history.whodidwhat.S.E. ? – user46925 Sep 12 '16 at 11:14

Dirac's discovery of the quantization of the magnetic charge is distinct from the Aharonov-Bohm effect. These effects depend on different topological properties of the manifold on which a charged particle moves. The Aharonov-Bohm effect appears on manifolds with a nonvanishing first cohomology group $H^1(M)$, while the Dirac quantization condition takes place when the second cohomology group $H^2(M)$is nontrivial. Please see, for example, V.P. Nair's lecture notes section 4 (pages 15-19).

In fact, these two effects combine to constitute all possible quantization types of a charged particle moving on a manifold.

It is true (in both cases) that when a charged particle moves on a trajectory $\Gamma$ on the manifold $M$, its wave function acquires a geometric phase of

$$e^{\frac{ie}{\hbar c} \int_{\Gamma}\mathbf{A}. d\mathbf{r}}$$

However, the consequences are different. In the first case (Aharonov-Bohm), the vector potential is proportional to a closed but nonexact one form $\alpha$ on the manifold.

$$A = \frac{\Phi}{2\pi} \alpha$$

This is the famous vector potential of the Aharonov-Bohm effect whose corresoponding magnetic field vanishes ($\Phi$ is the magnetic flux).

$$B = dA = \frac{\Phi}{2\pi} d\alpha = 0$$

When the particle moves on a loop dual to this form in homology, i.e., a noncontractible loop, the result (by the Stokes theorm) will not depend on which loop we take as a representative:

$$\int_{\Gamma_1}\mathbf{A}. d\mathbf{r} - \int_{\Gamma_2}\mathbf{A}. d\mathbf{r} = \int_S \mathbf{B}.d\mathbf{S} = 0$$

where $S$ is the area enclosed by $\Gamma_1$ and $\Gamma_2$.

Moreover, in the special case where $M=S^1$ (the circle), in which

$$\alpha = d\theta$$,

(The angle along the circumference).

In this case, the acquired phase is:

$$\frac{e \Phi}{\hbar c}$$

This means that two fluxes differing by $\frac{2 \pi\hbar c}{e}$ are physically equivalent.

In the second case when: $H^2(M)$is nontrivial, choosing the a closed trajectory to lie on a noncontractible two dimensional surface $\Omega$, allows us to compute the flux through the two halfs of the surface $\Omega_{\pm}$ whose boundaries is the closed trajectory:

$$\Phi_{\pm} = \int_{\Gamma} A_{\pm} = \pm \int_{\Omega_{\pm}} B$$.

Thus the phase difference acquired by the wave function when working using the upper half and the lower half coordinates is

$$\frac{e}{\hbar c}(\int_{\Omega_{+}}B + \int_{\Omega_{-}} B) = \frac{e}{\hbar c} \int_{\Omega} B$$

In the case of a monopole

$$\int_{\Omega} B = 4 \pi g$$

(The magnetic charge).

The phase difference should be an integer multiple of $2 \pi$ since it is physically equivalent to work on the lower or the upper halves. Thus we get the Dirac's quantization condition:

$$\frac{eg}{\hbar c} = \frac{n}{2}$$

Please observe also that in the present case the geometric phase is not topological since the magnetic field is nonvanishing.

• What do you mean by saying "magnetic field is nonvanishing"? Even far away from the magnetic charge? It does not matter because the quantization is not due to presence of $g$! – Vladimir Kalitvianski Sep 11 '16 at 11:34
• @ Vladimir Kalitvianski I meant not identically zero as in the Aharonov-Bohm case – David Bar Moshe Sep 11 '16 at 14:39
• But in both cases the field $B$ outside solenoid is identically zero and in both cases it is not zero within the solenoid. – Vladimir Kalitvianski Sep 11 '16 at 15:55
• @ Vladimir Kalitvianski but $B$ in the monopole case is constant on the surface of the sphere and if you consider the whole 3-space it behaves according the inverse square law. Its flux throug a finite surface does not vanish. – David Bar Moshe Sep 11 '16 at 16:07
• But we do not consider the field $B(\mathbf{r})$ of the monopole! Dirac's quantization of $g$ is derived from scattering from a solenoid field. – Vladimir Kalitvianski Sep 11 '16 at 16:42