Why Can't We Fix the Gauge in the Magnetic Aharonov-Bohm Effect?

[LarryS]

I am struggling with understanding the magnetic (solenoidal) version of the Aharonov-Bohm effect.

The magnetic vector potential, A, is defined as any vector field whose curl is the magnetic force field, B. This definition makes A not unique.

The presence of the solenoid makes the region of A not simply-connected. Somehow, that prevents us from choosing (fixing the gauge) one unique A for the entire region and this causes a phase difference between electrons that travel one way around the solenoid versus those that travel the other way.

I understand, in the electrostatic version of the AB effect why we cannot fix the gauge for the electrostatic potential (only simply-connected regions can be conservative), but why, in the above magnetic version can we not fix the gauge and why does that cause the phase difference/interference pattern?

Thanks in advance.

 

[Paul Colby]

Are you certain that it is not possible to choose a gauge in the multi connected case? If we choose $A = f(r) \hat{z}$ for a wire, then $B=f'(r) \bar{r}\times \hat{z}$ which yields a perfectly okay $B$-field. Is my $A$ not in a "fixed" gauge?

 

[vanhees71]

Of course you can choose a gauge, and any gauge you want. The only thing entering the AB effect is, the integral appearing in the phase factor
$$\Phi=\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{A}=\int_F \mathrm{d}^2 \vec{F} \cdot (\vec{\nabla} \times \vec{A}) = \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B},$$
which is the magnetic flux through (any) surface with the boundary $\partial F$ and this is a gauge-invariant quantity, only depending on the magnetic field.

It's understandable also by a very beautiful topological argument as a non-integrable phase factor:

Wu, T. T., Yang, C. N.: Concept of nonintegrable phase factors and global formulation of gauge fields, Phys. Rev. D 12, 3845, 1975

Also see the QM textbook by Sakurai.

 

[LarryS]

Of course you can choose a gauge, and any gauge you want. The only thing entering the AB effect is, the integral appearing in the phase factor
$$\Phi=\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{A}=\int_F \mathrm{d}^2 \vec{F} \cdot (\vec{\nabla} \times \vec{A}) = \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B},$$
which is the magnetic flux through (any) surface with the boundary $\partial F$ and this is a gauge-invariant quantity, only depending on the magnetic field.

It's understandable also by a very beautiful topological argument as a non-integrable phase factor:

Wu, T. T., Yang, C. N.: Concept of nonintegrable phase factors and global formulation of gauge fields, Phys. Rev. D 12, 3845, 1975

Also see the QM textbook by Sakurai.

Trying to understand your math. I see Stokes Theorem in combination with the definition of the magnetic vector potential. Must your surface F contain the "hole" caused by presence of the long solenoid?

 

[DrDu]

This whole non-connectedness is mathematical idealisation. For some reason I never quite understood mathematicians love to exclude the only region of physically relevance from their range of definition and then start to discuss what possibilities are left.
Physically, you have a more or less localized magnetic field in the center and you are integrating over all space. Whether it is due to a whisker or a coil pair is irrelevant.

 

[vanhees71]

In the case of the AB effect the point is that you get an non-integrable phase factor (see the above cited paper by Wu and Yang). It's most easily derived via the path integral. The point is that although the particles move in a (idealized) field-free region but still the magnetic field in the region, which is forbidden for the particles, has an effect in terms of this phase, i.e., measurable interference effects (as in the double-slit experiment).

@referframe of course the "allowed region" of the particle must be multiply connected, because otherwise the $\vec{\nabla} \times \vec{A}=0$ would imply via Poincare's lemma that $\vec{A}=-\vec{\nabla} \chi$ for some scalar field $\chi$, and the phase would vanish. Then any closed curve encircling a forbidden region, where $\vec{B} \neq 0$ encircles a surface containing this region, and thus the magnetic flux doesn't vanish. Which surface you choose is also independent due to Gauss's theorem, because you can take two different surfaces with the same boundary and make a closed surface out of it by adding the cylinder-like parts in the region, where $\vec{B}=0$. Then you use Gauss's theorem and $\vec{\nabla} \cdot \vec{B}=0$ to show that the surface integral along this closed surface is 0 and thus that the magnetic flux is the same for the two surfaces with the common boundary line.

 

[LarryS]

In the case of the AB effect the point is that you get an non-integrable phase factor (see the above cited paper by Wu and Yang). It's most easily derived via the path integral. The point is that although the particles move in a (idealized) field-free region but still the magnetic field in the region, which is forbidden for the particles, has an effect in terms of this phase, i.e., measurable interference effects (as in the double-slit experiment).

@referframe of course the "allowed region" of the particle must be multiply connected, because otherwise the $\vec{\nabla} \times \vec{A}=0$ would imply via Poincare's lemma that $\vec{A}=-\vec{\nabla} \chi$ for some scalar field $\chi$, and the phase would vanish. Then any closed curve encircling a forbidden region, where $\vec{B} \neq 0$ encircles a surface containing this region, and thus the magnetic flux doesn't vanish. Which surface you choose is also independent due to Gauss's theorem, because you can take two different surfaces with the same boundary and make a closed surface out of it by adding the cylinder-like parts in the region, where $\vec{B}=0$. Then you use Gauss's theorem and $\vec{\nabla} \cdot \vec{B}=0$ to show that the surface integral along this closed surface is 0 and thus that the magnetic flux is the same for the two surfaces with the common boundary line.

Thanks, I found a free version of that paper by Wu and Yang. It seems that an oversimplified answer would be that Stoke's Theorem fails if the curl of the enclosed vector field is zero and the domain is not simply-connected.

 

[vanhees71]

Stokes's theorem doesn't fail. How do you come to that conclusion? It's just that in a multiply connected domain $\vec{\nabla} \times \vec{A}=0$ does NOT imply that the line integral over $\vec{A}$ along a closed path vanishes.

The most simple example is the potential vortex. In cylindrical coordinates it's given by
$$\vec{A}=-\vec{\nabla} \varphi = -\frac{1}{r} \vec{e}_{\varphi}=-\frac{-y \vec{e}_x+x \vec{e}_y}{r^2}.$$
It's singular at $\rho=0$, i.e., along the $z$ axis, i.e., you have a multiply connected domain, and any closed curve around the $z$ axis gives
$$\int_{C} \mathrm{d} \vec{x} \cdot \vec{A}=-(\phi_1-\phi_0)=-2 \pi,$$
because for a closed curve (running once around the $z$ axis) you have $\phi_0=-\pi$ and $\phi_1=+\pi$. Of course, the curl vanishes everywhere around the $z$ axis. On the $z$ axis it has a $\delta$-distribution like singularity.

To see this we regularize the field by writing
$$\vec{A}_{\epsilon}=-(-y \vec{e}_x+x \vec{e}_y)/(r^2+\epsilon^2).$$
Now we have
$$\vec{\nabla} \times \vec{A}_{\epsilon}=-\frac{2 \epsilon^2}{(r^2+\epsilon^2)^2}.$$
Now take the surface integral over any plane $F$ parallel zu the $xy$ plane. Since $\mathrm{d} \vec{F}=\mathrm{d}r \, \mathrm{d} \varphi \, r \vec{e}_z$ you get
$$\int_{F} \mathrm{d} \vec{F} \cdot (\vec{\nabla} \times \vec{A}_{\epsilon})=-2 \pi.$$
Letting $\epsilon \rightarrow 0$ tells you that you have, written in Cartesian coordinates,
$$\vec{\nabla} \times \vec{A}=-2 \pi \delta(x) \delta(y) \vec{e}_z.$$
You cannot write this in cylinder coordinates, because cylinder coordinates have a singularity along the $z$ axis, which is the support of the above $\delta$ distribution.