Aharonov-Bohm topological explanation

1. Dec 2, 2012

TrickyDicky

In trying to get the Aharonov-effect right I've found something that I'm not sure how to sort out.
Briefly put my understanding of the effect is that it shows something that cannot be explained by classical physics in the sense that makes observable a classical EM global gauge transformation that shouldn't be observable within classical EM, where only the effects of the fields are observable but not the effects of the potentials. But this is not the case in QM where potentials are physical, and so it was theoretically predicted and later experimentally confirmed that charged particles going thru a a region where a magnetic field is negligible show a shift in their diffraction pattern caused just by the vector potential and varying with the flux thru the solenoid.
Now there is no question about where the shift comes from because the pattern is different between a close to zero magnetic field and a potential. So in practice there is no need to really make the magnetic field vanishing with an ideal infinite solenoid, which is great because otherwise the experiment would be impossible.

The explanation of the effect as given in topological terms is that the presence of the solenoid inside the loop-like disposition of the electrons makes the topology of the space nontrivial, and this is what causes the potential to be observable. Specifically the presence of a string defect makes the winding number around it observable. This is because R^3 space minus an infinite line makes the space no longer simply-connected. In practice the experimental setup obviously doesn't use an infinite solenoid, since it is considered that the space in wich the electrons are confined, kind of a torus-like, is also not simply-connected and that is what matters.

Now to the part I find difficult, it is my understanding that the topological requirement for the existence of a vector potential in the presence of a magnetic field when the gaus law for magnetism holds is that the space must be contractible,and contractibility implies simply-connectedness, so if the experimental space set up for the A-B effect to show up must be non-simply connected, it would seem like the very condition for the existence of the vector potential is absent.
Any hints?

2. Dec 2, 2012

Staff: Mentor

Magnetic field lines are closed - in a real experiment, they are closed somewhere and one beam part passes through the field loop. This way, space is really not simply connected if you are not allowed to "cross" the magnetic field. The vector potential does not have that limitation, it is defined on our real R^3-space which is simply connected.

3. Dec 2, 2012

TrickyDicky

Yes

Let's concentrate for a moment on the theoretical explanation of the effect, that is idealized, but should work in principle. The space is R^3-R, so not simply connected, this is the space where the magnetic field is switched on, but in such space the vanishing divergence of B doesn't imply a vector potential to begin with.

4. Dec 2, 2012

Staff: Mentor

That is not our universe. The vector potential is calculated in our universe, which is R^3 (neglecting time and possible extra dimensions ;)).

5. Dec 2, 2012

TrickyDicky

I'm not talking about "our universe" but rather a part of it "the Aharonov-Bohm effect universe". :P
You mean the magnetic field thru the solenoid in the A-B effect experiments is in simply connected space while while the electrons in the experiment are not?

6. Dec 2, 2012

Staff: Mentor

In R^3-R, you need a different way to define electromagnetism. Luckily we don't live in such a universe.

The region where curl(A)=0 is not a simply connected space (as this equation is not true where you have magnetic field).

7. Dec 2, 2012

TrickyDicky

Are you hinting that Maxwell equations are not necessarily right in this set up?
Yeah, noone disputes that, but that happens to be the nontrivial topology in the experimental set up space and used to explain the potential being observable.

My specific question was: is the magnetic field used in the A-B effect experimental set up in a simply connected space or in a non-symply connected one?

Last edited: Dec 2, 2012
8. Dec 2, 2012

Staff: Mentor

You can formulate alternative equations in other spaces, or even get different field configurations with the same equations.

The magnetic field is part of the R^3-space.

The volume with curl(A)=0 is not simply connected, but A itself is in a simply connected space (R^3).

9. Dec 2, 2012

TrickyDicky

Ok, but then I don't understand how the electrons wavefunctions, that surely are confined to the not symply connected space by definition of the experiment, are influenced by the vector potential A, that as you say is confined to the simply connected space.

10. Dec 2, 2012

strangerep

IMHO, all these explanations involving simply-connectedness obscure the important point...

The AB effect involves a line integral of the potential around a closed loop. This quantity is gauge invariant (unlike the potential at any given point), hence a candidate as a physical observable. But this line integral of the potential is equal (by Stokes Theorem) to the integral of the curl of the potential (i.e., the field) over the entire surface enclosed by the loop. Since the field is not zero everywhere on this surface, the line integral comes out nonzero.

The message from the AB effect (imho) is that there are other electrodynamic physical observables detectable by quantum interference which were not accessible classically. This emphasis is a bit different from the usual glib phrases like "the potential is unphysical classically, but physical in QM", which is subtly incorrect because only gauge invariant quantities can be physical.

11. Dec 2, 2012

tom.stoer

The loop integral

$$\oint A$$

is a gauge invariant quantity; and therefore it is (formally) an observable in classical Maxwell theory!

The question is whether one can find a classical experiment to measure this observable ;-)

12. Dec 2, 2012

strangerep

Yes, that's what I meant by "accessible classically".

13. Dec 3, 2012

The_Duck

Maybe I'm just ignorant about this, but I thought this was the requirement for the existence of a magnetic *scalar* potential? It seems trivially true that we can have a vector potential A such that curl(A)=B even if the space is not simply connected. I mean, suppose we set up an infinite solenoid, and construct its A field, and then remove the solenoid volume from our space. A is still a perfectly good vector potential, satisfying curl(A)=B everywhere, even though our space is no longer simply connected.

14. Dec 3, 2012

TrickyDicky

You mean that contrary to what is usually stated the AB effect is a classical effect that just happens to only have quantum mechanical confirmation so far?
If that were the case I don't understand the need to justify its observation thru topological defects.

15. Dec 3, 2012

TrickyDicky

Thinking it over, my topological problem holds in the same way for classical EM. So saying that in classical EM the vector potential is formally observable doesn't solve my issue with the simply versus non simply connected disposition.

16. Dec 3, 2012

tom.stoer

Not a classical effect, b/c using only the A-field doesn't make it an effect; the observable is classical, the measuring device = the interfering particle is quantum mechanical. So in a sense yes, it's a quantum mechanical confirmation of the classical configuration. And I am not saying that the vector potential is an observable, but that the integral is an observable! Which other condition but local gauge invariance do you need for an observable?

I think the topological interpretation is still relevant. The A-field has an U(1) gauge symmetry so for every infinite line S we can define a winding number wS[A] in the fundamental group π1. There are large, discrete gauge transformations labelled by this Z.

This is related to the fact that you can't simply gauge away the A-field b/c

$$\oint_S A \to \oint_S \,{}^\chi\!A = \oint_S (A + d\chi)$$

where the subscrips S does not label the integration contour but the line S for which we calculate the winding number, and where the second term must vanish due to periodicity. In order to see that more clearly one must not study

$$A \to A + d \chi$$

(where χ becomes discontinuous for large gauge trf's)

but

$$A \to U(A + d)U$$
$$U = \exp(-i\chi)$$

where U itself is periodic. This kind of reasoning is valid mathematically even w/o thinking about the physical source of the gauge field

17. Dec 3, 2012

TrickyDicky

It is a requirement for any potential if you want to derive its existence from the existence of a magnetic field and the Maxwell law that says the divergence of B is 0 to conclude that B=∇XA, this is the opposite to what you were saying, if you start with the magnetic potential A, you automatically have that its curl=B without topological requirements.

18. Dec 3, 2012

TrickyDicky

Yes, see my #15.

19. Dec 3, 2012

TrickyDicky

I can't see why must one introduce U, It doesn't seem mathematically valid due precisely to the topological defect.

20. Dec 3, 2012

tom.stoer

It's exactly the other way round: U is the valid mathematical object to be defined on a loop, not χ!

Consider

$$\chi(θ) = 2\pi θ$$

on a circle S1. This function χ violates periodicity on S1 so in a sense it's ill-defined.

But

$$U(θ) = \exp(- 2\pi iθ)$$

is perfectly well-defined.

The problem is not the definition of U but the definition of χ using a logarithm!

Note that in non-abelian gauge theories non-infinitesimal gauge transformations must always be defined using this U, the phase is never sufficient. The reason is that we have an U(N) gauge symmetry, not u(N). It's the gauge group that matters, not the algebra. Only in the case N=1 with U(1) it seems that the exponential is not required, but using the exponential makes the topological consideration much more clear!

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook