Recognitions:

Aharonov-Bohm topological explanation

I am not sure whether I get this point (which was not addressed to me, I guess) "Integrating around the closed loop independently of path requires the trivial topology".

The phase shift measured is

$$\Delta\phi \sim \oint_C A$$

This does neither require trivial topology (or Stokes theorem to calculate the flux) nor any B-field.

Let's cite Wikipedia:

 The Aharonov–Bohm effect ... is a quantum mechanical phenomenon in which an electrically charged particle is affected ... despite being confined to a region in which both the magnetic field B and electric field E are zero. The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wavefunction ...
This is the AB-effect!

The next sentence refers to the very specific case of solenoids, but this is only one possible experimental setup, not the general case:
 The most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, ...
So all you need is

$$\oint_C A \neq 0$$
 See #65
 There seems to be some confusing communication regarding "path dependence." Probably we can all agree on the following unambiguous statements: 1. In a simply connected space where curl(A) = 0 everywhere, line integrals of the A field are path-independent. 2. In a non-simply-connected space where curl(A) = 0 everywhere, line integrals of the A field may be path-dependent. TrickyDicky, I have the sense that you are arguing as follows: "People claim that the line integral of A depends on the winding number around the solenoid, because the space R^3 \ (solenoid) is not simply connected. But this is nonsense. The actual physical space is simply connected. Therefore the line integral of A cannot be path dependent. In particular, it's nonsense to claim that the line integral can depend on something topological like a winding number when the physical space has no nontrivial topology to speak of." Correct me if that's a misrepresentation; it's just a guess at what you are getting at. In any case, the response is that the two cases above neglected the actual physical case: 3. In a simply connected space where curl(A) is NOT everywhere zero, line integrals of the A field may be path dependent. If the physical space is simply connected, what justification do we have for talking about non-simply-connected spaces and reasoning topologically? The point is that if you work in the simply connected space, but don't consider line integrals that intrude into the solenoid, you get the same results for line integrals as if you had worked in the non-simply-connected space. Therefore whatever topology has to say about line integrals in the non-simply-connected space also applies to the same line integrals in the simply connected space. I think this addresses another point you seem to be making, namely, "People talk about this non-simply-connected space, but simultaneously use Stokes' theorem, which does not hold in such a space, to discuss the magnetic flux." Only in the simply connected space can you apply Stokes' theorem. Only in the non-simply-connected space do you have nontrivial topology and winding numbers. But since you get the same line integrals and the same physics either way, we feel comfortable talking about Stokes theorem and nontrivial topology at the same time, even though strictly speaking we are imagining different spaces in each case. We're just using different mathematical tools to get the same answer.

 Quote by The_Duck There seems to be some confusing communication regarding "path dependence." Probably we can all agree on the following unambiguous statements: 1. In a simply connected space where curl(A) = 0 everywhere, line integrals of the A field are path-independent. 2. In a non-simply-connected space where curl(A) = 0 everywhere, line integrals of the A field may be path-dependent.
Agreed.
 Quote by The_Duck TrickyDicky, I have the sense that you are arguing as follows: "People claim that the line integral of A depends on the winding number around the solenoid, because the space R^3 \ (solenoid) is not simply connected. But this is nonsense. The actual physical space is simply connected. Therefore the line integral of A cannot be path dependent. In particular, it's nonsense to claim that the line integral can depend on something topological like a winding number when the physical space has no nontrivial topology to speak of." Correct me if that's a misrepresentation; it's just a guess at what you are getting at. In any case, the response is that the two cases above neglected the actual physical case:
Thanks for trying to understand my point. Your interpretation is not exactly what I was getting at but not so far either. I can see the difference between our actual space and the space of the experiment, and see the logic behind the path dependence of the R^3 \ (solenoid) space of the experiment.
 Quote by The_Duck 3. In a simply connected space where curl(A) is NOT everywhere zero, line integrals of the A field may be path dependent.
But the electrons are always in the non-simply connected space, no?

 Quote by The_Duck I think this addresses another point you seem to be making, namely, "People talk about this non-simply-connected space, but simultaneously use Stokes' theorem, which does not hold in such a space, to discuss the magnetic flux."
This is closer to my point, yes.
 Quote by The_Duck Only in the simply connected space can you apply Stokes' theorem. Only in the non-simply-connected space do you have nontrivial topology and winding numbers. But since you get the same line integrals and the same physics either way, we feel comfortable talking about Stokes theorem and nontrivial topology at the same time, even though strictly speaking we are imagining different spaces in each case. We're just using different mathematical tools to get the same answer.
I'm just not sure you get the same line integrals and therefore the same physics if the Stokes theorem doesn't apply to the space of the experiment.
I don't know, I guess I'll just study it better, I might as well just leave it here.
Thanks everyone.
 Recognitions: Science Advisor thanks, I think this should contribute to clarification; my arguments seemed to be - at least partially - distracting. Regarding the_duck's the last paragraph, a) simply connected space and application of Stokes' theorem and b) non-simply-connected space i.e. nontrivial topology and winding numbers: that was the reason for me to cite Wikipedia. I wanted to make clear that they are talking about two entirely different things, namely a non-vanishing loop integral over A (which is the general case for the AB effect) and the solenoid case which is a rather special experimental setup.
 This might help understand my point too: http://unapologetic.wordpress.com/20...gnetism-texts/
 Recognitions: Science Advisor I am not sure if I get your point; are you saying that the existence of A satisfying B = rot A is ruled out b/c the space may not be contractable? Or what exactly is your conclusion derived from http://unapologetic.wordpress.com/20...gnetism-texts/ ?

 Quote by tom.stoer I am not sure if I get your point; are you saying that the existence of A satisfying B = rot A is ruled out b/c the space may not be contractable?
Yes, not so much as rule out, but certainly it makes one wonder what's going on.

Recognitions:
 Quote by TrickyDicky Yes, not so much as rule out, but certainly it makes one wonder what's going on.
The problem is that "existence" has not yet been well defined.

What topology tells us is that the existence of a global A-field and a global B-field is indeed problematic; that's the reason we have to use R³ \ R instead of R³. A- and B-field are not well-behaved on R³.

This can be seen by using the A-field outside the solenoid

$$A = \frac{\Phi}{2\pi r^2}(-y,x,0) = \frac{\Phi}{2\pi}\nabla\theta$$

and taking the limit

$$r \to 0; \Phi = \text{const.}$$

Physically this seems to be unproblematic b/c the electrons only "feel" the flux as a physical observable via the phase

$$e^{i\Phi}$$

which remains constant. But the A-field as defined above and the gauge function

$$U = e^{-i\Phi\theta / 2\pi}$$

from which it can be derived as

$$A^\prime = 0 \to A = U^\dagger\,(A + i\nabla)\,U$$

cannot be defined globally; r=0 has to be excluded from our configuration space, i.e. from the base manifold, which results in R³ \ R.

That means that we can define

$$F = dA = 0$$

locally but not globally.

$$dA = 0$$

means that A is locally flat i.e. has vanishing curvature 2-form F (i.e. vanishing el. mag. field strength, i.e. E=0, B=0) locally.

In other words A is pure gauge locally i.e. A ~ A' = 0 but not globally.

What Stokes' theorem (and cohomology) tells us is that

$$\oint_C A$$

does only depend on the homotopy class of the loop C w.r.t. r=0, i.e. it depends only on the winding number

$$w[C] = n = 0,\pm 1,\pm 2, ...$$

of the loop C around the r=0.

Using Stokes' theorem naively

$$\int_{\partial M} \omega = \int_M d\omega\;\;\to\;\; \int_{C} A = \int_D \nabla\times A = \int_D B = 0$$

one would derive a vanishing magnetic flux. But this is wrong due to the fact that the extension of A from the loop C to the disc D is not allowed due to the above mentioned singularity in r=0. Therefore the naive application of Stokes theorem is wrong, but this is due to the r.h.s. (= the surface integral) not due to the l.h.s.

Physically one can save the theorem by explaining the B-field via the solenoid, i.e. via replacing the A-field and the B-field by the solenoid solution on a disc D. Mathematically one can save the story by using the locally flat A-field outside and vanishing B-field!! Instead one has to cut out the disc D with shrinking radius such that one arrives at R³ \ R where the A-field is well-defined.

That means that the physical solenoid can be replaced by the non-trivial base manifold R³ \ R.

Now the question remains what is physically meaningful. At first glance one could argue in favor of the solenoid with non-vanishing B-field, but even this may be obscure: iff (if and only if) we know that the A-field has been prepared using a solenoid everything is fine. But we (and the electrons) are not able to penetrate the solenoid, so there is no physical mechanism to distinguish between the "physical solenoid" and the "R³ \ R".

What I am saying is that if somebody prepares an impenetrable solenoid with locally flat but non-zero A-field outside w/o telling you any details, you are not able to distinguish between the following two scenarios:
a) the impenetrable solenoid contains a non-vanishing B-field
b) the impenetrable solenoid wraps a topological singularity
 Tom, I do not understand how you can write solenoidal field as a gradient in that way. Is the angle $\theta$ in your writings a multiple-valued function of position? Otherwise we get sudden jump of A at $\theta = 0$, which makes the loop integral zero...
 Recognitions: Science Advisor I am not sure whether I understand you problem. It's true that θ is in [0,2π[ and that therefore θ itself is not a "global coordinate". But that doesn't matter b/c all physical quantities are well-behaved: The first definition of A is fine. The loop integral becomes a multipliation by the length i.e. by 2π for constant and is OK. The gauge transformation has been introduced via U which is periodic in θ and therefore well-defined. Therefore the gradient acting on U is well-defined, too; acting on θ may be problematic, but this is due to θ, not due to the gradient θ itself is not a globally well-defined coordinate, and in principle one would have to use a "cut R²" with several charts using different angle coordinates; but as usual in physics I am a bit sloppy and avoid all this stuff.
 OK, now I get it, your prescription gives correct solenoidal A without delta function in A. I mistakenly thought that we would get delta function in A due to chain rule and jump in theta. But now I realize the chain rule does not apply for $e^{i\theta}$, so it is OK. Tricky Dicky: I read the passage in the book by Nash&Sen and also sec. 2.2 in D. J. Thouless, Topological quantum numbers in nonrelativistic physics, (1998) which Wikipedia cites and was surprised that they do not say more than that topology is somehow relevant to BA shift. They do not elaborate on that. Furthermore, the papers on BA effect I have seen in PR do not seem to claim that considerations of topology are necessary. So can you give some reference to work where the authors really show there is some connection ?
 Recognitions: Science Advisor It depends on what you call the AB effect ;-) According to Wikipedia The Aharonov–Bohm effect ... is a quantum mechanical phenomenon in which an electrically charged particle is affected ... despite being confined to a region in which both the magnetic field B and electric field E are zero. The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wavefunction ... Following that line of reasoning one may conclude that the effect is due to locally but not globally flat gauge field which indicates a non-trivial vector bundle structure, so in this general setup it's a topological effect. http://www.encyclopediaofmath.org/in...haronov_effect http://iopscience.iop.org/0143-0807/9/3/007 But physically one could simply study the most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, ... which does by no means require any topological reasoning. As you may have observed I think that topology of gauge field and vector bundles is more fundamental than one specific experimental setup (just as quantization of action is a more fundamental principle than the black body radiation). I hope the above mentioned reference clarifies some of the topological concepts.

 Quote by tom.stoer It depends on what you call the AB effect ;-) According to Wikipedia The Aharonov–Bohm effect ... is a quantum mechanical phenomenon in which an electrically charged particle is affected ... despite being confined to a region in which both the magnetic field B and electric field E are zero. The underlying mechanism is the coupling of the electromagnetic potential with the complex phase of a charged particle's wavefunction ... Following that line of reasoning one may conclude that the effect is due to locally but not globally flat gauge field which indicates a non-trivial vector bundle structure, so in this general setup it's a topological effect. http://www.encyclopediaofmath.org/in...haronov_effect http://iopscience.iop.org/0143-0807/9/3/007
"A phenomenon in which the topological non-triviality of a gauge field is measurable physically [a1]. Moreover, this topological non-triviality, which can be expressed as a number , say, is a global topological invariant and so is not expressible by a local formula; this latter point being in contrast to a simpler topological invariant such as the dimension of the underlying space, which is deducible locally."
I think we all agree with this paragraph, the main reason of the need for globality being IMO that homotopy relations always contain global information as it is well known.
In the next line is where I find my conundrum rather than with the effect itself:
"To demonstrate this effect physically, one arranges that a non-simply-connected region of space has zero electromagnetic field . This electromagnetic field is related to the gauge field by the usual relation F=dA"
The usual relation F=dA for a non-simply connected region is not granted, that is what I mean by my doubts about the existence of the A-field in this particular not simply connected set up(see my link "A short rant..." from the Unapologetic mathematician to check this).
Here in the 5th paragraph:
"If we flip over to the language of differential forms, we know that the curl operator on a vector field corresponds to the operator $\alpha\mapsto *d\alpha$ on 1-forms, while the gradient operator corresponds to $f\mapsto df$. We indeed know that *ddf=0 automatically — the curl of a gradient vanishes — but knowing that $d\alpha=0$ is not enough to conclude that $\alpha=df$ for some f. In fact, this question is exactly what de Rham cohomology is all about!"
In our case F is a 2-form and A is a one-form. See also the wikipedia page "Closed and exact differential forms" to see that a closed form like F=0 needs a simply connected space to imply F=dA.

 Quote by tom.stoer But physically one could simply study the most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, ... which does by no means require any topological reasoning.
I don't know why you say the solenoid effect, that is an example of the AB effect, "by no means require any topological reasoning". I don't think your distinction between a topological general effect and a non-topological special magnetic case is founded. See this: http://rugth30.phys.rug.nl/quantummechanics/ab.htm

Recognitions:
 Quote by TrickyDicky "To demonstrate this effect physically, one arranges that a non-simply-connected region of space has zero electromagnetic field . This electromagnetic field is related to the gauge field by the usual relation F=dA" The usual relation F=dA for a non-simply connected region is not granted, that is what I mean by my doubts about the existence of the A-field in this particular not simply connected set up
The relation F=dA is granted locally but not globally, so F and A may required patching, cutting out singularities etc. In the case of the A-field as described above one has to remove r=0. On this R³ / R the relation F=dA=0 is valid (so "not globally" means "not on R³ but on R³ / R").

 Quote by TrickyDicky I don't know why you say the solenoid effect, that is an example of the AB effect, "by no means require any topological reasoning". I don't think your distinction between a topological general effect and a non-topological special magnetic case is founded.
In the solenoid case one has a finite disc with non-vanishing B-field; one has a regular solution inside the solenoid; one can use Stokes' theorem w/o any problem. In the case with a point-like singularity at r=0 there is a vanishing B-field; there is an A-field with dA=0 in R³ / R, so the explanation via a B-field is invalid b/c it's zero. The solenoid case can be realized experimentally; this is not possible for the R³ / R case. Therefore this distinction is relevant.

 Quote by tom.stoer The relation F=dA is granted locally but not globally, so F and A may required patching, cutting out singularities etc. In the case of the A-field as described above one has to remove r=0.On this R³ / R the relation F=dA=0 is valid (so "not globally" means "not on R³ but on R³ / R").
But you need it to be granted globally, at least at the global region where electrons are influenced by the A-field so their wave function is shifted.The homotopy from R^3 to R^3/R always carries global information.
In any case I don't think is granted locally either for the R^3/R case and you havent explained why.

 Quote by tom.stoer In the solenoid case one has a finite disc with non-vanishing B-field; one has a regular solution inside the solenoid; one can use Stokes' theorem w/o any problem. In the case with a point-like singularity at r=0 there is a vanishing B-field; there is an A-field with dA=0 in R³ / R, so the explanation via a B-field is invalid b/c it's zero. The solenoid case can be realized experimentally; this is not possible for the R³ / R case. Therefore this distinction is relevant.
The solenoid case should work just like the idealized case, you guarantee the simply- connectedness just by making the electrons unable to traverse the solenoid. So you are removing r=0 just the same.

Recognitions: