# Aharonov-Bohm topological explanation

by TrickyDicky
Tags: aharonovbohm, explanation, topological
P: 3,043
 Quote by strangerep And mine is (partly) that regarding the AB setup as a plane with a point (or small disc) removed (hence not simply connected) is an unphysical idealization. The space of the (physical) experiment does not have a hole. It has a region where curl A is nonzero. The topological arguments are a distracting waste of time.

 You could try solving the Maxwell equations expressed in terms of just A and a source current j. I.e., for an AB-like current configuration, find A. Then integrate your A solution around a closed loop...
Integrating around the closed loop independently of path requires the trivial topology, so only in case you were right and topology was irrelevant would you have a point.
P: 3,043
 Quote by tom.stoer Let's summarize: - the B-field is neither required nor sufficient to explain the Aharonov-Bohm effect ...
What do you mean is not required, there is no effect without a magnetic flux inside the solenoid, is there?
 Sci Advisor P: 5,451 I think it doesn't make sense to repeat everything umpteen times; the B-field is a derived quantity; the A-field is fundamental and sufficient; the A-field is pure gauge locally, so no B-field is assoiciated with it; the B-field is zero outside; the electron can't "look" into the solenoid and therefore does not see any B-field; the magnetic flux is not the same as the magnetic field, but it's a loop-integral over the A-field, the flux around the loop does not require a B-field inside b/c the electron cannot distinguish between the B-field in the solenoid and a pure gauge w/o any B-field ... As long as you insist on using the B-field you will never be able to understand the AB effect, local gauge theory with global gauge effects and other consequences.
P: 5,451
 Quote by TrickyDicky Integrating around the closed loop independently of path requires the trivial topology, so only in case you were right and topology was irrelevant would you have a point.
No!

Integrating around closed loops tells you something regarding topology, winding number etc. The integral is NOT path-independent in general but only path-indep. for homotopic paths. That's one key lesson: the magnetic flux IS a topological quantity related to the first homotopy group of the base manifold.
 P: 3,043 So do we agree at least that the solenoid cross section acts as a hole of the space of the paths of the electrons? If so, how does the independent of path closed line integral of A work? It is usually stated there must be no holes for the Green and Stokes theorems to hold. Is this wrong?
P: 3,043
 Quote by tom.stoer I think it doesn't make sense to repeat everything umpteen times; the B-field is a derived quantity; the A-field is fundamental and sufficient; the A-field is pure gauge locally, so no B-field is assoiciated with it; the B-field is zero outside; the electron can't "look" into the solenoid and therefore does not see any B-field; the magnetic flux is not the same as the magnetic field, but it's a loop-integral over the A-field, the flux around the loop does not require a B-field inside b/c the electron cannot distinguish between the B-field in the solenoid and a pure gauge w/o any B-field ... As long as you insist on using the B-field you will never be able to understand the AB effect, local gauge theory with global gauge effects and other consequences.
I agree with all the points you list, that shows to me you are not really getting my point.
P: 3,043
 Quote by tom.stoer No! Integrating around closed loops tells you something regarding topology, winding number etc. The integral is NOT path-independent in general but only path-indep. for homotopic paths. That's one key lesson: the magnetic flux IS a topological quantity related to the first homotopy group of the base manifold.
Once again you start with a negative and then go on to say what I'm saying in other words rather than answering specific questions.
 Sci Advisor P: 5,451 TrickyDicky, I don't get your point b/c there seems to be no point at all! I do not start with the negative! I simply say that the integral is not path-independent in general (this is contary your statement). The path-dependence related to the first homotopy group is is one key factor in understanding the topology of the AB effect. So I do not simply repeat what you are saying "in other words", there is a fundamental difference And I do not answer specific questions b/c I don't see them. So what are your questions? (and what is missing in our explanations?)
P: 3,043
 Quote by tom.stoer TrickyDicky, I don't get your point b/c there seems to be no point at all!
mfb, strangerep and Jano L. got what I was saying perfectly well. For the last two the solution came from discarding the topology as relevant to the effect, an opinion I don't agree with.

 Quote by tom.stoer I do not start with the negative! I simply say that the integral is not path-independent in general (this is contary your statement). The path-dependence related to the first homotopy group is is one key factor in understanding the topology of the AB effect. So I do not simply repeat what you are saying "in other words", there is a fundamental difference
Well, then you dindn't understand my wording, I was precisely underlining that the path independence is NOT general but TOPOLOGY DEPENDENT, are you actually reading what I write?

 Quote by tom.stoer And I do not answer specific questions b/c I don't see them. So what are your questions? (and what is missing in our explanations?)
Have you not spotted posts #56 and #59 for instance?
P: 612
 Quote by TrickyDicky I was precisely underlining that the path independence is NOT general but TOPOLOGY DEPENDENT
For a particle that behaves symmetrically with respect to external fields, path independence would be broken if the curl of the particle's internal field (projected outward) were interrupted by an obstacle of some kind, right? Are there other obvious reasons for a breakdown of path independence?
 P: 3,043 This is how the magnetic AB effect is described in wikipedia just for reference: "The most commonly described case, sometimes called the Aharonov–Bohm solenoid effect, takes place when the wave function of a charged particle passing around a long solenoid experiences a phase shift as a result of the enclosed magnetic field, despite the magnetic field being negligible in the region through which the particle passes and the particle's wavefunction being negligible inside the solenoid. This phase shift has been observed experimentally. The magnetic Aharonov–Bohm effect can be seen as a result of the requirement that quantum physics be invariant with respect to the gauge choice for the electromagnetic potential, of which the magnetic vector potential A forms part. Electromagnetic theory implies that a particle with electric charge q travelling along some path P in a region with zero magnetic field B, but non-zero A (by $\mathbf{B} = 0 = \nabla \times \mathbf{A}$), acquires a phase shift $\varphi$, given in SI units by $\varphi = \frac{q}{\hbar} \int_P \mathbf{A} \cdot d\mathbf{x},$ Therefore particles, with the same start and end points, but travelling along two different routes will acquire a phase difference $\Delta \varphi$ determined by the magnetic flux $\Phi_B$ through the area between the paths (via Stokes' theorem and $\nabla \times \mathbf{A} = \mathbf{B})$, and given by: $\Delta\varphi = \frac{q\Phi_B}{\hbar}.$ In quantum mechanics the same particle can travel between two points by a variety of paths. Therefore this phase difference can be observed by placing a solenoid between the slits of a double-slit experiment (or equivalent). An ideal solenoid (i.e. infinitely long and with a perfectly uniform current distribution) encloses a magnetic field B, but does not produce any magnetic field outside of its cylinder, and thus the charged particle (e.g. an electron) passing outside experiences no magnetic field B. However, there is a (curl-free) vector potential A outside the solenoid with an enclosed flux, and so the relative phase of particles passing through one slit or the other is altered by whether the solenoid current is turned on or off. This corresponds to an observable shift of the interference fringes on the observation plane." End quote I'm only centering on this part of the quote: "but non-zero A (by $\mathbf{B} = 0 = \nabla \times \mathbf{A}$)", to make the addition that is not mentioned in the wikipedia that what is between parenthesis only hold when the space is simply connected, and that requirement is precisely the one demanded by the explanation of the effect that not be fulfilled. Is this apparent contradiction so hard to see?
P: 3,043
 Quote by PhilDSP For a particle that behaves symmetrically with respect to external fields, path independence would be broken if the curl of the particle's internal field (projected outward) were interrupted by an obstacle of some kind, right? Are there other obvious reasons for a breakdown of path independence?
I'm not sure what you mean by "behaves symmetrically with respect to external fields" and " internal field (projected outward)".
Anyway here the obvious reason for a breakdown of path independence is the nontrivial topology.
P: 5,451
 Quote by TrickyDicky ... then you didn't understand my wording, I was precisely underlining that the path independence is NOT general but TOPOLOGY DEPENDENT, are you actually reading what I write?
Where have you written this statement?

 Quote by TrickyDicky Have you not spotted posts #56 and #59 for instance?
I saw these posts and I already commented them:
In #56 you are asking "What do you mean is not required, there is no effect without a magnetic flux inside the solenoid" I said (quite often in the meantime) that there is a difference between the magnetic FIELD (which is zero outside and which is NOT required) and the magnetic FLUX as calculated via the line integral.

Please try to get the following statement: the AB effect does not require any B-field; it can be expressed purely in terms of the A-field; and the A-field is evaluated purely outside the solenoid where the B-field is zero. The electrons do not interact with an B-field but with an A-field!

In #59 you are asking "how does the independent of path closed line integral of A work?" I think we agree that there is no path independence in general.

OK; so what are you specific questions that have not been addressed? (sorry for insisting on these questions, but I want to avoid answering the wrong questions or addressing issues that are already clear to you)
P: 3,043
 Quote by tom.stoer Where did you wriute this?
Tom, it is the key to my argument so it is all over my posts but for instance "Integrating around the closed loop independently of path requires the trivial topology" meaning that it requires simply connectednes, in post #55.
P: 5,451
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

$$\oint_C A \neq 0$$
 P: 3,043 See #65
 P: 865 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.
P: 3,043
 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.

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