Aharonov-Bohm topological explanation

  • #51
TrickyDicky said:
My entire point was that the Stokes theorem requires simply-connectednes to hold.
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.
[...] I'm finding it paradoxical under the usual assumptions with respect to Maxwell equations, closed line integrals and Stokes theorem dependence on trivial topology and the effect itself dependence on nontrivial one, so far I haven't found a way out.
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...
 
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  • #52
Jano L. said:
Can you please explain what you did above? I am sorry, but I do not see relevance of your calculation.
I'll come back to it later
 
  • #53
Let's summarize:
- the B-field is neither required nor sufficient to explain the Aharonov-Bohm effect
- the B-field can be regarded as a derived entity, whereas the A-field is fundamental
- this holds not only in QM but already in classical electrodynamics (Maxwell's equiations can be formulated w/o using the B-field)
- it is wrong (although it is often claimed) that the A-field is unphysical or unobservable; the loop integral represents a (gauge invariant) classical observable

strangerep said:
... regarding the AB setup as a plane with a point removed is an unphysical idealization. The space of the (physical) experiment does not have a hole. ... The topological arguments are a distracting waste of time.
Not really; I agree that physically the situation must be described using the solenoid rather than the punctured plane; but for the calculation of the phase shift there is no difference; the electron can't distinguish between the "physical" and the "topological" scenario b/c it can't penetrate the solenoid. Therefore the topological consideration is relevant.
 
  • #54
Don't the mechanics of "spin" enter into consideration here also? A rotating particle containing an extended charge or sub-charge will generate a magnetic dipole protruding from both ends of the rotation axis. But if the rotation is complex, possibly consisting of several rotations about several axes, then magnetic monopoles or quasi-monopoles would exist in the microscopic realm. The fields would no longer be simply-connected.

There seems to be a history, starting from Heaviside, to allow for the possibility of the existence of magnetic monopoles in an initial, primitive invocation of the Maxwell equations for particular problems where those monopoles are removed prior to the finalization of a particular solution to the EM problem. Doing so could be interpreted as invoking SU(n) topology in the beginning and resolving it into U(1), couldn't it?
 
  • #55
strangerep said:
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.
Hmm, your opininion contradicts all the texts I've consulted.


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.
 
  • #56
tom.stoer said:
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?
 
  • #57
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.
 
  • #58
TrickyDicky said:
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.
 
  • #59
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?
 
  • #60
tom.stoer said:
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.
 
  • #61
tom.stoer said:
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.
 
  • #62
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?)
 
  • #63
tom.stoer said:
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.

tom.stoer said:
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?
tom.stoer said:
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?
 
  • #64
TrickyDicky said:
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?
 
  • #65
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 traveling 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 traveling 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?
 
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  • #66
PhilDSP said:
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.
 
  • #67
TrickyDicky said:
... 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?

TrickyDicky said:
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)
 
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  • #68
tom.stoer said:
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.
 
  • #69
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
 
  • #70
See #65
 
  • #71
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.
 
  • #72
The_Duck said:
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.
The_Duck said:
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.
The_Duck said:
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?

The_Duck said:
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.
The_Duck said:
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.
 
  • #73
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.
 
  • #76
tom.stoer said:
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.
 
  • #77
TrickyDicky said:
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
 
  • #78
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...
 
  • #79
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.
 
  • #80
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 ?
 
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  • #81
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/index.php/Bohm-Aharonov_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.
 
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  • #82
tom.stoer said:
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/index.php/Bohm-Aharonov_effect
http://iopscience.iop.org/0143-0807/9/3/007
I don't have access to your second reference, so let me follow line by line the encyclopedia of math one:
"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.

tom.stoer said:
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
 
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  • #83
TrickyDicky said:
"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").

TrickyDicky said:
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.
 
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  • #84
tom.stoer said:
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 haven't explained why.

tom.stoer said:
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.
 
  • #85
TrickyDicky said:
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 haven't explained why.

Come on! it is granted locally! you can see A (it has been written down ;-) and you can check dA=0 by explicit calculation of the curl; and it is granted globally on R³/R b/c the check works for all (x,y,z) when r=0 is excluded
 
  • #87
In case,if one transform to a frame which moves uniformly with respect to any original one it is possible to eliminate magnetic field,in that frame of course the so far gauge invariant integral is zero.In that case the effect comes from A0, by a phase like-q/h-(∫∅dt),negative in sign with original one.Now here no line integral,no stokes theorem,or I should say no topology?
 
  • #88
tom.stoer said:
Come on! it is granted locally! you can see A (it has been written down ;-) and you can check dA=0 by explicit calculation of the curl; and it is granted globally on R³/R b/c the check works for all (x,y,z) when r=0 is excluded

You should realize that you are taking for granted A everytime, if we are debating the existence of A, you cannot write dA=0 to beging with bc then you are already assuming it. You are affirming the consequent to quote the UM. ;)

Wrt to the trick of cutting out the singularity r=0, well it is a bit silly since then we get back to R^3, and it defeats any further consideration about the problem of a non simply connected space for the existence of an A-field.
 
  • #89
I guess my doubt is more related to topology and vector calculus than to the quantum mechanical effect itself since it involves a step previous to the effect proper, so I might take it to the Topology and geometry subforum.
 
  • #90
andrien said:
In case,if one transform to a frame which moves uniformly with respect to any original one it is possible to eliminate magnetic field,in that frame of course the so far gauge invariant integral is zero.In that case the effect comes from A0, by a phase like-q/h-(∫∅dt),negative in sign with original one.Now here no line integral,no stokes theorem,or I should say no topology?
Can you show us the calculation? From what I understand you start in the A°=0 gauge in the rest frame and then transform to an inertial frame which moves parallel to the solenoid.
 
  • #91
TrickyDicky said:
You should realize that you are taking for granted A everytime, if we are debating the existence of A, you cannot write dA=0 to beging with bc then you are already assuming it. You are affirming the consequent to quote the UM. ;)
TrickyDicky, it becomes a little bit annoying!

I have written down (explicitly) an A-field valid for r>0. So this A-field certainly does exist on R³/R. And of course you can (and should) calculate rot A which again is perfectly valid for r>0.

What else do you need to be convinced that A does exist ?
 
  • #92
tom.stoer said:
Can you show us the calculation? From what I understand you start in the A°=0 gauge in the rest frame and then transform to an inertial frame which moves parallel to the solenoid.

http://books.google.co.in/books?id=f9viHUx3H5gC&pg=SA15-PA8&dq=vector+potential+and+quantum+mechanics+feynman+lectures+vol.+2&hl=en#v=onepage&q=vector%20potential%20and%20quantum%20mechanics%20feynman%20lectures%20vol.%202&f=false
15.9 onwards,However it is not written here that one can eliminate magnetic field,but it is possible to do so.In very simple cases,whenever there is a motion of charge producing magnetic field,one can transform to an inertial frame in which this field is zero and only electric field remains.Still there is a possibility of vector potential,but since the gauge invariant integral only depends on curl of A,it is zero because of absence of magnetic field.The law given in the reference is directly a consequence of lagrangian of charged particle coupled to electromagnetic field,as you are already aware of it.
 
  • #93
tom.stoer said:
TrickyDicky, it becomes a little bit annoying!

I have written down (explicitly) an A-field valid for r>0. So this A-field certainly does exist on R³/R. And of course you can (and should) calculate rot A which again is perfectly valid for r>0.

What else do you need to be convinced that A does exist ?
Ok, Tom, it is obvious to me you still think I am not aware that it is obviously the A-field that produces the effect, the OP is about something different as others understood, but as I said it is in part my fault for choosing the wrong subforum.
 
  • #94
Sorry, I don't get it. It's not the wrong subforum but perhaps the wrong way asking questions
 
  • #95
I believe this is the typical case where "you can't have your cake and eat it too".
We have here a potential A-field that we've agreed that is not globally defined in the space of the AB effect, we have to cut out the origin where the solenoid is, leaving a singularity. The only problem being that by removing the origin we eliminate the justification for having the A-field at all, that is the EM field that is switched on in the experiment to obtain the shift in wave function pattern.
 
  • #96
TrickyDicky said:
We have here a potential A-field that we've agreed that is not globally defined in the space of the AB effect, we have to cut out the origin
Yes.

TrickyDicky said:
The only problem being that by removing the origin we eliminate the justification for having the A-field at all
No, sorry, it seems you understood nothing.
 
  • #97
tom.stoer said:
No, sorry, it seems you understood nothing.

You mean the A-field in the AB experiment is unrelated to the EM field?
 
  • #98
I mean that the EM-field is zero!
 
  • #99
tom.stoer said:
I mean that the EM-field is zero!
I 'm not referring to that, you still don't have a clue what I'm asking do you? Either that or you are pulling my leg. Just in case I'll try with a kindergarten explanation.
What has to happen in order to observe the phase shift in the interference pattern of the electrons? Hint: It's got to do with switching something on.
So we have a situation for the AB experimental setup in which something is switched off, where the EM-field is zero for the electrons and there's no A-field, and a situation in which something is switched on where the EM field is still zero for the electrons but the A-field is nonzero and produces a phase shift. See? We can relate that nonzero A-field with the switching on of something, I'll let you call that something however you want, but I think it's called an EM-field. The take away point is that the A-field is related with switching it on, fine so far?
What I want to underline is that I'm not doubting the effect, I'm only concerned about the usual explanation of it because it seems to mix two incompatible scenarios in a contradictory way, the R^3 scenario and the R^3/R scenario.
The situation before the switching on is compatible with both spaces, but the situation after is compatible with the R^3/R space only, however an effect that is only causally compatible with the activation of something that doesn't exist for the R^3/R space is produced. If that is ok for you guys , then fine. I won't bother about it anymore
 
  • #100
I wonder why in the quantum physics forum nobody has mentioned the non-locality of the effect. Probably that's all my quibble amounts to.
 

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