Potentials and the Aharonov–Bohm effect

By Jess Riedel June 24, 2014 May 6, 2017 Philosophy, Physics

[This post was originally “Part 1” of my HTTAQM series. However, it’s old, haphazardly written, and not a good starting point. Therefore, I’ve removed it from that series, which now begins with “Measurements are about bases”. Other parts are here: 1,2,3,4,5,6,7,8. I hope to re-write this post in the future.]

It’s often remarked that the Aharonov–Bohm (AB) effect says something profound about the “reality” of potentials in quantum mechanics. In one version of the relevant experiment, charged particles are made to travel coherently along two alternate paths, such as in a Mach-Zehnder interferometer. At the experimenter’s discretion, an external electromagnetic potential (either vector or scalar) can be applied so that the two paths are at different potentials yet still experience zero magnetic and electric field. The paths are recombined, and the size of the potential difference determines the phase of the interference pattern. The effect is often interpreted as a demonstration that the electromagnetic potential is physically “real”, rather than just a useful mathematical concept.

The magnetic Aharanov-Bohm effect. The wavepacket of an electron approaches from the left and is split coherently over two paths, L and R. The red solenoid in between contains magnetic flux $\Phi$ . The region outside the solenoid has zero field, but there is a non-zero curl to the vector potential as measured along the two paths. The relative phase between the L and R wavepackets is given by $\Theta = e \Phi/c \hbar$ .

However, Vaidman recently pointed out that this is a mistaken interpretation which is an artifact of the semi-classical approximation used to describe the AB effect. Although it is true that the superposed test charges experience zero field, it turns out that the source charges creating that macroscopic potential do experience a non-zero field, and that the strength of this field is dependent on which path is taken by the test charges. In fact, if one modifies the experiment so that the source charges experience zero field, while retaining a potential difference between the two paths, the AB effect disappears. (In this modification, the semi-classical approximation breaks down.) It is the non-zero field at the source charges—not merely the potential—which is responsible for the observed phase. As Vaidman puts it: “The core of the Aharonov-Bohm effect is the same as the core of quantum entanglement: the quantum wave function describes all systems together.” The paper is very clear and straightforward; I encourage you to read it.

If you read about the AB effect and thought you had learned something deep about the universe, you were wrong (like I was^a). The AB effect is definitely a sign of something important^b, but it doesn’t have to do with asigning a privileged position to the potential.

EDIT: After reading Wikipedia, it’s worth responding to the idea that potentials are more real quantum mechanically because, otherwise, we’d have to give up locality. The argument is that, in order to calculate the behavior of a particle while using only fields and not potentials, we need to consider the field in places that the particle never travels. But this is a bad argument because we already know that quantum mechanics exhibits non-local effects.

EDIT 2: It’s worth emphasizing that Vaidman is only arguing that potentials in quantum mechanics are just as real or fake as they were classically; he’s not disputing the usefulness or elegance of potentials. See my comment below for details.

EDIT 3: OK, let’s make this explicit. The Hamiltonian for a particle with charge $e$ in a classical electromagnetic field given by the potential $(\Phi,\vec{A})$ is

(1) $\begin{align*} H=\frac{1}{2m}(\vec{p}-e\vec{A})^2+e \Phi(\vec{x}), \end{align*}$

where $H$ , $\vec{x}$ , and $\vec{p}$ are operators. This is the semi-classical limit, where we can ignore the quantum state of the source charges and assume the potential they generate, $(\Phi,\vec{A})$ , is a pre-determined classical variable. (Importantly, the interpretation here is that the charge $e$ cannot have an effect on $(\Phi,\vec{A})$ .)

Let’s concentrate on the electric AB effect, so that $\vec{A}=0$ everywhere:

(2) $\begin{align*} H = \frac{\vec{p}^2}{2m}+e \Phi(\vec{x}). \end{align*}$

We have $\Phi=\Phi_0 \neq 0$ (a constant) in a region corresponding to the left arm of the interferometer, but $\Phi=0$ in a region corresponding to the right arm. Since the potential is constant within each arm, there is no field anywhere the electron will travel. When the electron is brought into a superposition of being located in each arm, the component of the wavefunction in the left arm picks up a phase $\theta = e \Phi_0 T$ during a time $T$ passing through the interferometer.

However, Vaidman points out that we can also choose to write down the full quantum dynamics of the source charges producing the field. In this case, there is no classical potential $(\Phi,\vec{A})$ . Rather, there is a wavefunction for the source charges, a wavefunction for the electron, and a Hamiltonian coupling them. We adopt Vaidman’s toy model of two source charges of equal strength $Q$ . In accordance with $\vec{A}=0$ , we assume the two source charges can be approximated as stationary so that the full Hamiltonian is

(3) $\begin{align*} H = \frac{\vec{p}^2}{2m} + \frac{\vec{k}_1^2}{2M} + \frac{\vec{k}_2^2}{2M} - k_e [\frac{eQ}{\vec{x}-\vec{r}_1} + \frac{eQ}{\vec{x}-\vec{r}_2} + \frac{Q^2}{\vec{r}_1-\vec{r}_2}] \end{align*}$

where $\vec{k}_i$ and $\vec{r}_i$ , (with $i=1,2$ ) are the momentum and position operators for the source charges, and where $k_e$ is Coulomb’s constant. If the source charges were located exactly at a positions $r_1$ and $r_2$ for all time, then they would generate the (classical) scalar potential $\Phi(x) = - k_e Q[1/(\vec{x}-\vec{kr}_1) + 1/(\vec{x}-\vec{r}_2)]$ . This would allow us to recover the AB effect using potentials as usual.

But now we are going to get rid of Hamiltonians and evolve our wavefunction using the quantum mechanical version of Newton’s second law. (This is necessary if we want to talk about forces in quantum mechanics without speaking of potentials. After all, replacing forces with potentials is essential to the Hamiltonian formulation.) For concreteness, let the source charges have Gaussian wavepackets. It is known when a Gaussian wavepackets travels in a potential that the centroid of the wavepacket follows the classical path for the potential (so long as the width of the wavepacket is small compared to the curvature of the potential). Following Vaidman, we model the source charges as approaching from infinity to within a distance $R$ of the (essentially stationary) electron in the left arm, travelling with nearly constant velocity v toward the electron during a time interval $T$ , and then retreating to infinity. Although the source charge is taken to follow this path because of a much stronger controlling force (e.g. the experimenter’s equipment), the electron-source interaction does affect the wavepacket of the source charge. In particular, when the electron is in the left arm the source charge gains kinetic energy

(4) $\begin{align*} \int_{-\infty}^{x-R} F(r) dr = -\int_{-\infty}^{x-R} \frac{eQ}{(r-x)^2} dr = \frac{eQ}{R} = \delta(\frac{M v^2}{2}) = M v \delta v \end{align*}$

by being pulled toward the electron; it does not aquire this kinetic energy when the electron is in the right arm. The source charge is taken to have a macroscopic mass $M$ so that the fractional change in the velocity very small and the velocity can be taken to be nearly constant during the interval $T$ . The distance the center of the source charge wavepacket shifts due to the electron in the left arm, compared to where it is when the electron is in the right arm, is $\delta x = T\delta v = e Q T/M v R$ . For a source charge with de Broglie wavelength $\lambda = h/Mv$ (much larger than $\delta x$ ), this leads to an overall phase shift $\theta_0 = 2 \pi \delta x / \lambda$ in the wavefunction of the source charges. That is, if the state of one of the source charges evolves to $| \psi_{\mathrm{right}}\rangle_1$ when the electron is in the right arm, the state of the source charge evolves to $| \psi_{\mathrm{left}}\rangle_1 = e^{i \theta_0}| \psi_{\mathrm{right}}\rangle_1$ . (And likewise for the second source charge.) This means the total state evolves from

(5) $\begin{align*} |\psi_{\varnothing} \rangle_1 & | \psi_{\varnothing} \rangle_2 \left[ |e_\mathrm{left} \rangle_e + |e_\mathrm{right} \rangle_e \right] = |\psi_{\varnothing} \rangle_1 | \psi_{\varnothing}\rangle_2 | e_\mathrm{left} \rangle_e + | \psi_{\varnothing} \rangle_1 | \psi_{\varnothing} \rangle_2 |e_\mathrm{right} \rangle_e \end{align*}$

(6) $\begin{align*} |\psi_{\mathrm{left}}\rangle_1 | \psi_{\mathrm{left}}\rangle_2 &|e_\mathrm{left} \rangle_e + | \psi_{\mathrm{right}}\rangle_1 | \psi_{\mathrm{right}}\rangle_2 |e_\mathrm{right} \rangle_e \\ & = | \psi_{\mathrm{right}}\rangle_1 | \psi_{\mathrm{right}}\rangle_2 \left[e^{i \Theta_{AB}} |e_\mathrm{left} \rangle_e + | e_\mathrm{right} \rangle_e \right] \end{align*}$

where $\Theta_{AB} = 2 \theta_0 = - 2 e Q T / R \hbar$ is the AB phase.

Footnotes

(↵ returns to text)

I can vaguely remember noting some thin wisps of confusion at the time I learned about it–somehow the argument made my queasy–but I put those aside and convinced myself that they were normal. This was a mistake. I now trust those confused feelings more, and I trust less other people who claim confidence in similar circumstances. (If I could trick myself, they probably can too.)↵
Even though the AB effect doesn’t tell us anything new about the “reality” of the electromagnetic potential, it turns out to be a neat prototypical example of using a non-classical detection methods to detect a classically undetectable phenomenon. As you might have guessed, I have some thoughts on it (still incomplete).↵

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8 Comments

Reader297
June 30, 2014 at 5:05 pm

I’m not quite sure that thinking in terms of real potentials is a “bad” way to think about the Aharonov-Bohm effect.
People are often nervous about the interpretation of gauge potentials, but there isn’t really anything mysterious or confusing about them except that they don’t have a one-to-one relationship with physically distinct configurations of the electromagnetic field.
A given configuration of an electromagnetic field is equivalent to an entire class of gauge potentials, namely, all gauge potentials (Phi,A) that yield the given E and B fields.
Although any one choice of (Phi,A) is unphysical, the entire class (Phi,A) is just as physical as specifying a configuration of the E and B fields. And the consequence is that we can use (Phi,A) directly in calculations, and if we don’t make any mistakes, then we’ll always find that any physical results, like the Aharonov-Bohm effect, will depend only on the entire class and not on our particular choice of (Phi,A).
Moreover, if we spontaneously break the gauge invariance of electromagnetism, say, by turning on a Higgs-like field, then (Phi,A) now do become related in a one-to-one sense with distinct physical configurations (they now satisfy a massive Proca equation), and we can treat (Phi,A) just like regular classical fields. So if (Phi,A) can be treated as physical when the mass m of the photons are nonzero, why is it totally unacceptable to regard it as physical (at least in an equivalence-class sense) when m=0?
This is all to say that I’m not sure there is any good reason to be suspicious of regarding classes of gauge potentials as physical, and using them directly to derive effects like Aharonov-Bohm.
Vaidman seems to be saying that there is an alternative way to derive the Aharonov-Bohm effect in certain experimental cases, namely, when the background E or B fields are produced by physical source charges that themselves experience backreactions from the traveling test particle. But just because an effect like Aharonov-Bohm can be derived in multiple ways doesn’t mean that one of those ways is illegitimate or unphysical. After all, the whole magic of gauge theories is that we can do things in different gauge choices and always get the same answer.
Several things confuse me about Vaidman’s argument, however. The primary one is that in his alternative derivation of the Aharonov-Bohm effect (say, in the magnetic case), he seems to ignore completely the Aharonov-Bohm phase shift directly experienced by the test particle, focusing solely on the phase shift of the source charges. (He later says that the phase shift of the source charges accounts for the supposed phase shift of the test particle.)
So is he saying that the usual argument that the test particle directly experiences a phase shift is wrong? Where does he claim that the usual argument break down? Is he claiming that the A.v term in the Hamiltonian H of the test particle is invalid? That A.v term gives a cumulative phase shift to the test particle. So he seems to be saying that we should ignore that A.v term and its consequent phase shift and include only a phase shift coming from the source charges.
Has he implicitly chosen an alternative gauge in which the A.v in the test particle’s Hamiltonian is somehow not important anymore?
Another thing that confuses me is that, in principle, we can get Aharonov-Bohm phase shifts even without solenoids. For example, if we consider a space with cylindrical topology, then there can exist static but nonzero gauge potentials even without sources. (These are called “flat” potentials.) With no source charges for the test particle to backreact on, there doesn’t seem to be any way to explain the Aharonov-Bohm effect on the test particle in terms of phase shifts just of source charges.
Indeed, one could imagine a less exotic example in general relativity. On a fixed but sufficiently curved background space, the effective value of pi can be much smaller than 3.14, small enough that circumferences of circles can be much smaller than their radii. In that case, the test particle can go all the way around the circumference of the circle (and presumably develop an Aharonov-Bohm phase) quickly enough that the test particle’s own field does not have enough time to reach the solenoid at the center of the circle and backreact on the source charges. So then does the test particle experience an Aharonov-Bohm phase or not?
Reply
- Jess Riedel
  July 1, 2014 at 8:01 am
  
  > I’m not quite sure that thinking in terms of real potentials is a “bad” way to think about the Aharonov-Bohm effect.
  Whoops, sorry for not being clear. I’m definitely *not* trying to say that thinking in terms of potentials is bad. The mistake I’m highlight is claiming that there is something about *quantum mechanics* that makes the potential more real than it was in classical mechanics. Rather, potentials in quantum mechanics are just as real (or fictitious) as they were classically.
  > People are often nervous about the interpretation of gauge potentials, but there isn’t really anything mysterious or confusing about them except that they don’t have a one-to-one relationship with physically distinct configurations of the electromagnetic field.
  Well, I basically agree, except to stress this: When something doesn’t have a one-to-one correspondence with physically distinct configurations, that’s a pretty good indication that it’s not real. If I claim that some marbles are “blerg” and some are “blorg” but that there are no experiments that will distinguish between them, you are inclined to think that the distinction between blerg and blorg isn’t real.
  There’s a Feynman quote about the importance of knowing several different mathematical descriptions of exactly the same physics. Besides being calculationally and conceptually useful, having multiple maps of the same territory can help you “get at” the platonic ideal that they are trying to capture. When you only have one map, it’s much easier to confuse features of the map for features of the territory.
  > Vaidman seems to be saying that there is an alternative way to derive the Aharonov-Bohm effect in certain experimental cases, namely, when the background E or B fields are produced by physical source charges that themselves experience backreactions from the traveling test particle. But just because an effect like Aharonov-Bohm can be derived in multiple ways doesn’t mean that one of those ways is illegitimate or unphysical.
  Vaidman’s claim is that, in real life, *all* background fields are ultimately produced by source charges. It’s just that, in traditional statements of the AB effect, the semi-classical limit (i.e. treating the external field as a classical variable) obscures this.
  Remember, the sort of argument that Vaidman is trying to rebut is “quantum mechanics shows that potentials are real, even though classically it was legitimate to think only in terms of forces”. This argument was based off of the *necessity* of the potential. But instead, if you include a full quantum mechanical treatment of everything, it’s still valid to think in terms of *local forces* and never write down a potential. That’s all Vaidman shows.
  Yes, potentials are very useful things and there are cases where calculating in terms of forces is very inconvenient. But that’s true classically.
  > Vaidman…seems to ignore completely the Aharonov-Bohm phase shift directly experienced by the test particle, focusing solely on the phase shift of the source charges.
  Phase shifts aren’t properties of individual systems, they are properties of the global state as a whole. To see this, just consider a Bell state on two qubits
  |00> + exp(i theta) |11>
  If we have a physical process that changes theta, which qubit “owns” the phase? Neither.
  > So is he saying that the usual argument that the test particle directly experiences a phase shift is wrong? …So he seems to be saying that we should ignore that A.v term and its consequent phase shift and include only a phase shift coming from the source charges.
  No, the original argument is perfectly valid. Again, Vaidman’s point isn’t that his is the proper way to think about AB effect, just that it’s a valid one. That’s all he needs to rebut the idea that potentials are more real in quantum mechanics.
  > …if we consider a space with cylindrical topology, then there can exist static but nonzero gauge potentials even without sources…With no source charges for the test particle to backreact on, there doesn’t seem to be any way to explain the Aharonov-Bohm effect on the test particle in terms of phase shifts just of source charges.
  This is an excellent point to bring up, and it shows why all this handwringing about “reality” (which appears purely philisophical) is important: descriptions that yield equivalent predictions in one regime may diverge in another. (I say more about this below.) I’d have to think about it more, but my inclination is that the discovery of non-zero potentials in a cylindrical spacetime would be knock-down evidence for the reality of potentials.
  But of course, those are just mathematical constructions and we’ve never actually found spacetimes with nontrivial topology at the fundamental level. (Constructing a toy system in the lab that behaves like this is not enough, just like constructing toy systems with variable speeds of light doesn’t reduce the fundamental importance of Lorentz symmetry.) And certainly, the existence of such constructions is perfectly possible classically, so this aspect really isn’t about quantum mechanics per se.
  Insofar as the AB effect led you to raise your expectation of finding potentials on fundamental cylindrical topologies in nature (because of the supposed new-found “reality” of the potential), then Vaidman’s work should reduce those expectation to back what you believed before you read about the AB effect.
  > On a fixed but sufficiently curved background space…the test particle can go all the way around the circumference of the circle…quickly enough that the test particle’s own field does not have enough time to reach the solenoid at the center of the circle and backreact on the source charges. So then does the test particle experience an Aharonov-Bohm phase or not?
  Yes, an AB phase is measurable in that case.
  Here, I think you’re conflating two issues: (1) the reality of potentials versus forces, and (2) the fact that (when relativistic effects are important) the finite speed of light means you can’t just integrate out the electromagnetic field and replace it with forces that act instantaneously from a distance. (Of course, you can still integrate out the retarded fields and use “retarded forces”.)
  Classically, we already know that an object can emit radiation and recoil well before the emitted radiation interacts with another object. This is certainly evidence that the radiation is real, but it doesn’t distinguish between fields and potentials. Likewise, a particle’s wavefunction can be split over two paths in an AB experiment in curved spacetime in such a way that it produces a field which applies a force to the sources charges in a way that depends on the path taken, and this can produce a global phase that can be inferred before the field has a chance to chance to act on the source charges (at least in one frame). But that doesn’t mean the potential is more real than the field.
  It’s worth noting that (2) is an excellent example of how philosophical disputes on interpretation can have real-world implications when new experimental regimes become accessible. Before special relativity, any debate about the inelegance of instantaneous action-at-a-distance could have been dismissed as mere philosophy. If someone (e.g. Newton) insisted that gravity must, for reasons of beauty or physical intuition, be carried by a physical field, others could have pointed out this was experimentally indistinguishable from instantaneous force laws. But of course, in relativistic settings the gravitational field takes on a life of its own, vindicating Newton’s intuition.
  Reply
  - Reader297
    July 2, 2014 at 1:03 am
    
    I think I might push back slightly on the claim that quantum mechanics doesn’t make potentials more real. Are photons real in quantum theory? If so, then a coherent state of photons |0> + |p,mu> + (1/2!)|p,mu,p,mu> + (1/3!)|p,mu,p,mu,p,mu> + … defines a value for the potentials (indeed, it essentially gives a plane wave with polarization mu), and the ambiguity in the value of the potentials arises from the fact that the Hilbert space itself has to be defined as an appropriate quotient space due to the missing longitudinal polarization of the photon. (For a massive photon, one doesn’t need to do this, and the Hilbert space isn’t a quotient space.)
    Of course, one can define a Hilbert space for massless photons that isn’t a quotient space by working in a non-Lorentz-covariant gauge like Coulomb gauge. In Coulomb gauge, the vector potential has a concrete physical interpretation as the momentum per unit charge, just like Phi has an interpretation as the energy per unit charge.
    You write “When something doesn’t have a one-to-one correspondence with physically distinct configurations, that’s a pretty good indication that it’s not real.”
    That statement makes me a little nervous. Wave-functions (or state vectors) also do not have a one-to-one correspondence with physically distinct configurations, as they are only defined up to overall phase. So are state vectors real (in the ontological sense, not the complex variables sense) or not? A lot of people, including Vaidman, would say they are! (Although, I suppose instrumentalists would say they aren’t.)
    For that matter, electric and magnetic fields don’t have a one-to-one correspondence with physically-distinct configurations either—we can always rotate or Lorentz-boost them!
    And, again, the claim wasn’t ever that that a single choice of (Phi,A) is real. The claim is that it’s the entire equivalence class of (Phi,A)’s corresponding to a specific E,B that is real. Why can’t an equivalence class correspond to a real thing? After all, the standard way to avoid the ambiguity about state vectors being ambiguous due to overall phase factors is to declare the “state ray” to be real, but state rays are nothing other than equivalence classes of state vectors that differ only in overall phase. And the standard way to avoid the ambiguity about electric and magnetic fields is to say that we work with equivalence classes consisting of all values of the fields related by rotations and Lorentz transformations.
    So when you say “you are inclined to think that the distinction between blerg and blorg isn’t real,” I agree. No one claims that the distinction between two members (Phi,A) and (Phi’,A’) of the same equivalence class is real. But that’s different from saying that the entire equivalence class is real.
    You write “Vaidman’s claim is that, in real life, *all* background fields are ultimately produced by source charges. It’s just that, in traditional statements of the AB effect, the semi-classical limit (i.e. treating the external field as a classical variable) obscures this.”
    There is an old provenance to this way of thinking — that is, that all background fields are always produced by source charges. It was a significant motivation of folks like Jefimenko to try to formulate electromagnetism entirely in terms of sources. But this approach is not as general as the full theory of electromagnetism, and therefore limits the generality of electromagnetism in important ways.
    Photons are particles in their own right, and so there’s no reason why there can’t be primordial photons left over from the Big Bang just like primordial particles of all other kinds. So why can’t there be background fields left over from the early universe that don’t arise from sources?
    I accept that “Phase shifts aren’t properties of individual systems[;] they are properties of the global state as a whole.” My question was where the usual argument for the Aharonov-Bohm phase breaks down. When we plug the A.v term into a path integral for the test particle, for example, the Aharonov-Bohm phase just comes right when computing the path integral. In Vaidman’s treatment, he seems to be ignoring this calculation and focusing solely on the source charges. How does he justify ignoring the contribution from the test particle’s path integral? That is, why doesn’t he get *twice* the usual Aharonov-Bohm phase, namely, the usual contributing from the test particle, and his claimed additional contribution from the perturbation of the source charges?
    I’m also a little bit concerned with Vaidman’s use of approximations. The usual Aharonov-Bohm calculation is as exact as it gets. It’s a sharp calculation, whether done with Hamiltonians or with path integrals. But in Vaidman’s calculation, there is a lot of hand-waving with, e.g., de Broglie wavelengths. How trustworthy are these claims?
    And I have to apologize, but I don’t quite follow this paragraph or its relevance to my question:
    “Classically, we already know that an object can emit radiation and recoil well before the emitted radiation interacts with another object.
    This is certainly evidence that the radiation is real, but it doesn’t
    distinguish between fields and potentials. Likewise, a particle’s
    wavefunction can be split over two paths in an AB experiment in curved spacetime in such a way that it produces a field which applies a force to the sources charges in a way that depends on the path taken, and this can produce a global phase that can be inferred before the field has a chance to chance to act on the source charges (at least in one frame). But that doesn’t mean the potential is more real than the field.”
    My question is that according to the standard derivation of the Aharonov-Bohm effect, the phase appears instantly when the test particle’s wave-function recombines at the interferometer, whereas if we use Vaidman’s argument in my curved-space example, then the source may not have gotten phase-shifted by that time.
    Reply
    - Jess Riedel
      July 4, 2014 at 9:06 am
      
      > Are photons real in quantum theory?… the Hilbert space itself has to be defined as an appropriate quotient space due to the missing longitudinal polarization of the photon…Wave-functions (or state vectors) also do not have a one-to-one correspondence with physically distinct configurations, as they are only defined up to overall phase. So are state vectors real (in the ontological sense, not the complex variables sense) or not?
      I’m happy to bite the bullet on this one and say that wavefunctions themselves are not as real as, say, a rock. Certainly, wavefunction defy many attempts to independently confirm their various parts (such as the overall phase) in a suspicious way that is baked into the theory itself. (As opposed to the interior of the sun, which is difficult to probe only in practice). But I’m not sure there’s anything useful being done by separating our list of mathematical constructs into “real” and “not-real” columns, just like I don’t think there’s much accomplished be debating whether “justice” is “real”.
      I’m more interested in the misleading implications people get when they think of the wavefunction like a rock or a baseball. I expect future theories to overturn our view of the wavefunctions, especially the redundant bits, while I expect our view of rocks to stay pretty much the same. That’s all.
      > For that matter, electric and magnetic fields don’t have a one-to-one correspondence with physically-distinct configurations either—we can always rotate or Lorentz-boost them!
      Actually, this isn’t really true. If there is a rock in front of me, then — in my coordinates –a spatial translation of the rock a meter to the right is *not* the same thing. One refers to the rock actually in front of me, and the other refers to a counterfactual situation.
      If Bob gives a different description of the rock because he is passing by me at 0.5 c, that description is relative to his coordinates. And here I agree that there are two descriptions, with respect to two different coordinate systems, and the rock is best thought of as the equivalence class rather than either of the descriptions. But actually, that gets into questions about observers that would take us too far afield.
      > Why can’t an equivalence class correspond to a real thing?
      Yep, I’m mostly OK with this, except to say that I think you should slightly downgrade your confidence in reality than it would otherwise be. If your best description is an equivalence class, I think you should be slightly more likely to expect the theory to someday be replaced with one more concrete. But that’s a view I’m not confident in.
      > So why can’t there be background fields left over from the early universe that don’t arise from sources?
      Again, I think you’re are mixing up two different distinctions: between fields and potentials, and between photons and charges. The thrust of Vaidman’s paper is that we can always speak of fields rather than potentials (just like we could classically), so we it’s false to claim that quantum mechanics grants new-found reality to the potential over and above what it had classically. The question of integrating out photons to get charges is a separate one that Vaidman does not address. (In fact, he couldn’t do so easily here since he is working in a c=infinity limit. He’s pretty much stuck without photons.)
      > My question was where the usual argument for the Aharonov-Bohm phase breaks down.
      He’s not claiming it does because he’s not objecting to the traditional calculation. The traditional calculation is fine. He’s just doing a parallel calculation in a different framework, which is more awkward but just as valid, to show that one can live without potentials. There’s only one phase, but there are multiple ways of calculating it.
      So I guess maybe what you’re asking is why you can’t follow the old AB argument *within Vaidman’s framework*. And I think the reason is just that you can’t assume there is a potential *and* that there is a separate mechanism for transmitting the instantaneous action at a distance. If you infer a global phase by looking at the force applied to the source charges, then you can’t simultaneously add a phase from the potential. You’d be double counting.
      Likewise, you can’t reason that (1) a rock falling a certain distance D picks up the difference E from the gravitational potential V *and* that (2) the gravitational force F acts on the rock over the distance D giving it an additional amount of energy E, for a total increase of energy 2E. You can use F or V, but not both.
      > But in Vaidman’s calculation, there is a lot of hand-waving with, e.g., de Broglie wavelengths. How trustworthy are these claims?
      I am very confident. I’d be willing to wager cash that they would hold up to a more exhaustive analysis. I think, unfortunately, that the only contentious points are interpretational.
      > …I don’t quite follow this paragraph or its relevance to my question:…
      Yes sorry, that wasn’t very clear at all, so let me try to rephrase. That paragraph was trying to separate the two questions I think you were mixing up: fields versus potentials, and photons versus charges. As I said Vaidman is addressing the first, not the second.
      The form of Vaidman’s argument does not require signal propagation from the test particle to the source charges. It’s just that, in the non-relativistic limit in which he works, it’s easiest to integrate out the electromagnetic field and just work with charges that experience instantaneous action at a distance. Therefore, if he wants to show that fields are sufficient, and that potentials are not necessary, he must show that a force is actually applied to the source charges. (I guess he could have emphasized that the only necessary part is that the field was *produced* by the test particle, and de-emphasized that the field acted on the source charge, but in the c=infinity limit these events happen simultaneously.)
      If a full relativistic treatment were to be given, the source charges would not need to be discussed. Let’s take your nice modification where there is strong spatial curvature, and no signal can travel from the device to the source charges and back to the device in the lifetime of the experiment. In this case we would work only with two quantum systems: the test particle and the electromagnetic *field*. (But we would still never touch an electromagnetic potential.) What we would find is that there was global phase picked up by the joint field/test-particle state, depending on which path was taken by the test particle.
      Reply
      - Reader297
        July 4, 2014 at 6:06 pm
        
        Good points. So you’re saying that the source charges are a bit of a red herring. In the c less than infinity regime, we can talk about the phase shift in the emitted radiation rather than of the source charges. Thanks.
        You also write: “So I guess maybe what you’re asking is why you can’t follow the old AB argument *within Vaidman’s framework*. And I think the reason is just that you can’t assume there is a potential *and* that there is a separate mechanism for transmitting the instantaneous action at a distance.”
        Why not? Other than that it would mess up Vaidman’s argument by leading to double counting?
        “If you infer a global phase by looking at the force applied
        to the source charges, then you can’t simultaneously add a phase from the potential. You’d be double counting.”
        That was my original point. I argued that there’s something wrong with Vaidman’s approach, because it leads to double counting.
        “Likewise, you can’t reason that (1) a rock falling a certain distance D picks up the difference E from the gravitational potential V *and* that (2) the gravitational force F acts on the rock over the distance D giving it an additional amount of energy E, for a total increase of energy 2E. You can use F or V, but not both.”
        That’s correct, but there’s a reason for that — there is no formalism you can possibly write down in which both F and V appear simultaneously. You have to express F as minus the gradient of V, but then there’s no F anymore.
        My question about Vaidman’s approach is that I’d like an explanation for the analogous argument. He simply ignores what’s going on with the A.v term in the test particle’s Lagrangian. But why? What’s the analogy of writing F equals minus the gradient of V that eliminates the A.v term and turns it into a phase shift of the source charges (or emitted fields) and thereby proves that Vaidman isn’t hiding a double counting?
        Reply
        
        Jess Riedel
        August 1, 2014 at 5:41 pm
        
        > What’s the analogy of writing F equals minus the gradient of V that eliminates the A.v term and turns it into a phase shift of the source charges (or emitted fields) and thereby proves that Vaidman isn’t hiding a double counting?
        Sorry for the delay. Answering this (see new edit above) required me to go back and pick through Vaidman’s paper–i.e. actual work–and I didn’t have the time to do so until a recent flight.
        I did the electric version. Hopefully the mechanism is clear. I’ll leave it to you if you want to repeat the exercise to see exactly what happens with the A.v term in the magnetic version.
        Reply
Peter Morgan
August 11, 2014 at 5:29 pm

A different way to address the reality of EM observables is through QFT and gauge invariance, without reference to charges at all. The EM potential, $\hat A(x)$ , is an operator-valued distribution; an observable is constructed by smearing with a test function $f(x)$ , $\hat A_f=\inthat A_mu(x)f_\mu(x)\mathrm{d}^4 x$ , which is an inner product of 1-forms, $(\hat A,f)$ . In order for this to be an invariant object under gauge transformations $\hat A\mapsto \hat A+d\theta$ , we require $\delta f=0$ , in which case, on Minkowski space, we can write $f=\delta U$ , so that gauge invariant observables are
$(\hat A,\delta U)=(\mathrm{d}\hat A,U)=(\hat F,U).$
One has to worry a little about boundary conditions at infinity, but AFAICT not much. Effectively, $U$ is a 2-form potential for the test function $\delta U$ . The linear duality that is a required feature of the construction of observables in QFT makes gauge invariance a constraint that makes the gauge invariant observable $(\hat A,\delta U)$ the same as $(\mathrm{d}\hat A,U)=(\hat F,U)$ . Edit: No LaTeX. Sadness.
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- admin
  April 22, 2016 at 4:13 pm
  
  (The above comment was edited to fix LaTeX.)
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