5
$\begingroup$

Mathematically viewing, the Aharonov-Bohm experiment shows that the magnetic field creates a connection with a nonzero holonomy on a multiply-connected domain. This means that there isn't a state function that is continuous on that domain. This isn't a problem if we regard the state function as an element of $L^2$, since the elements of $L^2$ need not be continuous, only square-integrable. But the state function must be unambiguous up to a phase factor, and this is my problem. If we omit continuity on a zero-measure subset, then picking two different zero-measure subsets, the functions belonging to them will not differ by a constant phase factor.

A bit more details

The experimental setting described in this Wikipedia article looks like this:

enter image description here

The configuration space of the electron is the physical space minus the space occupied by solid objects like the wall (containing the slits) and the solenoid. The states of the electron are square-integrable functions on this multiply-connected domain. In the case when there is no current in the wire of the solenoid, let this state function be $\psi_0$, and let this state be $\psi_A$ when we switch the current on. Choosing a circle $c$ around the solenoid, let the restriction of $\psi_0$ to $c$ be $\psi_{0,c}$ and the restriction of $\psi_A$ to $c$ be $\psi_{A,c}$. Let's define $\Delta\psi:= \psi_{A,c}/\psi_{0,c}$.

Let's pick a point $p$ on the circle and take a parametrization $\gamma_p:[0,1]\to c: \lambda\mapsto \gamma_p(\lambda)$ of $c$ so that $\gamma(0)=\gamma(1)=p$. Then, according to the quantum theory of the A-B effect, $$\lim_{\lambda\to 1} \Delta\psi\circ\gamma(\lambda)=\Delta\psi\circ\gamma(0)e^{i\frac{q}{\hbar c}\oint_c A}\tag 1$$

Since $\gamma(0)=\gamma(1)=p$, $\Delta\psi\circ\gamma(1)=\Delta\psi\circ\gamma(0)=\Delta\psi(p)$, so $\Delta\psi$ isn't continuous at $p$,$\!\phantom{|}^{[*]}$ except when $\frac{q}{\hbar c}\oint_c A=2n\pi\ (n\in\mathbb N)$. This condition should imply the quantizedness of the flux in the solenoid, but in the Aharonov-Bohm effect, flux isn't quantized. In the discountinuous case, picking another point $q$ on $c$, we get a function that isn't continuous at $q$. It's not a problem that choosing different points on $c$ we get different state functions, but if these functions represent the same quantum state, then they should differ only by a complex number of modulus 1 as factor. But this is not the case, as seen in the following illustration (from $q$ to $p$, the factor is $1$ while on the rest, the factor is different from $1$). enter image description here

What is the resolution of this problem?

Edit

I am increasingly inclined to think that the above contradiction cannot be resolved within the framework of standard quantum mechanics. Is there a widely accepted alternative mathematical model of quantum mechanics that can describe exactly the Aharonov-Bohm experiment?

Perhaps the bundle formalism of quantum mechanics

Or the Segal quantization?

Or the Coherent foundations?

Edit 2

It seems that the problem is that the configuration space $Q$ of this quantum mechanical system is multiple-connected and standard quantum mechanics using the Hilbert space of states $L^2(Q)$ works only on simply connected $Q$. A solution could be to take a subspace of $L^2(\tilde Q)$ instead, where $\tilde Q$ is a $U(1)$ bundle on $Q$ (see Balachandran et al, Classical Topology and Quantum States), or the universal cover of $Q$. But I don't see the solution clearly yet.

Footnote

$\!\phantom{|}^{[*]}$ This property is seemingly similar to the case of the action of the Galilean group on the projective Hilbert space $P(H)$, where there isn't a continuous lift to the Hilbert space $H$ (regarded as an $U(1)$-bundle on $P(H)$. But here, $\psi$ is a function on the base space, not a section of a $U(1)$-bundle. The phase of $\psi$ is definite in each point. The phase freedom is to multiply the whole $\psi$ with a complex number of modulus 1, not to multiply its values point-by-point. The usual heuristic workaround of this problem is that we say that the particle is dragged along some pathes and meantime, it "picks a phase". But the notion of "dragging a particle" is alien to quantum machanics. In this case, where is the state function in a given time instance? I think, a more clear approximation would be to handle the $x$ coordinate of the particle classically, as Beltrametti and Cassinelli does in the case of the double slit experiment and to use wave functions defined on the planes perpendicular to the $x$-axis (see here). Or the other possibility, to make a kind of quantum mechanics in which the values of the state functions aren't complex numbers, but fibers of a $U(1)$-bundle. Of course, there is the Feynman path integral formalism, but currently I'm looking for a solution inside standard quantum mechanics, or at least, which is more close to the standard quantum mechanics than the path integrals.

$\endgroup$
7
  • $\begingroup$ Possible duplicate / closely related physics.stackexchange.com/q/420932/226902 $\endgroup$
    – Quillo
    Commented Mar 3 at 9:13
  • $\begingroup$ @Quillo I don't really understand that. Everybody talks about an $U(1)$ bundle which is of course there, but its trivialization (the "gauge") is not arbitrary. Since state functions have definite complex values at every point, the "gauge freedom" is restricted to multiplying by a constant complex number (i.e the whole $\psi$) of modulus 1. This number cannot depend on the space coordinates! This is quite different from that that if we consider a process $t\mapsto \psi(t)$ then it is equivalent to the process $t\mapsto \alpha(t)\psi(t)$. with any function $\alpha:\mathbb R\mapsto U(1)$. $\endgroup$
    – mma
    Commented Mar 3 at 12:04
  • $\begingroup$ > Since state functions have definite complex values at every point, the "gauge freedom" is restricted to multiplying by a constant complex number (i.e the whole ψψ) of modulus 1.This number cannot depend on the space coordinates! This is not true - psi function in quantum theory satisfying the Schroedinger equation implies it has gauge freedom in the sense any gauge transformation $A\to A'=A+\nabla \chi$ ($\chi$ being a function of spatial coordinates) induces change in the psi function $\psi\to\psi' = \psi e^{i\frac{q\chi}{\hbar}}$.Thus the phase factor can depend on particle coordinates. $\endgroup$ Commented Mar 8 at 23:52
  • $\begingroup$ It is hard to see why the discontinuity in $\psi$ somewhere around the solenoid is necessary. If we start at zero time with zero current in the solenoid, $\psi(t=0)$ obviously can be chosen a continuous function of particle coordinates. Then by increasing the current slowly, $\psi(t)$ will evolve continuously in time, and thus will remain a continuous function of particle coordinates. $\endgroup$ Commented Mar 8 at 23:56
  • 1
    $\begingroup$ @Quillo Yes, now I see, my question is practically a duplicate of that. However, I don't understand its accepted answer at all. I wrote there in a comment what are my problems with it. $\endgroup$
    – mma
    Commented Mar 12 at 4:54

1 Answer 1

2
$\begingroup$

All the talk about extending the formulation of QM (which is a good thing to understand) is in this case a red herring. Even if you insist on an unsophisticated view of the wavefunction as a class of functions $M\to\mathbb C,$ considered equivalent up to disagreement on null sets, the problem is just in your logic. You have correctly deduced that, by considering the wavefunction on a circle around the solenoid, the wavefunction must be discontinuous somewhere. You've also correctly stated that, in this system, the same physical state can be represented by multiple wavefunctions that don't just differ by a constant phase, but also can differ by how they distribute a jump in phase in the form of discontinuities around a circle. But this is not necessarily a problem (at least, not at the mathematical level). The theorem that quantum states differing by a constant phase represent the same physical state is only an implication in one direction. Having physically identical states only differ by a constant phase would be nice but is not necessary for the logical consistency of a formulation.

In classical electrodynamics, we know the gauge transformations $A_\mu\to A_\mu+\partial_\mu\alpha$ modify mathematical descriptions of a system without changing the physical state. You are basically hitting on the fact that, in QM+electrodynamics, the wavefunction also has a transformation rule that looks like $\psi\to e^{i\alpha}\psi.$ You can take a wavefunction that has a discontinuity at one point around the circle to a wavefunction that has the discontinuity somewhere else by making a gauge transformation with an $\alpha$ that is locally constant almost everywhere (thus $A_\mu=0$) except at a few points, where it contains jumps. (The infinite $A_\mu$ at the jumps serve to correct the derivative operator so that you get the correct kinetic momentum.) The "problem" you've identified is just a manifestation of gauge symmetry. As long as you are okay with having multiple (gauge-equivalent) descriptions of the same physical state, there is no problem.

$\endgroup$
6
  • $\begingroup$ I like this answer very much! However, it is not yet clear to me that the different wave functions, which differ in the location of discontinuity represent the same probability distribution, give the same transition probabilities, and give the same expected values for the physical quantities. $\endgroup$
    – mma
    Commented Mar 26 at 4:49
  • $\begingroup$ @mma That it's the same probability distribution should be obvious: $|e^{i\alpha}\psi|^2=|\psi|^2.$ The momentum operator $p=-i\nabla$ is gauge-dependent, so (even classically) it is unobservable. The actual observable, the kinetic momentum operator $P=p-A,$ transforms as $P\to P-\nabla\alpha$ and you can verify $(P-\nabla\alpha)(e^{i\alpha}\psi)=e^{i\alpha}P\psi,$ so gauge transforming and computing the kinetic momentum "commute". As $H$ only depends on $P$ and not $p,$ time evolution and also the transition amplitudes respect the gauge symmetry. $\endgroup$
    – HTNW
    Commented Mar 27 at 4:39
  • $\begingroup$ According to this comment, "If $\psi_1\neq \psi_2e^{i\varphi}$ there always exists a (bounded) hermitian operator with different expectation values. Hence, if you really mean "state", you should restrict the notion of observables, i.e. not every hermitian operator is an observable anymore." So, you should prove that all physically relevant hermitian operators have the same expectation values, not only the kinetic momentum operator. $\endgroup$
    – mma
    Commented Mar 28 at 9:57
  • $\begingroup$ @mma The physicals observables are the ones you can write as functions of the position $x$ and kinetic momentum $P$ (classically, the state of a particle of fixed mass is completely specified by $x$ and $P,$ and thus all observables are some function of $(x,P)$). Proving that $x$ and $P$ respect the gauge transformation is all that is essential to prove that all physical observables respect the gauge transformation. $\endgroup$
    – HTNW
    Commented Mar 28 at 14:06
  • $\begingroup$ OK, but what about transition probabilities? We talk about the $\psi_2=(1+\chi_S(z-1))\psi_1$ case (see here). In this case (taking $\psi$-s normed) $\langle\psi_2,\psi_2\rangle=1$ but $\langle\psi_2,\psi_1\rangle<1$. $\endgroup$
    – mma
    Commented Mar 29 at 7:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.