2
$\begingroup$

this is probably a very stupid question. First of all, I'm a mathematician so please try to use coordinate-free notations.

It's often used in quantum mechanics a wave function depending on a fixed trajectory. For instance, in How to derive the Aharonov-Bohm effect result?, $\psi (x, \gamma)$ depends on the path $\gamma$. I can understand this in terms path integral quantization (which is essentially perturbative and I don't like). However in the philosophical scope of quantum mechanics it seems unreasonable to fix a trajectory unless the wave function is collapsed at all possible times (we know the position at every instant). Note that the holonomy $\exp (-iq_e \int A)$ is gauge invariant, but $A$ (that's actually a cocyle $A =\{A_i, {g_{ij}}\}$ representing a gauge equivalence) need not to be $d$-closed (i.e., $dA_i = F_A$ may be non-zero), therefore it really depends on the path chosen (and not just on the endpoints). Maybe this condition is simply implicitly posed on the boundary conditions?

Let me be more precise. I will treat the Aharonov-Bohm effect in order to deal with something more concrete. For simplicity, let's take the space-time $X = \mathbb{R} \times \mathbb{R}^3$ together with a charge at $(t,0)$, i.e., $$dF_A = q_m\delta_{(t, 0)}$$ and a particle of charge $q_e$ moving through fixed paths $\gamma_i$, $i = 1, 2$ with the same endpoints.

What's the meaning of $\psi (x, \gamma_i)$ in this case? Is the wave function really collapsed at every instant (i.e., do we always know the position)?

If the above questions cannot be answered without propagators or path integrals, then is the Aharonov-Bohm effect a strictly perturbative phenomenum? If no, how to derive the Aharonov-Bohm effect non-perturbatively?

By deriving the Aharonov-Bohm effect I mean obtaining a phase difference that would lead to the Dirac charge quantization condition. For instance, in the path integral formulation the phase difference is given by $\exp (-iq_e \int_{W_e} A)$ (where $W_e$ is a loop through the solenoid) and this leads to $q_m q_e \in 2\pi \mathbb{Z}$ if one wants the solenoid to be undetectable.

For sake of completeness, I will add a possible solution of the Dirac charge quantization condition that does not makes the Aharonov-Bohm effect visible in any meaningful way, so that maybe someone can include the Aharonov-Bohm effect in this context.

A possible solution (to the Dirac charge quantization and not to the Aharonov-Bohm effect) is given at page 14-16 of http://www.maths.ed.ac.uk/~jmf/Teaching/Lectures/EDC.pdf , by declaring that $\chi = \log (g)$ for some gauge transformation $g$, that is the two open charts $U_{i} = \{(x, y, z) \in \mathbb{R}^3| \ (-1)^i z > 0 \}$ covering $\mathbb{R}^3\setminus 0$ together with the potentials $A_{-}$ and $A_{+}$ defines a $U(1)$-bundle.

Thanks in advance.

$\endgroup$
0
$\begingroup$

The demonstration of Aharonov-Bohm effect doesn't require the trajectories to be fixed. Usually diagrams showing fixed trajectories are simplified to two different trajectories to demonstrate the main issue at hand. Consider the following case, there is a magnetic field in a circular region in two dimensions which vanishes outside the radius of the circle. Even though the field vanishes outside the circular region, the magnetic vector potential doesn't.

Now in your model, you constrain the charge to move along the two fixed trajectories. In nature, that doesn't happen, and and the trajectories take all possible paths to reach the detector, where a pattern is produced due to the phase factor along the trajectories which is the path integral of the vector potential. For simplicity, usually two paths are depicted as they give a reasonably good treatment of the underlying phenomena, but that is not how it works and therefore the phase factor is not dependent on only the two paths, but it is essentially proportional to the path integral of the vector potential over all possible paths (which can be further simplified). And therefore the wavefunction is not collapsed at all points as in the OP but only at the detector.

$\endgroup$
  • $\begingroup$ Thanks for your answer. Well, but this relies on path integral quantization as I have already mentioned. How can you write the wave function from the Schroedinger equation directly assuming all the possible paths without going to a perturbative ill defined method? Scattering problems are very perturbative and I try to avoid this stuff. $\endgroup$ – user40276 Mar 5 '16 at 6:25
0
$\begingroup$

Consider the following setup for electromagnetic field:

$$\phi(x,y)=U_0\theta(R^2-x^2-y^2),$$ $$\vec A(x,y)=\frac{A_0}2\begin{cases} \left(y,-x,0\right)&x^2+y^2\le R^2\\ \left(\frac{R^2y}{x^2+y^2},-\frac{R^2x}{x^2+y^2},0\right)&\mathrm{otherwise} \end{cases},$$ where $\theta(x)$ is Heaviside theta, and $U_0$ is assumed to be very large — such that the electron would never get inside (may be infinity, i.e. homogeneous Dirichlet boundary condition at $x^2+y^2=R^2$), and $R$ is the radius of a solenoid.

If you propagate an electron wave so that it would go in the direction of the solenoid, the result of the interference on the side opposite to the source would depend on $A_0$, even though $\vec B=\nabla\times\vec A$ vanishes whenever $x^2+y^2>R^2$, and the wavefunction is nonzero only there. This is the essence of the Aharonov-Bohm effect. All the talk about pairs of fixed paths in this topic is merely a simplification to aid analysis.

$\endgroup$
  • $\begingroup$ Thanks for your answer. I don't get what boundary condition you're putting on the Schroedinger equation to require that the electron goes in the direction of the solenoid. $\endgroup$ – user40276 Mar 5 '16 at 9:39
  • $\begingroup$ The overall boundary condition is boundedness (since the space is infinite). For electron to go in the direction of the solenoid, you use an initial condition, not a boundary condition, e.g. $\psi(x,y,t_0)=\exp(ix)\exp(\frac1{x+2})\theta(-(x+2))$ or any nicer. You just need to make sure the wave packet or what you have moves (mostly) in the needed direction and initially avoids having nonzero values in the location of the solenoid (or its value there can be neglected). $\endgroup$ – Ruslan Mar 5 '16 at 10:15
  • $\begingroup$ Ops! Of course, initial value condition. Sorry but I still cannot understand how can you guarantee that the electron does not enter in a position where $B$ is non-zero. Furthermore, since I'm having some difficulties with the calculus notation (it has been like 4 years since the last time I used these notations, now I've got too used to differential forms), could you tell me what's the phase difference that you're getting from your derivation (are you really getting the holonomy? But in this case, the path must fixed?). My objective was to get the phase difference usually ... $\endgroup$ – user40276 Mar 6 '16 at 17:56
  • $\begingroup$ obtained when deriving the Dirac charge quantization condition. I will edit my question to clarify myself. $\endgroup$ – user40276 Mar 6 '16 at 17:57
  • $\begingroup$ The electron won't enter there because $U_0$ is taken to be large enough, up to infinity — in this limit case you can just put a homogeneous Dirichlet condition at the boundary of the solenoid, and thus the electron will automatically never appear past the border (even if it does, it won't affect the solution at all, as in infinite potential well). $\endgroup$ – Ruslan Mar 6 '16 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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