Is the bundle of the Aharonov-Bohm effect like the tangent bundle of a cylinder or like the tangent bundle of a truncated cone

aharonov-bohmdifferential-geometryelectromagnetismgauge-theoryquantum mechanics

Both are trivial bundles, and the natural (metric) connection is flat (curvature-free) for both. The difference between them is that the holonomy of the tangent bundle of the cylinder is trivial while the holonomy of the tangent bundle of the truncated cone isn't. This means that in the case of the cylinder, the horizontal lift of any closed curve to any point of the bundle is closed, while in the case of the truncated cone, the horizontal lifts of nonshrinkable closed curves are not closed. To put it plainly, the parallel transport of any tangent vector around a closed curve in the case of the cylinder returns always to its starting point, while in the case of the truncated cone, this is true only for those curves that don't go around the hole.enter image description here

One proposed solution to the issue of the non-existing global wave function in the Aharonov-Bohm experiment is the replacement of the wave function with a global horizontal section of an $U(1)$ bundle and getting different local wave functions for different patches in a local trivialization. But this presupposes the existence of global horizontal non-shrinkable sections of the bundle, like in the case of the cylinder. However, the Aharonov-Bohm effect is explained sometimes as a nontrivial holonomy of the (flat) connection due to the magnetic field inside the solenoid. This would be like the truncated cone. But in this case, there is no global horizontal section of the bundle, and we can't repair this with any gauge (trivialization).

So, which is the correct picture? In the first case, there is no problem, but there isn't any holonomy. In the second case, there is the much-talked-about holonomy, but in this case, the trivialization (gauge) doesn't solve anything.

Best Answer

Requiring (the derivative of) the wavefunction to be horizontal w.r.t. the $U(1)$-connection is absurd. If you just consider a free particle in space moving in the absence of any field at all (so space is $M=\mathbb{R}^3,$ the principal bundle is the trivial one $P=U(1)\times M,$ and the wavefunction is a section of the associated and also trivial vector bundle $B=M\times\mathbb{C}$ and therefore an ordinary function $\psi:M\to\mathbb{C}$), requiring the wavefunction to be "horizontal" means requiring it to be constant. After all, the connection tells you how to take the derivative of the wavefunction, so "horizontal" means $\frac{d\psi}{dx}=0.$ That's not a reasonable condition!

The only time I could see a (globally) horizontal wavefunction being appropriate for describing the Aharanov-Bohm effect is in the following situation: one particle is confined to a finite ring around a cylindrical solenoid. The flux in the solenoid is a multiple of the flux quantum $h/q$ (perhaps the flux is zero). Then the ground state of the particle is a horizontal section of the appropriate associated bundle. If the flux is not set to a multiple of the flux quantum, then a globally horizontal wavefunction is indeed impossible. This isn't a problem: the ground state is just not horizontal anymore. In particular, the single-valuedness of the wavefunction is not a priori related to the connection and is never in question. (What might happen is someone chooses coordinate charts to make components of the connection vanish. Then the wavefunction will have "multiple" values at the chart overlaps. But the wavefunction is still single-valued! You are just describing its single value at each point in multiple coordinate systems.)

That out of the way, I don't know where the bundles presented in the question come from. I will just describe the bundles and connection in the Aharanov-Bohm effect from first principles. Both bundles involved in the classic Aharanov-Bohm configuration (i.e. a particle moving around an infinite cylindrical solenoid) are trivial. Let $M=\mathbb{R}^3\mathbin\backslash\text{cylinder}$ be the space around the solenoid. The relevant (and only, I believe) principal $U(1)$-bundle on $M$ is the trivial one: $P=U(1)\times M.$

The space of wavefunctions is constructed by first replacing the $U(1)$ fibers of $P$ with $\mathbb{C}$ fibers through the associated bundle construction. This doesn't require the connection, and you get another boring trivial bundle, (isomorphically) $B=M\times\mathbb{C}.$ The wavefunction will be a section $\psi\in\Gamma(B)$ (or, rather, an $L^2$ class of sections of $B$). There is no obstruction to a global section here. In fact, as the bundle is trivial, the wavefunction is an ordinary global function $\psi:M\to\mathbb C,$ like in free space.

The vector potential is interpreted as a principal $U(1)$-connection on $P,$ i.e. a $\mathfrak{u}(1)$-valued $U(1)$-equivariant generator-preserving 1-form on $P.$ As $P$ is a trivial bundle, such an object does always exist. At zero field strength, we have the Maurer-Cartan connection on $P$ (which arises essentially by differentiating the action of $U(1)$ on $P$): $\omega_{MC}=d\phi,$ where $\phi$ is the angular coordinate around $U(1).$ If you restrict attention to a ring around the solenoid instead of the full space $M,$ $P$ restricts to a torus, and the "horizontal" subspaces (of $TP$) picked out by $\omega_{MC}$ are just toroidal circles (i.e. lines of "latitude"). When the solenoid is powered, the connection is instead $\omega=\omega_{MC}+A,$ where the vector potential $A$ is interpreted as a $\mathfrak{u}(1)$-valued 1-form on $M$ (not $P$, and with no further restrictions!). Specifically, letting $\theta$ be the azimuthal coordinate on $M,$ we can take $A=\frac{R^2B}{2}\,d\theta,$ where $\pi R^2B$ is the magnetic flux through the solenoid. The horizontal spaces picked out by $\omega$ on each torus are now "tilted".

Note there is no need for the "integral curves" of $\omega$ (i.e. the spiral paths in $P$ traced out by parallel transporting points on $P$ around circles in $M$) to close. What's important is there exist global choices of "horizontal", both twisted and untwisted, not globally horizontal sections. In particular, the connection imposes no restrictions on $\psi$ nor vice versa. The transport of a point of the wavefunction around the solenoid is allowed to cause a nontrivial phase shift, and will do so if the flux $\pi R^2B$ is not a multiple of the flux quantum $h/q.$

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