Aharonov-Bohm effect: exact solution for scattering problems

I'm reading the paper Significance of Electromagnetic Potentials in the Quantum Theory by Aharonov and Bohm (Phys. Rev. 115, 485 (1959); eprint). In the paragraph exact solution for scattering problems they solve the wave equation outside the magnetic field region, in cylindrical coordinates:,

$$\left[\dfrac{\partial^2}{\partial r^2} + \dfrac{1}{r}\dfrac{\partial}{\partial r} + \dfrac{1}{r^2}\left(\dfrac{\partial}{\partial \theta}-\dfrac{ie\phi}{ch}\right)^2+k^2\right]\psi=0,$$

where $\vec{k}$ is the wave vector of the electron and a gauge is chosen such that $A_r=0$ and $A_{\theta}=\phi/2\pi r$. I assume that $\phi$ is the flux of the solenoid and $e,c,h$ are the usual constants.

I'm guessing that this is a Schrödinger equation? Can someone help me find a derivation and help me understand the meaning of this problem?

Starting purely classically

So the Lagrangian for a particle in an electromagnetc field in Gaussian units is just $$L(\vec r, \vec v) = \frac12 m~ \vec v \cdot \vec v + \frac qc~\vec A(\vec r)\cdot \vec v - q~\varphi(\vec r).$$This generates the Lorentz force law with its Euler-Lagrange equations of motion, given a scalar potential $\varphi$ and a vector potential $\vec A.$

However those Euler-Lagrange equations have an interesting property in how this derivation comes about. The interesting property is, the "generalized momentum" for the particle is supposed to be $p_i = \partial L/\partial v_i$ and the "generalized force" in that direction is $F_i = \partial L/\partial r_i$ and the equations of motion just say that $\frac{d}{dt} p_i = F_i.$ Well, the only "force" term here is that scalar potential; the vector potential actually comes in by changing what the equation thinks "momentum" means; the equation thinks that "momentum" is $p_i = m v_i + (q/c)~A_i(\vec r).$ The full effect of this vector potential therefore happens in the $p_i$ stage, keeping in mind that we're talking about a total derivative on the left-hand-side:$$\frac{d p_i}{dt} =m \dot v_i + q\dot A_i/c + \frac qc~\big(\nabla A_i\big) \cdot \frac{d\vec r}{dt} = \frac{q}{c} \vec v \cdot \Big(\partial_i \vec A\Big) - q~\partial_i \varphi.$$ Just to complete the argument if you've never seen it before, we defined $B = \nabla \times A$ and $E = -c^{-1}~\dot A -\nabla\varphi$ originally to eliminate two Maxwell equations, and if we carefully work out the BAC-CAB identity in this context we have that $\big(\vec v \times (\nabla \times \vec A)\big)_i =\vec v \cdot (\partial_i \vec A) - (\vec v\cdot \nabla)A_i.$ So the above equation ends up as precisely the Lorentz force law in Gaussian units, $$m ~\vec a = q~\vec E + \frac qc \vec v \times \vec B.$$Furthermore, there is no assumption enforced by this Lagrangian about what gauge we're using for $\varphi$ and $\vec A.$

This also means that the classical Hamiltonian, which is $-L + \sum_i p_i v_i,$ becomes just $$H(\vec r, \vec v) = \frac 12 m~\vec v \cdot \vec v + q~\varphi(\vec r),$$ as the $q~\vec A \cdot \vec v/c$ terms cancel out. This has of course the nice interpretation in terms of conservation of energy, but it kind of looks weird because at first glance you might say "wait, does that mean there's no magnetic forces?" And there must be magnetic forces. In fact the Hamilton equations of motion require taking derivatives with respect to $p_i$ and not $v_i$ and this has to be the same as the canonical momentum derived above, so a better description for the sake of the equations of motion must be,$$H(\vec r, \vec p) = \frac 1{2m} \left(\vec p - \frac qc \vec A(\vec r)\right)^2 + q~\varphi(\vec r).$$ Now the equations of motion $\dot r_i = \partial H/\partial p_i,~~ \dot p_i = -\partial H/\partial r_i$ get you the Lorentz force law again.

Quantization

Those equations of motion are also of critical importance when we're talking about quantizing this Hamiltonian, but roughly speaking the procedure is just the same $\vec p \mapsto -i~\hbar\nabla$ procedure and hence,$$\hat H = \frac{1}{2m}\Big(-i\hbar \nabla - \frac qc \vec A\Big)^2 + q \varphi.$$Pulling $-i\hbar$ out of the parentheses if $\varphi = 0$ and writing $q=-e$ gives the Schrödinger equation as$$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\left(\nabla + \frac{i e}{\hbar} \vec A\right)^2 \Psi.$$ (The paper might have a sign error, or I might have above -- but either way if it does, it's not a particularly important one.) Most of the rest of this becomes the standard undergraduate exercise of deriving expressions for divergence, gradient, etc. in cylindrical coordinates.

Now Aharonov and Bohm have chosen the Coulomb gauge where $\nabla \cdot A = 0$ and therefore for these purposes $\nabla$ and $A$ are basically commuting operators. That the field is zero requires $\nabla\times A = 0$ and therefore in combination we have $\nabla \times (\nabla \times A) = \nabla(\nabla \cdot A) - \nabla^2 A = 0 - \nabla^2 A = 0$ so they have solved $\nabla^2 A = 0$ for a nontrivial solution $\vec A = c~\hat \theta / r,$ and since this, as mentioned, essentially commutes with the derivative operators one can just bring it into the $\theta$-component of divergence which is normally $r^{-2}~\partial^2\Psi/\partial\theta^2,$ to get $r^{-2} (\partial/\partial\theta + iec/\hbar)^2\Psi.$ Combined with other parts of the $\nabla^2$ expansion in cylindrical coordinates, one gets the expression above.

In normal classical physics $\vec A$ is thought to be a handy calculation tool, perhaps, but not as "real" as $\vec B$ and $\vec E$ are. What Aharonov and Bohm derived was a fantastic quantum effect of $\vec A$ which persists even if $E = B = 0.$ In other words, if electrons can travel two ways around a "loop" which encloses some $B$-field flux somewhere therein, then even if that $B$ field is totally zeroed out on the actual loop track, the electrons' wavefunctions interfering at the end of the track might see interference effects which have to be interpreted either as the $A$ field acting locally or you have to look at the $B$-field globally. Since physicists like local descriptions this forces treating the $A$-field as "more real" on the ring.
Essentially traversing the two branches of the loop each wave $e^{i(\omega t + \phi)}$ picks up a different phase $\phi_L \ne \phi_R$ when $A\ne 0$ and this is what causes interesting Aharonov-Bohm effects.
• @RubenVanBelle I mean it can also be $-k^2\Psi$ if you want to move that term to the right-hand-side; any equation $A = B$ can be rephrased as $A-B=0.$ Jul 31, 2017 at 22:23