# Momentum eigenfunction for non-vanishing vector potential

I'm reading a section in a textbook about the Aharonov-Bohm effect, which claims that the eigenfunction (outside the solenoid, so $$\operatorname{curl} A=0$$) for the canonical momentum operator $$\pi = \frac{\hbar}{i}\nabla - \frac{q}{c}A$$ is $$\Psi = \exp(i k_0\cdot x) \exp\left(\frac{iq}{\hbar c}\int_{x_0}^x A\cdot dx\right).$$

I don't understand why this holds. Doesn't this imply that $$A$$ can be expressed as the gradient of a potential: $$A = \nabla \int_{x_0}^x A\cdot dx$$ On the other hand $$A$$ is clearly not a conservative vector field, since $$\int_C A = \Phi_\text{mag} \neq 0,$$ for any closed $$C$$ around the solenoid. This is then in contradiction to the former statement as gradient fields are always conservative. Whats going wrong here?

• Sorry but I don't understand your question. Are you asking why $\pi=\frac{\hbar}{i}\nabla-\frac{q}{c}A$? Are you asking why the rotational of A is 0? A complete different thing? Oct 2, 2018 at 20:30
• Why is the given wavefunction a eigenfunction of the momentum operator? Oct 2, 2018 at 20:32
• Oh, easy then. Act with the momentum operator of that given wavefunction, and check that you get "one number times the same wavefunction". Oct 2, 2018 at 20:33
• but this requires $$A = \nabla \int_{x_0}^xA\cdot x$$ right? Oct 2, 2018 at 20:35
• Rather, it involves a gradient and an additional term, and the gradient acts on an exponential, and not the integral directly. Oct 2, 2018 at 20:39

The formula for $$\psi$$ in you textbook is only correct if there is only one path for the integral. This will be the case if you are considering a particle trapped on a ring, but in the case of a particle moving freely outside a thin solenoid threaded by $$\alpha$$ units of flux the actual Bohm-Aharonov wavefunction is much more complicated: $$\psi(r,\theta) \propto e^{i\pi |1-\alpha|/2}\sum_{m=-\infty}^{\infty} e^{i m\theta} J_{|m-\alpha|}(kr),$$ where $$J_m$$ is the Bessel function.