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?

  • $\begingroup$ 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? $\endgroup$
    – FGSUZ
    Oct 2, 2018 at 20:30
  • $\begingroup$ Why is the given wavefunction a eigenfunction of the momentum operator? $\endgroup$ Oct 2, 2018 at 20:32
  • $\begingroup$ Oh, easy then. Act with the momentum operator of that given wavefunction, and check that you get "one number times the same wavefunction". $\endgroup$
    – FGSUZ
    Oct 2, 2018 at 20:33
  • $\begingroup$ but this requires $$A = \nabla \int_{x_0}^xA\cdot x$$ right? $\endgroup$ Oct 2, 2018 at 20:35
  • $\begingroup$ Rather, it involves a gradient and an additional term, and the gradient acts on an exponential, and not the integral directly. $\endgroup$
    – FGSUZ
    Oct 2, 2018 at 20:39

1 Answer 1


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.


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