2
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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