Does the gauge field, $\left(\frac{y}{x^2+y^2},-\frac{x}{x^2+y^2}\right)=\nabla\arctan(x/y)\equiv\nabla f$ has effect on Schrödinger equation


Since we can perform a gauge transformation $A'\rightarrow A+\nabla f,\psi'\rightarrow e^{if}\psi$ to gauge out the field and lead to the equation below:



1 Answer 1


I am assuming that all charges of interest in the problem are $1$.

You can gauge out the field provided that you can find some $f$ such that $\nabla f=A$. Now, in your case, there is no such $f$ because the $f=\rm{arctan}(x/y)$ is not well defined when $x=y=0$. This is why the gauge field is not trivial.

More precisely, $f$ exists everywhere except at the origin. At the origin $B=-\nabla \times A=2\pi \delta(\bf{r}-0)$ (this you can get by applying Stokes theorem). Thus the gauge field configuration corresponds to an infinitesimally thin flux tube through the center carrying $2\pi$ flux.

However, if you want those solutions of the Schrodinger equation that vanish at the origin (Have not checked this, I think self-adjointness of the domain+operator would require that this is true), then you can restrict the Schrodinger equation to multiply connected domain $\mathbb{R}^2-\{0\}$, in which $f$ exists. Here the gauge field can be removed after a transformation.

You might find the following paper interesting (Byers and Yang PRL 1961): http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.7.46

and this wiki entry: https://en.wikipedia.org/wiki/Byers-Yang_theorem

Also pay attention to the periodicity of $f$ in the azimuthal direction. The flux at the center is an integer multiple of $2\pi$ and therefore $f$ is single valued. If the flux were not single valued, you would have ended up with the same Schrodinger equation, but the solutions $\psi$ will have to satisfy twisted boundary conditions due to the non-periodicity of $f$ in the azimuthal direction.


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