Polymer quantization, singularity resolution, and the $1∕{\mathit{r}}^2$ potential

We present a polymer quantization of the $-\lambda ∕r^2$ potential on the positive real line and compute numerically the bound state eigenenergies in terms of the dimensionless coupling constant $\lambda$. The singularity at the origin is handled in two ways: first, by regularizing the potential and adopting either symmetric or antisymmetric boundary conditions; second, by keeping the potential unregularized but allowing the singularity to be balanced by an antisymmetric boundary condition. The results are compared to the semiclassical limit of the polymer theory and to the conventional Schrödinger quantization on $L_2\left({\mathbb{R}}_{+}\right)$. The various quantization schemes are in excellent agreement for the highly excited states but differ for the low-lying states, and the polymer spectrum is bounded below even when the Schrödinger spectrum is not. We find, as expected, that for the antisymmetric boundary condition the regularization of the potential is redundant: the polymer quantum theory is well defined even with the unregularized potential and the regularization of the potential does not significantly affect the spectrum.

Figure 1

(Color online) The solid (red) line is the regulated $-1∕x^2$ potential ($\lambda =1$, $\mu =1$). The dashed (blue) line is the unregulated potential.

Figure 2

(Color online) The lowest two energy levels as a function of $\lambda$ for the unregulated potential with antisymmetric boundary conditions.

Figure 3

(Color online) The lowest two energy levels as a function of $\lambda$ for the regulated potential with antisymmetric boundary conditions.

Figure 4

(Color online) $\ln \left(E_n\right)$ vs $n$ for the unregulated potential with antisymmetric boundary conditions with linear fits $(R^2=1)$.

Figure 5

(Color online) $\ln \left(E_n\right)$ vs $n$ for the regulated potential with antisymmetric boundary conditions with linear fits $(R^2=1)$.

Figure 6

(Color online) $\beta$ vs $\alpha$ with $\mu =1$ for the unregulated potential with antisymmetric boundary conditions with a linear fit $(R^2=1)$.

Figure 7

(Color online) $\beta$ vs $\alpha$ with $\mu =1$ for the regulated potential with antisymmetric boundary conditions with a linear fit $(R^2=1)$.