Comment on 'The paradoxical zero reflection at zero energy'

We have read with interest the article by Ahmed et al [1] dealing with the anomalous threshold effect in the one-dimensional scattering problem. The authors discuss the reflection probability $R(E)$, which normally is unity at zero energy, i.e., $R(0)=1$. However the exception, or anomaly, occurs when the interaction, which the scattering particle is subject to, supports a zero-energy bound state, in particular a half-bound state, i.e., a non-normalizable zero-energy bound state. The authors consider the cases when $R(0)=0$ or $R(0)\ll 1$. We would like to supplement the authors' conclusions by pointing out that the anomaly includes the range of values for the reflection coefficient $0\leqslant R(E)\lt 1$ when there is a half-bound state and E is zero or very small.

In the original discussion of the threshold anomaly in one-dimension, Senn [2] already pointed out that R(0) can have any value between 0 and 1. See, for example, Senn's figure 1 [2]. Nogami and Ross [3] considered two classes of anomaly due to half-bound states: class I has $R(0)=0$ for symmetric potentials, and class II with $0\lt R(0)\lt 1$ for asymmetric potentials. Kiers and van Dijk [4] reached the same conclusions from the analysis of N-channel one-dimensional systems with general symmetric and asymmetric interactions. de Bianchi also discussed the threshold behaviour [5].

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To illustrate these features, which are true in general³ , we use the well-known example of a two-delta function potential,

$$\begin{eqnarray}&&V(x)=\displaystyle \frac{{{\rm{\hslash }}}^{2}}{2\mu }[\lambda \delta (x+a)+\widetilde{\lambda }\delta (x-a)],\end{eqnarray} \tag{ 1 }$$

where μ is the (reduced) mass of the particle(s) and $a\ne 0$. In the formulation below we use results, reduced to the single channel case, from section III.B.2 of [4]. The wave function has asymptotic form

$$\begin{eqnarray}\psi (k,x)=\left\{\begin{array}{ll}{{\rm{e}}}^{{\rm{i}}{k}{x}}+\rho (k){{\rm{e}}}^{-{\rm{i}}{k}{x}},\ \ & x\leqslant -a\\ \tau (k){{\rm{e}}}^{{\rm{i}}{k}{x}}, & x\geqslant a,\end{array}\right.\end{eqnarray} \tag{ 2 }$$

where $E={{\rm{\hslash }}}^{2}{k}^{2}/(2\mu )$, and $\rho (k)$ and $\tau (k)$ are the reflection and transmission amplitudes, respectively. The reflection amplitude is

$$\begin{eqnarray}&&\rho (k)=[\lambda {{\rm{e}}}^{-4{\rm{i}}{k}{a}}+(2{\rm{i}}{k}+\lambda ){(2{\rm{i}}{k}-\widetilde{\lambda })}^{-1}\widetilde{\lambda }]{{\rm{\Gamma }}}^{-1}(k,a;\lambda ,\widetilde{\lambda }){\lambda }^{-1},\end{eqnarray} \tag{ 3 }$$

where

$$\begin{eqnarray}&&{\rm{\Gamma }}(k,a;\lambda ,\widetilde{\lambda })=(2{\rm{i}}{k}-\lambda ){\lambda }^{-1}{{\rm{e}}}^{-2{\rm{i}}{k}{a}}-{(2{\rm{i}}{k}-\widetilde{\lambda })}^{-1}\widetilde{\lambda }{{\rm{e}}}^{2{\rm{i}}{k}{a}}.\end{eqnarray} \tag{ 4 }$$

The reflection probability is $R(E)=| \rho (k){| }^{2}$. It can be shown that the bound-state energies occur at the pure imaginary poles of reflection amplitude, so that the condition for these energies is

$$\begin{eqnarray}&&{\rm{\Gamma }}({\rm{i}}\alpha ,a;\lambda ,\widetilde{\lambda })=0,\end{eqnarray} \tag{ 5 }$$

where $E=-{{\rm{\hslash }}}^{2}{\alpha }^{2}/(2\mu )$. By setting $\alpha =0$ in (5), we have the condition on $\lambda ,\widetilde{\lambda }$ and a such that the system presents a half-bound state. The condition is the simple relation

$$\begin{eqnarray}&&\widetilde{\lambda }=-\displaystyle \frac{\lambda }{1+2a\lambda }\ \ \ \mathrm{or}\ \ \ \displaystyle \frac{1}{\lambda }+\displaystyle \frac{1}{\widetilde{\lambda }}-2a=0,\end{eqnarray} \tag{ 6 }$$

which is equivalent to equation (6.16) of [3].

Let us first consider the reflection amplitude at zero energy which is a real quantity. For graphing purposes we choose units so that the basic unit of length is a and of energy is ${{\rm{\hslash }}}^{2}/(2\mu {a}^{2})$. In other words, both quantities are set to unity. In figure 1 we plot $\rho (0)$ as a function λ when relation (6) is satisfied, so that for each value of λ the potential supports a half-bound state. Since $\rho (k)$ is complex when $k\ne 0$, we plot $R(E)$ as a function of λ for the half-bound cases for two energies, E = 0 and 0.25, in figure 2. Different (small) values of E yield similar curves. It should be noted that $\rho (0)$ is a discontinuous function of the potential strength when it passes through the critical point at which $\alpha =0$ [6].

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From both figures it is clear that when there is a half-bound state, the reflection probability at threshold has a range $0\leqslant R(0)\lt 1$. At $\lambda =-1$, $\widetilde{\lambda }=-1$, and the symmetric potential obtains, and $R(0)=0$. When $\lambda =0$ the potential is zero, and $R(E)$ vanishes. When $\lambda =-1/2$, $\widetilde{\lambda }=\infty$ and the potential is an infinite barrier. Then R(0) has the normal value of unity. From (6) it is clear that the antisymmetric potential, i.e., $\lambda =-\widetilde{\lambda }$, occurs only when $\lambda =0$. For all other λ the potential is asymmetric and $0\lt R(0)\lt 1$. In conclusion it is appropriate to speak of a threshold anomaly when $0\leqslant R(0)\lt 1$, rather than just $R(0)=0$ or $R(0)\ll 1$.