Bound state in the continuum

Potential and wave function corresponding to zero energy, for the first case of the Wigner-Von Neumann BIC

Let us consider the function ${\displaystyle \psi =r^{\alpha }{\sin r^{\beta }}}$. While the integral ${\displaystyle \int _{0}^{\infty }4\pi r^{2}|\psi (r)|^{2}dr=\int _{0}^{\infty }4\pi r^{2\alpha +2}{\sin ^{2}r^{\beta }}dr}$ must be finite, then considering the behavior when ${\displaystyle r=0}$, we get that ${\displaystyle 2\alpha +\beta +2>-1}$, then considering the behavior when ${\displaystyle r=\infty }$, we get ${\displaystyle 2\alpha +2<-1}$. The regularity ${\displaystyle V(r)}$ in ${\displaystyle r\neq 0}$ requires ${\displaystyle 2\alpha +\beta +1=0}$. Finally, we get ${\displaystyle \alpha =-2,\ \beta =3}$.

Assuming ${\displaystyle \psi (r)={\frac {\sin r^{3}}{r^{2}}}}$, then the potential will be equal to (discarding the irrelevant multiplier ${\displaystyle {h^{2}}/{8\pi ^{2}m}}$):

${\displaystyle V(r)={\frac {2}{r^{2}}}-9r^{4}}$

The eigenfunction and the potential curve are shown in the figure. It seems that the electron will simply roll off the potential and the energy will belong to the solid spectrum, but there is a stationary orbit with ${\displaystyle E=0}$.

In the work [4] is given the following interpretation: this behavior can be understood from an analogy with classical mechanics (considerations belong to Leo Szilard). The motion of a material point in the potential ${\displaystyle V={\frac {2}{r^{2}}}-9r^{4}}$ is described by the following equation:

${\displaystyle {{\frac {m}{2}}\left({\frac {dr}{dt}}\right)^{2}+V(r)=\mathrm {Const} }}$

${\displaystyle {{\frac {dr}{dt}}={\sqrt {{\frac {2}{m}}\mathrm {Const} -{\frac {4}{m}}{\frac {1}{r^{2}}}+{\frac {18}{m}}r^{4}}}}}$

It's easy to see that when ${\displaystyle r\rightarrow \infty }$, ${\displaystyle {\frac {dr}{dt}}\rightarrow \infty }$, so the asymptotic is

${\displaystyle {\frac {dr}{dt}}={\frac {3{\sqrt {2}}}{\sqrt {m}}}r^{2};\ \ \ \ \ \ \ {\frac {1}{r}}={\frac {3{\sqrt {2}}}{\sqrt {m}}}\left(t_{0}-t\right),\ \ }$ ${\displaystyle r={\frac {\sqrt {m}}{3{\sqrt {2}}\left(t_{0}-t\right)}}}$

that is, for a finite time ${\displaystyle t=t_{0}}$ the point goes to infinity. The stationary solution ${\displaystyle \psi (r)}$ means that the point returns from infinity again, that it is as if it is reflected from there and starts oscillating. The fact that ${\displaystyle \psi (r)}$ at ${\displaystyle r=\infty }$ tends to zero follows from the fact that it rolls down a large potential slide and has an enormous speed and therefore a short lifetime. And since the whole oscillatory process (from ${\displaystyle r_{min}}$ to infinity and back) is periodic, it is logical that this quantum mechanical problem has a stationary solution.

(a) Potential and wave function (on an arbitrary scale along the vertical axis) corresponding to zero energy, for the second case of the Wigner von Neumann SSC, ${\displaystyle A=1.5}$ (b) ${\displaystyle A=15}$.

Let's move on to the second example, which can no longer be interpreted from such considerations.

First of all, we take a function ${\displaystyle \psi ={\frac {\sin r}{r}}}$, then ${\displaystyle V=-1}$. These are divergent spherical waves, since the energy ${\displaystyle E=0}$ is greater than the potential ${\displaystyle V=-1}$, the classical kinetic energy remains positive. The wave function belongs to a continuous spectrum, the integral ${\displaystyle \int _{0}^{\infty }4\pi r^{2}|\psi (r)|^{2}dr=4\pi \int _{0}^{\infty }\sin ^{2}rdr}$ diverges. Let's try to change the wave function so that the quadratic integral converges and the potential varies near -1.

Consider the following ansatz:

${\displaystyle \psi ={\frac {\sin r}{r}}f(r)}$

If the function ${\displaystyle f(r)}$ is continuous, and at ${\displaystyle r\rightarrow \infty }$ the asymptotic is ${\displaystyle r^{\alpha },\ \ \alpha <-1/2}$ then the integral is finite. The potential would then be equal (with the corrected arithmetical error in the original article):[5]

${\displaystyle V=-1+2\operatorname {ctg} r{\frac {f^{\prime }(r)}{f(r)}}+{\frac {f^{\prime \prime }(r)}{f(r)}}.}$

In order for the potential to remain near -1, and at ${\displaystyle r\rightarrow \infty }$ tend to -1, we must make the functions ${\displaystyle \operatorname {ctg} r{\frac {f^{\prime }(r)}{f(r)}},\ {\frac {f^{\prime \prime }(r)}{f(r)}}}$ small and at ${\displaystyle r\rightarrow \infty }$ tend to zero.

In the first case, also ${\displaystyle {\frac {f^{\prime }(r)}{f(r)}}}$ should vanish for ${\displaystyle \operatorname {ctg} r=\infty }$, namely for ${\displaystyle r=0,\pi ,\ 2\pi ,\ 2\pi ,\dots }$, that is for ${\displaystyle \sin r=0}$. This is the case when ${\displaystyle f(r)=\int _{0}^{r}\sin ^{2}rdr={\frac {r}{2}}-{\frac {1}{4}}\sin 2r}$ or any other function of this expression.

Let assume ${\displaystyle f(r)=[A^{2}+({\frac {r}{2}}-{\frac {1}{4}}\sin 2r)^{2}]^{-1}}$, where ${\displaystyle A}$ is arbitrary (here ${\displaystyle f(r)}$ tends to ${\displaystyle r^{\alpha },\ \ \alpha <-1/2}$ when ${\displaystyle r\rightarrow \infty }$). Then

${\displaystyle \psi ={\frac {\sin r}{r(A^{2}+(2r-\sin 2r)^{2})}}.}$

The expression for the potential is cumbersome, but the graphs show that for ${\displaystyle r\rightarrow \infty }$ the potential tends to -1. Furthermore, it turns out that for any ${\displaystyle \varepsilon >0}$ one can choose such an A that the potential is between ${\displaystyle -1-\varepsilon }$ and ${\displaystyle -1+\varepsilon }$. We can see that the potential oscillates with period ${\displaystyle \pi }$ and the wave function oscillates with period ${\displaystyle 2\pi }$. It turns out that all reflected waves from the "humps" of such a potential are in phase, and the function is localized in the center, being reflected from the potential by a mechanism similar to the reflection from a Bragg mirror.