Bound state in the continuum

Wigner-Von Neumann BICs

Bound states in the continuum were first predicted in 1929 by Eugene Wigner and John von Neumann. Two potentials were described, in which BICs appear for two different reasons.

In this work, a spherically symmetric wave function is first chosen so as to be quadratically integrable over the entire space. Then a potential is chosen such that this wave function corresponds to zero energy.

The potential is spherically symmetric, then the wave equation will be written as follows:

${\displaystyle -\frac{h^2}{8{\pi }^{2}m}\mathrm{\Delta }\psi +(V(r)-E)\psi =0,}$

the angle derivatives disappear, since we limit ourselves to considering only spherically symmetric wave functions:

${\displaystyle \mathrm{\Delta }=\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{x}^2}+\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{y}^2}+\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{z}^2}=\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{r}^2}+\frac{2}{r}\frac{\mathrm{\partial }}{\mathrm{\partial }r},}$

For ${\displaystyle E=0}$ to be the eigenvalue for the spherically symmetric wave function ${\displaystyle \psi =\psi (r)}$, the potential must be

${\displaystyle V=\frac{h^2}{8{\pi }^{2}m}\left(\frac{{\psi }^{\mathrm{\prime }\mathrm{\prime }}}{\psi }+\frac{2{\psi }^{\mathrm{\prime }}}{r\psi }\right)}$.

We obtain the specific values ${\displaystyle \psi }$ and ${\displaystyle V}$ for which the BIC will be observed.

First case

Let us consider the function ${\displaystyle \psi =r^{\alpha }\sin r^{\beta }}$. While the integral ${\displaystyle {\int }_0^{\mathrm{\infty }}4\pi r^2|\psi (r){|}^2\text{d}r={\int }_0^{\mathrm{\infty }}4\pi r^{2\alpha +2}{\sin }^2{r}^{\beta }\text{d}r}$ 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 ${\displaystyle r=\mathrm{\infty }}$, we get ${\displaystyle 2\alpha +2<-1}$. The regularity ${\displaystyle V(r)}$ in ${\displaystyle r\ne 0}$ 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 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:

${\displaystyle \frac{m}{2}{\left(\frac{\text{d}r}{\text{d}t}\right)}^2+V(r)=\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}}$

${\displaystyle \frac{\text{d}r}{\text{d}t}=\sqrt{\frac{2}{m}\mathrm{C}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}-\frac{4}{m}\frac{1}{r^2}+\frac{18}{m}{r}^4}}$

It's easy to see that when ${\displaystyle r\to \mathrm{\infty }}$, ${\displaystyle \frac{\text{d}r}{\text{d}t}\to \mathrm{\infty }}$, 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 ${\displaystyle r=\mathrm{\infty }}$ 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.

Second case

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 ${\displaystyle {\int }_0^{\mathrm{\infty }}4\pi r^2|\psi (r){|}^2\text{d}r=4\pi {\int }_0^{\mathrm{\infty }}{\sin }^{2}r\text{d}r}$ 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 ${\displaystyle r\to \mathrm{\infty }}$ 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):

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

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

In the first case, also ${\displaystyle \frac{f^{\mathrm{\prime }}(r)}{f(r)}}$ should vanish for ${\displaystyle \mathrm{ctg}r=\mathrm{\infty }}$, namely for ${\displaystyle r=0,\pi ,2\pi ,2\pi ,\dots }$, that is for ${\displaystyle \sin r=0}$. This is the case when ${\displaystyle f(r)={\int }_0^r{\sin }^{2}r\text{d}r=\frac{r}{2}-\frac{1}{4}\sin 2r}$ 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 ${\displaystyle r\to \mathrm{\infty }}$). 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 ${\displaystyle r\to \mathrm{\infty }}$ the potential tends to -1. Furthermore, it turns out that for any ${\displaystyle \epsilon >0}$ one can choose such an A that the potential is between ${\displaystyle -1-\epsilon }$ and ${\displaystyle -1+\epsilon }$. 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.

Literature

• Hsu, Chia Wei; Zhen, Bo; Stone, A. Douglas; Joannopoulos, John D.; Soljačić, Marin (2016). "Bound states in the continuum". Nature Reviews Materials. 1 (9). Bibcode:2016NatRM...116048H. doi:10.1038/natrevmats.2016.48. hdl:1721.1/108400. S2CID 123778221.
• Koshelev, Kirill; Bogdanov, Andrey; Kivshar, Yuri (2020). "Engineering with Bound States in the Continuum". Optics and Photonics News. 31 (1): 38. Bibcode:2020OptPN..31a..38K. doi:10.1364/OPN.31.1.000038. S2CID 213414202.
• Azzam, Shaimaa I.; Kildishev, Alexander V. (2021). "Photonic Bound States in the Continuum: From Basics to Applications". Advanced Optical Materials. 9. doi:10.1002/adom.202001469. S2CID 228843072.
• Sadreev, Almas F. (2021). "Interference traps waves in an open system: Bound states in the continuum". Reports on Progress in Physics. 84 (5). arXiv:2011.01221. Bibcode:2021RPPh...84e5901S. doi:10.1088/1361-6633/abefb9. PMID 33730696.