• Source: Bound state in the continuum
    • A bound state in the continuum (BIC) is an eigenstate of some particular quantum system with the following properties:

      Energy lies in the continuous spectrum of propagating modes of the surrounding space;
      The state does not interact with any of the states of the continuum (it cannot emit and cannot be excited by any wave that came from the infinity);
      Energy is real and Q factor is infinite, if there is no absorption in the system.
      BICs are observed in electronic, photonic, acoustic systems, and are a general phenomenon exhibited by systems in which wave physics applies.
      Bound states in the forbidden zone, where there are no finite solutions at infinity, are widely known (atoms, quantum dots, defects in semiconductors). For solutions in a continuum that are associated with this continuum, resonant states are known, which decay (lose energy) over time. They can be excited, for example, by an incident wave with the same energy. The bound states in the continuum have real energy eigenvalues and therefore do not interact with the states of the continuous spectrum and cannot decay.


      Classification of BICs by mechanism of occurrence


      Source:


      = BICs arising when solving the inverse problem

      =


      Wigner-von Neumann's BIC (Potential engineering)


      The wave function of one of the continuum states is modified to be normalizable and the corresponding potential is selected for it.


      Hopping rate engineering


      In the tight binding approximation, the jump rates are modified so that the state becomes localized


      Boundary shape engineering


      Sources for BICs of different types, e.g. Fabry-Perot type are replaced by scatterers so as to create BIC of the same type.


      = BICs arising due to parameter tuning

      =


      Fabry-Perot BICs


      For resonant structures, the reflection coefficient near resonance can reach unity. Two such structures can be arranged in such a way that they radiate in antiphase and compensate each other.


      Friedrich-Wintgen BICs


      Two modes of the same symmetry of one and the same structure approach each other when the parameters of the structure are changed, and at some point an anti-crossing occurs. In this case, BIC is formed on one of the branches, since the modes as if compensate each other, being in antiphase and radiating into the same radiation channel.


      Single-resonance parametric BICs


      Occur when a single mode can be represented as a sum of contributions, each of which varies with the structure parameters. At some point, destructive interference of all contributions occurs.


      = Symmetry-protected BICs

      =
      Arise when the symmetry of the eigenstate differs from any of the possible symmetries of propagating modes in the continuum.


      = Separable BICs

      =
      Arise when the eigenvalue problem is solved by the Separation of Variables Method, and the wave function is represented, for example, as



      ψ
      =

      ψ

      1


      (
      x
      )

      ψ

      2


      (
      y
      )


      {\displaystyle \psi =\psi _{1}(x)\psi _{2}(y)}

      , where both multipliers correspond to localized states, with the total energy lying 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:








      h

      2



      8

      π

      2


      m



      Δ
      ψ
      +
      (
      V
      (
      r
      )

      E
      )
      ψ
      =
      0
      ,


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


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




      Δ
      =





      2





      x

      2





      +





      2





      y

      2





      +





      2





      z

      2





      =





      2





      r

      2





      +


      2
      r







      r



      ,


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


      For



      E
      =
      0


      {\displaystyle E=0}

      to be the eigenvalue for the spherically symmetric wave function



      ψ
      =
      ψ
      (
      r
      )


      {\displaystyle \psi =\psi (r)}

      , the potential must be




      V
      =



      h

      2



      8

      π

      2


      m




      (




      ψ





      ψ


      +



      2

      ψ






      r
      ψ




      )



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

      .
      We obtain the specific values



      ψ


      {\displaystyle \psi }

      and



      V


      {\displaystyle V}

      for which the BIC will be observed.


      = First case

      =

      Let us consider the function



      ψ
      =

      r

      α



      sin


      r

      β





      {\displaystyle \psi =r^{\alpha }{\sin r^{\beta }}}

      . While the integral






      0





      4
      π

      r

      2



      |

      ψ
      (
      r
      )


      |


      2



      d

      r
      =



      0





      4
      π

      r

      2
      α
      +
      2




      sin

      2




      r

      β




      d

      r


      {\displaystyle \int _{0}^{\infty }4\pi r^{2}|\psi (r)|^{2}{\text{d}}r=\int _{0}^{\infty }4\pi r^{2\alpha +2}{\sin ^{2}r^{\beta }}{\text{d}}r}

      must be finite, then considering the behavior when



      r
      =
      0


      {\displaystyle r=0}

      , we get that



      2
      α
      +
      β
      +
      2
      >

      1


      {\displaystyle 2\alpha +\beta +2>-1}

      , then considering the behavior when



      r
      =



      {\displaystyle r=\infty }

      , we get



      2
      α
      +
      2
      <

      1


      {\displaystyle 2\alpha +2<-1}

      . The regularity



      V
      (
      r
      )


      {\displaystyle V(r)}

      in



      r

      0


      {\displaystyle r\neq 0}

      requires



      2
      α
      +
      β
      +
      1
      =
      0


      {\displaystyle 2\alpha +\beta +1=0}

      . Finally, we get



      α
      =

      2
      ,

      β
      =
      3


      {\displaystyle \alpha =-2,\ \beta =3}

      .
      Assuming



      ψ
      (
      r
      )
      =



      sin


      r

      3




      r

      2






      {\displaystyle \psi (r)={\frac {\sin r^{3}}{r^{2}}}}

      , then the potential will be equal to (discarding the irrelevant multiplier





      h

      2




      /


      8

      π

      2


      m



      {\displaystyle {h^{2}}/{8\pi ^{2}m}}

      ):




      V
      (
      r
      )
      =


      2

      r

      2





      9

      r

      4




      {\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



      E
      =
      0


      {\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



      V
      =


      2

      r

      2





      9

      r

      4




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

      is described by the following equation:







      m
      2




      (




      d

      r



      d

      t



      )


      2


      +
      V
      (
      r
      )
      =

      C
      o
      n
      s
      t




      {\displaystyle {{\frac {m}{2}}\left({\frac {{\text{d}}r}{{\text{d}}t}}\right)^{2}+V(r)=\mathrm {Const} }}










      d

      r



      d

      t



      =




      2
      m



      C
      o
      n
      s
      t




      4
      m




      1

      r

      2




      +


      18
      m



      r

      4







      {\displaystyle {{\frac {{\text{d}}r}{{\text{d}}t}}={\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



      r




      {\displaystyle r\rightarrow \infty }

      ,







      d

      r



      d

      t







      {\displaystyle {\frac {{\text{d}}r}{{\text{d}}t}}\rightarrow \infty }

      , so the asymptotic is







      d
      r


      d
      t



      =



      3


      2




      m




      r

      2


      ;









      1
      r


      =



      3


      2




      m




      (


      t

      0



      t

      )

      ,




      {\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),\ \ }





      r
      =



      m


      3


      2



      (


      t

      0



      t

      )






      {\displaystyle r={\frac {\sqrt {m}}{3{\sqrt {2}}\left(t_{0}-t\right)}}}


      that is, for a finite time



      t
      =

      t

      0




      {\displaystyle t=t_{0}}

      the point goes to infinity. The stationary solution



      ψ
      (
      r
      )


      {\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



      ψ
      (
      r
      )


      {\displaystyle \psi (r)}

      at



      r
      =



      {\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




      r

      m
      i
      n




      {\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



      ψ
      =



      sin

      r

      r




      {\displaystyle \psi ={\frac {\sin r}{r}}}

      , then



      V
      =

      1


      {\displaystyle V=-1}

      . These are divergent spherical waves, since the energy



      E
      =
      0


      {\displaystyle E=0}

      is greater than the potential



      V
      =

      1


      {\displaystyle V=-1}

      , the classical kinetic energy remains positive. The wave function belongs to a continuous spectrum, the integral






      0





      4
      π

      r

      2



      |

      ψ
      (
      r
      )


      |


      2



      d

      r
      =
      4
      π



      0






      sin

      2



      r

      d

      r


      {\displaystyle \int _{0}^{\infty }4\pi r^{2}|\psi (r)|^{2}{\text{d}}r=4\pi \int _{0}^{\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:




      ψ
      =



      sin

      r

      r


      f
      (
      r
      )


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


      If the function



      f
      (
      r
      )


      {\displaystyle f(r)}

      is continuous, and at



      r




      {\displaystyle r\rightarrow \infty }

      the asymptotic is




      r

      α


      ,


      α
      <

      1

      /

      2


      {\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):




      V
      =

      1
      +
      2
      ctg

      r




      f




      (
      r
      )


      f
      (
      r
      )



      +




      f





      (
      r
      )


      f
      (
      r
      )



      .


      {\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



      r




      {\displaystyle r\rightarrow \infty }

      tend to -1, we must make the functions



      ctg

      r




      f




      (
      r
      )


      f
      (
      r
      )



      ,





      f





      (
      r
      )


      f
      (
      r
      )





      {\displaystyle \operatorname {ctg} r{\frac {f^{\prime }(r)}{f(r)}},\ {\frac {f^{\prime \prime }(r)}{f(r)}}}

      small and at



      r




      {\displaystyle r\rightarrow \infty }

      tend to zero.
      In the first case, also







      f




      (
      r
      )


      f
      (
      r
      )





      {\displaystyle {\frac {f^{\prime }(r)}{f(r)}}}

      should vanish for



      ctg

      r
      =



      {\displaystyle \operatorname {ctg} r=\infty }

      , namely for



      r
      =
      0
      ,
      π
      ,

      2
      π
      ,

      2
      π
      ,



      {\displaystyle r=0,\pi ,\ 2\pi ,\ 2\pi ,\dots }

      , that is for



      sin

      r
      =
      0


      {\displaystyle \sin r=0}

      . This is the case when



      f
      (
      r
      )
      =



      0


      r



      sin

      2



      r

      d

      r
      =


      r
      2





      1
      4


      sin

      2
      r


      {\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



      f
      (
      r
      )
      =
      [

      A

      2


      +
      (


      r
      2





      1
      4


      sin

      2
      r

      )

      2



      ]


      1




      {\displaystyle f(r)=[A^{2}+({\frac {r}{2}}-{\frac {1}{4}}\sin 2r)^{2}]^{-1}}

      , where



      A


      {\displaystyle A}

      is arbitrary (here



      f
      (
      r
      )


      {\displaystyle f(r)}

      tends to




      r

      α


      ,


      α
      <

      1

      /

      2


      {\displaystyle r^{\alpha },\ \ \alpha <-1/2}

      when



      r




      {\displaystyle r\rightarrow \infty }

      ). Then




      ψ
      =



      sin

      r


      r
      (

      A

      2


      +
      (
      2
      r

      sin

      2
      r

      )

      2


      )



      .


      {\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



      r




      {\displaystyle r\rightarrow \infty }

      the potential tends to -1.
      Furthermore, it turns out that for any



      ε
      >
      0


      {\displaystyle \varepsilon >0}

      one can choose such an A that the potential is between




      1

      ε


      {\displaystyle -1-\varepsilon }

      and




      1
      +
      ε


      {\displaystyle -1+\varepsilon }

      .
      We can see that the potential oscillates with period



      π


      {\displaystyle \pi }

      and the wave function oscillates with period



      2
      π


      {\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.


      Notes




      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.

    Kata Kunci Pencarian: