Exact scattering states in one-dimensional Hermitian and non-Hermitian potentials

Ruo-Lin Chai; Qiong-Tao Xie; Xiao-Liang Liu

doi:10.1088/1674-1056/ab928f

1. Introduction

In the past decades, the concept of reflectionless states has attracted extensive interest of research. On the one hand, several reflectionless potentials have been realized in engineered photonic lattices. [1 - 5] In the engineered photonic lattices, the temporal evolution of quantum systems can be mapped into the spatial propagation of light waves. These optical structures provide a controllable platform for studying the properties of the scattering states. By suitably modulating the coupling between the waveguides, reflectionless characteristics and supersymmetric scattering have been observed experimentally. [1, 2]

On the other hand, the concept of the reflectionless states has found wide applications in non-Hermitian systems with parity-time (PT) symmetry. [6 - 11] It has been shown that these non-Hermitian systems exhibit a rich variety of unique scattering properties without Hermitian counterparts. [12 - 15] Many interesting phenomena have been observed in different non-Hermitian systems, such as nonreciprocal light transmission in microcavities, [16] unidirectional reflectionlessness in optical metamaterials, [17] and coherent perfect absorption in coupled resonators. [18]

In the present work, our main aim is to investigate the scattering states in a type of one-dimensional potential and its non-Hermitian extension. We present the analytical solution for the scattering states in terms of Heun functions. It is found that for specially chosen parameter conditions, there exist an infinite number of the exact analytical solutions for the scattering states. In the Hermitian situation, the exact scatting states have zero reflection coefficients. In the non-Hermitian extension, we give an analytical demonstration for the unidirectional reflectionlessness.

2. Exact scattering states in a Hermitian potential

We begin with the following Schrödinger equation (2 m = ℏ = 1)

$$ \begin{eqnarray}-\displaystyle \frac{{{\rm{d}}}^{2}}{{\rm{d}}{x}^{2}}\psi (x)+V(x)\psi (x)=E\psi (x),\end{eqnarray}$$ (1)

where the potential V(x) is given as

$$ \begin{eqnarray}V(x)=\displaystyle \frac{{V}_{1}}{g+\cosh x}+\displaystyle \frac{{V}_{2}}{{(g+\cosh x)}^{2}}+\displaystyle \frac{{V}_{3}\sinh x}{{(g+\cosh x)}^{2}}.\end{eqnarray}$$ (2)

Here V_1,2,3 and g > –1 are real potential parameters. In the case of V₁ = V₃ = g = 0, the resulting potential is the well-known sech-squared potential which allows the exact solutions for the reflectionless states for V₂ = –n(n + 1) with integer n > 0.^[19] This reflectionless sech-squared potential has been realized in engineered photonic systems.^[1,2] In the case of g < 0 and V₁ = V₃ = 0, an exact transmission state was found in a Bose–Einstein condensate which is described by a nonlinear Schrödinger equation.^[20] In the case of V₃ = 0, the resulting potential is a double-well potential under certain parameter conditions, and exact bound states have been presented.^[21–24] Thus this potential (2) can be used to study various double-well systems, and find applications in systems of Bose–Einstein condensates. In this work, we mainly discuss the scattering states.

To solve the Schrödinger equation (1), we make the transformations

$$ \begin{eqnarray}z=\displaystyle \frac{{{\rm{e}}}^{-x}}{{{\rm{e}}}^{-x}-{r}_{1}},\,\,\,\,\psi (x)={z}^{{\lambda }_{1}x}{(z-1)}^{{\lambda }_{2}}{(z-a)}^{{\lambda }_{3}}\phi (z),\end{eqnarray}$$ (3)

with

r_{1} = \sqrt{g^{2} - 1} - g

a = r_{2} / (r_{2} - r_{1})

r_{2} = - \sqrt{g^{2} - 1} - g

λ_{1} = - λ_{2} = - \sqrt{- E}

, and

λ_{3} = 1 / 2 - \sqrt{1 + 4 V_{2} / (g^{2} - 1) + 4 V_{3} / \sqrt{g^{2} - 1}} / 2

. After a simple calculation, it is found that ϕ(z) obeys the Heun equation^[25,26]

$$ \begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}\phi }{{\rm{d}}{z}^{2}}+\left(\displaystyle \frac{\gamma }{z}+\displaystyle \frac{\delta }{z-1}+\displaystyle \frac{\varepsilon }{z-a}\right)\displaystyle \frac{{\rm{d}}\phi }{{\rm{d}}z}+\displaystyle \frac{\alpha \beta z-q}{z(z-1)(z-a)}\phi =0,\end{eqnarray}$$ (4)

where γ = 1 + 2λ₁, δ = 1 + 2 λ₂, α = λ₃ + λ₄, β = 1 + λ₃ – λ₄,

λ_{4} = 1 / 2 - \sqrt{1 + 4 V_{2} / (g^{2} - 1) - 4 V_{3} / \sqrt{g^{2} - 1}} / 2

, ε = α + β – δ – γ + 1 = 2λ₃, and

q = (V_{1} + V_{3}) / \sqrt{g^{2} - 1} + (1 + 2 λ_{1}) λ_{3}

. Mathematically, if

γ = 1 - 2 \sqrt{- E}

is not an integer, the Heun equation will have two linearly independent solutions near z = 0.^[25,26] For the scattering states with E > 0,

γ = 1 - 2 \sqrt{- E}

is complex, thus we obtain the two linearly independent solutions ϕ_1,2 (x) as follows:

$$ \begin{eqnarray}{\phi }_{1}(z) & = & Hl(a,q;\alpha,\beta,\gamma,\delta;z),\end{eqnarray}$$ (5)

$$ \begin{eqnarray}{\phi }_{2}(z) & = & {z}^{1-\gamma }Hl(a,q+(\varepsilon +\delta a)(1-\gamma );\\ & & \alpha -\gamma +1,\beta -\gamma +1,2-\gamma,\delta;z).\end{eqnarray}$$ (6)

Here

H l (a, q; α, β, γ, δ; z) = \sum_{n = 0}^{\infty} h_{n} z^{n}

is an infinite series known as the Heun function. The coefficients h_n are determined by the three-term recurrence relation: R_{n – 1}h_{n – 1} + P_nh_n + Q_{n + 1}h_{n + 1} = 0 with the initial conditions h₀ = 1 and h_–1 = 0. Here R_n = (n + α)(n + β), P_n = – q – n(n – 1 + γ)(1 + a)–n(aδ + ε), and Q_n = an(n – 1 + γ). In terms of the Heun functions, the analytical solutions for this potential are given as

$$ \begin{eqnarray}\begin{array}{lll}{\psi }_{1}(x) & = & {z}^{{\lambda }_{1}}{(z-1)}^{{\lambda }_{2}}{(z-a)}^{{\lambda }_{3}}Hl(a,q;\alpha,\beta,\gamma,\delta;z),\end{array}\end{eqnarray}$$ (7)

$$ \begin{eqnarray}\begin{array}{lll}{\psi }_{2}(x) & = & {z}^{-{\lambda }_{1}}{(z-1)}^{{\lambda }_{2}}{(z-a)}^{{\lambda }_{3}}\\ & & \times Hl(a,q+(\varepsilon +\delta a)(1-\gamma );\\ & & \alpha -\gamma +1,\beta -\gamma +1,2-\gamma,\delta;z).\end{array}\end{eqnarray}$$ (8)

To discuss the scattering states with the energy E = k² > 0, we need to discuss the asymptotic behavior of ψ_1,2(x). As x → ∞, one has z = e^−x/(e^−x – r₁)→ 0, thus ψ₁(x) = (–a)^λ₃/(r₁)^λ₁ e^ikx and ψ₂(x) = (r₁)^λ₁(–a)^λ₃ e^−ikx. We assume that a particle is incident from the left of the potential, thus ψ₁(x) is used to construct the analytical solution for the scattering states

$$ \begin{eqnarray}{\psi }_{{\rm{s}}}(x)=A{z}^{{\lambda }_{1}}{(z-1)}^{{\lambda }_{2}}{(z-a)}^{{\lambda }_{3}}Hl(a,q;\alpha,\beta,\gamma,\delta;z),\end{eqnarray}$$ (9)

where A is a constant to be determined. In order to obtain the transmission and reflection coefficients, we need to know the asymptotic behavior of ψ_s(x) as x → – ∞. However, this cannot be obtained easily. The reason is given as follows. The Heun function Hl(a,q;α,β,γ,δ; z) is only analytical in the range of |z| < min{a,1}. As x → – ∞, we have z = e^−x/(e^−x – r₁)→ 1. Hl(a,q; α,β,γ,δ; 1) is thus not analytical.^[25,26] This problem can be solved by constructing the solutions with good asymptotic behavior as x → – ∞.^[21] To discuss the asymptotic behavior as x → – ∞, we may make different transformations

$$ \begin{eqnarray}y=\displaystyle \frac{-{r}_{1}}{{{\rm{e}}}^{-x}-{r}_{1}},\,\,\,\,\psi (x)={y}^{-{\lambda }_{1}}{(y-1)}^{-{\lambda }_{2}}{(y-{a}_{1})}^{{\lambda }_{3}}\phi (y),\end{eqnarray}$$ (10)

with a₁ = r₁/(r₁ – r₂), and then obtain the Heun equation for ϕ(y) as follows:^[25,26]

$$ \begin{eqnarray}\displaystyle \frac{{{\rm{d}}}^{2}\phi }{{\rm{d}}{y}^{2}}+\left(\displaystyle \frac{{\gamma }_{1}}{y}+\displaystyle \frac{{\delta }_{1}}{y-1}+\displaystyle \frac{{\varepsilon }_{1}}{y-{a}_{1}}\right)\displaystyle \frac{{\rm{d}}\phi }{{\rm{d}}y}+\displaystyle \frac{{\alpha }_{1}{\beta }_{1}y-{q}_{1}}{y(y-1)(y-{a}_{1})}\phi =0,\end{eqnarray}$$ (11)

where γ₁ = 1 – 2λ₁, δ₁ = –2λ₂ + 1, ε₁ = 2λ₃, α₁ = 2λ₃, β₁ = 1, and

q_{1} = - V_{2} / \sqrt{g^{2} - 1} + (1 - 2 λ_{1}) λ_{3}

. For the scattering states with E > 0,

γ_{1} = 1 + 2 \sqrt{- E}

is not an integer, and thus the Heun equation supports two linearly independent solutions ϕ_3,4(x) as follows:

$$ \begin{eqnarray}\begin{array}{lll}{\phi }_{3}(y) & = & Hl({a}_{1},{q}_{1};{\alpha }_{1},{\beta }_{1},{\gamma }_{1},{\delta }_{1};y),\end{array}\end{eqnarray}$$ (12)

$$ \begin{eqnarray}\begin{array}{lll}{\phi }_{4}(y) & = & {y}^{1-{\gamma }_{1}}Hl({a}_{1},{q}_{1}+({\varepsilon }_{1}+{\delta }_{1}{a}_{1})(1-{\gamma }_{1});\\ & & {\alpha }_{1}-{\gamma }_{1}+1,{\beta }_{1}-{\gamma }_{1}+1,2-{\gamma }_{1},{\delta }_{1};y),\end{array}\end{eqnarray}$$ (13)

which leads to the analytical solutions

$$ \begin{eqnarray}\begin{array}{lll}{\psi }_{3}(x) & = & {y}^{-{\lambda }_{1}}{(y-1)}^{-{\lambda }_{2}}{(y-{a}_{1})}^{{\lambda }_{3}}\\ & & \times Hl({a}_{1},{q}_{1};{\alpha }_{1},{\beta }_{1},{\gamma }_{1},{\delta }_{1};y),\end{array}\end{eqnarray}$$ (14)

$$ \begin{eqnarray}\begin{array}{lll}{\psi }_{4}(x) & = & {y}^{{\lambda }_{1}}{(y-1)}^{-{\lambda }_{2}}{(y-{a}_{1})}^{{\lambda }_{3}}\\ & & \times Hl({a}_{1},{q}_{1}+({\varepsilon }_{1}+{\delta }_{1}{a}_{1})(1-{\gamma }_{1});\\ & & {\alpha }_{1}-{\gamma }_{1}+1,{\beta }_{1}-{\gamma }_{1}+1,2-{\gamma }_{1},{\delta }_{1};y).\end{array}\end{eqnarray}$$ (15)

In terms of the analytical solutions ψ_3,4, the analytical solution for the scattering states is given as

$$ \begin{eqnarray}{\psi }_{{\rm{s}}}(x)=A({C}_{1}{\psi }_{3}(x)+{C}_{2}{\psi }_{4}(x)),\end{eqnarray}$$ (16)

where the coefficients C_1,2 are constants to be determined. Clearly, in the common valid regions of ψ_1,3,4, we have

$$ \begin{eqnarray}{\psi }_{1}(x)={C}_{1}{\psi }_{3}(x)+{C}_{2}{\psi }_{4}(x).\end{eqnarray}$$ (17)

This leads to the explicit expressions for the coefficients C_1,2,

$$ \begin{eqnarray}{C}_{1}=\displaystyle \frac{W({\psi }_{1},{\psi }_{4})}{W({\psi }_{3},{\psi }_{4})},\,\,\,\,{C}_{2}=-\displaystyle \frac{W({\psi }_{1},{\psi }_{3})}{W({\psi }_{3},{\psi }_{4})}.\end{eqnarray}$$ (18)

Here W(ϕ,φ) = ϕ(x) dφ(x)/dx – φ(x)dϕ(x)/dx is the Wronskian of two functions ϕ(x) and φ(x). As x → – ∞, we have the asymptotic behavior

$$ \begin{eqnarray}\begin{array}{lll}{\psi }_{3}(x) & = & {(-{r}_{1})}^{-{\lambda }_{1}}{(-1)}^{-{\lambda }_{2}}{(-{a}_{1})}^{{\lambda }_{3}}{{\rm{e}}}^{{\rm{i}}kx},\end{array}\end{eqnarray}$$ (19)

$$ \begin{eqnarray}\begin{array}{lll}{\psi }_{4}(x) & = & {(-{r}_{1})}^{{\lambda }_{1}}{(-1)}^{-{\lambda }_{2}}{(-{a}_{1})}^{{\lambda }_{3}}{{\rm{e}}}^{-{\rm{i}}kx}.\end{array}\end{eqnarray}$$ (20)

If we set A = 1/(C₁(–r₁)^−λ₁(–1)^−λ₂(–a₁)^λ₃), we have

$$ \begin{eqnarray}{\psi }_{{\rm{s}}}(x)=A({C}_{1}{\psi }_{3}(x)+{C}_{2}{\psi }_{4}(x))={{\rm{e}}}^{{\rm{i}}kx}+r{{\rm{e}}}^{-{\rm{i}}kx},\end{eqnarray}$$ (21)

thereby resulting in the reflection and transmission coefficients

$$ \begin{eqnarray}r=\displaystyle \frac{{r}_{1}^{2{\lambda }_{1}}{C}_{2}}{{C}_{1}},\,\,\,\,t=\displaystyle \frac{{a}^{{\lambda }_{3}}}{{a}_{1}^{{\lambda }_{3}}{C}_{1}}.\end{eqnarray}$$ (22)

Therefore, the reflection and transmission coefficients are related to the Wronskians of the three analytical solutions ψ_1,3,4.

Besides the analytical solutions in terms of the Heun functions, there exist exact analytical solutions in terms of elementary functions, as studied in the following. It has been shown that for certain special conditions of

$$ \begin{eqnarray}\alpha,\beta =-N,\,\,\,\,N=0,1,2,\ldots,\end{eqnarray}$$ (23)

$$ \begin{eqnarray}{h}_{N+1}=0,\end{eqnarray}$$ (24)

the Heun function Hl(a,q; α,β,γ,δ; z) can be terminated as a polynomial in z.^[25,26] These terminated conditions allow us to obtain an infinite number of exact analytical solutions. From the condition α = λ₃ + λ₄ = –N in ψ_s(x), we have the parameter relation

$$ \begin{eqnarray}{V}_{2}=-\displaystyle \frac{N}{2}\left(\displaystyle \frac{N}{2}+1\right)(1-{g}^{2})+\displaystyle \frac{{V}_{3}^{2}}{{(N+1)}^{2}},\end{eqnarray}$$ (25)

with N ≥ 1. To give a simple form of the exact solutions, we first use the relation Hl(a,q; α,β,γ,δ; z) = (1 – z)^−αHl(a/(a – 1),(γα a – q)/(a – 1); α,α – δ + 1,γ,α – β + 1; z/(z – 1)) in ψ_s(x),^[26] and then set A = 1. Finally, the exact analytical solutions for the scattering states are given as

$$ \begin{eqnarray}{\psi }_{{\rm{s}}}^{N}(x)=\displaystyle \frac{{{\rm{e}}}^{{\rm{i}}kx}}{\sqrt{{({{\rm{e}}}^{-x}-{r}_{1})}^{N}{({{\rm{e}}}^{-x}-{r}_{2})}^{N}}}\displaystyle \sum _{n=0}^{N}{h}_{n}({E}_{{\rm{ex}}}^{N})\displaystyle \frac{{{\rm{e}}}^{-nx}}{{r}_{1}^{n}},\end{eqnarray}$$ (26)

where the exact energies

E_{ex}^{N}

are determined by h_{N + 1} = 0. We now discuss the asymptotic behavior of

ψ_{s}^{N} (x)

as x → ± ∞. After a simple analysis, we have

ψ_{s}^{N} (x) = e^{i k x}

as x → ∞, and

ψ_{s}^{N} (x) = (h_{N} / r_{1}^{N}) e^{i k x}

as x → – ∞. These results indicate that the exact analytical solutions

ψ_{s}^{N} (x)

correspond to the scattering states with the zero reflection coefficient. In the following, we present the exact results in the case of N = 1 as an example.

To give a simple explicit expression, we consider the case of V 3 = 0. Under the condition h 1 = 0, we obtain the simple parametric dependence of the exact energy

$$ \begin{eqnarray}{E}_{{\rm{ex}}}^{1}={k}^{2}=-\displaystyle \frac{1}{4}\displaystyle \frac{{(2{V}_{1}+g)}^{2}-1}{{g}^{2}-1}.\end{eqnarray}$$ (27)

It is valid for g > – 1 and g ≠ 1. Depending on the parameter value of g, we have different parameter regions for V₁, –(1 + g)/2 < V₁ < (1 – g)/2 for g > 1, and V₁ > (1 – g)/2 or V₁ < –(1 + g)/2 for –1 < g < 1. The corresponding scattering state takes the form

$$ \begin{eqnarray}\begin{array}{lll}{\psi }_{{\rm{s}}}^{1}(x) & = & \displaystyle \frac{{{\rm{e}}}^{{\rm{i}}kx}}{\sqrt{({{\rm{e}}}^{-x}-{r}_{1})({{\rm{e}}}^{-x}-{r}_{2})}}\\ & & \times \left(1+\left(g+\displaystyle \frac{2{V}_{1}}{1-2{\rm{i}}|k|}\right){{\rm{e}}}^{-x}\right).\end{array}\end{eqnarray}$$ (28)

It follows that we have

ψ_{s}^{1} (x) = e^{i k x}

as x → ∞, and

ψ_{s}^{1} (x) = (g + \frac{2 V_{1}}{1 - 2 i | k |}) e^{i k x}

as x → – ∞. We can verify the result of

| (g + \frac{2 V_{1}}{1 - 2 i | k |}) | = 1

easily. Therefore, the solution

ψ_{s}^{1} (x)

represents the reflectionless state.

E = E_{ex}^{1}

Figure 1.Profiles of the potential V(x) for (a) V₁ = –1.8, (b) V₁ = –1.5, (c) V₁ = –1.2 with g = 3, V₂ = 6, and V₃ = 0. In the right column, we plot the corresponding numerical results for the transmission probabilities T as a function of the incident energies E. The circles are for the exact results T = 1 at $E = E_{ex}^{1}$ .

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Figure 2.Profiles of the potential V(x) for (a) V₁ = –0.8, (b) V₁ = 0.6, (c) V₁ = 0.75, and (d) V₁ = 1.6 with g = V₃ = 0 and V₂ = –3/4. The corresponding transmission probabilities as a function of the incident energies E are plotted in the right column. The circles are for the exact results T = 1 at $E = E_{ex}^{1}$ .

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3. Exact scattering states in a complex potential

In this section, we study the case where the potential parameter V 3 is replaced by an imaginary parameter iV 3 . The resulting one-dimensional PT symmetric complex potential is given as

$$ \begin{eqnarray}\begin{array}{lll}V(x) & = & {V}_{{\rm{r}}}(x)+{\rm{i}}{V}_{{\rm{i}}}(x)\\ & = & \displaystyle \frac{{V}_{1}}{g+\cosh x}+\displaystyle \frac{{V}_{2}}{{(g+\cosh x)}^{2}}\\ & & +\displaystyle \frac{{\rm{i}}{V}_{3}\sinh x}{{(g+\cosh x)}^{2}}.\end{array}\end{eqnarray}$$ (29)

Here V₃ are real potential parameter. V_r(x) = V₁/(g + cosh x) + V₂/(g + cosh x)² and V_i(x) = V₃ sinh x/(g + cosh x)² are the real and imaginary parts of the complex potential. In general, the complex potential provides an effective description for open quantum systems. In atomic systems, the complex potential may emerge from the interaction of near resonant light with open atomic systems.^[27] In classical photonic systems, the complex potential may arise due to the complex index of refraction.^[28] The case of g = V₁ = 0 corresponds to the PT symmetric Scarf II potential which is exactly solvable.^[12–14] In the following, we shall show that there exist exact scattering states in this modified complex Scarf II potential.

V_{1} = - 2 k A \sqrt{1 - g^{2}}

$$ \begin{eqnarray}\psi (x)={{\rm{e}}}^{{\rm{i}}kx}{\left(\displaystyle \frac{{{\rm{e}}}^{-x}-{r}_{2}}{{{\rm{e}}}^{-x}-{r}_{1}}\right)}^{A}.\end{eqnarray}$$ (30)

After analyzing asymptotic behavior of ψ(x), we have ψ(x) = e^{i kx} as x → – ∞, and ψ(x) = (r₂/r₁)^A e^{i kx} as x → ∞. It is seen clearly that the reflection probability R = 0. Therefore, the obtained eigenfunction represents the reflectionless state.

On the other hand, with α = 0 and E = k 2, the Heun function in ϕ 2 (z) cannot be terminated as a polynomial. However, it is found that for g = 0 and A = n /2 with integer n, ϕ 2 (z) may be related to a hypergeometric function. For example, for the choice of A = 1/2 and g = 0, we obtain the eigenfunction of the form

$$ \begin{eqnarray}\begin{array}{lll}\psi (x) & = & {{\rm{e}}}^{-{\rm{i}}kx}{\left(\displaystyle \frac{{{\rm{e}}}^{-x}+{\rm{i}}}{{{\rm{e}}}^{-x}-{\rm{i}}}\right)}^{A}\\ & & \times (-1+2F(1,-2{\rm{i}}k,1-2{\rm{i}}k,-{{\rm{ie}}}^{x})).\end{array}\end{eqnarray}$$ (31)

Here F(α,β,γ,z) is the hypergeometric function.^[26] From the asymptotic behavior for the hypergeometric function,^[26] it follows that as x → – ∞, ψ(x) = – e^−ikx, and as x → ∞, ψ(x) = – i e^−ikx + i(2(i)^2ikπ k/sinh(2π k)) e^ikx. In this situation, the reflection probability R = 4π²k²/sinh²(2πk) is not zero. Therefore, our exact results reveal that with the same incident energy E = k², the reflection probability is sensitive to the direction of incidence. If a particle is incident from the left, its reflection probability is zero, and if it is incident from the right, its reflection probability is nonzero. This phenomena is called the unidirectional reflectionlessness which has been observed experimentally.^[17] In Fig. 3, we show the profiles of V_r,i(x) for four different values of k with A = 2 and g = 0. It is observed that as k is increased, V_r(x) is changed from the barrier to the well.

Figure 3.Profiles of the real (solid lines) and imaginary (dashed lines) parts of the generalized PT symmetric Scarf II potential for (a) k = –2, (b) k = –1, (c) k = –0.5, and (d) k = 0.5. The potential parameters are given as V₁ = –2kA, V₂ = –A², and V₃ = –A with A = 2 and g = 0.

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Finally, we discuss the case of g = 1 where the above exact analytical results are invalid. In fact, in the case of g = 1, we have r 1 = r 2 = -1 and thus a \to \infty. The confluent Heun equation is obtained from the Heun equation through a confluence process. [25, 26] The analytical solutions are then given by the confluent Heun functions. This will be discussed in detail in future. In the case of g = 1, we also obtain the exact scattering states. An example is given as follows. Here for brevity, we take V 1 = -2 kA, V 2 = - A 2, and V 3 = - A, where A is a real parameter. We obtain an exact scattering state

$$ \begin{eqnarray}\psi (x)={{\rm{e}}}^{{\rm{i}}kx}{{\rm{e}}}^{{\rm{i}} \frac{2A}{1+{{\rm{e}}}^{x}}}.\end{eqnarray}$$ (32)

This result indicates that as a particle is incident from the left-hand side of this complex potential, its reflection probability is zero. However, for the same parameter conditions, we cannot obtain the exact scattering states incident from the right-hand side of this complex potential.

4. Conclusions

In summary, we have presented an analytical solution for the scattering states in a type of one-dimensional potential in terms of Heun functions. The parameter conditions for the existence of the exact scattering states are derived analytically and confirmed numerically. In the Hermitian situation, it is shown that the obtained exact scattering states correspond to the reflectionless states. Depending on the chosen parameters, the potential may vary from the double well to the single well, the single barrier, and the double barrier. The exact reflectionless states also appear in these structures. In addition, we obtain some exact unidirectionally reflectionless states in a generalized PT-symmetric non-Hermitian Scarf II potential. Our analytical results may find applications in engineered photonic lattices.

Received: Apr. 12, 2020

Accepted: --

Published Online: Apr. 29, 2021

The Author Email: Qiong-Tao Xie (qiongtaoxie@yahoo.com)

DOI:10.1088/1674-1056/ab928f