Perturbation of the one-dimensional time-independent Schrödinger equation with a rectangular potential barrier

Lemma 2.1

Let I = ( a , b ) be an open interval and x 0 ∈ I , where − ∞ ≤ a < b ≤ + ∞ . Assume that r : I → ℂ is a continuous function and the characteristic equation, x 2 + β = 0 , has a pair of complex conjugate roots λ and μ with ℜ ( λ ) = ℜ ( μ ) = 0 and ℑ ( λ ) = − ℑ ( μ ) ≠ 0 . The following statement is true for all ε > 0 : For any twice continuously differentiable function y : I → ℂ satisfying the inequality

for all x ∈ I , there exists a solution y 0 : I → ℂ to the differential equation (2) with α = 0 and β > 0 such that

for all x ∈ I .

Lemma 2.2

Let ω be a positive real constant. Then

for all x ∈ ℝ .

Proof

For any x ∈ ℝ , we define n ( x ) = [ ω x π ] , where [ ω x π ] denotes the largest integer not exceeding the number ω x π . Then, the integer n ( x ) satisfies n ( x ) π ω ≤ x < ( n ( x ) + 1 ) π ω .

First, we assume that x ≥ 0 . Then

for any x ≥ 0 .

Now, let x < 0 . Since sin ( ω t ) is an even function with respect to t, we have

for any x < 0 .□

Theorem 3.1

Let E and V 0 be real numbers with 0 < E < V 0 . Given any ε > 0 , if a piecewise twice continuously differentiable function ψ : ℝ → ℂ satisfies the inequality

for all x ∈ ℝ \ { − c , c } , then there exists a piecewise twice continuously differentiable function ϕ : ℝ → ℂ , which is a solution to the one-dimensional time-independent Schrödinger equation (3) for all x ∈ ℝ \ { − c , c } , such that

Proof

It is to be proved that there exists a piecewise twice continuously differentiable solution ϕ : ℝ \ { − c , c } → ℂ to the Schrödinger equation (3) that satisfies inequality (6).

First, in view of Lemmas 2.1 and 2.2, there exists a twice continuously differentiable function ϕ 1 : ( − ∞ , − c ) → ℂ such that

and

for all x < − c , where we set ω 0 = 2 m E ℏ 2 .

Similarly, there exists a twice continuously differentiable function ϕ 3 : ( c , + ∞ ) → ℂ such that

and

for all x > c .

The solution ϕ 1 to (7) and the solution ϕ 3 to (9) are given by

where A 1 , B 1 , A 3 , and B 3 are some complex numbers.

Also, by [8, Theorem 2.2, Remark 2.3], we have a twice continuously differentiable function ϕ 2 : ( − c , c ) → ℂ such that

for all x ∈ ( − c , c ) . Indeed, ϕ 2 has the form

where A 2 and B 2 are some complex numbers, where we let ω 1 = 2 m ( V 0 − E ) ℏ 2 .

Let us define the function ϕ : ℝ → ℂ by

Then ϕ is a piecewise twice continuously differentiable solution to the one-dimensional time-independent Schrödinger equation (3). Moreover, it follows from (8), (10), and (12) that inequality (6) holds for all x ∈ ℝ \ { − c , c } .□

Theorem 3.2

Let E and V 0 be real numbers with 0 < E < V 0 . Assume that c ω 0 ≠ π 2 + n π for any n ∈ Z . Given ε > 0 , if a piecewise twice continuously differentiable function ψ : ℝ → ℂ satisfies inequality (5) for all x ∈ ℝ \ { − c , c } and if ψ is continuous, then there exists a piecewise twice continuously differentiable function ϕ : ℝ → ℂ , which is a solution to the one-dimensional time-independent Schrödinger equation (3) for all x ∈ ℝ \ { − c , c } , such that ϕ satisfies each of the following boundary conditions

1. ϕ is continuous on ℝ ;

2. ϕ ′ is continuous on ℝ

and such that

where B 1 and B 3 are given in (11), A 2 and B 2 are formulated in (13), and A 2 ∗ and B 2 ∗ are given in (21).

Proof

It is obvious that the proof of Theorem 3.1 is also valid for this theorem. We therefore use the proof of Theorem 3.1 as it is.

Moreover, we refer to (13) and consider the following function ϕ 2 ∗ : ( − c , c ) → ℂ given by

where A 2 ∗ and B 2 ∗ are some complex numbers. We obviously know that ϕ 2 ∗ is a particular solution to the Schrödinger equation (3) for − c < x < c .

Let us define the function ϕ : ℝ → ℂ by

Then ϕ is a piecewise twice continuously differentiable solution to the one-dimensional time-independent Schrödinger equation (3).

By the boundary condition (i), if we impose continuities on ϕ 1 , ϕ 2 ∗ , and ϕ 3 at x = − c and x = c , then it follows from (11) that

Moreover, by (11) and the boundary condition (ii), if we impose continuities on ϕ 1 ' , ϕ 2 ∗ ' , and ϕ 3 ' at x = − c and x = c , then we get

In the aforementioned four equations, if we perform a long calculation to eliminate A 2 ∗ and B 2 ∗ from them, we get the following:

where we set

If we substitute the expressions in (20) for A 1 and A 3 in (16) and (17), then we get

Furthermore, by using the last two equations, we obtain

So far, we have proved that the one-dimensional time-independent Schrödinger equation (3) has the piecewise twice continuously differentiable solution ϕ that satisfies the boundary conditions (i) and (ii).

Next, by the continuity of ψ at x = − c , it follows from (8), (11), and (15) that

Similarly, by the continuity of ψ at x = c , it follows from (10), (11), and (15) that

Finally, when − c < x < c , it follows from (12)–(15) and (21) that

where B 1 and B 3 are given in (11), A 2 and B 2 are formulated in (13), and A 2 ∗ and B 2 ∗ are given in (21).□

Corollary 3.3

Let E and V 0 be real numbers with 0 < E < V 0 . Assume that c ω 0 ≠ π 2 + n π for any n ∈ Z . Given ε > 0 , if a piecewise twice continuously differentiable function ψ : ℝ → ℂ satisfies inequality (5) for all x ∈ ℝ \ { − c , c } and if ψ is even and continuous, then there exists a piecewise twice continuously differentiable function ϕ : ℝ → ℂ , which is a solution to the one-dimensional time-independent Schrödinger equation (3) for all x ∈ ℝ \ { − c , c } , such that ϕ satisfies each of the following conditions

1. ϕ is continuous on ℝ ;

2. ϕ ′ is continuous on ℝ ;

3. ϕ is an even function

and such that

where A 2 is formulated in (13) and A 2 ∗ is given in (21).

Proof

As we see in (11), let ϕ 1 : ( − ∞ , − c ) → ℂ be the twice continuously differentiable function which satisfies the one-dimensional time-independent Schrödinger equation (7) and inequality (8) for all x < − c . Let us define the twice continuously differentiable function ϕ 3 : ( c , ∞ ) → ℂ by ϕ 3 ( x ) = ϕ 1 ( − x ) for all x > c . If we put z = − x , then we have ϕ 3 ( x ) = ϕ 1 ( z ) and

for all x > c and equivalently for all z < − c . In addition, it follows from (7) and the last equality that

for all x > c , i.e., ϕ 3 satisfies the one-dimensional time-independent Schrödinger equation (9) and inequality (10) for any x > c because of the evenness of ψ and by (8).

By (11), (14), and (20), we get

for all x > c . Hence, by (21) and the last equality, and by imposing the evenness on ψ and ϕ 2 ∗ , we obtain that A 1 = A 3 , B 1 = − B 3 , and A 2 ∗ = B 2 ∗ . In addition, we set

As we did in the proof of Theorem 3.2, we define the function ϕ : ℝ → ℂ by using formula (15). Then ϕ is a piecewise twice continuously differentiable solution to the one-dimensional time-independent Schrödinger equation (3) and ϕ satisfies inequality (23) and conditions (i), (ii), and (iii).□