Construction of the second trial function

Date: 2015-10-07; view: 455.

To construct the second trial function χ (x) introduced in (4.9), we define f^± (x) by

(4.47a)

and

(4.47b)

To retain flexibility it is convenient to impose only the boundary condition (4.11) first, but not (4.10); i.e., at x = 0

but leaving the choice of the overall normalization of χ⁺ (0) and χ⁻ (0) to be decided later. We rewrite the Schroedinger equations Figs. (4.12) and (4.13) in their equivalent forms

(4.49a)

and

(4.49b)

where

(4.50a)

and

(4.50b)

Because at x = 0, ⁺′(0) = ⁻′(0) = 0, in accordance with (4.5), we have, on account of (4.48),

So far, the overall normalization of f⁺ (x) and f⁻ (x) are still free. We may choose

(4.52)

From Figs. (4.6a), (4.47a), (4.49a) and (4.50a), we see that f⁺ (x) satisfies the integral equation (for x 0)

(4.53a)

Furthermore, from Figs. (4.6a) and (4.49a), we also have

(4.54a)

Likewise, f⁻ (x) satisfies (for x 0)

(4.53b)

and

(4.54b)

The function f⁺ (x) and f⁻ (x) will be solved through the iterative process described in Section 1. We introduce the sequences and for n = 1, 2, 3, … , with

(4.55a)

for x 0, and

(4.55b)

for x 0, where satisfies

(4.56a)

and

(4.56b)

Thus, Figs. (4.55a) and (4.55b) can also be written in their equivalent expressions

(4.57a)

for x 0, and

(4.57b)

for x 0. For n = 0, we set

(4.58)

through induction and by using Figs. (4.55a), (4.55b), (4.56a) and (4.56b), we derive all subsequent and . Because v_± (x) satisfies Figs. (4.44a), (4.44b) and (4.46), the Hierarchy theorem proved in Section 3 applies. The boundary conditions f⁺ (∞) = f⁻ (−∞) = 1, given by (4.52), lead to , in agreement with Figs. (4.55a) and (4.55b). According to Figs. (3.24), (3.25), (3.26) and (3.27) of Case (A) of the theorem, we have

(4.59a)

(4.59b)

(4.60a)

at all finite and positive x, and

(4.60b)

at all finite and negative x. Since

(4.61a)

and

(4.61b)

with both v_± (0) finite,

(4.62)

both exist. Furthermore, by using the integral equations Figs. (4.55a) and (4.55b) for and by following the arguments similar to those given in Section 5 of [3], we can show that

(4.63)

also exist. This leads us from the first trial function (x) given by (4.3) to f⁺ (x) and f⁻ (x) which are solutions of

(4.64a)

and

(4.64b)

with

(4.65)

and the boundary conditions at x = 0,

′(0)=f+′(0)=f-′(0)=0.

(4.66)

An additional normalization factor multiplying, say, f⁻ (x) would enable us to construct the second trial function χ (x) that satisfies Figs. (4.9), (4.10), (4.11), (4.12) and (4.13).