Construction of the first trial function

Date: 2015-10-07; view: 438.

We consider first the positive x region. Following Section 2.1, we begin with the usual perturbative power series expansion for

with

gS(x)=gS0(+)+S1(+)+g-1S2(+)+

(4.19)

and

E=gE0(+)+E1(+)+g-1E2(+)+

(4.20)

in which S_n (+) and E_n (+) are g-independent. Substituting Figs. (4.18), (4.19) and (4.20) into the Schroedinger equation (4.2) and equating both sides, we find

(4.21)

(4.22)

etc. Thus, (4.21) leads to

(4.23)

Since the left side of (4.22) vanishes at x = 1, so is the right side; hence, we determine

which leads to

Of course, the power series expansion Figs. (4.19) and (4.20) are both divergent. However, if we retain the first two terms in (4.19), the function

(4.26)

serves as a reasonable approximation of ψ (x) for x > 0, except when x is near zero. By differentiating (+), we find (+) satisfies

(T+V(x)+u+(x)) (+)=gE0(+) (+)

(4.27)

where

(4.28)

In order to construct the trial function (x) that satisfies the boundary condition (4.5), we introduce for x 0,

(4.29)

and

(4.30)

so that ⁺ (x) and its derivative ⁺′ (x) are both continuous at x = 1, and in addition, at x = 0 we have ⁺′ (0) = 0. For x 0, we observe that V (x) is invariant under

(4.31)

The same transformation converts ⁺ (x) for x positive to ⁻ (x) for x negative. Define

(4.32)

where

(4.33)

(4.34)

and

(4.35)

Both ⁻ (x) and its derivative ⁻′ (x) are continuous at x = −1; furthermore, ⁺ (x) and ⁻ (x) also satisfy the continuity condition Figs. (4.4) and (4.5), as well as the Schroedinger equation Figs. (4.6a) and (4.6b), with the perturbative potentials v₊ (x) and v_- (x) given by

(4.36a)

and

(4.36b)

in which u₊ (x) is given by (4.28),

(4.37)

(4.38a)

and

(4.38b)

In order that u₊ (x), be positive for x > 0 and u_- (x), positive for x < 0, we impose

(4.39)

in addition to the earlier condition λ > 0. From Figs. (4.28) and (4.37), we have

(4.40a)

and

(4.40b)

Likewise, from Figs. (4.38a) and (4.38b), we find

(4.41a)

and

(4.41b)

Furthermore, as x → ±1,

(4.42a)

and

(4.42b)

Thus, for x 0, we have

(4.43)

and, together with Figs. (4.36a) and (4.40a),

(4.44a)

for x positive. On the other hand for x 0, is not always positive; e.g., at x = 0,

which is positive for , but at x = −1+,

However, at x = −1, . It is easy to see that the sum can satisfy for x 0,

(4.44b)

To summarize: ⁺ (x) and ⁻ (x) satisfy the Schroedinger equation Figs. (4.6a) and (4.6b), with v_± (x) given by Figs. (4.36a) and (4.36b),

(4.45a)

and

(4.45b)

and the boundary conditions Figs. (4.4) and (4.5). In addition, v_± (x) satisfies

(4.46)

and the monotonicity conditions Figs. (4.7a) and (4.7b).