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Our second step is to regard χ (x) as a new trial function, which satisfies

with w (x) being a step function,

and

We see that w (x) is now monotonic, with

for the entire range of x from −∞ to +∞. The hierarchy theorem can be applied again, and that will lead from χ (x) to ψ (x), as we shall see.

4.1. Construction of the first trial function

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

with

and

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

etc. Thus, (4.21) leads to

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

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

where

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

and

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

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

where

and

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

and

in which u₊ (x) is given by (4.28),

and

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

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

and

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

and

Furthermore, as x → ±1,

and

Thus, for x 0, we have

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

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,

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),

and

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

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

4.2. Construction of the second trial function

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

and

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

and

where

and

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

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

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

Likewise, f⁻ (x) satisfies (for x 0)

and

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

for x 0, and

for x 0, where satisfies

and

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

for x 0, and

for x 0. For n = 0, we set

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

at all finite and positive x, and

at all finite and negative x. Since

and

with both v_± (0) finite,

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

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

and

with

and the boundary conditions at x = 0,

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).

4.3. Symmetric vs asymmetric potential

As we shall discuss, the general description leading from the trial function χ (x) to the final wave function ψ (x) that satisfies the Schroedinger equation (4.2) may be set in a more general framework. Decompose any potential V (x) into two parts

Next, extend the functions V_a (x) and V_b (x) by defining

Thus, both V_a (x) and V_b (x) are symmetric potential covering the entire x-axis. Let χ_a (x) and χ_b (x) be the ground state wave functions of the Hamiltonians T + V_a and T + V_b:

and

The symmetry (4.68) implies that

and at x = 0

Choose the relative normalization factors of χ_a and χ_b, so that at x = 0

The same trial function (4.9) for the specific quartic potential (4.1) is a special example of

with

In general, from Figs. (4.69a) and (4.69b) we see that χ (x) satisfies

Depending on the relative magnitude of E_a and E_b, we define, in the case of E_a > E_b

and

otherwise, if E_b > E_a, we set

and

Thus, we have either

at all finite x, or

at all finite x. A comparison between Figs. (4.9), (4.10), (4.11), (4.12), (4.13), (4.14), (4.15), (4.16) and (4.17) and (4.73)–(4.77a) shows that w (x) of (4.14) and the above differs only by a constant.

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