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In each of regions (I) and (III), we have r (x) > 0 from (3.62) and r′ (x) < 0 from (3.64). Since (f_m/f_l)′ is always negative by the assumption in (3.56), both terms inside the big parentheses of (3.63) are negative; hence the same (3.63) states that d²η/dξ² has the opposite sign from . From the sign of listed in Table 1, we see that

(3.76)

and

(3.77)

Within each region, η = D_m + 1 (x) and ξ = D_l + 1 (x) are both monotonic in x; therefore, η is a single-valued function of ξ and we can apply Lemma 2. In (I), at x = 0, both D_m + 1 (0) and D_l + 1 (0) are 0 according to (3.18), but their ratio is given by

(3.78)

Therefore,

(3.79)

Furthermore, from (3.76), in (I), it follows from Lemma 2, Case (i), the ratio η/ξ is a decreasing function of ξ. Since is >0 in (I), according to (3.59), we have

(3.80)

In (III), at x = ∞, both D_m + 1 (∞) and D_l + 1 (∞) are 0 according to (3.69). Their ratio is

which gives at x = ∞,

(3.81)

As x decreases from x = ∞ to x = x_l + 1, from (3.77) we have in (III). It follows from Lemma 2, Case (ii), η/ξ is an increasing function of ξ. Since is <0, because x > x_l + 1, we have

(3.82)

Thus, we prove Case (i) of Lemma 3. Case (ii) of Lemma 3 follows from Case (i) through the interchange of the subscripts m and l. Lemma 3 is then established. 

Lemma 4

Take any pair f_m (x) and f_l (x)(A) For the boundary condition f_n (∞) = 1,  if at all x > 0,

(3.83A)

therefore, if at all x > 0,

(3.84A)

(B) For the boundary condition f_n (0) = 1, if at all x > 0,

(3.83B)

therefore, if at all x > 0,

(3.84B)

Proof

Define

(3.85)

From (1.35) we see that

(3.86)

and

(3.87)

(A) In this case f_n (∞) = 1 for all n. Thus, at x = ∞, , , and their ratio

(3.88)

At the same point x = ∞, in accordance with (3.17), D_l + 1 (∞) = D_m + 1 (∞) = 0, but their ratio is, on account of w (∞) = 0 and (3.37) of Lemma 1,

(3.89)

in which the last inequality follows from the same assumption, if (f_m/f_l)′ < 0, shared by (3.37) of Lemma 1 and the present (3.83A) that we wish to prove. Thus, from (3.86), at x = ∞

(3.90)

As x decreases from ∞ to 0, increases from f_l + 1 (∞) = 1 to f_l + 1 (0) > 1, in accordance with Figs. (3.22) and (3.23A). On account of (3.56) of Lemma 3, we have (D_m + 1/D_l + 1)′ < 0, which when combined with (3.87) and leads to

(3.91)

Thus, by using Figs. (3.51) and (3.52) of Lemma 2, we have to be an increasing function of ; i.e.,

(3.92)

Because

(3.93)

and , we find

(3.94)

which establishes (3.83A). Through the interchange of the subscripts m and l, we also obtain (3.84A).

Next, we turn to Case (B) with the boundary condition f_n (0) = 1 for all n. Therefore at x = 0,

(3.95)

Furthermore from Figs. (3.16) and (3.18), we also have and D_m + 1 (0) = D_l + 1 (0) = 0, with their ratio given by

(3.96)

From (3.37) of Lemma 1, we see that if (f_m/f_l)′ < 0, then and therefore

(3.97)

(3.98)

Thus,

(3.99)

Analogously to (3.53), define

(3.100)

therefore

(3.101)

From (3.56) of Lemma 3, we know that if (f_m/f_l)′ < 0 then (D_m + 1/D_l + 1)′ < 0, which leads to

(3.102)

From (3.100), we have

(3.103)

and therefore at x = 0, because of (3.99),

L(0)<0.

(3.104)

Combining Figs. (3.102) and (3.104), we derive

(3.105)

Multiplying (3.100) by , we have

(3.106)

Because and L (x) are both negative, it follows then

which gives (3.83B) for Case (B), with the boundary condition f_n (0) = 1. Interchanging the subscripts m and l, (3.83B) becomes (3.84B), and Lemma 4 is established. 

We now turn to the proof of the theorem stated in Figs. (3.24), (3.25), (3.26), (3.27), (3.28), (3.29), (3.30), (3.31) and (3.32).
Proof of the hierarchy theorem. When n = 0, we have

f0(x)=1.

(3.107)

From Figs. (3.20), (3.21) and (3.22), we find for n = 1

(3.108)

and therefore

(f1/f0)′<0.

(3.109)

In Case (A), by using (3.83A) and by setting m = 1 and l = 0, we derive (f₂/f₁)′ < 0; through induction, it follows then (f_n + 1/f_n)′ < 0 for all n. From Lemma 1, we also find for all n. Thus, Figs. (3.24), (3.25), (3.26) and (3.27) are established.

In Case (B), by using Figs. (3.109) and (3.83B), and setting m = 1 and l = 0, we find (f₂/f₁)′ > 0, which in turn leads to (f₃/f₂)′ < 0, … , and Figs. (3.31) and (3.32). Inequalities Figs. (3.28), (3.29) and (3.30) now follow from Figs. (3.37) and (3.38) of Lemma 1. The Hierarchy Theorem is proved.

Assuming that w (0) is finite, we have for any n

(3.110)

Therefore, each of the monotonic sequences

and

converges to a finite limit . By following the discussions in Section 5 of [4], one can show that each of the corresponding monotonic sequences of f_n (x) also converges to a finite limit f (x). The interchange of the limit n → ∞ and the integrations in (3.13A) completes the proof that in Case (A) the limits and f (x) satisfy

(3.111A)

As noted before, the convergence in Case (A) can hold for any large but finite w (x), provided that w′ (x) is negative for x > 0. In Case (B), a large w (x) may yield a negative f_n (x), in violation of (3.23B) . Therefore, the convergence does depend on the smallness of w (x). One has to follow discussions similar to those given in [3] to ensure that the limits and f (x) satisfy

(3.111B)

4. Asymmetric quartic double-well problem

The hierarchy theorem established in the previous section has two restrictions: (i) the limitation of half-space x   0 and (ii) the requirement of a monotonically decreasing perturbative potential w (x). In this section, we shall remove these two restrictions.

Consider the specific example of an asymmetric quadratic double-well potential

(4.1)

with the constant λ > 0. The ground state wave function ψ (x) and energy E satisfy the Schroedinger equation

(T+V(x))ψ(x)=Eψ(x),

(4.2)

where , as before. In the following, we shall present our method in two steps: We first construct a trial function  (x) of the form

(4.3)

At x = 0,  (x) and ′ (x) are both continuous, given by

(0)= +(0)= -(0)

(4.4)

and

′(0)= +′(0)= -′(0)=0,

(4.5)

with prime denoting , as before. As we shall see, for x > 0, the trial function  (x) =  ⁺ (x) satisfies

(4.6a)

with

(4.7a)

whereas for x < 0,  (x) =  ⁻ (x) satisfies

(4.6b)

with

(4.7b)

Furthermore, at x = ±∞

v+(∞)=v-(-∞)=0.

(4.8)

Starting separately from ⁺ (x) and ⁻ (x) and applying the hierarchy theorem, as we shall show, we can construct from  (x) another trial function

(4.9)

with χ (x) and χ′ (x) both continuous at x = 0, given by

χ(0)=χ+(0)=χ-(0)

(4.10)

and

χ′(0)=χ+′(0)=χ-′(0)=0.

(4.11)

In addition, they satisfy the following Schroedinger equations

(4.12)

and

(4.13)

From V (x) given by (4.1) with λ positive, we see that at any x > 0, V (x) > V (−x); therefore, E⁺ > E⁻.