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Hierarchy Theorem (A) With the boundary condition f_n (∞) = 1, we have for all n

(3.24)

and

(3.25)

Thus, the sequences and {f_n (x)} are all monotonic, with

(3.26)

and

1<f1(x)<f2(x)<f3(x)<

(3.27)

at all finite and positive x.

(B) With the boundary condition f_n (0) = 1, we have for all odd n = 2m + 1 an ascending sequence

(3.28)

but for all even n = 2m, a descending sequence

(3.29)

furthermore, between any even n = 2m and any odd n = 2l + 1

(3.30)

Likewise, at any x > 0, for any even n = 2m

(3.31)

whereas for any odd n = 2l + 1

(3.32)

Remarks:

1. The validity of Eqs. Figs. (3.24) and (3.25) for the boundary condition f_n (∞) = 1 was established in [4]. The validity of Eqs. Figs. (3.28), (3.29), (3.30), (3.31) and (3.32) for the boundary condition f_n (0) = 1 is the new result of this paper, which we shall establish.

2. As we shall also show, the lowest eigenvalue E of the Hamiltonian T + V is the limit of the sequence {E_n} with

(3.33)

Thus, the boundary condition f_n (∞) = 1 yields a sequence, in accordance with (3.26),

E1>E2>E3> >E,

(3.34)

with each member E_n an upper bound of E, similar to the usual variational iterative sequence.

3. On the other hand, with the boundary condition f_n (0) = 1, while the sequence of its odd members n = 2l + 1 yields a similar one, like (3.34), with

E1>E3>E5> >E,

(3.35)

its even members n = 2m satisfy

E2<E4<E6< <E.

(3.36)

It is unusual to have an iterative sequence of lower bounds of the eigenvalue E. Together, these sequences may be quite efficient to pinpoint the limiting E.

The proof of the above generalized hierarchy theorem depends on several lemmas that are applicable to both boundary conditions: (A) f_n (∞) = 1 and (B) f_n (0) = 1; these lemmas will be established first, and then followed by the proof of the theorem.

Lemma 1

For any pair f_m (x) and f_l (x)

(3.37)

and

(3.38)

Proof

According to (3.14)

(3.39)

Also by definition (3.15),

(3.40)

Their difference gives

(3.41)

From (3.14),

(3.42)

Let x_l+1 be defined by (3.19). Multiplying (3.41) by f_l (x_l+1) and (3.42) by f_m (x_l+1) and taking their difference, we have

(3.43)

in which the unsubscripted x acts as a dummy variable; thus [f_m (x)] means [f_m] and [f_m (x_l+1)] means f_m (x_l+1) · [1], etc.

(i) If (f_m (x)/f_l (x))′ < 0, then for x < x_l + 1

(3.44)

In addition, since w′ (x) < 0 and , we also have for x < x_l+1

(3.45)

Thus, the function inside the square bracket on the right-hand side of (3.43) is positive for x < x_l+1. Also, the inequalities Figs. (3.44) and (3.45) both reverse their signs for x > x_l+1. Consequently, the right-hand side of (3.43) is positive definite, and so is its left side. Therefore, on account of Figs. (3.23A) and (3.23B), (3.37) holds.

(ii) If (f_m (x)/f_l (x))′ > 0, we see that for x < x_l+1, (3.44) reverses its sign but not (3.45). A similar reversal of sign happens for x > x_l+1. Thus, the right-hand side of (3.43) is now negative definite and therefore . Lemma 1 is proved. 

The following lemma was already proved in [4]. For the convenience of the readers, we also include it in this paper. Let

η=η(ξ)

(3.46)

be a single-valued differentiable function of ξ in the range between a and b with

0 a ξ b

(3.47)

and with

η(a) 0.

(3.48)

Lemma 2

(i) The ratio η/ξ is a decreasing function of ξ for a < ξ < b if

(3.49)

and

(3.50)

(ii) The ratio η/ξ is an increasing function of ξ for a < ξ < b if

(3.51)

and

(3.52)

Proof

Define

(3.53)

to be the Legendre transform L (ξ). We have

(3.54)

and

(3.55)

Since (3.49) says that L (a)   0 and (3.50) says that for a < ξ < b, these two conditions imply L (ξ) < 0 for a < ξ < b, which proves (i) in view of (3.55). The proof of (ii) is the same, but with inequalities reversed. 

Lemma 3

For any pair f_m (x)  and f_l (x)

(i) if over all x > 0,

(3.56)

and (ii) if over all x > 0,

(3.57)

Proof

From Figs. (3.17) and (3.18), we have

(3.58)

and

(3.59)

Define

(3.60)

In any local region of x where , we can regard η = η (ξ) through η (x) = η (ξ (x)). Hence, we have

(3.61)

where

(3.62)

and

(3.63)

where

(3.64)

(i) If (f_m/f_l)′ < 0, from Lemma 1, we have

(3.65)

From w′ (x) < 0 and the definition of x_m + 1 and x_l + 1, given by (3.19), we have

xm+1<xl+1,

(3.66)

(3.67)

We note that from Figs. (3.17) and (3.18) D_m + 1 (x) and D_l + 1 (x) are both positive continuous functions of x, varying from at x = 0,

Dm+1(0)=Dl+1(0)=0

(3.68)

to at x = ∞

Dm+1(∞)=Dl+1(∞)=0

(3.69)

with their maxima at x_m + 1 for D_m + 1 (x) and x_l + 1 for D_l + 1 (x), since in accordance with Figs. (3.58), (3.59) and (3.67),

(3.70)

From Figs. (3.64) and (3.65), we see that r′ (x) is always <0. Furthermore, from (3.62), we also find that the function r (x) has a discontinuity at x = x_l + 1. At x = 0, r (0) satisfies

(3.71)

As x increases from 0, r (x) decreases from r (0), through

r(xm+1)=0,

(3.72)

to −∞ at x = x_l + 1−; r (x) then switches to +∞ at x = x_l + 1+, and continues to decrease as x increases from x_l + 1+. At x = ∞, r (x) becomes

(3.73)

It is convenient to divide the positive x-axis into three regions:

(3.74)

Table 1 summarizes the signs of , , r, and r′ in these regions. Assuming (f_m/f_l)′ < 0 we shall show separately the validity of (3.56), (D_m + 1/D_l + 1)′ < 0, in each of these three regions.

Table 1.

The signs of , , , , r (x), and r′ (x) in the three regions defined by (3.74), when

Region					r (x)	r′ (x)
I	>0	>0	>0	>0	>0	< 0
II	<0	>0	<0	>0	<0	<0
III	<0	<0	<0	<0	>0	<0

Since

(3.75)

D_m + 1 (x) is decreasing and D_l + 1 (x) is increasing; it is clear that (3.56) holds in (II).