Hierarchy theorem and its generalization

Date: 2015-10-07; view: 522.

In this section, we restrict our discussions to a one-dimensional problem, in which the potential V (x) is an even function of x, as in the example given in Section 2.2. The Schroedinger equation (1.9) becomes

(3.1)

with ψ (x) as its ground state wave function, E the ground state energy, and ′ denoting , as before. For the one-dimensional problem, the trial function (x) satisfies

(3.2)

as in (1.11); therefore (3.1) can be written as

(3.3)

in which

and

(3.5)

as before. Throughout this section, we assume

(3.6)

hence, we need only to consider

x 0.

(3.7)

Furthermore, as in the example of the symmetric quartic double-well potential given in Section 2.2, we assume w (x) to satisfy

(3.8)

and

Therefore, w (x) is positive for x positive. Otherwise, the shape of w (x) can be arbitrary. The Schroedinger equation (3.1) will be solved through the iterative steps Figs. (1.34), (1.35), (1.36), (1.37), (1.38), (1.39), (1.40), (1.41), (1.42) and (1.43), using the sequences

(3.10)

for the energy difference , and the sequence

for the ratio f (x) = ψ (x)/ (x) with, for n = 0,

In this section, we differentiate two sets of sequences, labeled A and B, satisfying different boundary conditions:

or

Thus, in accordance with Figs. (1.42) and (1.43), we have in Case (A)

(3.13A)

whereas in Case (B)

(3.13B)

In both cases, is determined by the corresponding f_{n − 1} (x) through Figs. (1.38) and (1.40); i.e.,

(3.14)

in which [F] of any function F (x) is defined to be

(3.15)

Eqs. Figs. (1.35) and (1.37) give

(3.16)

Likewise, Figs. (1.38) and (1.39) lead to

(3.17)

which, on account of (1.40), is identical to

(3.18)

These two expressions of D_n (x) are valid for both Cases (A) and (B). Let x_n be defined by

(3.19)

Since w′ (x) < 0, (3.19) has one and only one solution, with negative for x > x_n and positive for x < x_n. Thus, if

for all x > 0, we have from Figs. (3.17) and (3.18)

and therefore, on account of (3.16),

(3.22)

In terms of the electrostatic analog introduced in Section 1, through Figs. (1.26), (1.27), (1.28) and (1.29), one can form a simple physical picture of these expressions. Represent D_n (x) by the standard flux of lines of force. Because the dielectric constant κ (x) = ² (x) is zero at x = ∞, so is the displacement field. Hence, D_n (∞) = 0; therefore each line of force must terminate at a finite point. Since the electric charge density is , the total electric charge to the right of x is

It must also be the negative of the flux D_n (x) passing through the same point x: i.e.,

which gives (3.17). In the whole range from x = 0 to ∞, the total electric charge is zero; therefore, we have

Furthermore, at any point x > 0, the total charge from the origin to the point x is

which must also be the negative of the above Q_n (x), and therefore the same as D_n (x); that leads to (3.18). From (3.19) and w′ (x) < 0, one sees that the charge distribution σ_n (x) is negative for x > x_n, 0 at x = x_n, and positive for 0 < x < x_n. Correspondingly, moving from x = ∞ towards the left, the displacement field increases from D_n (∞) = 0 to D_n (x) > 0, reaching its maximum at x = x_n, then as x further decreases, so does D_n (x), and finally reaches D_n (0) = 0 at x = 0. In Case (A), because of f_n (∞) = 1, (3.22) leads to

Since for n = 0, f₀ (x) = 1, Figs. (3.20) and (3.23A) are valid for n = 1; by induction these expressions also hold for all n; in Case (A), their validity imposes no restriction on the magnitude of w (x). In Case (B) we assume w (x) to be not too large, so that (3.13B) is consistent with

and therefore

As we shall see, these two boundary conditions (A) and (B) produce sequences that have very different behavior. Yet, they also share a number of common properties.

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

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

(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,

to at x = ∞

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

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).

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),

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

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

(3.108)

and therefore

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)