Symmetric vs asymmetric potential

Date: 2015-10-07; view: 444.

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

(4.67)

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

(4.68)

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

(4.70)

and at x = 0

(4.71)

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

(4.73)

with

(4.74)

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

(4.75)

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

(4.76a)

and

(4.77a)

otherwise, if E_b > E_a, we set

(4.76b)

and

(4.77b)

Thus, we have either

(4.78a)

at all finite x, or

(4.78b)

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.

As in (4.2), ψ (x) is the ground state wave function that satisfies

which can also be written in the same form as (1.14)

(4.80)

with

(4.81)

Here, unlike (1.32), V (x) can now also be asymmetric. Taking the difference between ψ (x) times (4.75) and χ (x) times (4.80), we derive

(4.82)

Introduce

in which f (x) satisfies

(4.84)

On account of Figs. (4.82) and (4.83), the same equation can also be written as

(4.85)

Eq. (4.80) will again be solved iteratively by introducing

with ψ_n and its associated energy determined by

(4.87)

and

(4.88)

In terms of f_n (x), we have

(4.89)

On account of (4.88), we also have

(4.90)

and

(4.91)

For definiteness, let us assume that

in Figs. (4.69a) and (4.69b); therefore and , in accordance with (4.76a). Start with, for n = 0,

we can derive {E_n} and {f_n (x)}, with

(4.94)

by using the boundary conditions, either

(4.95)

or

(4.96)

It is straightforward to generalize the Hierarchy theorem to the present case. As in Section 3, in Case (A), the validity of the Hierarchy theorem imposes no condition on the magnitude of . But in Case (B) we assume to be not too large so that (4.91) and the boundary condition f_n (−∞) = 1 is consistent with

for all finite x. From the Hierarchy theorem, we find in Case (A)

E1>E2>E3>

(4.96)

and

1 f1(x) f2(x) f3(x) ,

(4.97)

while in Case (B)

E1>E3>E5>

(4.98)

E2<E4<E6<

(4.99)

1 f1(x) f3(x) f5(x)

(4.100)

and

1 f2(x) f4(x) f6(x) .

(4.101)

A soluble model of an asymmetric square-well potential is given in Appendix A to illustrate these properties.

5. The N-dimensional problem

The N-dimensional case will be discussed in this section. We begin with the electrostatic analog introduced in Section 1. Suppose that the (n − 1)th iterative solution f_{n − 1} (q) is already known. The nth order charge density σ_n (q) is

(5.1)

in accordance with Figs. (1.23) and (1.24). Likewise, from Figs. (1.26) and (1.29) the dielectric constant κ of the medium is related to the trial function (q) by

κ(q)= 2(q)

(5.2)

and the nth order energy shift is determined by

(5.3)

In the following we assume the range of w (q) to be finite, with

and

0 w(q) Wmax.

(5.5)

Introduce

(5.6)

where δ (w (q) − W) is Dirac's δ-function, W is a constant parameter and the integrations in Figs. (5.3) and (5.6) are over all q-space. Similarly, for any function F (q), we define

(5.7)

In the N-dimensional case, the generalization of [F], introduced by (3.15), is

(5.8)

In terms of F^av (W), (5.8) can also be written as

(5.9)

Thus from Figs. (5.1) and (5.3) we have

(5.10)

the n-dimensional extension of (3.14).

Following Figs. (1.27) and (1.28), the nth order electric field is and the displacement field is

(5.11)

The corresponding Maxwell equation is

·Dn=σn.

(5.12)

Eqs. Figs. (5.11) and (5.12) determine f_n except for an additive constant, which can be chosen by requiring

(5.13)

Therefore,

fn(q) 1.

(5.14)

As in the one-dimensional case discussed in Section 3, (5.10) gives the same condition of fine energy tuning at each order of iteration. It is this condition that leads to convergent iterative solutions derived in Section 3. We now conjecture that

(5.15)

and

(5.16)

also hold in higher dimensions. Although we are not able to establish this conjecture, in the following we present the proofs of the N-dimensional generalizations of some of the lemmas proved in Section 3.

Lemma 1

For any pair f_m(q) and f_l(q) if at all W within the range (5.5),

(5.17)

and

(5.18)

Proof

For any function , define

(5.19)

Thus for any function F (q), we have

[F(q)]= Fav(W) ;

(5.20)

therefore,

(5.21)

and

(5.22)

By setting the subscript n in (5.10) to be m + 1, we obtain

(5.23)

Also by definition (5.19),

(5.24)

The difference of Figs. (5.23) and (5.24) gives

(5.25)

From (5.10) and setting the subscript n to be l + 1, we have

(5.26)

Regard and as two constant parameters. Multiply (5.25) by , (5.26) by and take their difference. The result is

(5.27)

analogous to (4.43).

(i) If , then for

(5.28)

Thus, the function inside the bracket in (5.21) is positive, being the product of two negative factors, and . Also, when , these two factors both reverse their signs. Consequently (5.17) holds.

(ii) If , we see that for , (5.28) reverses its sign, and therefore the function inside the bracket in (5.27) is now negative. The same negative sign can be readily established for . Consequently, (5.18) holds and Lemma 1 is established.

Lemma 2

Identical to Lemma 2 of Section 3.

In order to establish the N-dimensional generalization of Lemma 3 of Section 3, we define

(5.29)

Because of (5.3), Q_n (W) is also given by

(5.30)

We may picture that the entire q-space is divided into two regions

(5.31)

and

(5.32)

with Q_n (W) the total charge in I, which is also the negative of the total charge in II. By using Figs. (5.1) and (5.7), we see that

(5.33)

Lemma 3

For any pair f_m (q) and f_l (q) if at all W within the range (5.5)

(5.34)

(5.35)

Proof

Note that Figs. (5.34) and (5.35) are very similar to Figs. (3.56) and (3.57). As in (3.60), define

(5.36)

From (5.33), we have

(5.37)

and

(5.38)

Therefore,

(5.39)

where

(5.40)

Furthermore,

(5.41)

where

(5.42)

analogous to Figs. (3.61), (3.62), (3.63) and (3.64).

According to (5.30), at W = 0

and according to (5.29), at W = W_max

From (5.37), we see that the derivative is positive when , zero at , and negative when . Likewise, from (5.38), is positive when , zero at and negative when . Their ratio determines .

(i) If , from Lemma 1, we have

(5.45)

and therefore, on account of (5.42)

(5.46)

At W = 0,

(5.47)

As W increases, so does r (W). At , r (W) has a discontinuity, with

(5.48)

and

(5.49)

As W increases from , r (W) continues to increase, with

(5.50)

and

(5.51)

It is convenient to divide the range 0 < W < W_max into three regions:

(5.52)

(5.53)

(5.54)

Assuming , we shall show separately in these three regions.

In region B, Q_{l + 1} is decreasing, but Q_{m + 1} is increasing. Clearly,

(5.55)

In region A, , r (W) is positive according to Figs. (5.47) and (5.48) and is always >0 from (5.46). Therefore from (5.41),

(5.56)

In region C, , but r (W) and are both positive. Hence,

(5.57)

Within each region, η = Q_{m + 1} (W) and ξ = Q_{l + 1} (W) are both monotonic in W; therefore η is a single-valued function of ξ and we can apply Lemma 2 of Section 3.

In region A, at W = 0 both Q_{m + 1} (0) and Q_{l + 1} (0) are 0 according to (5.43), but their ratio is given by

(5.58)

Therefore

(5.59)

Furthermore, from (5.56), . It follows from Lemma 2 of Section 3, the ratio η/ξ is an increasing function of ξ. Since

(5.60)

we also have

(5.61)

In region C, at W = W_max, both Q_{m + 1} (W_max) and Q_{l + 1} (W_max) are 0 according to (5.44). Their ratio is

(5.62)

which gives at W = W_max

(5.63)

As W decreases from W_max to in region C, since , we have

(5.64)

Furthermore, from (5.57), in region C. It follows from Lemma 2 of Section 3, the ratio η/ξ is a decreasing function of ξ, which together with (5.64) lead to

(5.65)

Thus, we prove Case (i) of Lemma 3. Case (ii) of Lemma 3 follows from Case (i) by an exchange of the subscripts m and l. Lemma 3 is then proved.

So far, the above lk and lk are almost identical copies of lk and lk of Section 3, but now applicable to the N-dimensional problem. Difficulty arises when we try to generalize Lemma 4 of Section 3.

It is convenient to transform the Cartesian coordinates q₁, q₂, … , q_N to a new set of orthogonal coordinates:

with

w· βi=0

(5.67)

and

(5.68)

where i or j = 1, 2, … , N − 1. Introducing

(5.69)

(5.70)

In terms of the new coordinates, the components of D_n are

(5.71)

Its divergence is

(5.72)

Combining (5.12) with (5.30), we have

(5.73)

therefore,

(5.74)

in which the integration is along the surface

From Figs. (5.11) and (5.71), it follows that

(5.76)

In terms of curvilinear coordinates, (5.7) can be written as

(5.77)

Substituting (5.76) into (5.74), we find

(5.78)

Because , (5.78) can also be written as

(5.79)

Here comes the difficulty. While the above Lemma 3 transfers relations between to those between Q_{m + 1}/Q_{l + 1}, the latter is

(5.80)

which is quite different from . This particular generalization of the lemmas in higher dimensions fails to establish the Hierarchy Theorem.

For the one-dimensional case discussed in Section 3, we have w′ < 0 and x 0; consequently (5.80) is . Therefore, Lemma 4 of Section 3 can also be established by using (5.80), and the proof of the Hierarchy Theorem can be completed.