Appendix A. A soluble example

Date: 2015-10-07; view: 662.

In this appendix, we consider a soluble model in which the potential V (x) of (4.67) is

(A.1)

with W² > μ² and

Following (4.68), we introduce two symmetric potentials:

(A.3)

with, for x 0,

(A.4)

and

(A.5)

so that (A.1) can also be written as

(A.6)

Let ψ (x), χ_a (x), and χ_b (x) be, respectively, the ground state wave functions of

and

For |x| > γ, since V (x) = ∞, we have

For |x| < γ, these wave functions are of the form

(A.10)

(A.11)

and

(A.12)

By substituting these solutions to the Schroedinger equations Figs. (A.7), (A.8) and (A.9), we derive

(A.13)

(A.14)

and

(A.15)

The continuity of ψ′/ψ at x = ±α relates

and

with β given by (A.2). In the following, we assume the barrier heights W and

(A.18)

to be much larger than k and p; therefore, the wave function ψ is mostly contained within the two square wells; i.e., kβ and pβ are both near π. We write

(A.19)

and expect and θ to be small. Likewise, introduce

(A.20)

The explicit forms of these angles can be most conveniently derived by recognizing the separate actions of two related small parameters: one proportional to the inverse of the barrier height

(Wβ)-1 1

(A.21)

and the other

denoting the much smaller tunneling coefficient.

To illustrate how these two effects can be separated, let us consider first the determination of θ_b given by (A.20). The continuity of at x = ±α gives

From (A.15), we also have

(A.24)

Although the two small parameters Figs. (A.21) and (A.22) are not independent, their effects can be separated by introducing p_∞ and q_∞ that satisfy

and

(A.26)

Physically, p_∞ and q_∞ are the limiting values of p_b and q_b when the distance 2α between the two wells → ∞, but keeping the shapes of the two wells unchanged. Hence Figs. (A.23) and (A.25). Let

From (A.25), we may expand θ_∞ in terms of successive powers of (Wβ)⁻¹:

(A.28)

which determines both p_∞ and q_∞. By substituting

into (A.23) and using Figs. (A.24), (A.25), (A.26), (A.27) and (A.28), we determine

(A.30)

Likewise, the continuity of at x = ±α gives

with

(A.32)

As in (A.25), we introduce k_∞ and that satisfy

(A.33)

and

(A.34)

Similar to Figs. (A.27) and (A.28), we define

(A.35)

and derive

(A.36)

As in Figs. (A.29) and (A.30), we find to be given by

(A.37)

with

(A.38)

To derive similar expressions for θ and of (A.19), we first note that the transformation

brings Figs. (A.23) and (A.17), provided that we also change

and therefore

Since according to (A.1), the asymmetry of V (x) is due to the term in the positive x region, it is easy to see that

as will also be shown explicitly below. Thus, from (A.29) and through the transformations Figs. (A.39) and (A.40), we derive

where

with ν₁ given by (A.30). Likewise, we note that the transformation

brings Figs. (A.31) and (A.16), provided that we also change

and therefore

(A.45)

Here, we must differentiate three different situations:

(A.46)

and

In Case (i), when

e-2q∞(α-δ) 1,

(A.47)

from (A.37) and through the transformations given by Figs. (A.44) and (A.45), we find

(A.48)

where

(A.49)

with given by (A.38). According to Figs. (A.13) and (A.19), we have

(A.50)

which leads to

(A.51)

Since in accordance from Figs. (A.28) and (A.36), we find

(A.52)

and

(A.53)

Thus, the left side of (A.51) is dominated by its first term, μ²β². Since θ₁ and are exponentially small, we can neglect in (A.51). In addition, because θ_∞ and are much smaller than 2π, (A.51) can be reduced to

(A.54)

which gives the dependence of δ on μ². It is important to note that an exponentially small μ² can produce a finite δ. For δ < α, at x = δ we have, in accordance with (A.10)

which gives the minimum of ψ (x). The wave function ψ (x) has two maxima, one for each potential well.

In Case (ii), α = δ and (A.16) gives cot kβ = 0, and takes on the critical value with

(A.56)

In Case (iii), , and ψ (x) has only one maximum.

As in Figs. (4.73) and (4.75) we introduce χ (x) through

(A.57)

so that

(A.58)

in which, same as Figs. (4.76a) and (4.77a),

(A.59)

and

(A.60)

with E_a and E_b given by Figs. (A.14) and (A.15). Since according to Figs. (A.4) and (A.5), V_a (x) V_b (x), we have

therefore,

(A.62)

Write the Schroedinger equation (A.7) in the form (4.80):

(A.63)

with

(A.64)

As in (4.82), we have

(A.65)

In all subsequent equations, we restrict the x-axis to

|x| γ,

(A.66)

and set ψ (x), χ (x) positive. Define

We have, as in Figs. (4.84) and (4.85),

(A.68)

or, on account of (A.65), the equivalent form

(A.69)

The derivation of f (x) is given by

where

(A.71)

or equivalently,

(A.72)

In order to satisfy (A.65) and by using (A.59), we see that

(A.73)

Thus, Figs. (A.71), (A.72) and (A.73) give

(A.74)

The positivity of h (x) gives

(A.75)