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Introduce

We note that from (A.82),

and from Figs. (A.79) and (A.83),

Consequently, the three small energy parameters in (A.84) are related by

From e^-2q_∞^δ 1 and (A.76) , we see that

in accordance with Figs. (A.79) and (A.84). To understand the role of the parameter λ, we may start with the definition of V_b (x), given by (A.5), keep the parameters β = γ − α and fixed, but let the spacing 2α between the two potential wells approach ∞; in the limit 2α → ∞, we have E_b → E_∞. Thus, λ = E_∞ − E_b is the energy shift due to the tunneling between the two potential wells located at x < −α and x > α in V_b (x).

There is an alternative definition for λ, which may further clarify its physical significance. According to (A.3), V_b (x) is even in x; therefore, its eigenstates are either even or odd in x. In (A.9), χ_b (x) is the ground state of T + V_b (x), and therefore it has to be even in x. The corresponding first excited state χ_od is odd in x; it satisfies

We may define λ by

and regard Figs. (A.85) and (A.86) both as approximate expressions, as we shall see.

Multiplying (A.9) by χ_od (x) and (A.90) by χ_b (x), then taking their difference we derive

From (A.12), we may choose the normalization of χ_b so that

Correspondingly,

with

As in Figs. (A.25) and (A.26), q_od and p_od are determined by

and

At x = 0, we have

Integrating (A.92) from x = 0 to x = γ, we find

From Figs. (A.27), (A.28) and (A.29), we see that

Likewise, we can also show that

Thus, q_od q_b W, and the integral in (A.99) is

Since q_od W, we derive from (A.91)

in agreement with (A.86).

We are now ready to introduce the two-level model. We shall approximate the Hamiltonian T + V (x), T + V_a (x), and T + V_b (x) of Figs. (A.7), (A.8) and (A.9) by the following three 2 × 2 matrices:

and

with ψ, χ_a, and χ_b as their respective ground states which satisfy

The negative sign in the off-diagonal matrix element −λ in Figs. (A.104), (A.105) and (A.106) is chosen to make

simulating the evenness of χ_a (x) and χ_b (x). Likewise, the analog of χ_od is the excited state of h_b, with

and

It is straightforward to verify that

where

When , we have

in agreement with (A.88).

Next, we wish to examine the relation between the two-level model and the soluble square-well example when λ is . Assume, instead of (A.76),

Hence, in the square-well example, (A.83),

and (A.86),

remain valid; on the other hand, Figs. (A.54) and (A.78) now lead to

Thus, the above expressions for E_∞ − E and λ give

Together with (A.116), this shows that the soluble square-well example yields

in agreement with (A.113) given by the two-level model.

In both the square-well problem and the simple two-level model, we can also examine the limit, when . In that case, (A.113) gives

which leads to

in agreement with the exact square-well solution. Furthermore, if we include the first-order correction in O (μ²), (A.115) gives

As we shall discuss, for the exact square-well solution, (A.117) is also valid. Thus, the simple two-level formula (A.113) may serve as an approximate formula for the exact square-well solution over the entire range of .

A.2. Square-well example (Cont.)

We return to the soluble square-well example discussed in Appendix A.1. As before, ψ (x) is the ground state of T + V (x) with energy E, which is determined by the Schroedinger equation (A.7). Likewise, χ (x) is the trial function given by (A.57); i.e., the ground state of with eigenvalue Eˆ₀ = E_a, in accordance with Figs. (A.58), (A.59) and (A.60). From Figs. (A.59) and (A.65), we see that the energy difference

satisfies

where

and

Before we discuss the iterative sequence that approaches , as n → ∞, it may be instructive to verify (A.119) by evaluating the integrals Figs. (A.120) and (A.121) directly. Choose the normalization convention of ψ and χ so that at x = γ

From Figs. (A.10), (A.11) and (A.12) and (A.57) we write

By directly evaluating the integral ∫χ (x)ψ (x) dx, we can readily verify that for γ x 0

and for −γ x 0,

Both relations can also be inferred from the Schroedinger equations Figs. (A.7) and (A.58). Setting x = 0 and taking the sum (A.125) + (A.126), we derive

which, on account of Figs. (A.13), (A.14), (A.15), (A.16), (A.17) and (A.18), leads to the expression for the energy shift , in agreement with (A.119).

Next, we proceed to verify directly that f (x) = ψ (x)/χ (x) satisfies the integral equation (A.68). With the normalization choice (A.122), we find at x = γ, since ψ (γ) = χ (γ) = 0,

which gives the constant in the integral equation. The same equation (A.68) can also be cast in an equivalent form:

where (x|G|z) is the Green’s function that satisfies

For x < z, (x|G|z) is given by

where

is the irregular solution of the same Schroedinger equation (A.58), satisfied by χ (x). That is,

Consequently, over the entire range −γ < x < γ

According to Figs. (A.11), (A.12) and (A.57), we have

where A and B are constants given by

Since in (A.128), there are only single integrations of the products χ (z)ψ (z) and , one can readily verify that f (x) satisfies the integral equation, and therefore also its equivalent form (A.68).

A.3. The iterative sequence

The integral equation (A.68), or its equivalent form (A.128), will now be solved iteratively by introducing

As in Figs. (4.87), (4.88) and (4.89), f_n (x) and its associated energy are determined by

and

When n = 0, we set

Introduce

and

From (A.59) and

we derive

and

For n = 1, we have from Figs. (A.139), (A.140) and (A.141),

and

For small μ², since E_a−E_b and M₀−N₀ are both O (μ²), we find

in agreement with (A.117), given by the simple two-level formula.

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