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A.1. A two-level modelDate: 2015-10-07; view: 491. Before discussing the iterative solutions for f (x) and , it may be useful to first extract some essential features of the soluble square-well example. Let us first concentrate on Case (i) of (A.46), with the parameters α and δ satisfying
We shall also neglect (Wα)−1 or (Wβ)−1, when compared to 1. Thus, from Figs. (A.27) and (A.28), we have
in addition, from Figs. (A.30) and (A.38) we find
From (A.54), we have
On account of Figs. (A.15), (A.20), (A.27) and (A.29),
which, for
gives
On the other hand, from Figs. (A.13), (A.19), (A.42) and (A.43), we see that
Thus, under the condition (A.76), we find
As we shall see, these inequalities can be understood in terms of a simple two-level model. 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 Vb (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 Eb → E∞. Thus, λ = E∞ − Eb is the energy shift due to the tunneling between the two potential wells located at x < −α and x > α in Vb (x). There is an alternative definition for λ, which may further clarify its physical significance. According to (A.3), Vb (x) is even in x; therefore, its eigenstates are either even or odd in x. In (A.9), χb (x) is the ground state of T + Vb (x), and therefore it has to be even in x. The corresponding first excited state χod 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 χod (x) and (A.90) by χb (x), then taking their difference we derive
From (A.12), we may choose the normalization of χb so that
Correspondingly,
with
As in Figs. (A.25) and (A.26), qod and pod 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, qod qb W, and the integral in (A.99) is
Since qod 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 + Va (x), and T + Vb (x) of Figs. (A.7), (A.8) and (A.9) by the following three 2 × 2 matrices:
and
with ψ, χa, and χb 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 χa (x) and χb (x). Likewise, the analog of χod is the excited state of hb, 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 .
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