Asymmetric quartic double-well problem

Date: 2015-10-07; view: 587.

The hierarchy theorem established in the previous section has two restrictions: (i) the limitation of half-space x 0 and (ii) the requirement of a monotonically decreasing perturbative potential w (x). In this section, we shall remove these two restrictions.

Consider the specific example of an asymmetric quadratic double-well potential

(4.1)

with the constant λ > 0. The ground state wave function ψ (x) and energy E satisfy the Schroedinger equation

where , as before. In the following, we shall present our method in two steps: We first construct a trial function (x) of the form

(4.3)

At x = 0, (x) and ′ (x) are both continuous, given by

(0)= +(0)= -(0)

(4.4)

and

′(0)= +′(0)= -′(0)=0,

(4.5)

with prime denoting , as before. As we shall see, for x > 0, the trial function (x) = ⁺ (x) satisfies

(4.6a)

with

(4.7a)

whereas for x < 0, (x) = ⁻ (x) satisfies

(4.6b)

with

(4.7b)

Furthermore, at x = ±∞

Starting separately from ⁺ (x) and ⁻ (x) and applying the hierarchy theorem, as we shall show, we can construct from (x) another trial function

(4.9)

with χ (x) and χ′ (x) both continuous at x = 0, given by

and

In addition, they satisfy the following Schroedinger equations

(4.12)

and

(4.13)

From V (x) given by (4.1) with λ positive, we see that at any x > 0, V (x) > V (−x); therefore, E⁺ > E⁻.

Our second step is to regard χ (x) as a new trial function, which satisfies

with w (x) being a step function,

(4.15)

and

(4.16)

We see that w (x) is now monotonic, with

w′(x) 0

(4.17)

for the entire range of x from −∞ to +∞. The hierarchy theorem can be applied again, and that will lead from χ (x) to ψ (x), as we shall see.