4.3. Symmetric vs asymmetric potential

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:

(T+Va(x))χa(x)=Eaχa(x)

(4.69a)

and

(T+Vb(x))χb(x)=Ebχb(x).

(4.69b)

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

χa(0)=χb(0).

(4.72)

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

(T+V(x))ψ(x)=Eψ(x),

(4.79)

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

ψ(x)=χ(x)f(x),

(4.83)

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

ψn(x)=χ(x)fn(x)

(4.86)

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

Ea>Eb

(4.92)

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

f0(x)=1,

(4.93)

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

fn(x)>0

(4.95)

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

E1>E2>E3>

(4.96)

and

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

(4.97)

while in Case (B)

E1>E3>E5>

(4.98)

E2<E4<E6<

(4.99)

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

(4.100)

and

1f2(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.