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(2.19)

(2.20)

(2.21)

etc. These solutions give the convenient normalization convention at q = 0,

S(0)=0

and

e-S(0)=1.

(2.22)

Remarks:

(i) As an example, consider an N-dimensional harmonic oscillator with

(2.23)

From (2.2), one sees that the Hamilton–Jacobi equation (2.9) is for a particle moving in a potential given by

(2.24)

Thus, for any point q ≠ 0 the classical trajectory of interest is simply a straight line connecting the origin and the specific point, with the action

(2.25)

The corresponding energy is, in accordance with (2.10),

(2.26)

By using (2.8), one can readily show that E₁ = E₂ =   = 0 and S₁ = S₂ =   = 0. The result is the well-known exact answer with the ground state wave function for the Schroedinger equation (2.4) given by

(2.27)

and the corresponding energy

(2.28)

(ii) From this example, it is clear that the above expression Figs. (2.6), (2.7) and (2.8) is not the well-known WKB method. The new formalism uses −v (q) as the potential for the Hamilton–Jacobi equation, and its “classical” trajectory carries a 0+ energy; consequently, unlike the WKB method, there is no turning point along the classical trajectory, and the formalism is applicable to arbitrary dimensions.

2.2. Trial function for the quantum double-well potential

To illustrate how to construct a trial function, consider the quartic potential in one dimension with degenerate minima:

(2.29)

An alternative form of the same problem can be obtained by setting so that the Hamiltonian becomes

(2.30)

where

(2.31)

This shows that the dimensionless (small) expansion parameter is related to ; as it turns out, the relevant parameter is its square. In the following, we shall take a = 1 so that the expansion parameter is 1/g; in the literature [5], [6], [7], [8], [9], [10], [11], [12], [13] and [14] one often finds the assumption 2ga = 1 (placing the second minimum of the potential at q = 1/g) so that reduces to g and the anharmonic potential appears as (1/2)q² (1−gq)². Then g appears with positive powers instead of negative, but the coefficients of the power series are the same as with our form of the potential, apart from the overall factor 2ga.

For the above potential (2.29), the Schroedinger equation (2.4) is (with a = 1)

(2.32)

where, as before, ψ (x) = e^−gS (x) is the ground state wave function and E its energy. Using the expansions Figs. (2.6) and (2.7) and following the steps Figs. (2.8) and (2.10), and Figs. (2.15), (2.16), (2.17), (2.18), (2.19), (2.20) and (2.21), we find the well-known perturbative series

(2.33)

and

(2.34)

Both expansions S = S₀ + g⁻¹S₁ + g⁻²S₂ +  and E = gE₀ + E₁ + g⁻¹E₂ +  are divergent, furthermore, at x = −1 and for n   1, each S_n (x) is infinite. The reflection x → −x gives a corresponding asymptotic expansion S_n (x) → S_n (−x), in which each S_n (−x) is regular at x = −1, but singular at x = +1. We note that for g large, the first few terms of the perturbative series (with (2.33) for x positive and the corresponding expansion S_n (x) → S_n (−x) for x negative) give a fairly good description of the true wave function ψ (x) whenever ψ (x) is large (i.e., for x near ±1). However, for x near zero, when ψ (x) is exponentially small, the perturbative series becomes totally unreliable. This suggests the use of first few terms of the perturbative series for regions whenever ψ (x) is expected to be large. In regions where ψ (x) is exponentially small, simple interpolations by hand may already be adequate for a trial function, as we shall see. Since the quartic potential (2.29) is even in x, so is the ground state wave function; likewise, we require the trial function  (x) also to satisfy  (x) =   (−x). At x = 0, we require

(2.35)

To construct  (x), we start with the first two functions S₀ (x) and S₁ (x) in (2.33). Introduce, for x   0,

(2.36)

and

(2.37)

In order to satisfy (2.35), we define

(2.38)

Thus, by construct ′ (0) = 0,  (x) is continuous everywhere, for x from −∞ to ∞, and so is its derivative.

By differentiating ₊ (x) and  (x), we see that they satisfy

(T+V+u) +=g +

(2.39)

and

(T+V+w) =g ,

(2.40)

where

(2.41)

and

w(x)=w(-x)

(2.42)

with, for x   0

(2.43)

where

(2.44)

Note that for g > 1, gˆ (x) is positive, and has a discontinuity at x = 1. Furthermore, for x positive both u (x) and gˆ (x) are decreasing functions of x. Therefore, w (x) also satisfies for x > 0,

w′(x)<0,

(2.45)

a property that is very useful in our proof of convergence which will be discussed in the next section.

3. Hierarchy theorem and its generalization

In this section, we restrict our discussions to a one-dimensional problem, in which the potential V (x) is an even function of x, as in the example given in Section 2.2. The Schroedinger equation (1.9) becomes

(3.1)

with ψ (x) as its ground state wave function, E the ground state energy, and ′ denoting , as before. For the one-dimensional problem, the trial function  (x) satisfies

(3.2)

as in (1.11); therefore (3.1) can be written as

(3.3)

in which

U(x)=V(x)+w(x)

(3.4)

and

(3.5)

as before. Throughout this section, we assume

(3.6)

hence, we need only to consider

x 0.

(3.7)

Furthermore, as in the example of the symmetric quartic double-well potential given in Section 2.2, we assume w (x) to satisfy

(3.8)

and

w(∞)=0.

(3.9)

Therefore, w (x) is positive for x positive. Otherwise, the shape of w (x) can be arbitrary. The Schroedinger equation (3.1) will be solved through the iterative steps Figs. (1.34), (1.35), (1.36), (1.37), (1.38), (1.39), (1.40), (1.41), (1.42) and (1.43), using the sequences

(3.10)

for the energy difference , and the sequence

f1(x),f2(x),…,fn(x),…

(3.11)

for the ratio f (x) = ψ (x)/  (x) with, for n = 0,

f0(x)=1.

(3.12)

In this section, we differentiate two sets of sequences, labeled A and B, satisfying different boundary conditions:

or

Thus, in accordance with Figs. (1.42) and (1.43), we have in Case (A)

(3.13A)

whereas in Case (B)

(3.13B)

In both cases, is determined by the corresponding f_n − 1 (x) through Figs. (1.38) and (1.40); i.e.,

(3.14)

in which [F] of any function F (x) is defined to be

(3.15)

Eqs. Figs. (1.35) and (1.37) give

(3.16)

Likewise, Figs. (1.38) and (1.39) lead to

(3.17)

which, on account of (1.40), is identical to

(3.18)

These two expressions of D_n (x) are valid for both Cases (A) and (B). Let x_n be defined by

(3.19)

Since w′ (x) < 0, (3.19) has one and only one solution, with negative for x > x_n and positive for x < x_n. Thus, if

fn-1(x)>0

(3.20)

for all x > 0, we have from Figs. (3.17) and (3.18)

Dn(x)>0

(3.21)

and therefore, on account of (3.16),

(3.22)

In terms of the electrostatic analog introduced in Section 1, through Figs. (1.26), (1.27), (1.28) and (1.29), one can form a simple physical picture of these expressions. Represent D_n (x) by the standard flux of lines of force. Because the dielectric constant κ (x) =  ² (x) is zero at x = ∞, so is the displacement field. Hence, D_n (∞) = 0; therefore each line of force must terminate at a finite point. Since the electric charge density is , the total electric charge to the right of x is

It must also be the negative of the flux D_n (x) passing through the same point x: i.e.,

which gives (3.17). In the whole range from x = 0 to ∞, the total electric charge is zero; therefore, we have

Furthermore, at any point x > 0, the total charge from the origin to the point x is

which must also be the negative of the above Q_n (x), and therefore the same as D_n (x); that leads to (3.18). From (3.19) and w′ (x) < 0, one sees that the charge distribution σ_n (x) is negative for x > x_n, 0 at x = x_n, and positive for 0 < x < x_n. Correspondingly, moving from x = ∞ towards the left, the displacement field increases from D_n (∞) = 0 to D_n (x) > 0, reaching its maximum at x = x_n, then as x further decreases, so does D_n (x), and finally reaches D_n (0) = 0 at x = 0. In Case (A), because of f_n (∞) = 1, (3.22) leads to

fn(0)>fn(x)>fn(∞)=1.

(3.23A)

Since for n = 0, f₀ (x) = 1, Figs. (3.20) and (3.23A) are valid for n = 1; by induction these expressions also hold for all n; in Case (A), their validity imposes no restriction on the magnitude of w (x). In Case (B) we assume w (x) to be not too large, so that (3.13B) is consistent with

and therefore

fn(0)=1>fn(x)>fn(∞)>0.

(3.23B)

As we shall see, these two boundary conditions (A) and (B) produce sequences that have very different behavior. Yet, they also share a number of common properties.