Supersymmetry and cosmology

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Cosmology now provides unambiguous, quantitative evidence for new particle physics. I discuss the implications of cosmology for supersymmetry and vice versa. Topics include: motivations for supersymmetry; supersymmetry breaking; dark energy; freeze out and WIMPs; neutralino dark matter; cosmologically preferred regions...



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Supersymmetry and cosmology

Jonathan L. Feng,

Department of Physics and Astronomy, University of California, Irvine, CA 92697, United States

Received 23 February 2004;  accepted 29 September 2004.  Available online 18 December 2004.


Cosmology now provides unambiguous, quantitative evidence for new particle physics. I discuss the implications of cosmology for supersymmetry and vice versa. Topics include: motivations for supersymmetry; supersymmetry breaking; dark energy; freeze out and WIMPs; neutralino dark matter; cosmologically preferred regions of minimal supergravity; direct and indirect detection of neutralinos; the DAMA and HEAT signals; inflation and reheating; gravitino dark matter; Big Bang nucleosynthesis; and the cosmic microwave background. I conclude with speculations about the prospects for a microscopic description of the dark universe, stressing the necessity of diverse experiments on both sides of the particle physics/cosmology interface.

Article Outline

1. Introduction 

2. Supersymmetry essentials 

2.1. A new spacetime symmetry 

2.2. Supersymmetry and the weak scale 

2.3. The neutral supersymmetric spectrum 

2.4. R-Parity 

2.5. Supersymmetry breaking and dark energy 

2.6. Minimal supergravity 

2.7. Summary

3. Neutralino cosmology 

3.1. Freeze out and WIMPs 

3.2. Thermal relic density 

3.2.1. Bulk region 

3.2.2. Focus point region 

3.2.3. A funnel region 

3.2.4. Co-annihilation region

3.3. Direct detection 

3.4. Indirect detection 

3.4.1. Positrons 

3.4.2. Photons 

3.4.3. Neutrinos

3.5. Summary

4. Gravitino cosmology 

4.1. Gravitino properties 

4.2. Thermal relic density 

4.3. Production during reheating 

4.4. Production from late decays 

4.5. Detection 

4.5.1. Energy release 

4.5.2. Big Bang nucleosynthesis 

4.5.3. The cosmic microwave background

4.6. Summary

5. Prospects 

5.1. The particle physics/cosmology interface 

5.2. The role of colliders 

5.3. Synthesis 

5.4. Summary



1. Introduction

Not long ago, particle physicists could often be heard bemoaning the lack of unambiguous, quantitative evidence for physics beyond their standard model. Those days are gone. Although the standard model of particle physics remains one of the great triumphs of modern science, it now appears that it fails at even the most basic level—providing a reasonably complete catalog of the building blocks of our universe.

Recent cosmological measurements have pinned down the amount of baryon, matter, and dark energy in the universe [1] and [2]. In units of the critical density, these energy densities are







implying a non-baryonic dark matter component with


where h  0.71 is the normalized Hubble expansion rate. Both the central values and uncertainties were nearly unthinkable even just a few years ago. These measurements are clear and surprisingly precise evidence that the known particles make up only a small fraction of the total energy density of the universe. Cosmology now provides overwhelming evidence for new particle physics.

At the same time, the microscopic properties of dark matter and dark energy are remarkably unconstrained by cosmological and astrophysical observations. Theoretical insights from particle physics are therefore required, both to suggest candidates for dark matter and dark energy and to identify experiments and observations that may confirm or exclude these speculations.

Weak-scale supersymmetry is at present the most well-motivated framework for new particle physics. Its particle physics motivations are numerous and are reviewed in Section 2. More than that, it naturally provides dark matter candidates with approximately the right relic density. This fact provides a strong, fundamental, and completely independent motivation for supersymmetric theories. For these reasons, the implications of supersymmetry for cosmology, and vice versa, merit serious consideration.

Many topics lie at the interface of particle physics and cosmology, and supersymmetry has something to say about nearly every one of them. Regrettably, spacetime constraints preclude detailed discussion of many of these topics. Although the discussion below will touch on a variety of subjects, it will focus on dark matter, where the connections between supersymmetry and cosmology are concrete and rich, the above-mentioned quantitative evidence is especially tantalizing, and the role of experiments is clear and promising.

Weak-scale supersymmetry is briefly reviewed in Section 2 with a focus on aspects most relevant to astrophysics and cosmology. In Sections 3 and 4 the possible roles of neutralinos and gravitinos in the early universe are described. As will be seen, their cosmological and astrophysical implications are very different; together they illustrate the wealth of possibilities in supersymmetric cosmology. I conclude in Section 5 with speculations about the future prospects for a microscopic understanding of the dark universe. 

2. Supersymmetry essentials

2.1. A new spacetime symmetry

Supersymmetry is an extension of the known spacetime symmetries [3]. The spacetime symmetries of rotations, boosts, and translations are generated by angular momentum operators Li, boost operators Ki, and momentum operators Pμ, respectively. The L and K generators form Lorentz symmetry, and all 10 generators together form Poincare symmetry. Supersymmetry is the symmetry that results when these 10 generators are further supplemented by fermionic operators Qα. It emerges naturally in string theory and, in a sense that may be made precise [4], is the maximal possible extension of Poincare symmetry.

If a symmetry exists in nature, acting on a physical state with any generator of the symmetry gives another physical state. For example, acting on an electron with a momentum operator produces another physical state, namely, an electron translated in space or time. Spacetime symmetries leave the quantum numbers of the state invariant—in this example, the initial and final states have the same mass, electric charge, etc.

In an exactly supersymmetric world, then, acting on any physical state with the supersymmetry generator Qα produces another physical state. As with the other spacetime generators, Qα does not change the mass, electric charge, and other quantum numbers of the physical state. In contrast to the Poincare generators, however, a supersymmetric transformation changes bosons to fermions and vice versa. The basic prediction of supersymmetry is, then, that for every known particle there is another particle, its superpartner, with spin differing by 1/2.

One may show that no particle of the standard model is the superpartner of another. Supersymmetry therefore predicts a plethora of superpartners, none of which has been discovered. Mass degenerate superpartners cannot exist—they would have been discovered long ago—and so supersymmetry cannot be an exact symmetry. The only viable supersymmetric theories are therefore those with non-degenerate superpartners. This may be achieved by introducing supersymmetry-breaking contributions to superpartner masses to lift them beyond current search limits. At first sight, this would appear to be a drastic step that considerably detracts from the appeal of supersymmetry. It turns out, however, that the main virtues of supersymmetry are preserved even if such mass terms are introduced. In addition, the possibility of supersymmetric dark matter emerges naturally and beautifully in theories with broken supersymmetry. 

2.2. Supersymmetry and the weak scale

Once supersymmetry is broken, the mass scale for superpartners is unconstrained. There is, however, a strong motivation for this scale to be the weak scale: the gauge hierarchy problem. In the standard model of particle physics, the classical mass of the Higgs boson receives quantum corrections (see Fig. 1). Including quantum corrections from standard model fermions fL and fR, one finds that the physical Higgs boson mass is


where the last term is the leading quantum correction, with λ the Higgs-fermion coupling. Λ is the ultraviolet cutoff of the loop integral, presumably some high scale well above the weak scale. If Λ is of the order of the Planck scale 1019 GeV, the classical Higgs mass and its quantum correction must cancel to an unbelievable 1 part in 1034 to produce the required weak-scale mh. This unnatural fine-tuning is the gauge hierarchy problem.

Fig. 1. Contributions to the Higgs boson mass in the standard model and in supersymmetry.

In the supersymmetric standard model, however, for every quantum correction with standard model fermions fL and fR in the loop, there are corresponding quantum corrections with superpartners and . The physical Higgs mass then becomes


where the terms quadratic in Λ cancel, leaving a term logarithmic in Λ as the leading contribution. In this case, the quantum corrections are reasonable even for very large Λ, and no fine-tuning is required.

In the case of exact supersymmetry, where , even the logarithmically divergent term vanishes. In fact, quantum corrections to masses vanish to all orders in perturbation theory, an example of powerful non-renormalization theorems in supersymmetry. From Eq. (6), however, we see that exact mass degeneracy is not required to solve the gauge hierarchy problem. What is required is that the dimensionless couplings λ of standard model particles and their superpartners are identical, and that the superpartner masses be not too far above the weak scale (or else even the logarithmically divergent term would be large compared to the weak scale, requiring another fine-tuned cancellation). This can be achieved simply by adding supersymmetry-breaking weak-scale masses for superpartners. In fact, other terms, such as some cubic scalar couplings, may also be added without re-introducing the fine-tuning. All such terms are called “soft,” and the theory with weak-scale soft supersymmetry-breaking terms is “weak-scale supersymmetry.” 

2.3. The neutral supersymmetric spectrum

Supersymmetric particles that are electrically neutral, and so promising dark matter candidates, are shown with their standard model partners in Fig. 2. In supersymmetric models, two Higgs doublets are required to give mass to all fermions. The two neutral Higgs bosons are Hd and Hu, which give mass to the down-type and up-type fermions, respectively, and each of these has a superpartner. Aside from this subtlety, the superpartner spectrum is exactly as one would expect. It consists of spin 0 sneutrinos, one for each neutrino, the spin 3/2 gravitino, and the spin 1/2 Bino, neutral Wino, and down- and up-type Higgsinos. These states have masses determined (in part) by the corresponding mass parameters listed in the top row of Fig. 2. These parameters are unknown, but are presumably of the order of the weak scale, given the motivations described above.


Fig. 2. Neutral particles in the supersymmetric spectrum. M1, M2, μ, , and m3/2 are unknown weak-scale mass parameters. The Bino, Wino, and down- and up-type Higgsinos mix to form neutralinos.

The gravitino is a mass eigenstate with mass m3/2. The sneutrinos are also mass eigenstates, assuming flavor and R-parity conservation (see Section 2.4). The spin 1/2 states are differentiated only by their electroweak quantum numbers. After electroweak symmetry breaking, these gauge eigenstates therefore mix to form mass eigenstates. In the basis the mixing matrix is


where cW≡cosθW, sW≡sinθW, and β is another unknown parameter defined by tanβ≡Hu/Hd, the ratio of the up-type to down-type Higgs scalar vacuum expectation values (vevs). The mass eigenstates are called neutralinos and denoted χ ≡ χ1χ2χ3χ4 , in order of increasing mass. If M1  M2, |μ|, the lightest neutralino χ has a mass of approximately M1 and is nearly a pure Bino. However, for M1  M2  |μ|, χ is a mixture with significant components of each gauge eigenstate.

Finally, note that neutralinos are Majorana fermions; they are their own anti-particles. This fact has important consequences for neutralino dark matter, as will be discussed below. 

2.4. R-Parity

Weak-scale superpartners solve the gauge hierarchy problem through their virtual effects. However, without additional structure, they also mediate baryon and lepton number violation at unacceptable levels. For example, proton decay p → π0e+ may be mediated by a squark as shown in Fig. 3.


Fig. 3. Proton decay mediated by squark.

An elegant way to forbid this decay is to impose the conservation of R-parity Rp ≡ (−1)3(BL) + 2S, where B, L, and S are baryon number, lepton number, and spin, respectively. All standard model particles have Rp = 1, and all superpartners have Rp = −1. R-parity conservation implies ΠRp = 1 at each vertex, and so both vertices in Fig. 3 are forbidden. Proton decay may be eliminated without R-parity conservation, for example, by forbidding B or L violation, but not both. However, in these cases, the non-vanishing R-parity violating couplings are typically subject to stringent constraints from other processes, requiring some alternative explanation.

An immediate consequence of R-parity conservation is that the lightest supersymmetric particle (LSP) cannot decay to standard model particles and is therefore stable. Particle physics constraints therefore naturally suggest a symmetry that provides a new stable particle that may contribute significantly to the present energy density of the universe. 

2.5. Supersymmetry breaking and dark energy

Given R-parity conservation, the identity of the LSP has great cosmological importance. The gauge hierarchy problem is no help in identifying the LSP, as it may be solved with any superpartner masses, provided they are all of the order of the weak scale. What is required is an understanding of supersymmetry breaking, which governs the soft supersymmetry-breaking terms and the superpartner spectrum.

The topic of supersymmetry breaking is technical and large. However, the most popular models have “hidden sector” supersymmetry breaking, and their essential features may be understood by analogy to electroweak symmetry breaking in the standard model.

The interactions of the standard model may be divided into three sectors (see Fig. 4). The electroweak symmetry breaking (EWSB) sector contains interactions involving only the Higgs boson (the Higgs potential); the observable sector contains interactions involving only what we might call the “observable fields,” such as quarks q and leptons l; and the mediation sector contains all remaining interactions, which couple the Higgs and observable fields (the Yukawa interactions). Electroweak symmetry is broken in the EWSB sector when the Higgs boson obtains a non-zero vev: h → v. This is transmitted to the observable sector by the mediating interactions. The EWSB sector determines the overall scale of EWSB, but the interactions of the mediating sector determine the detailed spectrum of the observed particles, as well as much of their phenomenology.


Fig. 4. Sectors of interactions for electroweak symmetry breaking in the standard model and supersymmetry breaking in hidden sector supersymmetry breaking models.

Models with hidden sector supersymmetry breaking have a similar structure. They have a supersymmetry breaking sector, which contains interactions involving only fields Z that are not part of the standard model; an observable sector, which contains all interactions involving only standard model fields and their superpartners; and a mediation sector, which contains all remaining interactions coupling fields Z to the standard model. Supersymmetry is broken in the supersymmetry breaking sector when one or more of the Z fields obtains a non-zero vev: Z → F. This is then transmitted to the observable fields through the mediating interactions. In contrast to the case of EWSB, the supersymmetry-breaking vev F has mass dimension 2. (It is the vev of the auxiliary field of the Z supermultiplet.)

In simple cases where only one non-zero F vev develops, the gravitino mass is


where M ≡ (8πGN)−1/2  2.4 × 1018 GeV is the reduced Planck mass. The standard model superpartner masses are determined through the mediating interactions by terms such as


where cij and ca are constants, and λa are superpartners of standard model fermions and gauge bosons, respectively, and Mm is the mass scale of the mediating interactions. When Z → F, these terms become mass terms for sfermions and gauginos. Assuming order one constants,


In supergravity models, the mediating interactions are gravitational, and so Mm  M. We then have


and . In such models with “high-scale” supersymmetry breaking, the gravitino or any standard model superpartner could in principle be the LSP. In contrast, in “low-scale” supersymmetry breaking models with Mm  M, such as gauge-mediated supersymmetry breaking models,


, and the gravitino is necessarily the LSP.

As with electroweak symmetry breaking, the dynamics of supersymmetry breaking contributes to the energy density of the vacuum, that is, to dark energy. In non-supersymmetric theories, the vacuum energy density is presumably naturally instead of its measured value meV4, a discrepancy of 10120. This is the cosmological constant problem. In supersymmetric theories, the vacuum energy density is naturally F2. For high-scale supersymmetry breaking, one finds , reducing the discrepancy to 1090. Lowering the supersymmetry breaking scale as much as possible to gives , still a factor of 1060 too big. Supersymmetry therefore eliminates from 1/4 to 1/2 of the fine-tuning in the cosmological constant, a truly underwhelming achievement. One must look deeper for insights about dark energy and a solution to the cosmological constant problem. 

2.6. Minimal supergravity

To obtain detailed information regarding the superpartner spectrum, one must turn to specific models. These are motivated by the expectation that the weak-scale supersymmetric theory is derived from a more fundamental framework, such as a grand unified theory or string theory, at smaller length scales. This more fundamental theory should be highly structured for at least two reasons. First, unstructured theories lead to violations of low energy constraints, such as bounds on flavor-changing neutral currents and CP-violation in the kaon system and in electric dipole moments. Second, the gauge coupling constants unify at high energies in supersymmetric theories [5], and a more fundamental theory should explain this.

From this viewpoint, the many parameters of weak-scale supersymmetry should be derived from a few parameters defined at smaller length scales through renormalization group evolution. Minimal supergravity [6], [7], [8], [9] and [10], the canonical model for studies of supersymmetry phenomenology and cosmology, is defined by five parameters:


where the most important parameters are the universal scalar mass m0 and the universal gaugino mass M1/2, both defined at the grand unified scale MGUT  2 × 1016 GeV. In fact, there is a sixth free parameter, the gravitino mass



As noted in Section 2.5, the gravitino may naturally be the LSP. It may play an important cosmological role, as we will see in Section 4. For now, however, we follow most of the literature and assume the gravitino is heavy and so irrelevant for most discussions.

The renormalization group evolution of supersymmetry parameters is shown in Fig. 5 for a particular point in minimal supergravity parameter space. This figure illustrates several key features that hold more generally. First, as superpartner masses evolve from MGUT to Mweak, gauge couplings increase these parameters, while Yukawa couplings decrease them. At the weak scale, colored particles are therefore expected to be heavy, and unlikely to be the LSP. The Bino is typically the lightest gaugino, and the right-handed sleptons (more specifically, the right-handed stau ) are typically the lightest scalars.


Fig. 5. Renormalization group evolution of supersymmetric mass parameters. From [11].

Second, the mass parameter is typically driven negative by the large top Yukawa coupling. This is a requirement for electroweak symmetry breaking: at tree-level, minimization of the electroweak potential at the weak scale requires


where the last line follows for all but the lowest values of tan β, which are phenomenologically disfavored anyway. Clearly, this equation can only be satisfied if . This property of evolving to negative values is unique to ; all other mass parameters that are significantly diminished by the top Yukawa coupling also experience a compensating enhancement from the strong gauge coupling. This behavior naturally explains why SU(2) is broken while the other gauge symmetries are not. It is a beautiful feature of supersymmetry derived from a simple high energy framework and lends credibility to the extrapolation of parameters all the way up to a large mass scale like MGUT.

Given a particular high energy framework, one may then scan parameter space to determine what possibilities exist for the LSP. The results for a slice through minimal supergravity parameter space are shown in Fig. 6. They are not surprising. The LSP is either the the lightest neutralino χ or the right-handed stau . In the χ LSP case, contours of gaugino-ness




are also shown. The neutralino is nearly pure Bino in much of parameter space, but may have a significant Higgsino mixture for m0  1 TeV, where Eq. (15) implies |μ|  M1.


Fig. 6. Regions of the (m0M1/2) parameter space in minimal supergravity with A0 = 0, tanβ=10, and μ > 0. The lower shaded region is excluded by the LEP chargino mass limit. The stau is the LSP in the narrow upper shaded region. In the rest of parameter space, the LSP is the lightest neutralino, and contours of its gaugino-ness Rχ (in percent) are shown. From [12].

There are, of course, many other models besides minimal supergravity. Phenomena that do not occur in minimal supergravity may very well occur or even be generic in other supersymmetric frameworks. On the other hand, if one looks hard enough, minimal supergravity contains a wide variety of dark matter possibilities, and it will serve as a useful framework for illustrating many points below. 

2.7. Summary

• Supersymmetry is a new spacetime symmetry that predicts the existence of a new boson for every known fermion, and a new fermion for every known boson.

• The gauge hierarchy problem may be solved by supersymmetry, but requires that all superpartners have masses at the weak scale.

• The introduction of superpartners at the weak scale mediates proton decay at unacceptably large rates unless some symmetry is imposed. An elegant solution, R-parity conservation, implies that the LSP is stable. Electrically neutral superpartners, such as the neutralino and gravitino, are therefore promising dark matter candidates.

• The superpartner masses depend on how supersymmetry is broken. In models with high-scale supersymmetry breaking, such as supergravity, the gravitino may or may not be the LSP; in models with low-scale supersymmetry breaking, the gravitino is the LSP.

• Among standard model superpartners, the lightest neutralino naturally emerges as the dark matter candidate from the simple high energy framework of minimal supergravity.

• Supersymmetry reduces fine tuning in the cosmological constant from 1 part in 10120 to 1 part in 1060 to 1090, and so does not provide much insight into the problem of dark energy.

3. Neutralino cosmology

Given the motivations described in Section 2 for stable neutralino LSPs, it is natural to consider the possibility that neutralinos are the dark matter [13], [14] and [15]. In this section, we review the general formalism for calculating thermal relic densities and its implications for neutralinos and supersymmetry. We then describe a few of the more promising methods for detecting neutralino dark matter. 

3.1. Freeze out and WIMPs

Dark matter may be produced in a simple and predictive manner as a thermal relic of the Big Bang. The very early universe is a very simple place—all particles are in thermal equilibrium. As the universe cools and expands, however, interaction rates become too low to maintain this equilibrium, and so particles “freeze out.” Unstable particles that freeze out disappear from the universe. However, the number of stable particles asymptotically approaches a non-vanishing constant, and this, their thermal relic density, survives to the present day.

This process is described quantitatively by the Boltzmann equation


where n is the number density of the dark matter particle χ, H is the Hubble parameter, σAv is the thermally averaged annihilation cross-section, and neq is the χ number density in thermal equilibrium. On the right-hand side of Eq. (18), the first term accounts for dilution from expansion. The n2 term arises from processes that destroy χ particles, and the term arises from the reverse process , which creates χ particles.

It is convenient to change variables from time to temperature,


where m is the χ mass, and to replace the number density by the co-moving number density


where s is the entropy density. The expansion of the universe has no effect on Y, because s scales inversely with the volume of the universe when entropy is conserved. In terms of these new variables, the Boltzmann equation is


In this form, it is clear that before freeze out, when the annihilation rate is large compared with the expansion rate, Y tracks its equilibrium value Yeq. After freeze out, Y approaches a constant. This constant is determined by the annihilation cross-section σAv. The larger this cross-section, the longer Y follows its exponentially decreasing equilibrium value, and the lower the thermal relic density. This behavior is shown in Fig. 7.


Fig. 7. The co-moving number density Y of a dark matter particle as a function of temperature and time. From [16].

Let us now consider WIMPs—weakly interacting massive particles with mass and annihilation cross-section set by the weak scale: . Freeze out takes place when



Neglecting numerical factors, neq  (mT)3/2em/T for a non-relativistic particle, and H  T2/M. From these relations, we find that WIMPs freeze out when


Since , WIMPs freeze out with velocity v  0.3.

One might think that, since the number density of a particle falls exponentially once the temperature drops below its mass, freeze out should occur at T  m. This is not the case. Because gravity is weak and M is large, the expansion rate is extremely slow, and freeze out occurs much later than one might naively expect. For a m  300 GeV particle, freeze out occurs not at T  300 GeV and time t  10−12 s, but rather at temperature T  10 GeV and time t  10−8 s.

With a little more work [17], one can find not just the freeze out time, but also the freeze out density


A typical weak cross-section is


corresponding to a thermal relic density of Ωh2  0.1. WIMPs therefore naturally have thermal relic densities of the observed magnitude. The analysis above has ignored many numerical factors, and the thermal relic density may vary by as much as a few orders of magnitude. Nevertheless, in conjunction with the other strong motivations for new physics at the weak scale, this coincidence is an important hint that the problems of electroweak symmetry breaking and dark matter may be intimately related. 

3.2. Thermal relic density

We now want to apply the general formalism above to the specific case of neutralinos. This is complicated by the fact that neutralinos may annihilate to many final states: , W+W, ZZ, Zh, hh, and states including the heavy Higgs bosons H, A, and H±. Many processes contribute to each of these final states, and nearly every supersymmetry parameter makes an appearance in at least one process. The full set of annihilation diagrams is discussed in [18]. Codes to calculate the relic density are publicly available [19].

Given this complicated picture, it is not surprising that there are a variety of ways to achieve the desired relic density for neutralino dark matter. What is surprising, however, is that many of these different ways may be found in minimal supergravity, provided one looks hard enough. We will therefore consider various regions of minimal supergravity parameter space where qualitatively distinct mechanisms lead to neutralino dark matter with the desired thermal relic density. 

3.2.1. Bulk region

As evident from Fig. 6, the LSP is a Bino-like neutralino in much of minimal supergravity parameter space. It is useful, therefore, to begin by considering the pure Bino limit. In this case, all processes with final state gauge bosons vanish. This follows from supersymmetry and the absence of 3-gauge boson vertices involving the hypercharge gauge boson.

The process through a t-channel sfermion does not vanish in the Bino limit. This process is the first shown in Fig. 8. This reaction has an interesting structure. Recall that neutralinos are Majorana fermions. If the initial state neutralinos are in an S-wave state, the Pauli exclusion principle implies that the

New ways to solve the Schroedinger equation 

R. Friedberg and T.D. Lee

aDepartment of Physics, Columbia University, New York, NY 10027, USA
bChina Center of Advanced Science and Technology (CCAST) (World Laboratory), P.O. Box 8730, Beijing 100080, People’s Republic of China
cRIKEN BNL Research Center (RBRC), Brookhaven National Laboratory, Upton, NY 11973, USA

Received 27 July 2004;  accepted 5 August 2004.  Available online 15 December 2004.


We discuss a new approach to solve the low lying states of the Schroedinger equation. For a fairly large class of problems, this new approach leads to convergent iterative solutions, in contrast to perturbative series expansions. These convergent solutions include the long standing difficult problem of a quartic potential with either symmetric or asymmetric minima.

Article Outline

1. Introduction 

2. Construction of trial functions 

2.1. A new formulation of perturbative expansion 

2.2. Trial function for the quantum double-well potential

3. Hierarchy theorem and its generalization 

4. Asymmetric quartic double-well problem 

4.1. Construction of the first trial function 

4.2. Construction of the second trial function 

4.3. Symmetric vs asymmetric potential

5. The N-dimensional problem 

Appendix A. A soluble example 

A.1. A two-level model 

A.2. Square-well example (Cont.) 

A.3. The iterative sequence


1. Introduction

Quantum physics is largely governed by the Schroedinger equation. Yet, exact solutions of the equation are relatively few. Besides lattice and other numerical calculations, we rely mostly on perturbative expansions. Such expansion quite often leads to a divergent series with zero radius of convergence, as in quantum electrodynamics, quantum chromodynamics, and problems involving tunneling and instantons. In a series of previous papers [1], [2], [3] and [4] we have presented a new approach to solve the low lying states of the Schroedinger equation. In the special case of one-dimensional problems, this new approach leads to explicit convergent iterative solutions, in contrast to perturbative series expansions. These convergent solutions include the long standing difficult problem [5], [6], [7], [8], [9], [10], [11], [12], [13] and [14] of a quartic potential with symmetric minima.

In this paper, we discuss some additional results bearing on the new method. In the one-dimensional case, we show that by changing the boundary condition to be applied at each iteration, we can obtain a convergent alternating sequence for the ground state energy and wave function instead of the monotonic sequence found before [4]. This result will be spelled out later in this section and proved in Section 3. We also find that the asymmetric quartic double-well potential can be treated by an extension of the procedure used previously for the symmetric case. This extension is treated in Section 4.

In addition, we have begun the exploration of higher dimensional problems along the same line. Although the same kind of iterative procedure can be set up, the linear inhomogeneous equation to be solved at each step cannot now be reduced to simple quadratures, as was done for one dimension. However, it is of interest that this equation is identical in form to an electrostatic analog problem with a given position-dependent dielectric constant media; at each nth iteration, there is an external electrostatic charge distribution determined by the (n − 1)th iterated solution, as we shall discuss in this section.

Consider the Schroedinger equation



where H is the Hamiltonian operator, ψ the wave function, and E its energy. For different physics problems, H assumes different forms. For example, for a system of n non-relativistic particles in three dimensions, H may be written as


where x stands for x1x2, … , x3n the coordinate components of these n particles, V (x) is the potential function, Cij are constants, and p1p2, … , p3n are the momentum operators satisfying the commutation relation



(Throughout the paper, we set Planck’s constant  = 1.) For a relativistic field theory, the Hamiltonian usually takes on a different form. Let Φ (r) be a scalar boson field at a three-dimensional position vector r, and Π (r) be the corresponding conjugate momentum operator. In this case we may write



with Π (r) and Φ (r′) satisfying the commutation relation



In both cases, the dependence of H on the momentum operators pi and Π (r) are quadratic. Consequently, they can be brought into an identical standard form. In the above case of a system of non-relativistic particles, through a linear transformation



the Hamiltonian (1.2) can be written in the standard form




Likewise for the relativistic boson field Hamiltonian (1.4), we can use the Fourier-components of Φ (r) and Π (r) as the set {xi} and {pi}. Through a similar transformation (1.6), the field Hamiltonian (1.4) can also be brought into the standard form Figs. (1.7) and (1.8), but with the number of variables N = ∞. All our subsequent discussions will start from the Schroedinger equation in this standard form Figs. (1.7) and (1.8). Furthermore, in this paper, we shall limit our discussions only to the ground state. In order to solve


where q stands for the set {qi}, we proceed as follows:

1. Construct a good trial function  (q). A rather efficient way to find such trial functions is given in the next section.

2. By differentiating , we define


in which the constant E0 may be determined by, e.g., setting the minimum value of U (q) to be zero. Thus,  (q) satisfies a different Schroedinger equation


Define w (q) and by





The original Schroedinger equation (1.9) can then be written as


Multiplying this equation on the left by  (q) and (1.11) by ψ (q), we find their difference to be


The integration of its left-hand side over all space is zero, which yields


3. The above equation (1.14) will be solved iteratively by considering the sequences


that satisfy





As in Figs. (1.15) and (1.16), we multiply (1.11) by ψn and (1.18) by ; their difference gives


and therefore


As we shall show, for many interesting problems


in contrast to the perturbative series expansion using w (q) as the perturbation. The key difference lies in the above expression (1.21) of , which is a ratio, with both its numerator and denominator depending on the (n − 1)th iterative solution ψn − 1.

4. There exists a simple electrostatic analog problem for the iterative equation (1.18). Assuming that ψn − 1 (q) has already been solved, we can determine through (1.21). The right-hand side of (1.20), defined by


is then a known function. Introduce



In terms of fn (q), the nth order iterative equation (1.20) becomes


Consider a dielectric medium with a dielectric constant dependent on q, given by



Interpret σn (q) as the external electrostatic charge distribution, the electrostatic potential, the electrostatic field, and


the corresponding displacement vector field. Thus (1.25) becomes



the Maxwell equation for this electrostatic analog problem.

At infinity,  (∞) = 0. In accordance with Figs. (1.26) and (1.27), we also have Dn (∞) = 0. Hence the integration of (1.28) leads to the total external electrostatic charge to be also zero; i.e.,


which is the same result given by (1.21) for the determination of . Because the dielectric constant κ (q) in this analog problem is zero at q = ∞, the dielectric media becomes a perfect dia-electric at ∞. Thus, the equation of zero total charge, given by (1.29), may serve as a much simplified model of charge confinement, analogous to color confinement in quantum chromodynamics.

We note that (1.25) can also be derived from a minimal principle by defining


Because of (1.29), the functional I (fn (q)) is invariant under



Since the quadratic part of I (fn (q)) is the integral of the positive definite , the curvature of I (fn (q)) in the functional space fn (q) is always positive. Hence, I (fn (q)) has a minimum, and that minimum determines a unique electrostatic field , as we shall see. To establish the uniqueness, let us assume two different fn, both satisfy (1.25), with the same κ = 2 and the same σn; their difference would then satisfy (1.25) with a zero external charge distribution. For σn = 0, the minimum of I (fn (q)) is clearly zero with the corresponding fn = 0. To derive fn (q) from fn, there remains an additive constant at each iteration. As we shall show, this arbitrariness allows us the freedom to derive different types of convergent series.

To illustrate this freedom, let us consider a one-dimensional problem in which we may replace the variables {qi} by a single x. Furthermore, for this discussion, let us assume the potential V (x) to be an even function, with



(a condition that will be relaxed in our later analysis). The evenness of V (x) requires ψ (x) = ψ (−x) and therefore also  (x) =  (−x). Thus, we need only to consider the half-space



Eqs. Figs. (1.24), (1.27) and (1.28) can be written now as











same as before. Throughout the paper, ′ denote .

From (1.36) and Dn (∞) = 0, we have


and, since σn (x) is even in x, we have from (1.29),


It follows then from (1.34) and Figs. (1.38), (1.39) and (1.40),


which lead to, through (1.35),




Consider first the case that w (x) in (1.38) is positive and satisfies


The hierarchy theorem that will be proved in Section 3 states that if w (x) satisfies (1.44) then the iterative solution of (1.42) with the boundary condition


gives a convergent monotonic sequence , where for all n,




likewise, the sequence f0 (x) = 1, f1 (x), f2 (x), … is also monotonic and convergent at any x  0 with





Furthermore, the convergence of Figs. (1.47) and (1.49) can hold for arbitrarily large but finite w (x). A result that is surprising, but pleasant.

On the other hand, if instead of (1.45), we impose a different boundary condition, one given by


then instead of (1.47), we have for all odd n = 2m + 1 an ascending sequence


however, for the even n = 2m series, we have a descending sequence


furthermore, between any even n = 2m and any odd n = 2l + 1, we have


Since according to (1.13), is the nth order iteration towards


each odd member in (1.51) gives an upper bound of E, whereas each even member in (1.52) leads to a lower bound of E. Both sequences approach the correct as n → ∞, one from above and the other from below. For the boundary condition (1.50), our proof of convergence requires a condition on the magnitude of w (x). Still this is quite a remarkable result.

In Section 2, we discuss the details of how to construct a good trial function  (q) for the N-dimensional problem. Section 3 gives the proof of the hierarchy theorem for the one-dimensional problem in which V (x) = V (−x) is an even function of x and the potential-difference function w (x) is assumed to satisfy (1.44); i.e., w′ (x) < 0 for x > 0. The extension to the asymmetric case V (x) ≠ V (−x) is discussed in Section 4. The hierarchy theorem is also applicable to Mathieu’s equation, which has infinite number of maxima and minima. In Appendix A, we give a soluble example in one dimension.

In dimensions greater than 1, at each iteration Eq. (1.21) gives a fine tuning of the energy, just like the one-dimensional problem. Hence, there are good reasons to expect our approach to yield convergent solutions in any higher dimension. In Section 5, we formulate an explicit conjecture to this effect. We describe an attempt to prove this conjecture by generalizing the steps used to prove the hierarchy theorem in one dimension. The attempt fails at present because the proof of one of the lemmas does not appear to generalize in higher dimension.

The present paper represents the synthesis and generalization of results, some of which have appeared in our earlier publications [1], [2], [3] and [4]. The function Dn introduced in this paper is identical to the function hn used in [4].

2. Construction of trial functions

2.1. A new formulation of perturbative expansion

In many problems of interest, perturbative expansion leads to asymptotic series, which is not the aim of this paper. Nevertheless, the first few terms of such an expansion could provide important insight to what a good trial function might be. For our purpose, a particularly convenient way is to follow the method developed in [1] and [2]. As we shall see, in this new method to each order of the perturbation, the wave function is always expressible in terms of a single line-integral in the N-dimensional coordinate space, which can be readily used for the construction of the trial wave function.

We begin with the Hamiltonian H in its standard form (1.7). Assume V (q) to be positive definite, and choose its minimum to be at q = 0, with



Introduce a scale factor g2 by writing



and correspondingly



Thus, the Schroedinger equation (1.9) becomes


where, as before, q denotes q1q2, … , qN and the corresponding gradient operator. Hence S (q) satisfies


Considering the case of large g, we expand






Substituting Figs. (2.6) and (2.7) into (2.5) and equating the coefficients of gn on both sides, we find


etc. In this way, the second-order partial differential equation (2.5) is reduced to a series of first-order partial differential equations (2.8). The first of this set of equations can be written as


As noted in [1], this is precisely the Hamilton–Jacobi equation of a single particle with unit mass moving in a potential “−v (q)” in the N-dimensional q-space. Since q = 0 is the maximum of the classical potential energy function −v (q), for any point q ≠ 0 there is always a classical trajectory with a total energy 0+, which begins from q = 0 and ends at the other point q ≠ 0, with S0 (q) given by the corresponding classical action integral. Furthermore, S0 (q) increases along the direction of the trajectory, which can be extended beyond the selected point q ≠ 0, towards ∞. At infinity, it is easy to see that S0 (q) = ∞, and therefore the corresponding wave amplitude e-gS0(q) is zero. To solve the second equation in (2.8), we note that, in accordance with Figs. (2.1) and (2.2) at q = 0, . By requiring S1 (q) to be analytic at q = 0, we determine


It is convenient to consider the surface



its normal is along the corresponding classical trajectory passing through q. Characterize each classical trajectory by the S0-value along the trajectory and a set of N − 1 angular variables



so that each α determines one classical trajectory with






(As an example, we note that as q → 0, and therefore . Consider the ellipsoidal surface S0 = constant. For S0 sufficiently small, each classical trajectory is normal to this ellipsoidal surface. A convenient choice of α could be simply any N − 1 orthogonal parametric coordinates on the surface.) Each α designates one classical trajectory, and vice versa. Every (S0α) is mapped into a unique set (q1q2, … , qN) with S0  0 by construction. In what follows, we regard the points in the q-space as specified by the coordinates (S0α). Depending on the problem, the mapping (q1q2, … , qN) → (S0α) may or may not be one-to-one. We note that, for q near 0, different trajectories emanating from q = 0 have to go along different directions, and therefore must associate with different α. Later on, as S0 increases each different trajectory retains its initially different α-designation; consequently, using (S0α) as the primary coordinates, different trajectories never cross each other. The trouble-some complications of trajectory-crossing in q-space is automatically resolved by using (S0α) as coordinates. Keeping α fixed, the set of first-order partial differential equation can be further reduced to a set of first-order ordinary differential equation, which are readily solvable, as we shall see. Write



the second line of (2.8) becomes


and leads to, besides (2.10), also


where the integration is taken along the classical trajectory of constant α. Likewise, the third, fourth, and other lines of (2.8) lead to





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






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


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


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


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


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


and the corresponding energy


(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:


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




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)


where, as before, ψ (x) = egS (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




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 Sn (x) is infinite. The reflection x → −x gives a corresponding asymptotic expansion Sn (x) → Sn (−x), in which each Sn (−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 Sn (x) → Sn (−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


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




In order to satisfy (2.35), we define


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











with, for x  0




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



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


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


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


in which





as before. Throughout this section, we assume


hence, we need only to consider



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





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


for the energy difference , and the sequence

f1(x),f2(x), ,fn(x),


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



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


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


whereas in Case (B)


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


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


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


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


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


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


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



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



and therefore, on account of (3.16),


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 Dn (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, Dn (∞) = 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 Dn (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 Qn (x), and therefore the same as Dn (x); that leads to (3.18). From (3.19) and w′ (x) < 0, one sees that the charge distribution σn (x) is negative for x > xn, 0 at x = xn, and positive for 0 < x < xn. Correspondingly, moving from x = ∞ towards the left, the displacement field increases from Dn (∞) = 0 to Dn (x) > 0, reaching its maximum at x = xn, then as x further decreases, so does Dn (x), and finally reaches Dn (0) = 0 at x = 0. In Case (A), because of fn (∞) = 1, (3.22) leads to



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



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.

Hierarchy Theorem (A) With the boundary condition fn (∞) = 1, we have for all n




Thus, the sequences and {fn (x)} are all monotonic, with





at all finite and positive x.

(B) With the boundary condition fn (0) = 1, we have for all odd n = 2m + 1 an ascending sequence


but for all even n = 2m, a descending sequence


furthermore, between any even n = 2m and any odd n = 2l + 1


Likewise, at any x > 0, for any even n = 2m


whereas for any odd n = 2l + 1



1. The validity of Eqs. Figs. (3.24) and (3.25) for the boundary condition fn (∞) = 1 was established in [4]. The validity of Eqs. Figs. (3.28), (3.29), (3.30), (3.31) and (3.32) for the boundary condition fn (0) = 1 is the new result of this paper, which we shall establish.

2. As we shall also show, the lowest eigenvalue E of the Hamiltonian T + V is the limit of the sequence {En} with


Thus, the boundary condition fn (∞) = 1 yields a sequence, in accordance with (3.26),



with each member En an upper bound of E, similar to the usual variational iterative sequence.

3. On the other hand, with the boundary condition fn (0) = 1, while the sequence of its odd members n = 2l + 1 yields a similar one, like (3.34), with



its even members n = 2m satisfy



It is unusual to have an iterative sequence of lower bounds of the eigenvalue E. Together, these sequences may be quite efficient to pinpoint the limiting E.

The proof of the above generalized hierarchy theorem depends on several lemmas that are applicable to both boundary conditions: (A) fn (∞) = 1 and (B) fn (0) = 1; these lemmas will be established first, and then followed by the proof of the theorem. 

Lemma 1

For any pair fm (x) and fl (x)





According to (3.14)


Also by definition (3.15),


Their difference gives


From (3.14),


Let xl+1 be defined by (3.19). Multiplying (3.41) by fl (xl+1) and (3.42) by fm (xl+1) and taking their difference, we have


in which the unsubscripted x acts as a dummy variable; thus [fm (x)] means [fm] and [fm (xl+1)] means fm (xl+1) · [1], etc.

(i) If (fm (x)/fl (x))′ < 0, then for x < xl + 1


In addition, since w′ (x) < 0 and , we also have for x < xl+1


Thus, the function inside the square bracket on the right-hand side of (3.43) is positive for x < xl+1. Also, the inequalities Figs. (3.44) and (3.45) both reverse their signs for x > xl+1. Consequently, the right-hand side of (3.43) is positive definite, and so is its left side. Therefore, on account of Figs. (3.23A) and (3.23B), (3.37) holds.

(ii) If (fm (x)/fl (x))′ > 0, we see that for x < xl+1, (3.44) reverses its sign but not (3.45). A similar reversal of sign happens for x > xl+1. Thus, the right-hand side of (3.43) is now negative definite and therefore . Lemma 1 is proved. 

The following lemma was already proved in [4]. For the convenience of the readers, we also include it in this paper. Let



be a single-valued differentiable function of ξ in the range between a and b with



and with



Lemma 2

(i) The ratio η/ξ is a decreasing function of ξ for a < ξ < b if




(ii) The ratio η/ξ is an increasing function of ξ for a < ξ < b if







to be the Legendre transform L (ξ). We have




Since (3.49) says that L (a)  0 and (3.50) says that for a < ξ < b, these two conditions imply L (ξ) < 0 for a < ξ < b, which proves (i) in view of (3.55). The proof of (ii) is the same, but with inequalities reversed.  

Lemma 3

For any pair fm (x)  and fl (x)

(i) if over all x > 0,


and (ii) if over all x > 0,



From Figs. (3.17) and (3.18), we have






In any local region of x where , we can regard η = η (ξ) through η (x) = η (ξ (x)). Hence, we have








(i) If (fm/fl)′ < 0, from Lemma 1, we have


From w′ (x) < 0 and the definition of xm + 1 and xl + 1, given by (3.19), we have




We note that from Figs. (3.17) and (3.18) Dm + 1 (x) and Dl + 1 (x) are both positive continuous functions of x, varying from at x = 0,



to at x = ∞



with their maxima at xm + 1 for Dm + 1 (x) and xl + 1 for Dl + 1 (x), since in accordance with Figs. (3.58), (3.59) and (3.67),


From Figs. (3.64) and (3.65), we see that r′ (x) is always <0. Furthermore, from (3.62), we also find that the function r (x) has a discontinuity at x = xl + 1. At x = 0, r (0) satisfies


As x increases from 0, r (x) decreases from r (0), through



to −∞ at x = xl + 1−; r (x) then switches to +∞ at x = xl + 1+, and continues to decrease as x increases from xl + 1+. At x = ∞, r (x) becomes


It is convenient to divide the positive x-axis into three regions:


Table 1 summarizes the signs of , , r, and r′ in these regions. Assuming (fm/fl)′ < 0 we shall show separately the validity of (3.56), (Dm + 1/Dl + 1)′ < 0, in each of these three regions.

Table 1.

The signs of , , , , r (x), and r′ (x) in the three regions defined by (3.74), when


r (x)

r′ (x)







< 0

















Dm + 1 (x) is decreasing and Dl + 1 (x) is increasing; it is clear that (3.56) holds in (II).

In each of regions (I) and (III), we have r (x) > 0 from (3.62) and r′ (x) < 0 from (3.64). Since (fm/fl)′ is always negative by the assumption in (3.56), both terms inside the big parentheses of (3.63) are negative; hence the same (3.63) states that d2η/dξ2 has the opposite sign from . From the sign of listed in Table 1, we see that




Within each region, η = Dm + 1 (x) and ξ = Dl + 1 (x) are both monotonic in x; therefore, η is a single-valued function of ξ and we can apply Lemma 2. In (I), at x = 0, both Dm + 1 (0) and Dl + 1 (0) are 0 according to (3.18), but their ratio is given by




Furthermore, from (3.76), in (I), it follows from Lemma 2, Case (i), the ratio η/ξ is a decreasing function of ξ. Since is >0 in (I), according to (3.59), we have


In (III), at x = ∞, both Dm + 1 (∞) and Dl + 1 (∞) are 0 according to (3.69). Their ratio is

which gives at x = ∞,


As x decreases from x = ∞ to x = xl + 1, from (3.77) we have in (III). It follows from Lemma 2, Case (ii), η/ξ is an increasing function of ξ. Since is <0, because x > xl + 1, we have


Thus, we prove Case (i) of Lemma 3. Case (ii) of Lemma 3 follows from Case (i) through the interchange of the subscripts m and l. Lemma 3 is then established.  

Lemma 4

Take any pair fm (xand fl (x)(A) For the boundary condition fn (∞) = 1,  if at all x > 0,


therefore, if at all x > 0,


(B) For the boundary condition fn (0) = 1, if at all x > 0,


therefore, if at all x > 0,





From (1.35) we see that




(A) In this case fn (∞) = 1 for all n. Thus, at x = ∞, , , and their ratio


At the same point x = ∞, in accordance with (3.17), Dl + 1 (∞) = Dm + 1 (∞) = 0, but their ratio is, on account of w (∞) = 0 and (3.37) of Lemma 1,


in which the last inequality follows from the same assumption, if (fm/fl)′ < 0, shared by (3.37) of Lemma 1 and the present (3.83A) that we wish to prove. Thus, from (3.86), at x = ∞


As x decreases from ∞ to 0, increases from fl + 1 (∞) = 1 to fl + 1 (0) > 1, in accordance with Figs. (3.22) and (3.23A). On account of (3.56) of Lemma 3, we have (Dm + 1/Dl + 1)′ < 0, which when combined with (3.87) and leads to


Thus, by using Figs. (3.51) and (3.52) of Lemma 2, we have to be an increasing function of ; i.e.,




and , we find


which establishes (3.83A). Through the interchange of the subscripts m and l, we also obtain (3.84A).

Next, we turn to Case (B) with the boundary condition fn (0) = 1 for all n. Therefore at x = 0,


Furthermore from Figs. (3.16) and (3.18), we also have and Dm + 1 (0) = Dl + 1 (0) = 0, with their ratio given by


From (3.37) of Lemma 1, we see that if (fm/fl)′ < 0, then and therefore





Analogously to (3.53), define




From (3.56) of Lemma 3, we know that if (fm/fl)′ < 0 then (Dm + 1/Dl + 1)′ < 0, which leads to


From (3.100), we have


and therefore at x = 0, because of (3.99),



Combining Figs. (3.102) and (3.104), we derive


Multiplying (3.100) by , we have


Because and L (x) are both negative, it follows then

which gives (3.83B) for Case (B), with the boundary condition fn (0) = 1. Interchanging the subscripts m and l, (3.83B) becomes (3.84B), and Lemma 4 is established. 

We now turn to the proof of the theorem stated in Figs. (3.24), (3.25), (3.26), (3.27), (3.28), (3.29), (3.30), (3.31) and (3.32).
Proof of the hierarchy theorem. When n = 0, we have



From Figs. (3.20), (3.21) and (3.22), we find for n = 1


and therefore



In Case (A), by using (3.83A) and by setting m = 1 and l = 0, we derive (f2/f1)′ < 0; through induction, it follows then (fn + 1/fn)′ < 0 for all n. From Lemma 1, we also find for all n. Thus, Figs. (3.24), (3.25), (3.26) and (3.27) are established.

In Case (B), by using Figs. (3.109) and (3.83B), and setting m = 1 and l = 0, we find (f2/f1)′ > 0, which in turn leads to (f3/f2)′ < 0, … , and Figs. (3.31) and (3.32). Inequalities Figs. (3.28), (3.29) and (3.30) now follow from Figs. (3.37) and (3.38) of Lemma 1. The Hierarchy Theorem is proved.

Assuming that w (0) is finite, we have for any n


Therefore, each of the monotonic sequences


converges to a finite limit . By following the discussions in Section 5 of [4], one can show that each of the corresponding monotonic sequences of fn (x) also converges to a finite limit f (x). The interchange of the limit n → ∞ and the integrations in (3.13A) completes the proof that in Case (A) the limits and f (x) satisfy


As noted before, the convergence in Case (A) can hold for any large but finite w (x), provided that w′ (x) is negative for x > 0. In Case (B), a large w (x) may yield a negative fn (x), in violation of (3.23B) . Therefore, the convergence does depend on the smallness of w (x). One has to follow discussions similar to those given in [3] to ensure that the limits and f (x) satisfy


4. Asymmetric quartic double-well problem

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


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


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






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




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




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


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






In addition, they satisfy the following Schroedinger equations




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,




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



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.

4.1. Construction of the first trial function

We consider first the positive x region. Following Section 2.1, we begin with the usual perturbative power series expansion for









in which Sn (+) and En (+) are g-independent. Substituting Figs. (4.18), (4.19) and (4.20) into the Schroedinger equation (4.2) and equating both sides, we find



etc. Thus, (4.21) leads to


Since the left side of (4.22) vanishes at x = 1, so is the right side; hence, we determine



which leads to



Of course, the power series expansion Figs. (4.19) and (4.20) are both divergent. However, if we retain the first two terms in (4.19), the function


serves as a reasonable approximation of ψ (x) for x > 0, except when x is near zero. By differentiating  (+), we find  (+) satisfies





In order to construct the trial function  (x) that satisfies the boundary condition (4.5), we introduce for x  0,




so that + (x) and its derivative +′ (x) are both continuous at x = 1, and in addition, at x = 0 we have +′ (0) = 0. For x  0, we observe that V (x) is invariant under


The same transformation converts + (x) for x positive to  (x) for x negative. Define







Both  (x) and its derivative ′ (x) are continuous at x = −1; furthermore, + (x) and  (x) also satisfy the continuity condition Figs. (4.4) and (4.5), as well as the Schroedinger equation Figs. (4.6a) and (4.6b), with the perturbative potentials v+ (x) and v- (x) given by




in which u+ (x) is given by (4.28),





In order that u+ (x), be positive for x > 0 and u- (x), positive for x < 0, we impose


in addition to the earlier condition λ > 0. From Figs. (4.28) and (4.37), we have




Likewise, from Figs. (4.38a) and (4.38b), we find




Furthermore, as x → ±1,




Thus, for x  0, we have


and, together with Figs. (4.36a) and (4.40a),


for x positive. On the other hand for x  0, is not always positive; e.g., at x = 0,

which is positive for , but at x = −1+,

However, at x = −1, . It is easy to see that the sum can satisfy for x  0,


To summarize: + (x) and  (x) satisfy the Schroedinger equation Figs. (4.6a) and (4.6b), with v± (x) given by Figs. (4.36a) and (4.36b),




and the boundary conditions Figs. (4.4) and (4.5). In addition, v± (x) satisfies


and the monotonicity conditions Figs. (4.7a) and (4.7b).

4.2. Construction of the second trial function

To construct the second trial function χ (x) introduced in (4.9), we define f± (x) by




To retain flexibility it is convenient to impose only the boundary condition (4.11) first, but not (4.10); i.e., at x = 0



but leaving the choice of the overall normalization of χ+ (0) and χ (0) to be decided later. We rewrite the Schroedinger equations Figs. (4.12) and (4.13) in their equivalent forms








Because at x = 0, +′(0) = ′(0) = 0, in accordance with (4.5), we have, on account of (4.48),



So far, the overall normalization of f+ (x) and f (x) are still free. We may choose


From Figs. (4.6a), (4.47a), (4.49a) and (4.50a), we see that f+ (x) satisfies the integral equation (for x  0)


Furthermore, from Figs. (4.6a) and (4.49a), we also have


Likewise, f (x) satisfies (for x  0)




The function f+ (x) and f (x) will be solved through the iterative process described in Section 1. We introduce the sequences and for n = 1, 2, 3, … , with


for x  0, and


for x  0, where satisfies




Thus, Figs. (4.55a) and (4.55b) can also be written in their equivalent expressions


for x  0, and


for x  0. For n = 0, we set


through induction and by using Figs. (4.55a), (4.55b), (4.56a) and (4.56b), we derive all subsequent and . Because v± (x) satisfies Figs. (4.44a), (4.44b) and (4.46), the Hierarchy theorem proved in Section 3 applies. The boundary conditions f+ (∞) = f (−∞) = 1, given by (4.52), lead to , in agreement with Figs. (4.55a) and (4.55b). According to Figs. (3.24), (3.25), (3.26) and (3.27) of Case (A) of the theorem, we have




at all finite and positive x, and


at all finite and negative x. Since




with both v± (0) finite,


both exist. Furthermore, by using the integral equations Figs. (4.55a) and (4.55b) for and by following the arguments similar to those given in Section 5 of [3], we can show that


also exist. This leads us from the first trial function  (x) given by (4.3) to f+ (x) and f (x) which are solutions of






and the boundary conditions at x = 0,



An additional normalization factor multiplying, say, f (x) would enable us to construct the second trial function χ (x) that satisfies Figs. (4.9), (4.10), (4.11), (4.12) and (4.13).

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


Next, extend the functions Va (x) and Vb (x) by defining


Thus, both Va (x) and Vb (x) are symmetric potential covering the entire x-axis. Let χa (x) and χb (x) be the ground state wave functions of the Hamiltonians T + Va and T + Vb:






The symmetry (4.68) implies that


and at x = 0


Choose the relative normalization factors of χa and χb, so that at x = 0



The same trial function (4.9) for the specific quartic potential (4.1) is a special example of




In general, from Figs. (4.69a) and (4.69b) we see that χ (x) satisfies


Depending on the relative magnitude of Ea and Eb, we define, in the case of Ea > Eb




otherwise, if Eb > Ea, we set




Thus, we have either


at all finite x, or


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



which can also be written in the same form as (1.14)




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





in which f (x) satisfies


On account of Figs. (4.82) and (4.83), the same equation can also be written as


Eq. (4.80) will again be solved iteratively by introducing



with ψn and its associated energy determined by




In terms of fn (x), we have


On account of (4.88), we also have




For definiteness, let us assume that



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



we can derive {En} and {fn (x)}, with


by using the boundary conditions, either




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 fn (−∞) = 1 is consistent with



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






while in Case (B)










A soluble model of an asymmetric square-well potential is given in Appendix A to illustrate these properties.

5. The N-dimensional problem

The N-dimensional case will be discussed in this section. We begin with the electrostatic analog introduced in Section 1. Suppose that the (n − 1)th iterative solution fn − 1 (q) is already known. The nth order charge density σn (q) is


in accordance with Figs. (1.23) and (1.24). Likewise, from Figs. (1.26) and (1.29) the dielectric constant κ of the medium is related to the trial function  (q) by



and the nth order energy shift is determined by


In the following we assume the range of w (q) to be finite, with








where δ (w (q) − W) is Dirac’s δ-function, W is a constant parameter and the integrations in Figs. (5.3) and (5.6) are over all q-space. Similarly, for any function F (q), we define


In the N-dimensional case, the generalization of [F], introduced by (3.15), is


In terms of Fav (W), (5.8) can also be written as


Thus from Figs. (5.1) and (5.3) we have


the n-dimensional extension of (3.14).

Following Figs. (1.27) and (1.28), the nth order electric field is and the displacement field is


The corresponding Maxwell equation is



Eqs. Figs. (5.11) and (5.12) determine fn except for an additive constant, which can be chosen by requiring





As in the one-dimensional case discussed in Section 3, (5.10) gives the same condition of fine energy tuning at each order of iteration. It is this condition that leads to convergent iterative solutions derived in Section 3. We now conjecture that




also hold in higher dimensions. Although we are not able to establish this conjecture, in the following we present the proofs of the N-dimensional generalizations of some of the lemmas proved in Section 3. 

Lemma 1

For any pair fm(q) and fl(q) if at all W within the range (5.5),





For any function , define


Thus for any function F (q), we have







By setting the subscript n in (5.10) to be m + 1, we obtain


Also by definition (5.19),


The difference of Figs. (5.23) and (5.24) gives


From (5.10) and setting the subscript n to be l + 1, we have


Regard and as two constant parameters. Multiply (5.25) by , (5.26) by and take their difference. The result is


analogous to (4.43).

(i) If , then for


Thus, the function inside the bracket   in (5.21) is positive, being the product of two negative factors, and . Also, when , these two factors both reverse their signs. Consequently (5.17) holds.

(ii) If , we see that for , (5.28) reverses its sign, and therefore the function inside the bracket   in (5.27) is now negative. The same negative sign can be readily established for . Consequently, (5.18) holds and Lemma 1 is established. 

Lemma 2

Identical to Lemma 2 of Section 3.

In order to establish the N-dimensional generalization of Lemma 3 of Section 3, we define


Because of (5.3), Qn (W) is also given by


We may picture that the entire q-space is divided into two regions




with Qn (W) the total charge in I, which is also the negative of the total charge in II. By using Figs. (5.1) and (5.7), we see that


Lemma 3

For any pair fm (q) and fl (q) if at all W within the range (5.5)




Note that Figs. (5.34) and (5.35) are very similar to Figs. (3.56) and (3.57). As in (3.60), define


From (5.33), we have












analogous to Figs. (3.61), (3.62), (3.63) and (3.64).

According to (5.30), at W = 0



and according to (5.29), at W = Wmax



From (5.37), we see that the derivative is positive when , zero at , and negative when . Likewise, from (5.38), is positive when , zero at and negative when . Their ratio determines .

(i) If , from Lemma 1, we have


and therefore, on account of (5.42)


At W = 0,


As W increases, so does r (W). At , r (W) has a discontinuity, with




As W increases from , r (W) continues to increase, with




It is convenient to divide the range 0 < W < Wmax into three regions:




Assuming , we shall show separately in these three regions.

In region B, Ql + 1 is decreasing, but Qm + 1 is increasing. Clearly,


In region A, , r (W) is positive according to Figs. (5.47) and (5.48) and is always >0 from (5.46). Therefore from (5.41),


In region C, , but r (W) and are both positive. Hence,


Within each region, η = Qm + 1 (W) and ξ = Ql + 1 (W) are both monotonic in W; therefore η is a single-valued function of ξ and we can apply Lemma 2 of Section 3.

In region A, at W = 0 both Qm + 1 (0) and Ql + 1 (0) are 0 according to (5.43), but their ratio is given by




Furthermore, from (5.56), . It follows from Lemma 2 of Section 3, the ratio η/ξ is an increasing function of ξ. Since


we also have


In region C, at W = Wmax, both Qm + 1 (Wmax) and Ql + 1 (Wmax) are 0 according to (5.44). Their ratio is


which gives at W = Wmax


As W decreases from Wmax to in region C, since , we have


Furthermore, from (5.57), in region C. It follows from Lemma 2 of Section 3, the ratio η/ξ is a decreasing function of ξ, which together with (5.64) lead to


Thus, we prove Case (i) of Lemma 3. Case (ii) of Lemma 3 follows from Case (i) by an exchange of the subscripts m and l. Lemma 3 is then proved. 

So far, the above lk and lk are almost identical copies of lk and lk of Section 3, but now applicable to the N-dimensional problem. Difficulty arises when we try to generalize Lemma 4 of Section 3.

It is convenient to transform the Cartesian coordinates q1q2, … , qN to a new set of orthogonal coordinates:

(q1,q2, ,qN)→(w(q),β1(q),…,βN-1(q))







where i or j = 1, 2, … , N − 1. Introducing



In terms of the new coordinates, the components of Dn are


Its divergence is


Combining (5.12) with (5.30), we have




in which the integration is along the surface



From Figs. (5.11) and (5.71), it follows that


In terms of curvilinear coordinates, (5.7) can be written as


Substituting (5.76) into (5.74), we find


Because , (5.78) can also be written as


Here comes the difficulty. While the above Lemma 3 transfers relations between to those between Qm + 1/Ql + 1, the latter is


which is quite different from . This particular generalization of the lemmas in higher dimensions fails to establish the Hierarchy Theorem.

For the one-dimensional case discussed in Section 3, we have w′ < 0 and x  0; consequently (5.80) is . Therefore, Lemma 4 of Section 3 can also be established by using (5.80), and the proof of the Hierarchy Theorem can be completed.


[1] R. Friedberg, T.D. Lee and W.Q. Zhao, IL Nuovo Cimento A 112 (1999), p. 1195. Abstract-INSPEC   | $Order Document 

[2] R. Friedberg, T.D. Lee and W.Q. Zhao, Ann. Phys. 288 (2001), p. 52. Abstract | Abstract + References | PDF (214 K) | MathSciNet 

[3] R. Friedberg, T.D. Lee, W.Q. Zhao and A. Cimenser, Ann. Phys. 294 (2001), p. 67. Abstract | Abstract + References | PDF (269 K) 

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Appendix A. A soluble example

In this appendix, we consider a soluble model in which the potential V (x) of (4.67) is


with W2 > μ2 and



Following (4.68), we introduce two symmetric potentials:


with, for x  0,




so that (A.1) can also be written as


Let ψ (x), χa (x), and χb (x) be, respectively, the ground state wave functions of








For |x| > γ, since V (x) = ∞, we have


For |x| < γ, these wave functions are of the form





By substituting these solutions to the Schroedinger equations Figs. (A.7), (A.8) and (A.9), we derive





The continuity of ψ′/ψ at x = ±α relates






with β given by (A.2). In the following, we assume the barrier heights W and


to be much larger than k and p; therefore, the wave function ψ is mostly contained within the two square wells; i.e., kβ and pβ are both near π. We write


and expect and θ to be small. Likewise, introduce


The explicit forms of these angles can be most conveniently derived by recognizing the separate actions of two related small parameters: one proportional to the inverse of the barrier height



and the other



denoting the much smaller tunneling coefficient.

To illustrate how these two effects can be separated, let us consider first the determination of θb given by (A.20). The continuity of at x = ±α gives



From (A.15), we also have


Although the two small parameters Figs. (A.21) and (A.22) are not independent, their effects can be separated by introducing p and q that satisfy





Physically, p and q are the limiting values of pb and qb when the distance 2α between the two wells → ∞, but keeping the shapes of the two wells unchanged. Hence Figs. (A.23) and (A.25). Let



From (A.25), we may expand θ in terms of successive powers of (Wβ)−1:


which determines both p and q. By substituting



into (A.23) and using Figs. (A.24), (A.25), (A.26), (A.27) and (A.28), we determine


Likewise, the continuity of at x = ±α gives





As in (A.25), we introduce k and that satisfy




Similar to Figs. (A.27) and (A.28), we define


and derive


As in Figs. (A.29) and (A.30), we find to be given by




To derive similar expressions for θ and of (A.19), we first note that the transformation



brings Figs. (A.23) and (A.17), provided that we also change

and therefore



Since according to (A.1), the asymmetry of V (x) is due to the term in the positive x region, it is easy to see that



as will also be shown explicitly below. Thus, from (A.29) and through the transformations Figs. (A.39) and (A.40), we derive






with ν1 given by (A.30). Likewise, we note that the transformation



brings Figs. (A.31) and (A.16), provided that we also change

and therefore


Here, we must differentiate three different situations:



In Case (i), when



from (A.37) and through the transformations given by Figs. (A.44) and (A.45), we find




with given by (A.38). According to Figs. (A.13) and (A.19), we have


which leads to


Since in accordance from Figs. (A.28) and (A.36), we find




Thus, the left side of (A.51) is dominated by its first term, μ2β2. Since θ1 and are exponentially small, we can neglect in (A.51). In addition, because θ and are much smaller than 2π, (A.51) can be reduced to


which gives the dependence of δ on μ2. It is important to note that an exponentially small μ2 can produce a finite δ. For δ < α, at x = δ we have, in accordance with (A.10)



which gives the minimum of ψ (x). The wave function ψ (x) has two maxima, one for each potential well.

In Case (ii), α = δ and (A.16) gives cot kβ = 0, and takes on the critical value with


In Case (iii), , and ψ (x) has only one maximum.

As in Figs. (4.73) and (4.75) we introduce χ (x) through


so that


in which, same as Figs. (4.76a) and (4.77a),




with Ea and Eb given by Figs. (A.14) and (A.15). Since according to Figs. (A.4) and (A.5), Va (x)  Vb (x), we have





Write the Schroedinger equation (A.7) in the form (4.80):




As in (4.82), we have


In all subsequent equations, we restrict the x-axis to



and set ψ (x), χ (x) positive. Define



We have, as in Figs. (4.84)