MODIFIED PRÜFER AND EFGP TRANSFORMS AND

DETERMINISTIC MODELS WITH DENSE POINT SPECTRUM

Yoram Last and Barry Simon^{∗}

Division of Physics, Mathematics, and Astronomy California Institute of Technology Pasadena, CA 91125

June 10, 1997

Abstract. We provide a new proof of the theorem of Simon and Zhu that in the region |E| < λ for a.e. energies, − \frac{d^{2}}{dx^{2}} + λ cos(x^{α}), 0 < α < 1 has Lyapunov behavior with a quasi-classical formula for the Lyapunov exponent. We also prove Lyapunov behavior for a.e. E ∈ [−2, 2] for the discrete model with V (j^{2}) = e^{j}, V (n) = 0 if n ∉ {1, 4, 9, … }. The arguments depend on a direct analysis of the equations for the norm of a solution.

§1. Introduction

In this paper, we will consider half-line Schrödinger operators

H_{θ} = −\frac{d^{2}}{dx^{2}} + V (x)  (1.1)

on L^{2}(0, ∞) with u(0) cos(θ) + u′(0) sin(θ) = 0 boundary conditions and the discrete analog on ℓ^{2}(ℤ^{+}), ℤ^{+} = {1, 2, 3, … },

(h_{α}u)(n) = \{ \begin{matrix}u(n + 1) + u(n − 1) + V (n)u(n) & n ≤ 2 \\ u(2) + (V (1) + α)u(1) & n = 1 \end{matrix}  (1.2)

where α plays the role of a boundary condition.

We are interested in models where H_{θ} has dense point spectrum in some interval [a, b]. By general instability results [3,7], this cannot happen for all θ but can and does for a.e. θ if [a, b] ⊂ spec(H_{θ}), and if for a.e. E ∈ [a, b], there is a solution −u′′ + V u = Eu which is

Typeset by AMS-TEX

∗ This material is based upon work supported by the National Science Foundation under Grant No. DMS-9401491. The Government has certain rights in this material.

L^{2} at infinity [4,15,16]. The first examples of such operators involved random V ’s where one proves dense point spectrum for a.e. V.

Examples which are deterministic were first found by Gordon [6] (also see [9]), who showed it for problems with very high and sparse but not too sparse barriers. Simon-Zhu [17] proved a similar result for slowly oscillating potentials like V (x) = λ cos(x^{α}); 0 < α < 1. Attention on the first class was focused by work of Simon-Spencer [15], and on the second by work of Behncke [1] and Stolz [18] — these authors showed the absence of a.c. spectrum.

Our goal here is to obtain dense pure point spectrum by direct control of the asymptotics of the transfer matrix T (0, x), defined by T (0, x)\binom{u′(0)}{u(0)} = \binom{u′(x)}{u(x)} for solutions of

−u′′ + V u = Eu  (1.3)

in the continuum case, and T (0, n)\binom{u(1)}{u(0)} = \binom{u(n+1)}{u(n)} for solutions of

u(n + 1) + u(n − 1) + V (n)u(n) = Eu(n)  (1.4)

in the discrete case. It follows from results of Ruelle [13] that if lim_{n→∞} \frac{1}{n} ln ∥T (0, n)∥ > 0, then there is an L^{2} solution (here we include existence and finiteness of the limit). The same idea works for other situations, for example, if lim_{n→∞} \frac{1}{n^{γ}} ln ∥T (0, n)∥ > 0 for any γ > 0; see [11].

For the case V (x) = λ cos(x^{α}), that lim_{n→∞} \frac{1}{n} ln ∥T (0, n)∥ exists for a.e. E and an explicit formula for the limit was found by Simon and Zhu [17]. In Section 6, we will prove

Theorem 1.1. Let V (x) = 1 + cos(x^{α}); 0 < α < 1. Let x_{n} = (2πn)^{1/2}; let a(n) = n^{(1−α)/2α}, and let {E_{j}^{(n)}}_{j=1}^{∞} be the eigenvalues of

−\frac{d^{2}}{dx^{2}} + V (x); u(x_{n−1}) = u(x_{n}) = 0

on L^{2}(x_{n−1}, x_{n}; dx) and let

Ā = ⋂_{k=1}^{∞} ⋃_{m=k}^{∞} ⋃_{j=1}^{∞}(E_{j}^{(n)} − e^{−a(n)}, E_{j}^{(n)} + e^{−a(n)})

(so that Ā is a G_{δ} dense in [0, ∞) of Lebesgue measure zero).

Let E ∈ (0, 2)\Ā. Then,

lim_{n→∞} \frac{1}{n} ln ∥T_{E}(0, n)∥ = \frac{1}{2π} ∫_{\{y|1+cos(y)≥E\}}(1 + cos(y) − E)^{1/2} dy.

Remarks. 1. The forbidden set Ā in [17] is larger; they conjecture that our Ā is the “right” one. One should be able to use WKB methods to describe Ā more completely.

2. It is known [1,18] that for E > 2, the limit exists and is zero.

3. V (x) can be replaced by any f(x^{α}) where f is any C^{2} periodic function with a finite number of critical points in each period.

Unlike Simon-Zhu [17], we will directly attack the transfer matrix by using a transformation idea. In the continuum case, we transform from u′(x), u(x) to R(x), θ(x) defined by (k = \sqrt{E}):

u′(x) = kR(x) cos(θ(x))  (1.5a)
u(x) = R(x) sin(θ(x)).  (1.5b)

Then (1.3) is equivalent to

\frac{d(θ(x))}{dx} = k − \frac{V (x)}{k} sin^{2}(θ(x))  (1.6)
\frac{d ln R}{dx} (x) = \frac{1}{2k} V (x) sin(2θ(x)).  (1.7)

In [10], together with A. Kiselev, we have shown how to exploit these formulas in a variety of spectral situations, and our main goal here is to show that they are useful in many tunnelling calculations.

In the classically forbidden region where V (x) > k^{2}, (1.6) tends to drive θ toward values where the left side vanishes, that is,

sin(θ) = ±\sqrt{\frac{k^{2}}{V (x)}} .  (1.8)

At such points,

\frac{1}{2k} V (x) sin(2θ(x)) = ±\sqrt{V (x) − k^{2}} .  (1.9)

The solutions of (1.8) where (1.9) has the plus sign are attracting, which is why R grows like exp ( + ∫ \sqrt{V (x) − k^{2}} ).

In the classically allowed region where V (x) < k^{2}, it is useful to define R, θ in a slightly different way. Define

k(x) = \sqrt{k^{2} − V (x)}  (1.10a)
u′(x) = k(x)R_{w}(x) cos(θ_{w}(x))  (1.10b)
u(x) = R_{w}(x) sin(θ_{w}(x))  (1.10c)

which we will call WKB-Prüfer variables. Then, R_{w}, θ_{w} obey

\frac{dθ_{w}(x)}{dx} = k(x) − \frac{k′(x)}{2k(x)} sin(2θ_{w}(x))  (1.11)
\frac{d ln R_{w}(x)}{dx} = −\frac{k′}{k} cos^{2}(θ_{w}(x)).  (1.12)

For later purposes, note that cos^{2}(u) = \frac{1}{2} + \frac{1}{2} cos(2u) suggests we define

R̃_{w}(x) = k(x)^{1/2}R_{w}(x)  (1.13)

in which case,

\frac{d ln R̃_{w}(x)}{dx} = −\frac{k′}{2k} cos(2θ_{w}(x)).  (1.14)

Equations (1.11)/(1.14) can be found, for example, in [2].

To the extent that purely oscillatory terms are unimportant because they average to zero, we have R̃_{w}(x) = constant, θ_{w}(x) = ∫^{x} k(y) dy so that u has the the WKB form,

k(x)^{−1/2} sin(θ_{0} + ∫_{x_{0}}^{x} k(y) dy).

As we will see in the appendix, this makes (1.11/1.14) ideal tools for WKB approximations in the classically allowed region.

For the discrete case, we need an analog of modified Prüfer variables, and these are provided by what we have called EFGP variables after contributions in [5,8,12]. Define R(n), θ(n) by

R(n) cos(θ(n)) = u(n) − cos(k)u(n − 1)  (1.15a)
R(n) sin(θ(n)) = (sin k)u(n − 1),  (1.15b)

where E ∈ (−2, 2) and k are related by E = 2 cos(k). Then

θ̄(n) ≡ θ(n) + k;  ν_{k}(n) ≡ −\frac{V (n)}{sin(k)}  (1.16)
cot(θ(n + 1)) = cot(θ̄(n)) + ν_{k}(n)  (1.17)
\frac{R(n + 1)^{2}}{R(n)^{2}} = 1 + ν_{k}(n) sin(2θ̄(n)) + ν_{k}(n)^{2} sin^{2}(θ̄(n)).  (1.18)

We will also need the following relation between θ and u:

\frac{u(n)}{u(n − 1)} = \frac{sin(θ̄(n))}{sin(θ(n))} .  (1.19)

For any θ_{0} ∈ [0, π), define R(x, θ_{0}) (resp. R(n, θ_{0})) to solve (1.6/1.7) (resp. (1.17/1.18)) with θ(0) = θ_{0} (resp. θ(1) = θ_{0}) and R(0) = 1 (resp. R(1) = 1). Then the behavior of any two R’s determines the growth of T in the sense that for any fixed k with k > 0 (resp. k ∈ (0, π)),

C_{1}(k)R(n, θ_{1}) ≤ ∥T (0, n)∥ ≤ C_{2}(k, θ_{1}, θ_{2}) max(R(n, θ_{1}), R(n, θ_{2})),

where C_{1} and C_{2} are constants independent of n and V. In particular, Proposition 1.2. If for θ_{1} ≠ θ_{2} (both in [0, π)) we have

lim_{n→∞} \frac{1}{n} ln R(n, θ_{1}) = lim_{n→∞} \frac{1}{n} ln R(n, θ_{2}) = γ,

then

lim_{n→∞} \frac{1}{n} ln ∥T (n, 0)∥ = γ.

As already mentioned, in the classically forbidden region, the basic equations push R to want to grow as exp (+∫ \sqrt{V (y) − E} dy) or else to decay as exp (−∫ \sqrt{V (y) − E} dy). In examples like cos(x^{α}), forbidden and allowed regions alternate. Our strategy will be to prove one of three possibilities occurs:

(i) All forbidden regions are decay regions for x sufficiently large. In that case, u will be in L^{2}.

(ii) All forbidden regions are growth regions for x sufficiently large. In that case, R

grows in the expected WKB manner. (iii) Arbitrarily far out, there will be a growing region followed by a decaying region.

In that case, we can cut off u at the centers of those forbidden regions and get a very good approximate eigenfunction, and so see that E ∈ Ā.

So if E ∉ Ā, either R grows in the expected way or u is L^{2}. Since at most one θ_{0} can lead to an L^{2} solution, we can always find two θ’s with the expected growth and so use Proposition 1.2.

In Section 2, we discuss a discrete model with sparse growing barriers for which

lim_{n→∞}

\frac{1}{n} lim ln ∥T (0, n)∥ > 0. This shows the use of EFGP variables. In Section 3, we discuss a model like cos(x^{α}) but where cos is replaced by a periodized step function. Sections 4–6 present our proof of Theorem 1.1. An appendix discusses bounded transfer matrices in the region E > 2 in the 1 + cos(x^{α}) model if α < \frac{1}{2}.

B.S. would like to thank M. Ben-Artzi for the hospitality of the Hebrew University where some of this work was done. Y.L. would like to thank J. Avron for the hospitality of the ITP at the Technion where some of this work was done.

§2. A Model of Simon-Spencer Type

In this section, we will study the following model on ℓ^{2}([1, ∞)),

(H_{α}u)(n) = u(n + 1) + u(n − 1) + V (n)u(n)  n ≥ 2
= u(n + 1) + αu(n)  n = 1

where

V (j^{2}) = e^{βj}  j ≥ 2
V (n) = 0  n ≠ 4, 9, 16, …

α plays the role of a boundary condition. β is a parameter, β > 0.

Define

A_{m} = ⋃_{j=1}^{m−1}(2 cos(πj/m) − e^{−\sqrt{m}} , 2 cos(πj/m) + e^{−\sqrt{m}} )

and let

Ā = ⋂_{k=1}^{∞} ⋃_{m=k}^{∞} A_{2m+1}

so Ā is a dense G_{δ} in [−2, 2] of Lebesgue measure zero.

We will the prove the following theorem:

Theorem 2.1. Suppose E ∉ Ā is in [−2, 2]. Then,

lim_{n→∞} \frac{1}{n} ln ∥T (n, 0)∥ = \frac{β}{2} .  (2.1)

For a.e. α, H_{α} has dense point spectrum in [−2, 2] with eigenfunctions decaying as e^{−βn/2}.

Remarks. 1. By “eigenfunctions decaying as e^{−βn/2},” we mean lim_{n→∞} ln(|u(n)|^{2} + |u(n + 1)|^{2})^{1/2}/n = −\frac{β}{2} .

2. Since \overline{lim} |V (n)| = ∞, the results of Simon-Spencer [15] imply σ_{ac}(H_{α}) = ∅. Gordon [6] and Kirsch-Molchanov-Pastur [9] proved that for some potentials of Simon-Spencer type (where the distances between the bumps are not too large), H has dense point spectrum for a.e. boundary condition. Their methods apply to the problem discussed here. Our method is different and identifies the set Ā and the Lyapunov exponent γ(E) = \frac{β}{2} .

3. A similar result holds if V ([j^{β}]) = e^{νj^{μ}} for any μ > 0 and β > 1 (here [j^{β}] is the greatest integer less than j^{β}). Then, lim_{n→∞} \frac{ln ∥T (n,0)∥}{n^{ζ}} = \frac{ν}{μ+1} where ζ = (μ + 1)/β. Where we use Ruelle’s result [13] in the argument below, one instead uses its extension in [11].

Proof. It obviously suffices to prove that for E ∉ Ā, lim \frac{1}{n} ln ∥T (n, 0)u_{θ_{0}}∥ = \frac{β}{2} for u_{θ_{0}} = (cos(θ_{0}), sin(θ_{0})) and at least two out of three values of θ_{0}, say, θ_{0} = 0, \frac{π}{4} , and \frac{π}{2} . Pick θ_{0} = 0 and let k be defined by E = 2 cos(k) and let θ(n), R(n) be the EFGP variables for this value of θ_{0} and E.

Assume that for j ≥ j_{0},

|θ̄(j^{2})| ≥ exp(−j^{2/3}).  (2.2)

We then have that by (1.18), R(n)^{2} is constant for n = j^{2} + 1, …, (j + 1)^{2} and jumps from n = j^{2} to n = j^{2} + 1. By (1.16) and (1.18),

\frac{R(j^{2} + 1)^{2}}{R(j^{2})^{2}} ≤ 1 + \frac{e^{βj}}{| sin k|} + \frac{e^{2βj}}{sin^{2}(k)}

≥ 1 + \frac{sin^{2} exp(−j^{2/3})}{sin^{2}(k)} e^{2βj} − \frac{e^{βj}}{sin(k)}

for j ≥ j_{0}. From these inequalities and ∑_{j=1}^{m} j = \frac{m(m+1)}{2}, one easily sees that lim_{n→∞}

\frac{ln R(j^{2}+1)}{j^{2}} = \frac{β}{2} and then

lim_{n→∞} \frac{ln R(n)}{n} = \frac{β}{2} .  (2.3)

We need to examine (2.2). We will prove that at least one of the following holds for E, θ_{0} fixed:

(i) (2.2) holds; or (ii) E ∈ Ā; or (iii)

∑_{n} ∥T (n, 0)u_{θ_{0}}∥^{2} < ∞.  (2.4)

If we prove this and E ∉ Ā, then for each of θ_{0} = 0, θ_{0} = \frac{π}{4} , and θ_{0} = \frac{π}{2} , one of (i) or (iii) must hold. Since (2.4) can hold for at most one θ_{0} (by constancy of the Wronskian), (2.2) must hold for at least two θ_{0}’s and so (2.3) holds for two θ_{0}’s, and thus (2.1) holds.

Once (2.1) holds, application of Ruelle’s theorem [13] implies that for E ∉ Ā, there exists an initial u_{E} so that lim ln ∥T (0, n)u_{E}∥/n = −\frac{β}{2} , and then the Simon-Wolff [16] method proves point spectrum for a.e. α (see, e.g., [4,14]).

Thus, we need only prove that one of the three alternatives (i)–(iii) above holds. Suppose neither (i) nor (ii) holds. We will prove that (iii) holds.

Since (ii) is assumed false, there exists j_{0} large so that Lemma 2.2 holds and so that E ∉ Ā_{2j−1} for j ≥ j_{0}. In particular, alternative (a) of Lemma 2.2 does not hold. Suppose j_{1} ≥ j_{0} and |θ̄(j_{1}^{2})| ≤ exp(−j_{1}^{2/3}). Since alternative (b) of Lemma 2.2 holds, we can iterate and see that (2.5) holds for j = j_{0} + 1, …, j_{1}.

If alternative (i) fails, there are j_{1}’s going to infinity with |θ̄(j_{1}^{2})| ≤ exp(−j_{1}^{2/3}), so (2.7) holds for all j ≥ j_{0}, and thus |R(n)| ≤ Ce^{−(α−ϵ)n/4} so (2.4) holds. □

Lemma 2.2. There is a j_{0} (depending only on k and β) so that if j ≥ j_{0} and |θ̄(j^{2})| ≤ exp(−j^{2/3}), then either

(a) For some ℓ ∈ \{1, …, 2j − 2\}, |E − 2 cos(\frac{πℓ}{2j−1})| ≤ e^{−\sqrt{2j−1}} , or

(b) |θ̄(j − 1)^{2})| ≤ exp(−(j − 1)^{2/3}) and

R(j^{2}) ≤ e^{−βj/2}R((j − 1)^{2}).  (2.5)

Proof. By (1.19),

|u(j^{2})| = |\frac{R(j^{2})}{sin k} sin(θ̄(j^{2}))| ≤ CR(j^{2}) exp(−j^{2/3}),  (2.6)

where C will be used to indicate a constant depending only on k. C can vary from formula to formula!

Let q = (j − 1)^{2}. If (recall R(n) = R(j^{2}) if q < n ≤ j^{2})

|u(q)| ≤ R(j^{2}) exp(−j^{2/3}),  (2.7)

then ũ ≡ u ↾ (j = q + 1, …, j^{2} − 1) is an extremely accurate trial function for H̃_{0}, the Jacobi matrix on (j = q + 1, …, j^{2} − 1), for by (2.6), (2.7)

∥(H̃_{0} − E)ũ∥^{2} ≤ CR(j)^{2} exp(−2j^{2/3})

while by an elementary estimate,

∥ũ∥^{2} ≥ CjR(j)^{2}.

Thus, by taking j_{0} large enough, we can be certain that

\frac{∥(H̃_{0} − E)ũ∥}{∥u∥} ≤ exp(−\sqrt{2j − 1} )

if j ≥ j_{0}. Since the eigenvalues of H̃_{0} are \{2 cos(\frac{πℓ}{2j−1})\}_{ℓ=1,...,2j−2}, we conclude if j ≥ j_{0} and (2.7) holds, then alternative (a) holds.

So suppose that (2.7) fails. Since u(q)^{2} + u(q + 1)^{2} ≥ CR(q + 1)^{2} = CR(j^{2}), we conclude that

|\frac{u(q)}{u(q + 1)}| ≥ C exp(−j^{2/3}).  (2.8)

Thus, using the eigenfunction equation,

|\frac{u(q − 1)}{u(q)}| ≥ e^{βj} − |E| − C exp(j^{2/3}) ≥ \frac{1}{2} e^{β(j−1)}  (2.9)

if j ≥ j_{0} and j_{0} is sufficiently large.

By (1.19), (2.9) implies

|\frac{sin(θ̄(q))}{sin(θ(q))}| ≤ 2e^{−β(j−1)}

so for j ≥ j_{0} with j_{0} large, we have |θ̄(q)| ≤ Ce^{−β(j−1)} ≤ exp(−(j − 1)^{2/3}) verifying one of the conclusions.

Moreover, by (2.8) and (2.9) (C is a constant whose value changes!),

R(j)^{2} = R(q + 1)^{2} ≤ C(u(q)^{2} + u(q − 1)^{2})
≤ Ce^{−2βj}[1 + exp(2j^{2/3})]u(j − 1)^{2}
≤ Ce^{−2βj}[1 + exp(2j^{2/3})]R(q)^{2}
≤ e^{−βj}R(q)^{2}

if j ≥ j_{0} with j_{0} large. This proves (2.5). □

§3. A Warm-up Problem

In this section, we treat an elementary tunnelling problem that, because V is piecewise constant, avoids some of the technicalities we will need in the cos(x^{α}) case. Throughout this section, let [x] ≡ greatest integer less than x, and define

f(x) = \frac{1}{2}(1 + (−1)^{[x]})
V (x) = f(x^{α})  (3.1)

with 0 < α < 1. Thus, f is 1 (resp. 0) for x in [0, 1)∪[2, 3)∪[4, 5) ⋯ (resp. [1, 2)∪[3, 4)∪ ⋯). V consists of regions 1, 2, …, where V is first 1, then 0, then 1, … and region n runs from (n − 1)^{1/α} to n^{1/α} and has approximate width α^{−1}n^{(1/α)−1} going to infinity. We will use Q_{1}, L_{1}, Q_{2}, L_{2}, … for the regions and |Q_{1}|, |L_{1}|, … for their widths. L_{n} is the n^{th} region where V is 0.

For each L, consider the potential W_{L} which is 0 on [0, L] and 1 for other x. Let e_{1}(L) < ⋯ < e_{m_{L}}(L) denote the eigenvalues of −\frac{d^{2}}{dx^{2}} + W_{L} of energy less than 1. A Sturm oscillation argument shows that m_{L} = \frac{L}{2π} + O(1) as L → ∞.

Define

A_{j} = ⋃_{k=1}^{m_{L_{j}}}(e_{k}(L_{j}) − e^{−\sqrt{|Q_{j}|}} , e_{k}(L_{j}) + e^{−\sqrt{|Q_{j}|}} )

and let

Ā = ⋂_{ℓ=1}^{∞} ⋃_{j=ℓ}^{∞} A_{j}

which is a dense G_{δ} of [0, 1] of Lebesgue measure zero.

Theorem 3.1. Let H_{θ} = −\frac{d^{2}}{dx^{2}} + V (x) with V given by (3.1) with 0 < α < 1 and θ boundary conditions at 0. Suppose E ∈ (0, 1)\Ā. Then

lim_{x→∞} \frac{1}{|x|} ln ∥T (0, x)∥ = \frac{1}{2} \sqrt{1 − E} .  (3.2)

Proof. Let Q_{i} = (x_{i}, y_{i}), L_{i} = (y_{i}, x_{i+1}). We will look at three values of θ_{0}^{(k)}, say, 0, \frac{π}{4} , \frac{π}{2} for k = 1, 2, 3 and let R_{k}(x), θ_{k}(x) be the solution of the usual modified Prüfer equations with θ(0) = θ_{0}^{(i)} and R_{k}(0) = 1. Our goal is to prove that if E ∈ (0, 1)\Ā, then for each k, either ∥T (0, x)u_{θ_{k}}∥ ∈ L^{2} or lim_{n→∞} \frac{1}{|x|} ln R_{k}(x) = \frac{1}{2} \sqrt{1 − E}. Since at most one k can have ∥T (0, x)u_{θ_{k}}∥ ∈ L^{2}, we conclude (3.2).

In each region L_{j}, T (y_{j}, x) is just the free transfer matrix at energy E and so ∥T (y_{j}, x)∥ ≤ C if x ∈ L_{j}. Thus, | ln R_{i}(x_{j+1}) − ln R_{i}(y_{j})| ≤ C and thus

lim_{n→∞} \frac{1}{x_{n+1}} ∑_{j=1}^{n} | ln R(x_{j+1}) − ln R(y_{j})| → 0

since x_{n+1} ∼ (2n)^{1/α} and thus \frac{n}{x_{n+1}} → 0. That means we only have to control the change of R in the tunnelling regions Q_{i}.

Define an angle η in (0, \frac{π}{2} ) by sin(η) = k so cos(η) = \sqrt{1 − E}. Then in the regions Q_{i}, (1.6) can be rewritten as

\frac{dθ}{dx} = \frac{1}{k} (sin^{2}(η) − sin^{2}(θ)).  (3.3)

The equation (3.3) with θ thought of as running mod 2π has a simple structure. There are four fixed points where sin^{2}(θ) = sin^{2}(η), viz. θ = ±η, π ± η. The fixed points at η and π + η are attracting, and the ones at −η and π − η are repelling. As x increases, θ moves away from the neighboring repeller and toward the neighboring attractor. For definiteness, we will talk about θ in the interval (−η, η) and suppose η < \frac{π}{2} , but a similar argument works for any other interval.

For x ∈ Q_{i},

\frac{1}{2k} V (x) sin(2η) = \frac{1}{k} sin(η) cos(η) = \sqrt{1 − E}

and one sees similarly that at the two attracting fixed points (2k)^{−1}V (x) sin(2θ) is \sqrt{1 − E}, and at the two repelling fixed points, it is −\sqrt{1 − E}, and for regions near −η, ln R decreases by (δx)\sqrt{1 − E}.

Fix ϵ small and define η_{1} by (2k)^{−1} sin(2η_{1}) = \sqrt{1 − E} − ϵ. Let ℓ_{0} be the length of x it takes a solution of (3.3) to run from −η_{1} to η_{1}. Consider the region (x_{k}, x_{k} + |Q_{k}|^{2/3}) at the start of Q_{k}. Suppose θ(x_{k}) ∈ (−η, η) in accordance with our simplifying assumption. If θ(x_{k} + |Q_{k}|^{2/3}) ≤ −η_{1}, then

ln R(x_{k} + |Q_{k}|^{2/3}) ≤ ln R(x_{k}) − (\sqrt{1 − E} − ϵ) |Q_{k}|^{2/3}.  (3.4)

If θ(x_{k} + |Q_{k}|^{2/3}) ≥ −η_{1}, then once x ≥ x_{k} + |Q_{k}|^{2/3} + ℓ_{0}, we have \frac{d ln R}{dx} ≥ \sqrt{1 − E} − ϵ, and so

ln R(y_{k}) ≥ ln R(x_{k}) + (\sqrt{1 − E} − ϵ)(|Q_{k}| − |Q_{k}|^{2/3} − ℓ_{0}) − (|Q_{k}|^{2/3} + ℓ_{0}).  (3.5)

Thus for such intervals, either (3.4) or (3.5) holds. For (η, −η +π) intervals (or if η > \frac{π}{2} ), we need to deal with \sqrt{1 − E} + ϵ instead of \sqrt{1 − E} − ϵ. The net result is that

|ln R(y_{k}) − ln R(x_{k}) − \sqrt{1 − E} |Q_{k}| | ≤ ϵ |Q_{k}| + Cℓ_{0} + C |Q_{k}|^{2/3}.  (3.6)

If we can show that (3.4) fails for large k, then for y large,

|ln R(y) − \frac{1}{2} \sqrt{1 − E} |y| | ≤ \frac{1}{2}ϵ|y| + o(y).  (3.7)

So, since ϵ is arbitrary, we obtain the desired result.

Suppose (3.4) holds. Go back to Q_{k−1} = (x_{k−1}, y_{k−1}). Again, for simplicity, suppose θ(y_{k−1}) ∈ (−η, η). If θ(y_{k−1} − |Q_{k}|^{2/3}) ≥ η_{1}, then

ln R(y_{k−1} − |Q_{k}|^{2/3}) ≤ ln R(y_{k−1}) − (\sqrt{1 − E} − ϵ)|Q_{k}|^{2/3}.  (3.8)

If θ(y_{k−1} − |Q_{k}|^{2/3}) ≤ η_{1}, then θ(x) ≤ −η_{1} for x_{k−1} ≤ x ≤ y_{k−1} − |Q_{k}|^{2/3} − ℓ_{0}, and (assuming k is so large that |Q_{k}| ≥ 2|Q_{k}|^{2/3} + ℓ_{0}) we conclude that (3.4) holds for k − 1 replacing k. Moreover, R(y_{k}) ≤ R(x_{k}) exp(−\frac{1}{2} \sqrt{1 − E} |Q_{k}|) (again for k large).

If (3.4) and (3.8) hold, we can smoothly cut off u at y_{k−1} − \sqrt{Q_{k}} and x_{k} + \sqrt{Q_{k}} and get a trial function for −\frac{d^{2}}{dx^{2}} + W_{L_{k}}, and so we see that |E − e_{j}(|L_{k}|)| ≤ e^{−\sqrt{Q_{k}}}. As in the last section, we see that one of the following holds:

(1) E ∈ Ā

(2) (3.5) holds for all large k (and so (3.7) holds)

(3) u ∈ L^{2}. As explained at the start of the proof, this suffices. □

§4. The Classically Allowed Region

In proving Theorem 1.1, we will break up [0, ∞) into three regions where V (x) ≤ E − ϵ_{0}, where V (x) ≥ E − ϵ_{0}, and where |V (x) − E| ≤ ϵ_{0}. Here ϵ_{0} is a parameter we will take to zero eventually, using the fact that we can show the contribution of the |V (x) − E| ≤ ϵ_{0} region to lim \frac{ln R(x)}{x} is bounded by Cϵ_{0}. In this section, we will control the contribution of the classically allowed region where V (x) ≤ E − ϵ_{0}. The goal will be to show that each oscillation of V contributes at most a constant C to ln R(x), so that, since x^{−1}# of oscillations → 0, the classically allowed region makes no contribution to γ (as it makes no contribution to the integral in Theorem 1.1).

Theorem 4.1. Fix 0 ≤ A ≤ B < E. Suppose that V is a C^{1} function on (a, b) so that there is a c ∈ (a, b) with

(i) V (x) is monotone decreasing on (a, c) and monotone increasing on (c, b).

(ii) A ≤ V (x) ≤ B on (a, b). Fix θ_{0} and let any R(x) solve (1.7) on [a, b] with θ(a) = θ_{0}. Then there is a constant C (depending only on A, B, E but not on V, a, b or θ_{0} ) so that

| ln R(a) − ln R(b)| ≤ C.

Remarks. 1. The proof shows that one can take

C = ln(E/(E − B)) + ln((E − A)/(E − B)).

12 Y. LAST AND B. SIMON

2. That V be piecewise monotone is convenient but not critical. In general, one gets

C = ln(E/(E − B)) + \frac{1}{2} ∫_{a}^{b}|\frac{d}{dx} ln(E − V (x))| dx.

Proof. We use what we called WKB-modified Prüfer variables, that is, we let k(x) = \sqrt{E − V (x)} and R_{w}(x)^{2} = u(x)^{2} + (u′(x))^{2}/k(x)^{2}. Then by (ii) and A ≥ 0,

R(x)^{2} ≤ R_{w}(x)^{2} ≤ \frac{E}{E − B} R(x)^{2}

so

| ln R(x) − ln R_{w}(x)| ≤ \frac{1}{2} ln(E/(E − B)).  (4.1)

By (1.12),

|\frac{d ln R_{w}(x)}{dx}| ≤ |\frac{k′(x)}{k(x)}|

so

| ln R_{w}(a) − ln R_{w}(b)| ≤ ∫_{a}^{b}|\frac{d}{dx} ln k(x)| dx

= ∫_{a}^{c} \frac{d}{dx} ln k(x) − ∫_{c}^{b} \frac{d}{dx} ln k(x)

= ln(\frac{k(c)}{k(a)}) + ln(\frac{k(c)}{k(b)})

since \frac{d}{dx}k ≥ 0 on (a, c) and \frac{d}{dx}k ≤ 0 on (c, b). So

| ln R_{w}(a) − ln R_{w}(b)| ≤ \frac{1}{2} × 2 ln(\frac{E − A}{E − B} ).  (4.2)

(4.1) and (4.2) prove the theorem. □

§5. The Classically Forbidden Region

Our goal in this section is to prove the following: Theorem 5.1. Let 0 < E < 2. Let V be the potential of Theorem 1.1. Let R(x), θ(x) be the solution of (1.6/1.7) for some θ_{0} with ∥T (x, 0)u_{θ_{0}}∥ ∉ L^{2}. Suppose E ∉ Ā. Then

lim_{x→∞} \frac{1}{x} ∫_{\{y|V (y)≥E+ϵ_{0};0≤y≤x\}} \frac{d}{dy} ln R(y)

= \frac{1}{2π} ∫_{\{y|1+cos(y)≥E+ϵ_{0};0≤y≤2π\}}(1 + cos(y) − E)^{1/2} dy. (5.1)

Proof. The argument is very similar to that in Section 3, so we will focus on the new elements. In \{y | V (y) ≥ E + ϵ_{0}\}, let k(x) ≡ \sqrt{V (x) − E} and η(x) = Arcsin(\sqrt{E/V (x)} ) so that (because of the ϵ_{0} cutoff) |k′(x)| ≤ Cx^{1−α} and |η′(x)| ≤ Cx^{1−α}. Notice that

\frac{1}{2k} V (x) sin(2η(x)) = k(x).  (5.2)

Pick ϵ ≤ \frac{1}{2} min(η(x), \frac{π}{2} − η(x)) ≡ ϵ_{1}. ϵ_{1} > 0 because of the ϵ_{0} cutoff and E > 0. We claim that

(a) If |θ − η(x)| < ϵ, then |\frac{1}{2k} V (x) sin(2θ) − k(x)| ≤ Dϵ

(b) k − \frac{V (x)}{k} sin^{2}(η(x) − ϵ) ≥ Y > 0 uniformly in x

(c) k − \frac{V (x)}{k} sin^{2}(η(x) + ϵ) ≤ −Y < 0 uniformly in x. Here D and Y are fixed ϵ independent (but they are ϵ_{0} dependent) non-zero constants.

(a) holds by (5.2). (b), (c) follow from the monotonicity of sin^{2} in (0, \frac{π}{2} ) and the condition ϵ ≤ ϵ_{1}.

We claim in any interval where V (x) ≥ E+ϵ_{0} and |x| is sufficiently large, as x increases, once x ∈ (η(x) − ϵ, η(x) + ϵ) ≡ I_{1}, it remains in that interval. For \frac{d}{dx}[θ(x) − η(x)] ≥ E − Cx^{1−α} at θ = η − ϵ and ≤ −E + Cx^{1−α} at θ = η + ϵ. Similarly, once θ leaves (−η − ϵ, η + ϵ) ≡ I_{2}, it stays outside it; and in a finite distance ℓ_{0}, it moves from anywhere outside I_{2} into I_{1} (or the interval (π + η − ϵ, π + η + ϵ)).

By mimicking the arguments in Section 3, we see that either E ∈ Ā or ∥T (x, 0)u_{θ_{0}}∥ ∈ L^{2} or else

lim_{x→∞} \frac{1}{x} ∫_{\{y|V (y)≥E+ϵ_{0};0≤y≤x\}}|\frac{d}{dy} ln R(y) − k(y)| ≤ Dϵ.

Since we can take ϵ to zero and

\frac{1}{x} ∫_{\{y|V (y)≥E+ϵ_{0};0≤y≤x\}} k(y) dy = \frac{1}{2π} ∫_{\{1+cos(y)≥E+ϵ_{0};0≤y≤2π\}}(1 + cos(y) − E)^{1/2} dy

the theorem is proven. □

§6. Putting It Together

Here we will prove Theorem 1.1. Suppose E ∉ Ā and θ_{0} is such that ∥T (x, 0)u_{θ_{0}}∥ ∉ L^{2}. Let R(x) solve (1.7) with θ(x) = θ_{0}. Fix ϵ_{0} < 0 and consider the three regions:

Z(1) : {x | V (x) ≤ E − ϵ_{0}}
Z(2) : {x | |V (x) − E| ≤ ϵ_{0}}
Z(3) : {x | V (x) ≥ E + ϵ_{0}}.

In Section 4, we proved that

\frac{1}{x} ∫_{Z(1)∩\{y≤x\}}(\frac{d}{dy} ln R(y)) dy → 0.

In Section 5, we proved that

\overline{lim} |\frac{1}{x} ∫_{Z(3)∩\{y≤x\}}(\frac{d}{dy} ln R(y)) dy − \frac{1}{2π} ∫_{\{y|1+cos(y)≥E;0≤y≤2π\}}(1−cos(y)−E)^{1/2} dy |

≤ D_{0}ϵ_{0}

for a constant D_{0}.

By (1.7), |\frac{d}{dy} ln R(y)| ≤ \frac{2}{2k}. Moreover, it is clear that \overline{lim} \frac{1}{x}|Z(2) ∩ \{y ≤ x\}| ≤ D_{1}ϵ_{0} for some constant D_{1}.

Thus, we have

\overline{lim} |\frac{1}{x} [ln R(x) − ln R(0)] − \frac{1}{2π} ∫_{\{y|1+cos(y)≤E;0≤y≤2π\}}(1 + cos(y) − E)^{1/2} dy | ≤ D_{2}ϵ_{0}

with D_{2} = D_{0} + \frac{D_{1}}{2k}. Taking ϵ_{0} to zero, we prove that

\frac{1}{x} ln R(x) → \frac{1}{2π} ∫_{\{y|1+cos(y)≥E;0≤y≤2π\}}(1 + cos(y) − E)^{1/2} dy.

Since at most one θ_{0} has ∥T (x, 0)u_{θ_{0}}∥ ∈ L^{2}, we see that \frac{1}{x} ln ∥T (x, 0)∥ has the required limit. □

Appendix: WKB Prüfer Variables and Bounded Transfer Matrices

In this appendix, we will show how to use WKB-Prüfer variables to show for E > 1, the transfer matrix for cos(x^{α}) potentials is bounded. This is a result of Behncke [1] and Stolz [18] whose proof is not unrelated. Their method is basically a variation of parameters, and this appendix reiterates the idea of [10] that modified Prüfer variables are often a suitable replacement for variation of parameters.

Recall the definition (1.13) for R̃_{w}(x) and θ̃_{w}(x). They obey

\frac{dθ_{w}}{dx} = k(x) + \frac{1}{2} \frac{k′}{kx} sin(2θ_{2}(x))  (A.1)
\frac{d ln R̃_{w}}{dx} = −\frac{k′}{2k} cos(2θ_{w}(x)).  (A.2)

Let V (x) = cos(x^{α}), with \frac{1}{2} < α < 1 and E > 1. Then k(x) = \sqrt{E − V (x)} > \sqrt{|E − 1|} is bounded away from zero. Moreover, we have for j = 0, 1, 2, … and |x| ≥ 1:

|\frac{d^{j}k(x)}{dx^{j}} | ≤ C_{j}(1 + |x|)^{−j(1−α)}  (A.3)
|\frac{d^{j}}{dx^{j}} (\frac{k′}{k} )| ≤ D_{j}(1 + |x|)^{−(j+1)(1−α)}.  (A.4)

In particular, for x large, \frac{dθ}{dx} ≥ \sqrt{|E − 1|}−D_{0}(1+|x|)^{−(1−α)} > 0. By (A.1) and (A.3/A.4), we see

|\frac{d^{2}}{dx^{2}} θ_{w}(x)| ≤ x^{−(1−α)}.  (A.5)

Integrate (A.2) to get (where x_{0} is picked so large that k(x) > δ > 0 for x > x_{0})

ln R̃_{w}(x) − ln R̃_{w}(x_{0}) = ∫_{x}^{x_{0}} −\frac{k′}{4k} \frac{1}{\frac{dθ_{w}}{dx}} \frac{d}{dx} (sin(2θ_{w}(x))) dx

and integrate by parts. By (A.4) and (A.5), the integrand bounded by (1 + |x|)^{−2(1−α)} is integrable, so R̃_{w}(x) is bounded.

Remarks. 1. One doesn’t gain anything by iterating the integration by parts because \frac{d^{j}}{dx^{j}} θ_{w}(x) only falls as (1 + |x|)^{−2(1−α)}.

2. One also sees by integrating by parts that θ_{w}(x) − ∫_{x_{0}}^{x} k(y) dy has a limit, and so one can prove there are WKB-type solutions.

3. By using higher-order modifications, it should be possible to accommodate 0 < α ≤ \frac{1}{2}.

4. All this proof requires, if one keeps track of the derivatives, is that V is C^{2} and (i) sup_{x} V (x) = V_{+} < ∞, inf_{x} V (x) > −∞ (ii) V′(x) → 0 as x → ∞ (iii) V′ ∈ L^{2}, V′′ ∈ L^{1}.

One obtains a bounded transfer matrix if E > V_{+}.

5. The point of this is that bounded transfer matrices imply purely a.c. spectrum [1,11,18].

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