RESONANCES IN ONE DIMENSION AND FREDHOLM DETERMINANTS

BARRY SIMON

Abstract. We discuss resonances for Schrödinger operators in wholeand half-line problems. One of our goals is to connect the Fredholm determinant approach of Froese to the Fourier transform approach of Zworski. Another is to prove a result on the number of antibound states — namely, in a half-line problem there are an odd number of antibound states between any two bound states.

1. Introduction

In this paper, we want to discuss resonances and antibound states in one dimension for Schrödinger operators H = H_{0} + V, where H_{0} is one of the following:

Case 1: −\frac{d^{2}}{dx^{2}} on L^{2}(ℝ)

Case 2: −\frac{d^{2}}{dx^{2}} with u(0) = 0 boundary conditions on L^{2}(0, ∞)

Case 3: −\frac{d^{2}}{dx^{2}} with u′(0) + hu(0) = 0 boundary condition on L^{2}(0, ∞)

Case 4: −\frac{d^{2}}{dx^{2}} + \frac{ℓ(ℓ+1)}{x^{2}} on L^{2}(0, ∞); ℓ = 1, 2, . . ..

We will often consider Case 2 as the ℓ = 0 case of Case 4. We will normally suppose V has compact support, although many of our results only require

∫ e^{a|x|}|V (x)| dx < ∞  (1.1)

for all a > 0 (and some only that (1.1) hold for suitable a > 0).

The operator H_{0} has spectrum [0, ∞) except for Case 3 with h < 0 which has a single eigenvalue at energy −h^{2}. That means that (H_{0} + κ^{2})^{−1} is a well-defined operator in the region Re κ > 0 (except for a pole at κ = −h in Case 3 with h < 0). Its integral kernel

(H_{0} + κ^{2})^{−1}(x, y) ≡ G_{0}(x, y; κ)

has an explicit formula in terms of exponential functions in Cases 1–3 and Bessel functions in Case 4. For example

G_{0}(x, y; κ) = e^{−κ|x−y|}/2κ  (Case 1)  (1.2)
G_{0}(x, y; κ) = e^{−κx_{>}} sinh(κx_{<})/κ (Case 2)  (1.3)

Date: July 31, 2000.

This material is based upon work supported by the National Science Foundation under Grant No. DMS-9707661. The Government has certain rights in this material.

where x_{<} = min(x, y) and x_{>} = max(x, y).

From these explicit formulae, G_{0}(x, y; κ) as a function of κ has an analytic continuation to the entire κ plane except for a simple pole at κ = 0 in Case 1 and at κ = −h, in Case 3 with h ≥ 0.

By the explicit formula for G_{0}, it can be seen that when (1.1) holds for all a > 0:

K(κ)(x, y) = V (x)^{1/2}G_{0}(x, y; κ)|V (y)|^{1/2}  (1.4)

is in L^{2}(ℝ × ℝ, dx dy) for all κ (except for the simple pole in Cases 1 and 3) and so defines an analytic Hilbert-Schmidt operator-valued function on ℂ (or ℂ\{0}, or ℂ\{−h}). In (1.4), V (x)^{1/2} is short for

V (x)^{1/2} = V (x)/|V (x)|^{1/2} if V (x) ≠ 0
= 0  if V (x) = 0.

K(κ) is called the Birman-Schwinger kernel.

It is well known (see, e.g., [7, 23] that Birman-Schwinger Principle. H_{0} + V has −κ^{2} as an eigenvalue (with Re κ > 0) if and only if K(κ) has eigenvalue −1.

With this in mind, one defines Definition. κ is called a resonance energy if Re κ < 0 and K(κ) has eigenvalue −1. The multiplicity of the resonance is the algebraic multiplicity of the eigenvalue. If Im κ = 0 as well, we call κ an antibound state.

Of course, −κ^{2} is really the energy, but since κ is the natural parameter, we will abuse terminology. This definition is the one of Froese [8, 9]. Melrose and Zworski and their school in numerous papers (e.g., [29, 31]) have defined resonance in terms of poles of suitable elements of the analytically contained S-matrix. It follows from our results below (Proposition 2.9 and 2.10) that the definitions agree and are equivalent to the solution of −u′′ + V u = −κ^{2}u with u(x) = e^{−κx} (a suitable Hankel function in Case 4) near +∞ obeys the boundary condition at zero in Cases 2 and 3, is e^{+κx} near −∞ in Case 1 and decreases as x ↓ 0 in Case 4.

While one has obviously that K(κ) is Hilbert-Schmidt, the following is true:

Theorem 1. K(κ) is trace class for all κ (except for the poles in Cases 1 and 3).

This is a result of Froese [8] for Cases 1 and 2, but it seems worthwhile to give an alternate proof that also works in Case 4 (Case 3 is an easy consequence of Case 2). We do this in Section 2. Once one has Theorem 1, it is natural to define

d(κ) = det(1 + K(κ))  (1.5)

in which case the resonances or bound states are precisely the zeros of d(κ) (or κd(κ) in Case 1 or (κ + h) d(κ) in Case 3). Determinants have also been used by Melrose [20] and Zworski [30] in their work on resonances.

Our main goals in this paper are the following:

(1) Zworski [29] proved his theorem on counting resonances by realizing the inverse of a suitable S-matrix element as a Laplace transform. Froese’s alternate proof directly analyzes the asymptotics of d(κ). In this paper, we link the two methods by showing the Fredholm determinant can be realized as a Laplace transform. This provides an alternate to using Melin’s theory [19].

(2) We want to present a new result on counting antibound states. During the final preparation of this manuscript, I received a preprint of Kargaev-Korotyaev [13] who independently found this result.

To be more specific, in Section 2, we will prove expansion formulae for the Fredholm determinants d(κ) in Cases 1 and 2 (this part also works for Cases 3 and 4, but we do not state these results explicitly since we will only handle Cases 1 and 2 in Section 3).

Theorem 2. In Case 2 (half-line with u(0) = 0 boundary conditions):

d(κ) = 1 + ∑_{n=1}^{∞} d_{n}(κ)  (1.6)

with

\begin{matrix}d_{n}(κ) = \\ ∫_{x_{0}≡0<x_{1}<···<x_{n}} V (x_{1}) … V (x_{n}) ∏_{j=1}^{n} \frac{sinh(κ(x_{j} − x_{j−1}))}{κ} e^{−κx_{n}} dx_{1} … dx_{n}. \end{matrix} (1.7)

Theorem 3. In Case 1 (H_{0} ≡ −\frac{d^{2}}{dx^{2}} on L^{2}(ℝ)):

d(κ) = 1 + ∑_{n=1}^{∞} d_{n}(κ)  (1.8)

with

\begin{aligned}d_{n}(κ) = & ∫_{x_{1}<···<x_{n}} V (x_{1}) … V (x_{n}) \\ & [ ∏_{j=2}^{n} \frac{sinh(κ(x_{j} − x_{j−1}))}{κ}] \frac{e^{−κ(x_{n}−x_{1})}}{2κ} dx_{1} … dx_{n}. \end{aligned}  (1.9)

Notes. 1. The product in (1.9) has n − 1 terms and is empty in case n = 1 so that

d_{1}(k) = \frac{1}{2κ} ∫ V (x) dx.  (1.10)

2. (1.9) implies that if we look at the problem of λV with a coupling constant λ added

κd(κ) = κ + λ ∫ V (x) dx + λ^{2}F (λ, κ)

with F analytic near λ = κ = 0. It follows that if ∫ V (x) dx < 0, there is a bound state for small λ with energy (−κ^{2}) given by −λ^{2}(∫ V (x) dx)^{2}+O(λ^{3}) and an antibound state at that energy if ∫ V (x) dx > 0. This recovers old results of Landau-Lifshitz [16, Section 45] (see also Simon [25]).

Once we have these expansions, we can use

\frac{sinh κy}{κ} = ∫_{0}^{y} e^{−2κα} e^{κy} dα

to obtain the following Laplace transform representations:

Theorem 4. In Case 2, let a = sup(supp(V)). Then

d(κ) = 1 + ∫_{0}^{a} t(α) e^{−2ακ} dα

for a suitable L^{1}-function t on [0, a] with a ∈ supp(t).

Theorem 5. In Case 1, let [a, b] be the convex hull of the support of V. Then

κd(k) = κ + \frac{1}{2} ∫_{a}^{b} V (x) dx + \frac{1}{2} ∫_{0}^{b−a} t(α) e^{−2ακ} dα,

where b − a ∈ supp(t).

Following Zworski, these Laplace transform formulae allow one to use Titchmarsh’s theorem [28] to obtain the result of Regge [24] and Zworski [29] on the density of resonances.

While we will directly write down these Laplace transforms, we could use a less direct result. As we will show in Section 2, the expansion in Theorem 2 lets us identify d(κ) with a Jost function, the value of the Jost solution at x = 0 (this is not a new result; it is due to Jost-Pais [12]). But Levin [17] has a Laplace transform formula for the Jost solution as used extensively by Marchenko [18] and that allows one to prove Theorem 4 from Theorem 2 (the estimate we use to show a ∈ supp(t) is in Marchenko’s book).

Our final results are on a different subject

Theorem 6. In a half-line problem (Cases 2, 3 or 4), suppose h has n bound states 0 < κ_{1} < ⋯ < κ_{n}. Then each interval (−κ_{j+1}, −κ_{j}) has an odd number of antibound states and, in particular, at least one antibound state. In particular, there are at least (n − 1) antibound states.

We will prove Theorems 1, 2, and 3 in Section 2, Theorems 4 and 5 in Section 3 and Theorem 6 in Section 4. An appendix has a result on finite determinants that we need in Section 2.

I would like to thank F. Gesteszy, M. Hitrik, K. Makarov, and A. Pushnitski for useful discussions.

2. Expansion of the Fredholm Determinant

In this section, we want to first prove that K(κ) is trace class for all κ ∈ ℂ (or ℂ\{0}), that is, prove Theorem 1. Then we present an expansion of the determinant, det(1 + K(κ)), in the ℓ = 0 case (Theorems 2 and 3).

We begin with the whole-line case, so

K(κ)(x, y) ≡ V (x)^{1/2} \frac{e^{−κ|x−y|}}{2κ} |V (y)|^{1/2}.

Since V has compact support, K(κ)(x, y) is L^{2} in ℝ × ℝ for all κ ∈ ℂ\{0} and the kernel is analytic. Thus,

Proposition 2.1. K(κ) is Hilbert-Schmidt for all κ ∈ ℂ\{0} and analytic in κ.

For Re κ > 0, we can write

K(κ) = V^{1/2}(x)(p^{2} + κ^{2})^{−1}|V (x)|^{1/2}.

It is a basic fact [26, Theorem 4.1] that if f, g ∈ L^{2}(ℝ), then f(p)g(x) is Hilbert-Schmidt with

∥f(p)g(x)∥_{2} ≤ (2π)^{−1/2}∥f∥_{2}∥g∥_{2}.  (2.1)

It follows immediately that

Proposition 2.2. If g, h ∈ L^{2}(ℝ) and f ∈ L^{1}(ℝ), then g(x)f(p)h(x) is trace class and

∥g(x)f(p)h(x)∥_{1} ≤ (2π)^{−1}∥g∥_{2}∥h∥_{2}∥f∥_{1}.

Note next that if | Arg κ| ∈ (\frac{π}{4} , \frac{π}{2} ),

∫ \frac{dρ}{|p^{2} + κ^{2}|} ≤ \frac{c}{| Im κ|} [1 + log (| Im κ|/ Re κ)],

so we have

Proposition 2.3. For | Arg κ| ∈ (\frac{π}{4} , \frac{π}{2} ), K(κ) is trace class with

∥K(κ)∥_{1} ≤ c∥V ∥_{1}[\frac{1}{| Im κ|} [1 + log (| Im κ|/ Re κ)]].  (2.2)

Next, we note that

(K(κ) − K(−κ))(x, y) = V^{1/2}(x) \frac{2 cosh(κ|x − y|)}{κ} V (y)^{1/2}

= V^{1/2}(x) \frac{2 cosh(κ(x − y))}{κ} |V (y)|^{1/2}

is a rank 2 operator. It follows that

Proposition 2.4. If Re(κ) > 0, K(κ) − K(−κ) is trace class with

∥K(κ) − K(−κ)∥_{1} ≤ 2 ∫ |V (x)| e^{2(Re κ)|x|} dx.

As a result

Proposition 2.5. If Re(κ) < 0, K(κ) is trace class with

∥K(κ)∥_{1} ≤ C(1 + log(| Im κ|)),  (2.3)

where C is a constant bounded on each half-annulus {κ | Re κ < 0, 0 < A < |κ| < B}.

Now given κ_{0} with κ_{0} = iα, α ∈ ℝ\{0}, write

K(κ_{0}) = ∫_{−π}^{π} K(κ_{0} + \frac{α}{2} e^{iθ}) \frac{dθ}{2π} .

Since ∥K(κ_{0} + \frac{α}{2} e^{iθ}∥_{1} has only logarithmic singularities at θ = ±\frac{π}{2} , we conclude

Proposition 2.6. K(κ) is trace class on ℂ\{0}.

This is just Theorem 1 in this case.

The proof of Theorem 1 in the ℓ(ℓ+1)/r^{2} case depends on an eigenfunction expansion for h_{ℓ}^{(0)}. Let

u_{ℓ}(k, x) = kx j_{ℓ}(kr),

where j_{ℓ} is a spherical Bessel function. Then [1],

(F_{ℓ}φ)(k) = \sqrt{\frac{2}{π}} ∫_{0}^{∞} u_{ℓ}(k, x)φ(x) dx  (2.4)

is a unitary map of L^{2}(ℝ, dx) to L^{2}(ℝ, dk), that is,

∫_{0}^{∞} |(F_{ℓ}φ)(k)|^{2} dk = ∫_{0}^{∞} |φ(x)|^{2} dx  (2.5)

and

(F_{ℓ}h_{ℓ}^{(0)}φ)(k) = k^{2}(F_{ℓ}φ)(k).  (2.6)

While the usual proof [26] that ∥f(p)g(x)∥_{2} = (2π)^{−1/2}∥f∥_{2}∥g∥_{2} comes from writing f(p) as a convolution operator, here is another proof. Let V be the Fourier transform and let M_{f} be multiplications by f(x). Then

f(p)g(x) = V^{−1}M_{f}V M_{g}

so since V^{−1} is unitary ∥f(p)g(x)∥_{2} = ∥M_{f}V M_{g}∥ but M_{f}V M_{g} has integral kernel:

(M_{f}V M_{g})(p, x) = f(p) \frac{e^{−ipx}}{\sqrt{2π}} g(x)

from which the Hilbert-Schmidt norm formula is immediate. Similarly,

Proposition 2.7. If f, g ∈ L^{2}(0, ∞), then f(\sqrt{h_{ℓ}^{(0)}} )g(x) is Hilbert-Schmidt and

∥f(\sqrt{h_{ℓ}^{(0)}} )g(x)∥_{2} ≤ C∥f∥_{2}∥g∥_{2}.  (2.7)

Proof. As above,

f(\sqrt{h_{ℓ}^{0}} )g(x) = F_{ℓ}^{−1}M_{f}F_{ℓ}M_{g},

so since F_{ℓ} is unitary (2.5), we have

∥f(\sqrt{h_{ℓ}^{(0)}} )g(x)∥_{2} = ∥M_{f}F_{ℓ}M_{g}∥

and the latter has integral kernel

f(k)\sqrt{\frac{2}{π}} u_{ℓ}(k, x)g(x)

so (2.7) follows from |u(k, x)| ≤ C_{1} (see [1]).

With (2.7) in hand, the proof of Theorem 1 is essentially identical to the proof in the whole-line case. The ℓ = 0 situation is identical to the u(0) = 0 boundary conditions for the half-line −\frac{d^{2}}{dx^{2}} case and then the general h bound condition situation follows from the fact that the difference of the resolvents is rank 1 when h changes. That completes the proof of Theorem 1.

Once we know K(κ) is trace class, we can form d(κ) = det(1 + K(κ)). We turn to the expansion of the Fredholm determinant. We will do this for a general trace class operator, A, with an integral kernel A(x, y) on ℝ (or [0, ∞)) of the form

A(x, y) = V (x)^{1/2}G_{0}(x, y)|V (y)|^{1/2},  (2.8)

where G_{0} has the form

G_{0}(x, y) = f_{−}(x_{<})f_{+}(x_{>}),  (2.9)

with x_{<} = min(x, y), x_{>} = max(x, y). We will show that

Proposition 2.8. Let A be a trace class operator of the form (2.8)/ (2.9). Then

\begin{matrix}det(1 + A) = & 1 + ∑_{n=1}^{∞} ∫_{x_{1}<···<x_{n}} V (x_{1}) … V (x_{n})f_{−}(x_{1})f_{+}(x_{n}) \\ ∏_{j=1}^{n−1}[f_{+}(x_{j+1})f_{−}(x_{j}) − f_{+}(x_{j})f_{−}(x_{j+1})] dx_{1} … dx_{n}. \end{matrix} (2.10)

Remarks. 1. We are vague about convergence issues since in the applications we make, they are trivial. In general, it certainly suffices that ∫ |V (x)| [|f_{+}(x)| + |f_{−}(x)| + 1]^{2} dx < ∞.

2. Theorems 2 and 3 are immediate corollaries of Proposition 2.8 given that

e^{−κx}(2κ)^{−1} e^{+κy} − e^{−κy}(2κ)^{−1} e^{+κx} = κ^{−1} sinh(κ(y − x))

and

e^{−κx}κ^{−1} sinh(κy) − e^{−κy}κ^{−1} sinh(κx) = κ^{−1} sinh(κ(y − x)).

3. While we are interested in the cases given by Theorems 2 and 3 because of the application in the next section, the proposition applies directly to the ℓ(ℓ + 1)/x^{2} example; f_{−} is a Bessel function (of imaginary argument) and f_{+} is a suitable Hankel function.

4. Jost-Pais [12] have a related result.

Proof. As shown in [26] (and essentially due to Fredholm)

det(1 + A) = 1 + ∑_{n=1}^{∞} \frac{1}{n!} ∫ A\binom{x_{1} … x_{n}}{x_{1} … x_{n}} dx_{1} … dx_{n},  (2.11)

where

A\binom{x_{1} … x_{n}}{y_{1} … y_{n}} = det([A(x_{i}, y_{j})]_{i,j=1,...,n}).

Since A\binom{x_{1}...x_{n}}{x_{1}...x_{n}} is symmetric in the x’s, we can remove the n! and integrate over x_{1} < ⋯ < x_{n}. In that case,

A\binom{x_{1} … x_{n}}{x_{1} … x_{n}} = V (x_{1}) … V (x_{n}) det(f_{−}(x_{min(i,j)})f_{+}(x_{max(i,j)})).

Determinants of this form are discussed in the appendix where it is shown that

det(f_{−}(x_{min(i,j)})f_{+}(x_{max(i,j)}))

= f_{+}(x_{n})f_{−}(x_{j}) ∏_{j=1}^{n−1}[f_{+}(x_{j+1})f_{−}(x_{j}) − f_{+}(x_{j})f_{−}(x_{j+1})],

which proves the proposition.

We close this section by using the expansions in Theorem 2 and 3 to identify d(κ) with quantities related to the Jost function. Consider first the half-line case. Let d(κ, x_{0}) be the d(κ)-function for the potential {V (x + x_{0})}_{x≥0}. Then, with

F (x, y; κ) = \frac{sinh(κ(y − x))}{κ} ,  (2.12)

we have that

d(κ,x_{o})

= 1 + ∑_{n=1}^{∞} ∫_{0<x_{1}<···<x_{n}} F (0, x_{1}; κ)V (x_{0} + x_{1})F (x_{1}, x_{i}; κ) … V (x_{0} + x_{n})

e^{−κx_{n}} dx_{1} … dx_{n}

= 1 + e^{κx_{0}} ∑_{n=1}^{∞} ∫_{x_{0}<x_{1}<···<x_{n}} F (x_{0}, x_{1}; κ)V (x_{1})F (x_{1}, x_{n}) … V (x_{n})

e^{−κx_{n}} dx_{1} … dx_{n}

by a change of variables. That means if we define

f(x, κ) = e^{−κx} d(κ, x),  (2.13)

then f obeys

\begin{matrix}f(x, κ) = & e^{−κx} + ∑_{n=1}^{∞} ∫_{x<x_{1}<···<x_{n}} F (x, x_{1}; κ)V (x_{1})F (x_{1}, x_{2}; κ) … V (x_{n}) \\ e^{−κx_{n}} dx_{1} … dx_{n}, \end{matrix} (2.14)

so that f obeys the integral equation

f(x, κ) = e^{−κx} + ∫_{x}^{∞} F (x, y; κ)V (y)f(y, κ) dy,  (2.15)

which implies that f obeys

−f′′ + V f = −κ^{2}f  (2.16)

with the boundary condition

f(x, κ) = e^{−κx}  if x ≥ a = sup[suppV].  (2.17)

Thus,

Proposition 2.9. f(x, κ) is the Jost solution, that is, solution of (2.16) which obeys (2.17). In particular, d(κ) is the Jost function, that is, f(x = 0, κ).

Remark. It is a known result that the Jost function is a Fredholm determinant. See Jost-Pais [12] and Simon [26]. Our proof is related to that of Jost-Pais. Similarly, Proposition 2.10 below is known; it follows by combining formulae in Newton [21].

There is a related expression for the whole-line case. First, we need some notation. Let [a, b] be the convex hull of the support of V. The Jost solutions f_{±}(x, κ) are the solutions of (2.16) that obey the boundary conditions

f_{±}(x, κ) = e^{∓κx}  for ± x ≥ |a| + |b|.  (2.18)

Given two C^{1} functions f, g, we define their Wronskian by

W (f, g)(x) = f′(x)g(x) − f(x)g′(x).  (2.19)

As usual, if f, g obey the same second-order differential equation, W is constant and we denote its value as W (f, g). Finally, we define the free Jost solutions

f_{±}^{(0)}(x, κ) = e^{∓κx}.

Proposition 2.10. In the whole-line case

κd(κ) = \frac{1}{2} W (f_{−}( · , κ), f_{+}( · , κ)).  (2.20)

Proof. By looking at the expansion (2.14) of the Jost solution f_{+}(x, κ), and the expansion of d(λ) given by Theorem 3, we see that

d(κ) = 1 + \frac{1}{2κ} ∫_{a}^{b} e^{κx}V (x)f_{+}(x, κ) dx  (2.21)

so

κd(κ) = κ + \frac{1}{2} ∫_{a}^{b} f_{−}^{(0)}(x, κ)V (x)f_{+}(x, κ) dx.  (2.22)

But since f_{+} = f_{+}^{(0)} for x > b,

κ = \frac{1}{2} W (f_{−}^{(0)}, f_{+})(b)

and

f_{−}^{(0)}(x, κ)V (x)f_{+}(x, κ) = −\frac{d}{dx} W (f_{−}^{(0)}, f_{+}),

so (2.22) shows that

κd(κ) = \frac{1}{2} W (f_{−}^{(0)}, f_{+})(a).

But for x ≤ a, f_{−}^{(0)} = f_{−}, so

κd(κ) = \frac{1}{2} W (f_{−}, f_{+})(a) = \frac{1}{2} W (f_{−}, f_{+})(x)

for all x.

Remarks. 1. Proposition 2.9 can be rephrased in a form close to (2.20). Namely, if u(x, κ) is the solution of (2.16) obeying u(0, κ) = 0, u′(0, κ) = 1, then d(κ) = W (u, f).

2. Similarly, in the ℓ(ℓ + 1)/x^{2} case, if f_{ℓ}(x, κ) is the solution of the equation

−u′′ + \frac{ℓ(ℓ + 1)}{x^{2}} u + V u = κ^{2}u,

which is given by kxh^{(1)}(κx) if x ≥ a = sup(supp(V)) and u_{ℓ} is the solution that obeys lim_{x↓0} \frac{u(x,r)}{kxj_{ℓ}(kx)} = 1, then

d(κ) = W (u, f).

In the next section, we will need an additional function and a relation between d and this new function. Let c(κ) be defined by

κc(κ) ≡ \frac{1}{2} W (f_{−}( · , κ), f_{+}( · , −κ)).  (2.23)

Proposition 2.11.

(i)  f_{−}( · , κ) = −c(κ)f_{+}( · , κ) + d(κ)f_{+}( · , −κ)  (2.24)
(ii)  d(κ)d(−κ) = 1 + c(κ)c(−κ)  (2.25)

Proof. (2.23) is a direct consequence of (2.20), (2.23) and

W (f_{+}( · , κ), f_{+}( · , −κ)) = −2κ

since f_{+}(x, κ) = e^{−κx} for x near +∞.

Writing (2.24) for κ and −κ and

W (f_{−}( · , κ), f_{−}( · , −κ) = 2κ

(since f_{−}(x, κ) = e^{κx} for x near −∞), we have that

2κ = −2κc(κ)c(−κ) + 2κd(κ)d(−κ).

which implies (2.25).

3. Determinants as Laplace Transforms

Our goal in this section is to prove Theorems 4 and 5. Given the expansion of Theorems 2 and 3, the argument is similar to part of the construction of the A-function in [27]. We will use

\frac{sinh κx}{κ} = ∫_{−x/2}^{x/2} e^{−2ℓκ} dℓ  (3.1)
= ∫_{0}^{x} e^{xκ}e^{−2ℓκ} dℓ.  (3.2)

We start with the half-line case. Using (1.7) and (3.2), we have that

d(κ) = 1 + ∑_{n=1}^{∞} d_{n}(κ)  (3.3)

d_{n}(κ) = ∫_{0<x_{1}<···<x_{n}; 0<ℓ_{1}<x_{1}, 0<ℓ_{2}<x_{2}−x_{1},...,0<ℓ_{n}<x_{n}−x_{n−1}} V (x_{1}) … V (x_{n})
e^{−2 ∑_{j=1}^{n} ℓ_{j}κ} dx_{1} … dx_{n} dℓ_{1} … dℓ_{n}  (3.4)
≡ ∫ t_{n}(α) e^{−2ακ} dα,  (3.5)

where

t_{n}(α) = ∫_{0<x_{1}<···<x_{n}} V (x_{1}) … V (x_{n})R_{n}(x_{1}, …, x_{n}; α) dx_{1} … dx_{n} (3.6)

with

\begin{matrix}R_{n}(x_{1}, …, x_{n}; α) \\ = ∫_{0<ℓ_{1}<x_{1}, 0<ℓ_{2}<x_{2}−x_{1},...,0<ℓ_{n}<(x_{n}−x_{n−1})} δ( ∑_{j=1}^{n} ℓ_{j} − α) dℓ_{1} … dℓ_{n}. \end{matrix} (3.7)

Lemma 3.1. (i) R_{n} = 0 if α > x_{n} (ii) |R_{n}| ≤ \frac{α^{n−1}}{(n−1)!}

(iii) For n ≥ 2, R_{n} is C^{1} and |\frac{∂R_{n}}{∂α} | ≤ 2 \frac{α^{n−2}}{(n−2)!} .

Proof. (i) By the inequalitites on ℓ_{j}, ∑ ℓ_{j} ≤ (x_{1} + (x_{2} − x_{1}) + ⋯ + (x_{n} − x_{n−1})) = x_{n}.

(ii) Clearly, |R_{n}| ≤ ∫_{0<ℓ_{1}, 0<ℓ_{2},...,0<ℓ_{n}} δ(∑_{j=1}^{n} ℓ_{j} − α) dℓ_{1} … dℓ_{n} = \frac{α^{n−1}}{(n−1)!} by induction.

(iii)

\frac{∂R_{n}}{∂α} = ∫_{...} δ′( ∑_{j=1}^{n} ℓ_{j} − α) dℓ_{1} … dℓ_{n}

= ∫_{...} δ( ∑_{j=1}^{n−1} ℓ_{j} − α) dℓ_{1} … dℓ_{n−1}

− ∫_{...} δ( ∑_{j=1}^{n−1} ℓ_{j} − α − (x_{n} − x_{n−1})) dℓ_{1} … dℓ_{n−1}.

Now use the estimation method of (ii).

Proposition 3.2. Let a = sup(supp(V)). (i)

d(κ) = 1 + ∫_{0}^{a} t(α) e^{−2ακ} dα  (3.8)

with t(α) a continuous function. (ii) t is an absolutely continuous function with

|t′(α) + V (α)| ≤ C ∫_{α}^{a} |V (y)| dy.  (3.9)

(iii) The convex hull of the support of δ(α) + t(α) is [0, a].

Proof. By (3.6) and (ii) of Lemma 3.1, we have that

|t_{n}(α)| ≤ \frac{α^{n−1}}{(n − 1)!} [ ∫_{α}^{a} |V (y)| dy]\frac{1}{(n − 1)!}( ∫_{0}^{a} |V (y)| dy)^{n−1}

so

t(α) = ∑_{n=1}^{∞} t_{n}(α)

converges and we can justify the interchange of sum and integral to obtain (3.8).

Similarly, by (iii) of Lemma 3.1, ∑_{n=2}^{∞} t_{n}(α) is C^{1} with derivative g(α) that obeys

|g(α)| ≤ ∑_{n=2}^{∞}(2α^{n−2})/(n − 2)! [ ∫_{α}^{a} V (y) dy]\frac{1}{(n − 1)!} ( ∫_{0}^{a} |V (y)| dy)

≤ C ∫_{α}^{a} |V (y)| dy.

By a direct calculation.

t_{1}(α) = ∫_{α}^{a} V (y) dy

so that (3.9) is proven.

Clearly, t(α) is supported on [0, a] and δ + t has 0 in its support. That means we need only show t is non-zero on each interval [a − ε, a]. Suppose t vanishes identically on such an interval. Then by (3.9),

|V (α)| ≤ C ∫_{α}^{a} |V (y)| dy

for α > a − ε. So for x ∈ [a − ε, a],

∫_{x}^{a} |V (y)| dy ≤ |a − x|C ∫_{x}^{a} |V (y)| dy

so V vanishes on [a − ε′, a) with ε′ = min(ε, \frac{1}{C}). Since a = sup(supp(V)) by hypothesis, this is a contradiction which shows that a ∈ supp(t).

This implies Theorem 4. Proposition 3.2 also makes it easy to construct potentials with an infinity of antibound states. The basic idea is close to those of Titchmarsh [28] on zeros of Laplace transforms and a similar analysis (but using Melin’s theory and on the whole line) has been made by Zworski [29]. Define intervals I_{1}, I_{2}, … ⊂ [0, 1] by I_{1} = [0, \frac{1}{2}], I_{2} = [\frac{1}{2}, \frac{3}{4}], …, I_{n} = [1 − \frac{1}{2^{n−1}} , 1 − \frac{1}{2^{n}}]. Let a_{n} = 2^{n} e^{−2^{n+1}} and

V (x) = (−1)^{n}a_{n} if x ∈ I_{n}.

It is easy to modify V to be C^{∞}. The infinity of oscillations of V is critical since if V has a definite sign near x = a, it is easy to see that t(α) has a definite sign, and so f(−κ) does for κ near infinity and thus there are only finitely many antibound states.

Since a_{n} is decreasing, |t′(α)+V (α)| ≤ C2^{−n}a_{n} by (ii) of Proposition 3.2. Thus for n large, t′(α) is very close to (−1)^{n+1}a_{n} on I_{n}. Thus

∫_{0}^{1} t′(α) e^{2ακ} dα ∼ ∑_{n=1}^{∞}(−1)^{n+1}a_{n}2^{−n} e^{2(1−2^{−n})κ}

∼ e^{2κ} ∑_{n=1}^{∞}(−1)^{n+1} exp(−2^{−n−1}κ − 2^{n+1}).

For κ_{m} = 2^{2(m+1)}, it is easy to see this sum is dominated by the term with n = m. Since ∫_{0}^{1} t′(α) e^{2ακ} dα ∼ −2κ ∫_{0}^{1} t(α) e^{2ακ} dα, we concluded that

(−1)^{m}f(−2^{2(m+1)}) > 0

for m large, and so f(−κ) has infinitely many zeros as κ → ∞.

Once one has a half-line potential with an infinity of antibound states, the same is true of whole-line problems with suitable even potentials.

14 B. SIMON

We now turn to the whole-line case and Theorem 5. We will use (2.21)

2κd(κ) = 2κ + ∫_{a}^{b} e^{κx}V (x)f_{+}(x, κ) dx,  (3.10)

as well as the following formula proven in a similar way:

2κc(κ) = ∫_{a}^{b} e^{κx}V (x)f_{+}(x, −κ) dλ  (3.11)
= ∫_{a}^{b} f_{−}(x, κ)V (x) e^{κx} dx.  (3.12)

Proposition 3.3. In the whole-line case: (i)

κd(κ) = κ + \frac{1}{2} ∫_{a}^{b} V (x) dx + \frac{1}{2} ∫_{0}^{b−a} t(α)e^{−2ακ} dα  (3.13)

for an L^{1}-function t on [0, b − a]. (ii)

κc(κ) = \frac{1}{2} ∫_{a}^{b} e^{2κα}s(α) dα,  (3.14)
where s(α) is a function in L^{1}(a, b).

(iii)

|s(α) − V (α)| ≤ C ∫_{a}^{α} |V (y)| dy
≤ C ∫_{α}^{b} |V (y)| dy.

(iv) a, b ∈ supp(s). Proof. (i) We begin with the extension of (3.8) which implies that

f_{+}(x, κ) = e^{−κx} + ∫_{x}^{b} t_{+}(α, x) e^{−2κα} e^{+κx}.  (3.15)

Plugging this into (3.10),

κd(κ) = κ + \frac{1}{2} ∫_{a}^{b} V (x) dx + \frac{1}{2} ∫_{a}^{b} dx ∫_{x}^{b} t_{+}(α, x) e^{−2κ(α−x)} dα

= κ + \frac{1}{2} ∫_{a}^{b} V (x) dx + \frac{1}{2} ∫_{0}^{b−a} ( ∫_{a}^{b−β} t_{+}(β + x, x) dx) e^{−2βκ} dβ, which is (3.13) if t(α) = ∫_{a}^{b−α} t_{+}(α + x, x) dx.

(ii) Plugging (3.15) into (3.11) yields κc(κ) = \frac{1}{2} ∫_{a}^{b} e^{2κx}V (x) dx + \frac{1}{2} ∫_{a}^{b} V (x) dx( ∫_{x}^{b} t_{+}(α, x) e^{2κα} dα) which is (3.14) if

s(α) = V (α) + ∫_{a}^{α} t_{+}(α, x)V (x) dx.  (3.16)

Using (3.12) instead of (3.11) yields

s(α) = V (α) + ∫_{α}^{b} t_{−}(α, x)V (x) dx.  (3.17)

(iii) This follows from (3.16)/(3.17) with

c = sup_{α,x} |t_{±}(α, x)|.

(iv) This follows as in the proof of (iii) of Proposition 3.2.

(3.13) is Theorem 5 if we prove that b − a ∈ supp(t). We give a proof of this fact that is part of Zworski’s proof [29] translated to our language; Froese [8] has a different proof.

Proposition 3.4. b − a ∈ supp(t).

Proof. Write κd(κ) = ∫ m(α) e^{−2ακ} dα and κc(κ) = ∫ n(α) e^{−2ακ} dα in distributional sense where

m(α) = \frac{1}{2}δ′(α) + (\frac{1}{2} ∫_{a}^{b} V (x) dx)δ(α) + \frac{1}{2} t(α)

and

n(α) = \frac{1}{2} s(−α).

By (2.25) and the uniqueness of inverse Laplace transforms

(m ∗ m̃)(α) = \frac{1}{4} δ′′(α) + (n ∗ ñ)(α),  (3.18)

where m̃(α) = m(−α) and ∗ is convolution. Since a, b ∈ supp(s), we have that a−b and b−a lie in supp[n∗ñ]. If supp[m] ⊂ [0, c], then supp[m∗m̃] ⊂ [−c, c]. Hence by (3.18), b − a ∈ supp[m].

4. Antibound States

Our goal in this section is to prove Theorem 6. We will provide three rather different proofs, two that I found and one supplied to me by G.M. Graf [11]. We will present them first in the case when u(0) = 0 boundary conditions with V of compact support, and then make some remarks about the other cases.

First Proof of Theorem 6. We begin by noting some properties of the Fredholm determinant d(κ):

(i) If κ ≠ 0, either d(κ) ≠ 0 or d(−κ) ≠ 0 (or both). This is true because if d(±κ) = 0, then the Jost solution f(x, ±κ) vanishes at x = 0, so the function solving −u′′ + qu = −κ^{2}u with u(0) = 0, u′(0) = 1 is αe^{∓κx} for x > a ≡ max supp(V). This cannot happen for both κ and −κ since u ≠ 0 for x large.

(2) If d(κ = 0) = 0, then d′(κ = 0) ≠ 0. This well-known result holds because \frac{∂}{∂κ}f(x, κ)|_{κ=0} and f(x, κ)|_{κ=0} both solve −u′′+qu = 0 and are equal to x and 1 for x > a. As with the proof of (i), it cannot happen that both d(κ = 0) = 0 (in which case the solution with u(0) = 0, u′(0) = 1 is a constant for x > a) and \frac{∂d}{∂κ}(κ = 0) = 0 (in which case u is equal to αx for x > a).

(iii) If κ_{n}(λ) are the bound state “energies” of −\frac{d^{2}}{dx^{2}} + λq, then κ_{n}(λ) is increasing in λ for λ in (L_{n}; ∞) where L_{n} is the infimum over those λ with n or more bound states. This follows from first-order perturbation theory (Feynman-Hellman theorem), which implies that \frac{∂(−κ_{n}^{2})}{∂λ} = ⟨φ, qφ⟩ ≤ −κ_{n}^{2}/λ < 0.

(iv) There are no zeros of f in the quadrants Re κ > 0, Im κ ≠ 0 since they would correspond to imaginary eigenvalues.

(v) Resonances (zeros with Re κ < 0, Im κ ≠ 0) occur in complex conjugate pairs.

This means as λ increase and we look at eigenvalues of −\frac{d^{2}}{dx^{2}} + λq, the only way bound states can change is by an antibound state turning into a bound state. However, antibound states can change due to a complex pair of resonances turning into a pair of antibound states or vice-versa.

Now imagine λ increasing past L_{1}. A single antibound state turns into a bound state κ_{1}(λ). For λ near L_{1}, there is exactly one bound state/antibound state near 0 since d′(L_{1}) ≠ 0 by (ii). In particular, there are no antibound states in (−κ_{1}(λ), 0). Since −κ_{1}(λ) cannot be an antibound state, no antibound states can pass along the real axis into or out of (−κ_{1}(λ), 0) for λ ∈ (L_{1}, L_{2}). Only pairs of resonances can produce pairs of antibound states. Thus, for λ ∈ (L_{1}, L_{2}), there are an even number of antibound states in (−κ_{1}(λ), 0). At λ = L_{2}, a single antibound state turns into a bound state leaving an odd number in (−κ_{1}(λ), −κ_{2}(λ)). As λ increases, antibound states only get added or subtracted in pairs, so the number stays odd. The argument is simple for each similar interval (−κ_{n}(λ), −κ_{n+1}(λ)) for λ > L_{n+1}.

Second Proof of Theorem 6. This proof has some connection with an old paper of Ciafaloni-Menotti [6], who discuss a related result concerning alternation as coupling constant is changed rather as κ is varied. We define

\frac{d(−κ)}{d(κ)} = e(κ).  (4.1)

If κ_{j} is the jth bound state 0 < κ_{n} < κ_{n−1} < ⋯ < κ_{1}, then d(κ_{j}) = 0 and d(−κ_{j}) ≠ 0 (as in (i) in the last proof). We will show that

lim_{κ→κ_{j}}(κ − κ_{j})e(κ) < 0.  (4.2)

It follows that e(κ_{j} + ε) < 0 and e(κ_{j−1} − ε) > 0 for ε small and so, by continuity, e(κ) has an odd number of zeros (counting multiplicity) in (κ_{j}, κ_{j−1}). But zeros of d(−κ) are the same as zeros of e(κ).

We will prove (4.2) by using a representation for e(κ) due to Froese [8, 9]. In this case

d(κ) = det(1 + K(κ))

K(κ)(x, y) = V (x)^{1/2} \frac{sinh(κx_{<})}{κ} e^{−κx_{>}}|V (y)|^{1/2}

so

[K(κ) − K(−κ)](x, y) = 2κ^{−1}V (x)^{1/2} sinh(κx) sinh(κy)|V (y)|^{1/2}

is rank 1; call it A(κ). Then

d(−κ) = d(κ) det(1 − (1 + K(κ))^{−1}A(κ))
= d(κ)[1 − Tr((1 + K(κ))^{−1}A(κ))]

so [8, 9]:

e(κ) = 1 − 2κ^{−1}⟨|V |^{1/2}φ_{0}, (1 + K(κ))^{−1}V^{1/2}φ_{0}⟩,  (4.3)

where φ_{0}(x) = sinh(κx).

Let L(κ) = V^{1/2}(h_{0} + V + κ^{2})^{−1}|V |^{1/2}. Then, by the resolvent formula:

L(κ) = K(κ) − K(κ)L(κ)

or

(1 + K(κ))^{−1} = 1 − L(κ).

Then by (4.3)

e(κ) = 1 − 2κ^{−1}⟨φ_{0}, V φ_{0}⟩ + 2κ^{−1}⟨V φ_{0}, (h_{0} + V + κ^{2})^{−1}V φ_{0}⟩.

It follows from this that

lim_{κ→κ_{j}}(κ − κ_{j})e(κ_{j}) = −2κ_{j}^{−1}(2κ_{j})^{−1}|⟨V φ_{0}, η_{j}⟩|^{2},

where η_{j} is the normalized eigenvector for H_{0} + V at energy −κ_{j}^{2}. Thus, the limit is non-positive. Since e has a pole, the limit is non-zero, and so (4.2) is proven.

Third Proof of Theorem 6 (Graf [11]). Let u(x, κ) solve the equation −u′′ + qu = −κ^{2}u with u(0) = 0, u′(0) = 1. As is well known, u(x, κ_{j}) has j − 1 zeros, so since u(x, κ) > 0 for x small, (−1)^{j−1}u(x, κ_{j}) > 0 for x near infinity. Since q has compact support, say [0, a],

u(x, κ) = α(κ) e^{−κx} + β(κ) e^{κx}, x ≥ a

and α and β are continuous since u(x, κ), u′(x, κ) are, and α, β can be expressed in terms of suitable u, u′ data. By the sign condition on u near infinity

(−1)^{j−1}α(κ_{j}) > 0,  (4.4)

where one has strict positivity since u cannot be identically zero on (a, ∞).

By the association of f and the Jost function, f(−κ) = 0 if and only if the solution which is e^{κx} near infinity vanishes at x = 0, that is, if and only if α(κ) = 0. By (4.4), α has an odd number of zeros in (κ_{j}, κ_{j−1}).

These proofs were stated for the case where H has u(0) = 0 boundary conditions, but each proof can be modified to handle the other half-line cases. For example, the third proof accommodates u′(0)+hu(0) = 0 boundary conditions by looking at the solution obeying that boundary condition and u(0) = 1 normalization. And it handles an \frac{ℓ(ℓ+1)}{x^{2}} term by replacing exponentials by suitable Bessel functions.

The proofs also accommodate potentials with superexponential decay with minor modification. For example, the second proof applies verbatim to such potentials.

An illuminating example is the Bargmann potential [5] with Jost function (suggested by Newton [22])

f(κ) = \frac{(κ − k_{1})(κ − k_{2})}{(κ + k_{3} + ik_{4})(κ + k_{3} − ik_{4})} ,

where k_{j} > 0 for j = 1, 2, 3, 4. Such a potential has no antibound states and seemingly violates Theorem 6. The point, of course, is that q is not superexponential in this case, but only decays as e^{−αx} with α = min(k_{1}, k_{2}, k_{3}). In general, Theorem 6 does hold for potentials with exponential decay, say, bounded by e^{−αx}, but only for intervals (−κ_{j}, −κ_{j−1}) with κ_{j} < α.

Appendix A. The Determinant of a Green’s Matrix

Let {a_{i}}_{i=1}^{N}, {b_{i}}_{i=1}^{N} be two sequences and let C be the N × N matrix

c_{ij} = a_{min(i,j)}b_{max(i,j)};  1 ≤ i, j ≤ N.  (A.1)

Our main goal in this appendix is to give a simple proof of the following theorem of Barrett-Feinsilver [3], which we used in Section 2:

Theorem A.1. Let C be a matrix given by (A.1). Then

det(C) = b_{N}(a_{N}b_{N−1} − a_{N−1}b_{N})(a_{N−1}b_{N_{2}} − a_{N−2}b_{N−1}) … (a_{2}b_{1} − a_{1}b_{2})a_{1}. (A.2)

Remarks. 1. Barrett-Feinsilver [3] actually state the theorem for matrices closely related to ones given by A.1 and express the result in terms of the c_{ij}, viz:

det(C) = ∏_{i=1}^{N−1}(c_{i,i}c_{i+1,i−1} − c_{i,i+1}c_{i+1,i})/ ∏_{i+1}^{N} c_{ii}.

2. The proof in [3] is combinatoric but has the advantage of generalizations in a variety of directions [2, 15, 4].

3. An idea similar to our proof in a related context appears in Jost-Pais [12].

4. Matrices of type (A.1) were dubbed Green’s matrices by Karlin [14] since they are a discrete analog of one-dimensional Green’s functions. Indeed, they occur in Section 2 in connection with free Green’s functions. And they arise as the inverses of symmetric tridiagonal matrices

J = (\begin{matrix}x_{1} & y_{1} … & 0 \\ y_{1} & … & 0 \\ ⋮ & ... & ⋮ \\ 0 & … y_{N−1} & x_{N} \end{matrix} )  (A.3)

in that there is a one-one correspondence realized by the inverse between invertible matrices of the form (A.1) and invertible matrices of the form (A.3). This is a theorem of Gantmacher-Krein [10] (see also Barrett [2]). Indeed, given an invertible (A.3), a can be determined by

a_{1} = 1; x_{1}a_{1} + y_{1}a_{2} = 0; y_{n−1}a_{n−1} + x_{n}a_{n} + y_{n}a_{n+1} = 0, n = 2, … N − 1 (A.4)

y_{N−1}b_{N−1} + x_{N}b_{n} = 0; y_{n−1}b_{n−1} + x_{n}b_{n} + y_{n}b_{n+1} = 0, n = 2, …, N − 1, (A.5)

where b is normalized by

y_{n}(a_{n+1}b_{1} − b_{n+1}a_{n}) = 1,  n = 1, …, N − 1.  (A.6)

[Note: (A.4)/(A.5) imply the left side of (A.6) is independent of n and it is non-zero if J is invertible.] Conversely, given a matrix (A.1), one defines y_{n} by (A.6). [Note: det(C) ≠ 0 means (a_{n−1}b_{n} −b_{n−1}a_{n}) ≠ 0 by Theorem A.1.] And then one defines x_{n} by (A.4) or (A.5).

Proof of Theorem A.1. We prove the result by induction in N. The result for N = 1 is obvious. If we can prove the result when b_{N−1} ≠ 0, it follows by continuity for all b_{N−1}, so suppose b_{N−1} ≠ 0. Consider the last row and column of C. It has a_{N}b_{N} in the corner and otherwise every element is a_{j}b_{N} for some j ∈ {1, …, N − 1}. It follows that

det(C) = a_{N}b_{N} det(C_{N−1}) + b_{N}^{2}F (a_{1}, …, a_{n−1}; b_{1}, … b_{N−1})

where C_{N−1} is the (N − 1) × (N − 1) Green’s matrix with the last row and column removed and F is some function of {a_{i}}_{i=1}^{N−1} and {b_{i}}_{i=1}^{N−1}.

Fix these values of a and b and think of a_{N} and b_{N} as variable and det(C) as a function of them. When a_{N} = a_{N−1} and b_{N} = b_{N−1}, then C has the identical rows, so det(C) = 0, that is,

a_{N−1}b_{N−1} det(C_{N−1}) + b_{N−1}^{2}F = 0

so, since b_{N−1} ≠ 0, F = −a_{N−1} det(C_{N−1})/b_{N−1} and thus

det(C) = b_{N}(a_{N}b_{N−1} − a_{N−1}b_{N}) det(C_{N−1})/b_{N−1}.  (A.7)

But by induction, we can assume (A.2) for det(C_{N−1}) and thus (A.7) implies (A.2) for det(C).

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Division of Physics, Mathematics, and Astronomy 253-37, California Institute of Technology, Pasadena, CA 91125-3700

   E-mail address: bsimon@caltech.edu