Trace formulas for Schr\"odinger operators on the half-line

We study the scattering problem for the Schr\"odinger equation on the half-line with Robin boundary condition at the origin. We derive an expression for the trace of the difference of the perturbed and unperturbed resolvent in terms of a Wronskian. This leads to a representation for the perturbation determinant and to trace identities of Buslaev-Faddeev type.


Introduction
Let H be the self-adjoint operator in L 2 [0, ∞) defined by where γ ∈ R. The potential V is real-valued and goes to zero at infinity (in some averaged sense).
Then H has continuous spectrum on the positive semiaxis and discrete negative spectrum, consisting of eigenvalues {λ j }. If V decays fast enough, then there are only finitely many negative eigenvalues.
The Hamiltonian H describes a one-dimensional particle restricted to the positive semiaxis. The parameter γ describes the strength of the interaction of the particle with the boundary. Negative γ corresponds to an attractive interaction and positive γ to a repulsive one.
In this paper we derive trace formulas for the negative eigenvalues of H. Formulas of this type first appeared in 1953 in the paper of Gel'fand and Levitan, [9], where some identities for the eigenvalues of a regular Sturm-Liouville operator were obtained. Later, also Dikiȋ studied similar formulas, see [8]. The next important contribution in this direction was made by Buslaev and Faddeev [6] in 1960. They studied the singular Sturm-Liouville operator on the half-line with Dirichlet boundary condition at the origin. Under some assumptions on the short range potential (i.e. integrable on (0, ∞) with finite first moment), they proved a series of trace identities. The second one in this series states that whereη(k) is the so-called limit phase and has a scattering theoretical nature. A more precise definition will be given later. This result was extended in 1997 by Rybkin to long-range potentials (nonintegrable on (0, ∞)), [2,15]. Analog formulas for charged particles were obtained already in 1972 by Yafaev [16]. Trace formulas for the whole line Schrödinger operator as well as their generalizations to the multidimensional case have already been studied extensively (see, e.g., the surveys [3,4,12]) and have great importance in applications. Numerous papers are devoted to the subject of inverse spectral problems for Schrödinger operators, where these formulas turn out to be very useful, see e.g. [1,5,7,10,11] and references therein. Another consequence are the well-known Lieb-Thirring inequalities, which in dimension one follow from the third Faddeev-Zakharov trace formula, see [18] and [14]. This formula was extended in [13] by Laptev and Weidl to systems of Schrödinger operators, which leads to sharp Lieb-Thirring inequalities in all dimensions.
Our goal is to prove the analog of the Buslaev-Faddeev trace formulas for the half-line Schrödinger operator with Robin boundary conditions (1.1). Thereby, we follow Yafaev's book "Mathematical Scattering Theory, Analytic Theory" [17], which contains complete proofs in the case of Dirichlet boundary condition. We aim to point out the differences arising from the Robin boundary conditions and to give an interpretation for them.
The outline of this paper is as follows. We consider the differential equation where ζ ∈ C and x > 0. We are concerned with two particular solutions of this equation, the regular solution ϕ and the Jost solution θ. The first one is characterized by the conditions ϕ(0, ζ) = 1, ϕ ′ (0, ζ) = γ, (1.4) and the latter one by the asymptotics θ(x, ζ) ∼ e iζx as x → ∞.
In section 2 we prove existence and uniqueness of the regular solution. The corresponding properties of the Jost solution are well-known. Further, we introduce a quantity w(ζ), which we call the Jost function. We emphasize that this function depends on γ and does not coincide with what is called the Jost function in the Dirichlet case. More precisely , w(ζ) is defined as the Wronskian of the regular solution and the Jost solution of (1.3). It turns out that Section 3 contains our first main result. Denoting the resolvents of the unperturbed and perturbed operators by R 0 (z) = (H 0 − z) −1 and R(z) = (H − z) −1 , respectively, we derive an expression for Tr(R(z) − R 0 (z)) in terms of the Jost function.
From this relation we get a representation for the perturbation determinant in terms of w(ζ). Section 4 deals with the asymptotic expansion of the perturbation determinant, which we shall use to derive trace identities in Section 5. For complex numbers s, we define the function Under some regularity and decay assumptions on the potential V we prove infinitely many trace identities. The analogous to (1.2) will be given by where η(k) is now the corresponding limit phase for the Robin boundary problem. We recall that if γ ≥ 0, then H 0 has purely absolutely continuous spectrum [0, ∞). If γ < 0, then H 0 has a simple negative eigenvalue −γ 2 and purely absolutely continuous spectrum on [0, ∞). Hence the first two terms on the left-hand side of (1.6), N j=1 |λ j | − M 1 (γ), correspond to the shift of the discrete spectrum between H and H 0 . Similarly, the last term on the left-hand side corresponds to the shift of the absolutely continuous spectrum. The trace formula (1.6) and its higher order analogs proved below relate this shift of the spectrum to the potential V .
Finally, in Theorem 5.5 we prove a trace formula of order zero. Namely, the so-called Levinson formula for the Schrödinger operator H with Robin boundary condition.

The regular solution and the Jost solution
In this section, we prove existence and uniqueness of the regular solution and recall some elementary results on the Jost solution. The γ-dependent Jost function is studied.
2.1. The associated Volterra equation and auxiliary estimates. Existence and uniqueness of the regular solution of (1.3) can be proved by using Volterra integral equations. For different boundary conditions, equation (1.3) is associated with different Volterra integral equations.
considered on locally bounded functions ϕ.
Proof. Suppose that equation (1.3) holds for ϕ. Then the equality is true. We integrate the right-hand side twice by parts. Taking into account boundary conditions (1.4), we see that the right-hand side equals ϕ(x, ζ)− cos(ζx)− γ ζ sin(ζx). Thus equation (2.1) follows. Conversely, assume that equation (2.1) holds. Then ϕ ∈ C 1 loc ([0, ∞)) and Therefore ϕ ′ is absolutely continuous and In the following Lemma it is proved that the regular solution ϕ(x, ζ) of (1.3) with boundary conditions (1.4) exists uniquely. For the case of Dirichlet boundary condition, this result was represented e.g. by Yafaev in [17].
From this equation, we obtain in the limit N → ∞ equation (2.1). To prove estimate (2.4), we consider Because of (2.8), the right-hand side in (2.11) is bounded from above bỹ If γ = 0, then we use the same estimates (2.7) and get successively, From this estimate it follows, that which proves estimate (2.5). The uniqueness of a bounded solution of equation (2.1) can be proved by contradiction. Suppose that ϕ 1 and ϕ 2 are two different solutions of equation (2.1). Then ϕ 1 − ϕ 2 satisfies the corresponding homogeneous equation and is bounded for an arbitrary n, by the righthand side of (2.8) and hence is zero. Therefore ϕ 1 = ϕ 2 .
2.2. The Jost solution and the Jost function. The so-called Jost solution, which was first studied by Jost, is important in scattering theory. This solution of equation (1.3) is characterized by the asymptotics θ(x, ζ) ∼ e iζx as x → ∞. It is proved, e.g., in [17], that under the assumption For any fixed x ≥ 0, the function θ(x, ζ) is analytic in ζ in the upper half-plane Im ζ > 0 and continuous in ζ up to the real axis with a possible exception of the point ζ = 0. Moreover, it satisfies the estimates where C depends on c and the value of the integral (2.12) only. We will need an analog of estimate (2.14) for the derivative of the Jost solution.
Lemma 2.3. Assume condition (2.12) and ζ = 0, Im ζ > 0. Then for the derivative of the solution θ(x, ζ) with asymptotics (2.13), the following estimate holds where K depends only on k and the value of the integral (2.12).
The proof of this Lemma follows closely the arguments of [17]. For the sake of completeness, we provide the necessary modifications in the appendix.
Next, we study some properties of the γ-dependent Jost function. Below we suppose that condition (2.12) is satisfied and that Im ζ ≥ 0. Setting x = 0 in (2.17), we see that This is the definition of w(ζ), that was used in the introduction. The Jost function w(ζ) is analytic in ζ in the upper halfplane Im ζ > 0 and is continuous in ζ up to the real axis, with a possible exception of the point ζ = 0. Moreover, it follows from (2.14) and (2.16) that Remark 2.5. Usually, in the literature the Wronskian w D of the Jost solution and the regular solution satisfying a Dirichlet boundary condition is called Jost function. In our case of Robin boundary condition (1.4), the Wronskian differs from the usual one and depends on γ. We emphasize, that for every γ ∈ R, the function w(ζ) grows linearly in ζ as |ζ| → ∞, whereas in the Dirichlet case we have for the corresponding Jost function w D (ζ) Our next goal is to give an integral representation for w(ζ).
We integrate the first integral in the right-hand side twice by parts and get Passing to the limit x → ∞ in the above equation and using (2.21), we arrive at (2.20) for Im ζ > 0. By continuity, (2.20) can be extended to the real axis.
As the Jost solution of equation (1.3) is unique, it follows that For real numbers k > 0 both Jost solutions θ(x, k) and θ(x, −k) of the equation are correctly defined and their Wronskian W {θ(·, k), θ(·, −k)} equals 2ik. Thus, they are linearly independent. In particular, we get from (2.23), It is useful to express the regular solution in terms of the Jost solutions as follows, Indeed, it is easy to verify that the right-hand side of (2.26) satisfies equation (2.24) and conditions (1.4). Now, we introduce the limit amplitude and phase shift for real values of k.
The functions a(k) and η(k) are called the limit amplitude and the limit phase, respectively.
These functions determine the asymptotics of the regular solution of the Schrödinger equation as x → ∞. Indeed, comparing (2.13) and (2.26), we find Thus, This asymptotic behavior should be compared with the exact expression for the solution ϕ 0 (x, ζ) of the equation −ϕ ′′ = ζ 2 ϕ satisfying the conditions (1.4), namely, Finally, we note, that w(k) = 0 for all k > 0.
(2.28) Indeed, if there was a number k such that w(k) = 0, then it would follow from relations (2.25) and (2.26), that ϕ(x, k) = 0 for all x.

A Trace formula and the perturbation determinant
We consider the Hamiltonian we denote the free Hamiltonian with the same boundary condition (1.4) but with V ≡ 0. The resolvents of H and H 0 are denoted by R(z) and R 0 (z), respectively.
In this section, we derive an expression for Tr (R(z) − R 0 (z)) in terms of the Jost function. From this relation we get a representation for the perturbation determinant.
It is a well-known fact, that R(z) can be constructed in terms of solutions ϕ(x, ζ) and θ(x, ζ) of equation (1.3) and their Wronskian (2.17). Suppose, that (2.12) holds. Then for all z such that Im z = 0 and w(ζ) = 0, the resolvent is the integral operator with kernel and R(x, y; z) = R(y, x; z). Moreover, the estimate holds. We note, that in the particular case V ≡ 0, the unperturbed resolvent R 0 (z) has the integral kernel 3) The self-adjoint operator H has discrete negative spectrum, which consists of negative eigenvalues λ j = (iκ j ) 2 , κ j > 0, which possibly accumulate at zero. It is important to note, that the zeros of the function w(ζ) and the eigenvalues of H are related as follows. Proof. First, assume that w(ζ) = 0 for Im ζ > 0. Then the Jost function θ(x, ζ) fullfills boundary conditions (1.4) and is in the space L 2 (R + ) because of (2.13). Thus θ(x, ζ) is an eigenfunction of the operator H corresponding to the eigenvalue λ = ζ 2 . Since H is self-adjoint, it follows that λ < 0. Conversely, assume that λ is an eigenvalue of H. Then its resolvent R(z) has a pole in λ. Therefore, it follows from (3.2) that w(ζ) = 0. As the resolvent of a self-adjoint operator has only simple poles, the zeros of w(ζ) are simple.
Proof. Since R − R 0 is a trace class operator and kernels of the operators R and R 0 are continuous functions, we have (R(y, y; z) − R 0 (y, y; z)) dy.
The following equation is true for any two arbitrary solutions of equation (1.3) Applying (3.7) to the regular solution ϕ(x, ζ) and the Jost solution θ(x, ζ), we get 2ζw(ζ) x 0 R(y, y; z) dy = 2ζ x 0 ϕ(y, ζ)θ(y, ζ) dy = ϕ ′ (y, ζ)θ(y, ζ) − ϕ(y, ζ)θ ′ (y, ζ) x 0 . (3.8) Note, that the contribution of the right-hand side in (3.8) for y = 0 is Consider now the case of potentials of compact support. Then, for all ζ ∈ C , we have θ(x, ζ) = e iζx for sufficiently large x. Further, in this case we can generalize (2.26) to complex ζ, namely For large x we have, Taking into account that, since Im ζ > 0, the terms containing e 2ixζ tend to zero as x → ∞ and using (3.10), we get for sufficiently large x Combining (3.9) and (3.11), we arrive at Finally, we conclude from (3.6) and (3.12), that lim x→∞ x 0 (R 0 (y, y; z) − R(y, y; z)) dy = 1 2ζ is well-defined. Here ρ(H 0 ) denotes the resolvent set of the operator H 0 . Furthermore, it can easily be verified that the perturbation determinant is related to the trace of the resolvent difference by

Thus, it follows from (3.4) that
Therefore, we conclude that D(z) = Cw( √ z)/(γ − i √ z). Because of the asymptotics (2.19), it follows that This is the sought after relation.

Low and High-energy asymptotics
Here we derive an asymptotic expansion of the perturbation determinant D(ζ) as |ζ| → ∞.
It follows from (4.2) that asymptotics (2.13) and the asymptotics are equivalent to each other. For an arbitrary N , the equality holds with the remainder satisfying the estimates for all x ≥ 0 and Im ζ ≥ 0, |ζ| ≥ c > 0. Here b 0 (x) = 1 and b n (x) are real C ∞ functions defined by the recurrent relation Further the following estimates hold, b (j) n (x) = O(x −n(ρ−1)−j ), j ∈ N 0 , x → ∞. Now, we can prove the asymptotic expansion of the perturbation determinant for |ζ| → ∞. as |ζ| → ∞, Im ζ ≥ 0. The coefficients d n are given by We emphasize that expansion (4.6) is understood in the sense of an asymptotic series.
Proof. It follows from (4.5) that (4.8) Applying the geometric series to the first sum in the right-hand side of (4.8) for |ζ| > γ, we conclude On the other hand, it is easy to see that Thus, setting x = 0 and combining (4.9) with (4.10), we arrive at (4.6).
Note, that because of (2.19), we have for |ζ| → ∞, Im ζ ≥ 0, Thus, we can fix the branch of the the function ln D by the condition ln D(ζ) → 0 as |ζ| → ∞. The following Corollary is an immediate consequence of Lemma 4.1.
Seperating in Corollary 4.2 the function ln D(k) into its real and imaginary part, we finally conclude that for k → ∞,

4.2.
Low-energy asymptotics. In this section we assume, that (4.14) We denote the regular solution for ζ = 0 by ϕ(x). This is the solution of the integral equation (2.1), This solution exists under condition (2.12). As shown in [17], the stronger condition (4.14) guarantees the existence of a Jost solution θ(x, ζ) at ζ = 0. For any fixed x ≥ 0, the Jost solution θ(x, ζ) is continuous as ζ → 0, Im ζ ≥ 0. Moreover, where C does not depend on ζ and x. The function θ(x) := θ(x, 0) = θ(x, 0) satisfies the equation and, as x → ∞, Indeed, asymptotics (4.17) follow from the integral equation (A.1) for ζ = 0, namely, One can also show that the Jost function w(ζ) is continuous as ζ → 0, Im ζ ≥ 0, and from (2.20) we get for ζ = 0 If (4.14) holds, then the integral in (4.18) is convergent, in view of the estimate following from (2.4). Moreover, we have w(0) = w(0). After these preliminaries we claim that the operator H has no zero eigenvalue. Indeed, the function defined by is a solution of equation (4.16) and is linearly independent of θ(x). Again, x 0 is an arbitrary point such that θ(x) = 0 for x ≥ x 0 . Further, Thus, the equation (4.16) does not have solutions, tending to zero at infinity, as claimed.
While the operator does not have a zero eigenvalue, it may have a so-called zero resonance.
Since the Jost function is the Wronskian of the Jost and the regular solution, the resonance condition means that ϕ is a multiple of θ and therefore that equation (4.16) has a bounded solution satisfying boundary condition (1.4).
The third and fourth term in the right-hand side of (4.22) are bounded from above by C|ζ|xe Im ζx . This follows in the same way as shown in [17], which only uses that ϕ(x) is bounded. It remains to give an estimate for the first and second term in (4.22), which we write as cos(ζx) − 1 = −2 sin 2 (ζx/2).
In order to prove that w 0 = 0, we use equation (4.15) to write as x → ∞. On the other hand, ϕ is proportional to θ, which satisfies (4.17). This shows that w 0 = 0, as claimed.

Trace identities
We now put the material from the previous sections together to prove our main result, namely, a family of trace formulas for the operator H. These identities provide a relation between the shift of the spectra between H and H 0 and quantities involving the potential V . The spectral shift consists of two parts, one coming from the discrete spectrum (expressed in terms of the eigenvalues of H and H 0 ) and the other one coming from the continuous spectrum (expressed in terms of the quantities η and a).
In this section we assume that ∞ 0 (1 + x)|V (x)| dx < ∞, which guarantees that H has only a finite number N of negative eigenvalues λ 1 , . . . , λ N . We recall that H 0 has a single negative eigenvalue −γ 2 if γ < 0 and no negative eigenvalues if γ ≥ 0. We also recall that M s (γ) was defined at the end of the introduction.
While we are mainly interested in trace formulas of integer and half-integer order, we prove a version of these formulas for every complex s with Re s > 0. We proceed by analytic continuation, where the starting point is the following proposition. Proof. Let Γ R,ε be the contour (with counterclockwise direction) which consists of the halfcircles C + R = {|ζ| = R, Im ζ ≥ 0} and C + ε = {|ζ| = ε, Im ζ ≥ 0} and the intervals (ε, R) and (−R, −ε). Figure 1. contour of integration The argument of ζ ∈ C is fixed by the condition 0 ≤ arg ζ ≤ π. We consider the integral The set of the singularities of the integrand is the set of zeros of the function w(ζ)/(γ − iζ).
Calculating the integral by residues, we see that for κ j = |λ j | 1/2 Since by Lemma 3.1, zeros iκ j of w(ζ) are simple, the residues work out to be e iπs κ 2s j . Hence, Next, we show that the integral over the semicircle C + r tends to zero as r → ∞ or r → 0. Integrating by parts, we see that Note, that we can choose ln(w(ζ)/(γ − iζ)) as a continuous function on C + r . If r → ∞, then this integral tends to zero for Re s < 1/2 because of (4.11). If r → 0, then the integral also tends to zero. Indeed, this follows from the fact that either w(0) = 0 or w(ζ) satisfies (4.23) with w 0 = 0. Therefore passing to the limits R → ∞ and ε → 0 in equality (5.4), we obtain that Integrating in the left-hand side by parts and taking into account relations (2.25) and (2.27), we obtain = 2s(e 2iπs − 1)F (s) − 2is(e 2iπs + 1)G(s).
In order to prove trace identities for arbitrary powers s ∈ C + , we need the analytic continuation of the functions F (s) and G(s) to the entire half-plane Re s > 0.
If n ≥ 1 and n − 1/2 < Re s < n + 1/2, then Proof. We can write the function F , given in Lemma 5.2 as follows The first integral in the right-hand side of equation (5.7) is an analytic function of s in the entire half-plane Re s > 0. The second integral is in view of (4.13) an analytic function of s in the strip 0 < Re s < n + 1. For Re s > n, we have Thus, the function F is an analytic function in the strip n < Re s < n + 1. Similarly, we split the integral in the right-hand side of (5.6). Note, that we now have for n ≥ 1 and Re s > n − 1/2, Therefore, it follows with analog arguments as for F that the function G, given in (5.6), is an analytic function of s in the strip n − 1/2 < Re s < n + 1/2. Theorem 5.3. Let estimates (4.1) and and for n ≥ 1, n ∈ N, 10) The coefficients ℓ n are given as in (4.12).
We compute the first four trace formulas.
Corollary 5.4. Let estimates (4.1) and Finally, we prove a trace formula of order zero for the operator H with boundary conditions (1.4). Such formulas are called in the literature the Levinson formula and relate the number of negative eigenvalues of H to the phase shift η.
Theorem 5.5. Suppose (4.14) and let N be the number of negative eigenvalues of the operator H with boundary condition (1.4). Then, the following formulas hold. For w(0) = 0, For w(0) = 0, Proof. We apply the argument principle to the function D(ζ) and the contour Γ R,ε given in Figure  1. We choose R and ε such that all of the N negative eigenvalues of H lie inside the contour Γ R,ε . Remember that if γ < 0, then H 0 has a simple negative eigenvalue −γ 2 . Thus, it follows with Lemma 3.1 that Note, that the branch of the function ln D(ζ) was fixed by the condition ln D(ζ) → 0 as |ζ| → ∞. Thus, we have ln D(ζ) = ln |D(ζ)| + iarg D(ζ). As for k ∈ R, arg D(k) = η(k) and η(−k) = −η(k), it follows from equation (5.13) that Note, that lim R→∞ var C + R arg D(ζ) = 0 because of (4.11). To compute lim ε→0 var C + ε arg D(ζ), we rewrite var C + ε arg D(ζ) = var C + ε arg w(ζ) − var C + ε arg(γ − iζ). The second term in the right-hand side of (5.15) depends on the sign of γ and turns out to be in the limit, This proves (2.15). If |ζ| ≥ k > 0 and condition (2.12) is satisfied, then estimate (2.16) follows from (2.15).