Coupling constant limits of Schrödinger operators with critical potentials

A family of Schrödinger operators, P(\lambda )=P_0+\lambda V, is studied in this paper. Here P_0=-\mathrm{\Delta }+f(x) with f(x)\sim \frac{1}{{|x|}^2} when |x| is large enough and V(x)=O({|x|}^{-2-ϵ}) for some ϵ>0. We show that each discrete eigenvalue of P(\lambda ) tends to 0 when λ tends to some {\lambda }_0. We get asymptotic behavior of the smallest discrete eigenvalue when λ tends to {\lambda }_0.

In this paper, we consider a family of Schrödinger operators P(\lambda ) which are the perturbation of P_0 in the form

on L^2({\mathbb{R}}^d), d\ge 2. Here P_0=-\mathrm{\Delta }+\frac{q(\theta )}{r^2}. (r,\theta ) are the polar coordinates on {\mathbb{R}}^d, and q(\theta ) is a real continuous function. V\le 0 is a non-zero continuous function satisfying

Here 〈x〉={(1+{|x|}^2)}^{1/2}. Let {\mathrm{\Delta }}_s denote the Laplace operator on the sphere {\mathbb{S}}^{d-1}. Assume that

If (2) holds, then P_0\ge 0 in L^2({\mathbb{R}}^d) (see [1]).

Under the assumption on V, we know that P(\lambda ) has discrete eigenvalues when λ is large enough, and each discrete eigenvalue tends to zero when λ tends to some {\lambda }_0 (see Section 2). We study the asymptotic behaviors of the discrete eigenvalues of P(\lambda ) in this paper. The asymptotic behaviors for Schrödinger operators with fast decaying potentials were studied by Klaus and Simon [2]. In [2], they studied the convergence rate of discrete eigenvalues of H(\lambda )=-\mathrm{\Delta }+\lambda V when \lambda \to {\lambda }_0. {\lambda }_0 is the value at which some discrete eigenvalue e_i(\lambda ) tends to zero. The main method they used in their paper is the Birman-Schwinger technique.

In order to use the Birman-Schwinger technique to P(\lambda ), we need to get the asymptotic expansion of {(P_0-\alpha )}^{-1} for α near zero, , which was studied by Wang [1]. In this paper, we first show that there exists some {\lambda }_0 such that when \lambda >{\lambda }_0, P(\lambda ) has discrete eigenvalues. Then, we define the Birman-Schwinger kernel K(\alpha ) for P(\lambda ) and find that there is one-to-one correspondence between the discrete eigenvalues of K(\alpha ) and the discrete eigenvalues of P(\lambda ). Hence, the asymptotic expansion of the discrete eigenvalue of P(\lambda ) can be got through the asymptotic expansion of the discrete eigenvalue of K(\alpha ). In our main results, we need to use that K(\alpha ) is a bounded operator from L^2 to L^2. To get that, we add a strong condition on V (i.e., {\rho }_0>6 in (1)). We show that K(\alpha ) is a family of compact operators converging to K(0) and obtain the asymptotic expansions of the discrete eigenvalues of K(\alpha ) by functional calculus. After that, the convergence rate of the smallest discrete eigenvalue of P(\lambda ) is obtained.

Here is the plan of our work. In Section 2, we recall some results of P_0 and define the Birman-Schwinger kernel K(\alpha ) for P(\lambda ). The relationship between the eigenvalues of these two kinds of operators is studied. In Section 3, we first study the asymptotic behavior of the discrete eigenvalues of K(\alpha ). Then the convergence rate of the smallest discrete eigenvalue of P(\lambda ) is obtained. We get the leading term and the estimate of the remainder term of the smallest discrete eigenvalue.

Let us introduce some notations first.

Notation The scalar product on L^2({\mathbb{R}}^{+};r^{d-1}dr) and L^2({\mathbb{R}}^d) is denoted by 〈\cdot ,\cdot 〉 and that on L^2({\mathbb{S}}^{d-1}) by (\cdot ,\cdot ). H^{r,s}({\mathbb{R}}^d), r\in \mathbb{Z}, s\in \mathbb{R}, denotes the weighted Sobolev space of order r with volume element {〈x〉}^{2s}dx. The duality between H^{1,s} and H^{-1,-s} is identified with the L^2 product. Denote H^{0,s}=L^{2,s}. Notation \mathcal{L}(H^{r,s},H^{r^{\mathrm{\prime }},s^{\mathrm{\prime }}}) stands for the space of continuous linear operators from H^{r,s} to H^{r^{\mathrm{\prime }},s^{\mathrm{\prime }}}. The complex plane ℂ is slit along positive real axis so that z^{\nu }=e^{\nu lnz} and lnz=ln|z|+iargz with are holomorphic there.

Assume that (r,\theta ) are the polar coordinates on {\mathbb{R}}^d. Then the condition

implies

in L^2({\mathbb{R}}^d) (see [1]).

Now, we recall some results on the resolvent and the Schrödinger group for the unperturbed operator P_0. Let

Denote

For \nu \in {\sigma }_{\mathrm{\infty }}, let n_{\nu } denote the multiplicity of {\lambda }_{\nu }={\nu }^2-\frac{{(d-2)}^2}{4} as the eigenvalue of -{\mathrm{\Delta }}_s+q(\theta ). Let {\phi }_{\nu }^{(j)}, \nu \in {\sigma }_{\mathrm{\infty }}, 1\le j\le n_{\nu } denote an orthogonal basis of L^2({\mathbb{S}}^{d-1}) consisting of eigenfunctions of -{\mathrm{\Delta }}_s+q(\theta ):

Let {\pi }_{\nu } denote the orthogonal projection in L^2({\mathbb{S}}^{d-1}) onto the subspace spanned by the eigenfunctions of -{\mathrm{\Delta }}_s+q(\theta ) associated with the eigenvalue {\lambda }_{\nu }, and let {\pi }_{\nu }^{(i)} denote the orthogonal projection in L^2({\mathbb{S}}^{d-1}) onto the eigenfunction {\phi }_{\nu }^{(i)}:

Denote for \nu \in {\sigma }_{\mathrm{\infty }}

Here {\nu }^{\mathrm{\prime }}=\nu -[\nu ], and [\nu ] is the largest integer which is not larger than ν. For \nu >0, let {[\nu ]}_{-} be the largest integer strictly less than ν. When \nu =0, set {[\nu ]}_{-}=0. Define {\delta }_{\nu } by {\delta }_{\nu }=1, if \nu \in {\sigma }_{\mathrm{\infty }}\cap \mathbb{N}, {\delta }_{\nu }=0, otherwise. One has [\nu ]={[\nu ]}_{-}+{\delta }_{\nu }.

The following is the asymptotic expansion for the resolvent R_0(z)={(P_0-z)}^{-1}.

Theorem 2.1 (Theorem 2.2 [1])

The following asymptotic expansion holds for z near zero with \mathrm{\Im }z>0:

in \mathcal{L}(-1,s;1,-s), s>2N+1. Here

Here

for , and

with P_{j,\nu }(\rho ) a polynomial in ρ of degree j:

First, we show that P(\lambda ) has discrete eigenvalues when λ is large enough. In fact, we need only to show that there exists a function \psi \in L^2({\mathbb{R}}^d) such that .

From the assumption on V, we know that there exists a point x_0\in {\mathbb{R}}^d such that V(x_0)={inf}_{x\in {\mathbb{R}}^d}V(x). Choose \delta >0 small enough such that for all x\in B(x_0,\delta ), . For \psi \in C_0^{\mathrm{\infty }}({\mathbb{R}}^d), \parallel \psi (x)\parallel =1, supp\psi \subset B(x_0,\delta ), one has

when λ is large enough, one has . This means that P(\lambda ) has discrete eigenvalues when λ is large enough.

P(\lambda ) has a continuous spectrum [0,\mathrm{\infty }) for \lambda \ge 0 because {lim}_{|x|\to \mathrm{\infty }}V(x) exists and equals zero (see [3]). We know that \sigma (P(0))=\sigma (P_0)=[0,\mathrm{\infty }). Hence, from the continuity of a discrete spectrum of P(\lambda ), we know that there exists some {\lambda }_0 such that when \lambda >{\lambda }_0, P(\lambda ) has eigenvalues less than zero, and when \lambda \le {\lambda }_0, \sigma (P(\lambda ))=[0,\mathrm{\infty }). So, P(\lambda ) has an eigenvalue at the bottom of its spectrum for \lambda >{\lambda }_0. In Section 3 (Proposition 3.1), we prove that e_1(\lambda ) is simple and the corresponding eigenfunction can be chosen to be positive everywhere. (There are many results about the simplicity of the smallest eigenvalue of the Schrödinger operator without singularity, but there is no result which can be used directly, because the potential we use in this paper has singularity at zero. Theorem XIII.48 [4] can treat the Schrödinger operator with the potential which has singularity at zero, but the positivity of potential is needed. Hence, we give this result.) From the discussion above and the continuity of a discrete spectrum, one has that e_1(\lambda ) tends to zero at some λ. The asymptotic behavior of e_1(\lambda ) is studied in this paper.

To study the eigenvalues of P(\lambda ), we first define a family of Birman-Schwinger kernel operators. Let

for z\notin \sigma (P_0), and

Then we have the following result.

Proposition 2.2 Let . Then

1. (a)

   Let

   \begin{array}{c}A=\{\psi \in L^2\left({\mathbb{R}}^d\right);(P(\lambda )-\alpha )\psi =0\},\hfill \\ B=\{\varphi \in L^2\left({\mathbb{R}}^d\right);K(\alpha )\varphi ={\lambda }^{-1}\varphi \}.\hfill \end{array}

Then {|V|}^{1/2} is injective from A to B, and {(P_0-\alpha )}^{-1}{|V|}^{1/2} is injective from B to A.

1. (b)

   The multiplicity of α as the eigenvalue of P(\lambda ) is exactly the multiplicity of {\lambda }^{-1} as the eigenvalue of K(\alpha ).

Proof (a) First, we prove that {|V|}^{1/2} is injective from A to B. Note that if \psi \in A, then

with \varphi ={|V|}^{1/2}\psi. And if \varphi =0, then

It follows that {|V|}^{1/2} is injective from A to B.

Next, we show that {(P_0-\alpha )}^{-1}{|V|}^{1/2} is injective from B to A. If \varphi \in B, then

And if \psi =0, then

It follows that {(P_0-\alpha )}^{-1}{|V|}^{1/2} is injective from B to A.

(b) From (a), one has dimA=dimB. This means that the multiplicity of α as the eigenvalue of P(\lambda ) is exactly the multiplicity of {\lambda }^{-1} as the eigenvalue of K(\alpha ). □

From the last proposition, we know that there exists one-to-one correspondence between the discrete eigenvalues of P(\lambda ) and the discrete eigenvalues of K(\alpha ). Hence, we can study the eigenvalues of K(\alpha ) first.

If P_0 and V are defined as above, we show that if P_0+V has the eigenvalue less than zero, then the smallest eigenvalue of P_0+V is simple. We use Theorems XIII.44, XIII.45 [4] to prove it.

Proposition 3.1 Suppose P_0+V has an eigenvalue at the bottom of its spectrum. Then this eigenvalue is simple and the corresponding eigenfunction can be chosen to be a positive function.

Proof Let 0\le \chi (t)\le 1 be a smooth nonincreasing function such that \chi (t)=1 if and \chi (t)=0 if t>2. Let {\chi }_n(t)=\chi (t/n). Set V_n={\chi }_n(r)\frac{q(\theta )}{r^2}+V-〈x〉, H_0=-\mathrm{\Delta }+〈x〉, H_n=H_0+V_n, H=P=P_0+V. From the proof of Theorem XIII.47 [4], we know that e^{-t{H}_0} is positivity preserving and \{e^{-t{H}_0}\}\cup L^{\mathrm{\infty }} acts irreducibly on L^2. Hence, by Theorem XIII.45 [4], if H_n converges to H and H-V_n converges to H_0 in the strong resolvent sense, then e^{-tH} is positivity preserving and \{e^{-tH}\}\cup L^{\mathrm{\infty }} acts irreducibly on L^2. By Theorems XIII.43 and XIII.44 [4], we can get the result. Since C_0^{\mathrm{\infty }}({\mathbb{R}}^d) is the core for all P_n and P, and for any \psi \in C_0^{\mathrm{\infty }}({\mathbb{R}}^d), V_n\psi \to (\frac{q(\theta )}{r^2}+V)\psi in L^2, then we have the necessary strong resolvent convergence by Theorem VIII.25(a) [5]. This ends the proof. □

Proposition 3.2 Assume that 0\notin {\sigma }_{\mathrm{\infty }}. P_0{F}_{0}u=u in H^{-1,s} for any u\in H^{-1,s}, s>1.

Proof If u\in H^{-1,s}, then F_{0}u\in H^{1,-s}. For any test function \varphi \in C_0^{\mathrm{\infty }}({\mathbb{R}}^n), we have 〈P_0{F}_{0}u,\varphi 〉=〈u,F_0{P}_0\varphi 〉. If 0\notin {\sigma }_{\mathrm{\infty }}, then we have {lim}_{z\to 0}{(P_0-z)}^{-1}=F_0 in H^{-1,s} for \mathrm{\Im }z>0. It follows 〈u,F_0{P}_0\varphi 〉={lim}_{z\to 0}〈u,{(P_0-z)}^{-1}{P}_0\varphi 〉={lim}_{z\to 0}〈u,\varphi -z{(P_0-z)}^{-1}\varphi 〉=〈u,\varphi 〉 because ϕ and P_0\varphi belong to H^{-1,s}. Hence, P_0{F}_{0}u=u in H^{-1,s}. □

Proposition 3.3 Assume that 0\notin {\sigma }_{\mathrm{\infty }}. K(\alpha ) is a compact operator for \alpha \le 0. And K(\alpha ) converges to K(0) in operator norm sense.

Proof For , K(\alpha )={|V|}^{1/2}{(P_0-\alpha )}^{-1}{|V|}^{1/2}. Since {(P_0-\alpha )}^{-1} is a bounded operator from L^2({\mathbb{R}}^d) to H^1({\mathbb{R}}^d), and V is a compact operator from H^1({\mathbb{R}}^d) to L^2({\mathbb{R}}^d), then V{(P_0-\alpha )}^{-1} is a compact operator on L^2({\mathbb{R}}^d). Using a similar method to that in Proposition 2.2, we can show that V{(P_0-\alpha )}^{-1} and K(\alpha ) have the same non-zero eigenvalues, and for the same eigenvalue e(\alpha ), the multiplicity of e(\alpha ) as the eigenvalue of V{(P_0-\alpha )}^{-1} and the multiplicity of e(\alpha ) as the eigenvalue of K(\alpha ) are the same. Hence, K(\alpha ) is a compact operator. Because

and if {\rho }_0>2, then {|V|}^{1/2}{R}_0^{(0)}{|V|}^{1/2}=o({|\alpha |}^{ϵ}) in L^2({\mathbb{R}}^d). Hence, K(\alpha )\to K(0) in operator norm sense as \alpha \to 0. This means that K(0) is a compact operator. □

Lemma 3.4 Suppose A_1, A_2 are two bounded self-adjoint operators on a Hilbert space H. Set

then |{\mu }_n(A_1)-{\mu }_n(A_2)|\le \parallel A_1-A_2\parallel.

Proof By the definition of {\mu }_n(A_i), one has

This ends the proof. □

Lemma 3.5 Suppose T(\alpha ) is a family of compact self-adjoint operators on a separable Hilbert space H, and T(\alpha )=T_0+o({|\alpha |}^{ϵ}) for α near zero. Set

Then:

1. (a)

   {\mu }_k(\alpha ) is an eigenvalue of T(\alpha ), and {\mu }_k(\alpha ) converges when \alpha \to 0. Moreover, if {\mu }_k(\alpha )\to {\mu }_k, then {\mu }_k is an eigenvalue of T_0.

2. (b)

   Suppose that E_0\ne 0 is an eigenvalue of T_0 of the multiplicity of m. Then there are m eigenvalues (counting multiplicity), E_k(\alpha ) (1\le k\le m), of T(\alpha ) near E_0. Moreover, we can choose \{{\varphi }_k(\alpha );1\le k\le m\} such that ({\varphi }_k(\alpha ),{\varphi }_j(\alpha ))={\delta }_{kj} (1\le k,j\le m), {\varphi }_k(\alpha ) is the eigenvector of T(\alpha ) corresponding to E_k(\alpha ) (E_k(\alpha )\to E_0), and {\varphi }_k(\alpha ) converges as \alpha \to 0. If {\varphi }_k(\alpha ) converges to {\varphi }_k, then {\varphi }_k is the eigenvector of T_0 corresponding to E_0.

Proof (a) By the min-max principle, we know that {\mu }_k(\alpha ) is an eigenvalue of T(\alpha ). By Lemma 3.4, one has

It follows that {\mu }_k(\alpha ) converges to the eigenvalue of T_0.

(b) Because T_0 is a compact operator and E_0\ne 0 is an eigenvalue of T_0, then E_0 is a discrete spectrum of T_0. Then there exists a constant \delta >0 small enough such that T_0 has only one eigenvalue E_0 in B(E_0,\delta ) (). For α small enough, T(\alpha ) has exactly m eigenvalues (counting multiplicity) in B(E_0,\delta ) because the eigenvalues of T(\alpha ) converge to the eigenvalues of T_0 by part (a) of lemma. Suppose the m eigenvalues, near E_0, of T(\alpha ) are E_1(\alpha ),E_2(\alpha ),\dots ,E_m(\alpha ), and the corresponding eigenvectors are {\psi }_1(\alpha ),{\psi }_2(\alpha ),\dots ,{\psi }_m(\alpha ) such that 〈{\psi }_k(\alpha ),{\psi }_j(\alpha )〉={\delta }_{kj}. Let

Then P_{\alpha }={\sum }_{k=1}^m〈\cdot ,{\psi }_k(\alpha )〉{\psi }_k(\alpha ). Let P_{\alpha }^{(k)}=〈\cdot ,{\psi }_k(\alpha )〉{\psi }_k(\alpha ), then P_{\alpha }={\sum }_{k=1}^m{P}_{\alpha }^{(k)}. For α near zero, one has

Let A=\{\varphi ;\parallel \varphi \parallel =1,\varphi \in Ran{P}_0\}. Let {\varphi }_k be an element in A such that \parallel \varphi -{\psi }_k(\alpha )\parallel acquires the minimum value. Then we have

and

In the last equality, we use the fact

and

It follows that \parallel P_{\alpha }^{(k)}{\varphi }_k-{\varphi }_k\parallel =O({|\alpha |}^{ϵ}). Let {\varphi }_k(\alpha )=\frac{P_{\alpha }^{(k)}{\varphi }_k}{\parallel P_{\alpha }^{(k)}{\varphi }_k\parallel }. Then 〈{\varphi }_k(\alpha ),{\varphi }_j(\alpha )〉=0 for k\ne j because P_{\alpha }^{(k)}{P}_{\alpha }^{(j)}=0 if k\ne j, and \parallel {\varphi }_k(\alpha )-{\varphi }_k\parallel \le \parallel P_{\alpha }^{(k)}{\varphi }_k-{\varphi }_k\parallel +\parallel (1-\frac{1}{\parallel P_{\alpha }^{(k)}{\varphi }_k\parallel })P_{\alpha }^{(k)}{\varphi }_k\parallel =O({|\alpha |}^{ϵ}). This ends the proof. □

Let and

Here, {\delta }_i=0 or 1, T_0\ge 0, T_i (1\le i\le m), T_r(\beta ) are compact operators, and T_r(\beta )=O({|\beta |}^{{\alpha }_m+ϵ}) for β near zero. Set

Then, by the min-max principle, e_s is an eigenvalue of T_0. Moreover, if e_s\ne 0, then e_s is a discrete eigenvalue of T_0 because T_0 is a compact operator. If e_s\ne 0 is an eigenvalue of T_0 of multiplicity m, without loss, we can suppose that e_s=e_{s+1}=\cdots =e_{s+m-1}. Then there exist exactly m eigenvalues (counting multiplicity), e_s(\beta ),e_{s+1}(\beta ),\dots ,e_{s+m-1}(\beta ), of T(\beta ) near e_s. By Lemma 3.5, we know that there exists a family of normalized eigenvectors \{{\varphi }_j(\beta );j=s,s+1,\dots ,s+m-1\} of T(\beta ) such that T(\beta ){\varphi }_j(\beta )=e_j(\beta ){\varphi }_j(\beta ), 〈{\varphi }_j(\beta ),{\varphi }_k(\beta )〉={\delta }_{jk} (j,k=s,s+1,\dots ,s+m-1), and {\varphi }_j(\beta ) (j=s,s+1,\dots ,s+m-1) converge as \beta \to 0. Suppose that {\varphi }_j(\beta ) converge to {\varphi }_j for all j such that e_j\ne 0. Then 〈{\varphi }_s,{\varphi }_j〉={\delta }_{sj}. \{{\varphi }_s\} can be extended to a standard orthogonal basis. Set

Then we have the following.

Lemma 3.6 T(\beta ), e_s are given as before. Then the eigenvalue of T(\beta ), e_j(\beta ) (j=s,s+1,\dots ,s+m-1) has the following form:

Here

Proof If e_s\ne 0, then e_s is the discrete eigenvalue of T_0. Suppose that the multiplicity of e_s is m, and suppose that e_s=e_{s+1}=\cdots =e_{s+m-1} as before. Hence, we can choose \delta >0 small enough such that there is only one eigenvalue e_s in . We know that e_j(\beta ) (j=s,s+1,\dots ,s+m-1) converge to e_s. It follows that if δ is small enough, there are exactly m eigenvalues (counting multiplicity) of T(\beta ) in B(e_s,\delta ) for β small. Set

Then

Since

then

Then

In particular,