Coexistence of spectra in rank-one perturbation problems.

R. del Rio, ^{∗} S. Fuentes ^{†} A. Poltoratski ^{‡}^{§}

Introduction.

In this note we shall study the behavior of different parts of the spectrum of an operator under small perturbations. Such problems became an object of active research in recent years mostly due to the applications to differential equations and mathematical physics. In such applications the operators are usually unitary or self-adjoint and acting in a separable Hilbert space.

Let A_{0} be a cyclic self-adjoint operator, φ its cyclic vector and μ the corresponding spectral measure. Denote by A_{λ} the rank one perturbations of A_{0}:

A_{λ} = A_{0} + λ(·, φ)φ, λ ∈ ℝ.  (1)

Let μ_{λ} be the spectral measures of A_{λ} corresponding to φ. The general problem related to such a family is to understand how the measures μ_{λ} change when one varies the parameter (the coupling constant) λ.

Such problems appear in many applications. For instance, if one considers a discrete Schrödinger operator and starts changing the potential at one of the points of the lattice one will obtain a family of self-adjoint rank-one perturbations. Our question then translates into the problem of predicting the changes in the dynamical properties of the quantum system caused by small perturbations in the potential field. Similarly, if one takes a Sturm-Liouville differential operator on the half-axis and begins varying the boundary condition at 0 one again obtains an example of a family A_{λ}. The same problem appears in many other areas including self-adjoint extensions of symmetric operators and spaces of pseudocontinuable functions in the unit disk. For more on these connections see [8], [16] and [13].

∗IIMAS-UNAM, Circuito Escolar, Ciudad Universitaria, 04510, México, D.F., México. E-mail: delrio@servidor.unam.mx

†Universidad Anahuac, Dpto Ciencias Basicas. Av. Univ. Anahuac s/n, Lomas Anahuac Huixquilucan, Edo. de Mexico, Mexico. E-mail:sergio.fuentes@anahuac.mx

‡Department of Mathematics, College Station, Texas 77843–3368, U.S.A. Email: alexei.poltoratski@math.tamu.edu

§A. P. was supported in part by the NSF grant DMS-9970151

Every measure μ_{λ} can be in a standard way decomposed into the sum of its absolutely continuous and singular parts. In quantum dynamics the spectrum of an operator represents the set of all admissible energies for the underlying quantum system. The spectral subspaces corresponding to the different parts of the spectral measure consist of vectors (states) displaying different dynamic behavior. It is, therefore, important to understand the interplay between different parts of μ_{λ} under small perturbations. In this paper we focus on the correlations between the absolutely continuous and singular spectrum.

By the Kato-Rosenblum theorem the absolutely continuous spectrum is invariant under such small perturbations. At the same time, examples obtained in recent years show that the singular spectrum is extremely unstable under the change of the parameter λ in particular in ”mixed” situations when both singular and absolutely continuous spectra are present. Such ”mixed” measures correspond to the systems admitting states with different dynamic properties, which seems to be an especially interesting case in many situations. The usual place for the singular spectrum of A_{λ} is the gaps of the absolutely continuous spectrum. With the change of λ the isolated eigenvalues of A_{λ} move inside their respective gaps and often dissappear when they hit the edge of the gap. Looking at such simple examples one may get an impression that in the case when there are no gaps in the a.c. spectrum, the dense singular spectrum, even if it originally existed, must be immediately destroyed when the parameter λ starts changing. In fact, most of the existing examples seem to support this intuitive model (see for instance the papers by Naboko [11] and Simon [17] where the singular spectrum exists only for a set of parameters of Lebesgue measure zero). However, the example obtained in [2] produced a family of operators with dense absolutely continuous and dense singular spectrum for all values of the parameter.

In many applications the perturbations occur randomly, i. e. the parameter λ is a random variable with certain distribution on the real line. If the problem is considered from such a point of view, one can say that the probability of coexistence of dense singular and absolutely continuous spectra in systems constructed in [11] and [17] is 0 (for any a. c. distribution of λ) and 1 for the example from [2].

The next natural question seems to be if the coexistence is possible for the sets of parameters other than the whole line or zero-sets. In the next section, with an addition of new analytic tools, we are able to complete the construction of [2] to obtain the following result.

Theorem 1 For any measurable set B ⊂ ℝ there exists a family of rank-one perturbations {A_{λ}}_{λ∈ℝ} such that A_{λ} has dense absolutely continuous and dense singular spectrum for almost every λ ∈ B and dense absolutely continuous (but no singular) spectrum for almost every λ ∈ B.

Although in probabilistic settings the sets of measure zero are negligible, it would be interesting to know if the words ”almost every” could be removed from the last statement. Such a problem requires a completely different approach. As an example, in this paper we show that there exists a family of operators such that the spectrum is dense absolutely continuous for all λ ∈ [0, 1] and dense absolutely continuous plus dense singular for all λ ∈ ℝ\[0, 1]. In this construction we utilize the Krein spectral shift function. While a similar technique may provide the answer for other simple sets B, the general question remains open.

1 The Main Proof

Every Pick function F (z) (an analytic function which takes the upper half-plane ℂ_{+} into itself) has an integral representation of the form

F (z) = a + bz + Kμ  (2)

where a, b ∈ ℝ, b ≥ 0, μ is a non negative Borel measure which satisfies

_{ℝ}^{∫} \frac{dμ(t)}{1 + t^{2}} < ∞

and Kμ is its Cauchy integral in the upper half-plane:

Kμ = _{ℝ}^{∫} [\frac{1}{t − z} − \frac{t}{1 + t^{2}} ] dμ(t).  (3)

The term bz can be viewed as the Cauchy integral of a point mass b at infinity. Conversely, any function of this form is an analytic map from the upper half-plane into itself, see for instance [15]. We will also denote by P μ the Poisson integral of μ:

P μ(x + iy) = ImKμ(x + iy) = _{ℝ}^{∫} \frac{y}{(x − t)^{2} + y^{2}} dμ(t).

Proof of Theorem 1

Part I.

Let B ⊂ ℝ be an arbitrary measurable set. In this part we construct a family A_{λ} such that the spectrum is absolutely continuous for a. e. λ ∈ B̄ = ℝ \ B and absolutely continuous plus singular for a. e. λ ∈ B. In the second part we will make all parts of the spectrum dense in ℝ. For any measurable set S we denote by |S| the measure of S with respect to the normalized Lebesgue measure dx/(1 + x^{2}).

Without loss of generality (abbreviated WLOG henceforth) 0 ∉ B. Denote E = −1/B = {x| − 1/x ∈ B}. Then there exist open sets E_{n} containing E such that E_{1} ⊃ E_{2} ⊃ ... and |E_{k} \ E| → 0.

If O ⊂ ℝ is an open set and ϵ is a positive number, we denote by Ω(O, ϵ) the open domain in the upper half-plane constructed in the following way. Let O = ∪I_{n} where I_{n} are disjoint open intervals. Denote by Q_{n} the open rectangle in the upper half-plane with base I_{n} and of height greater than ϵ. We will denote

Ω(O, ϵ) = (∪Q_{n}) ∪ {z|Im z > ϵ}.

Let {ϵ_{n}} be a decreasing sequence of positive numbers tending to 0 (to be chosen later). For each n consider the domain Ω(E_{n}, ϵ_{n}), see fig. 1.

Fig. 1

Put

Γ = ∩Ω(E_{n}, ϵ_{n}),

see fig. 2.

Fig. 2

It can be easily verified that Γ is an open domain. Hence by the Riemann Theorem there exists a conformal map ϕ from the upper halfplane to Γ. Since ϕ maps ℂ_{+} into itself, it is a Pick function allowing the representation (2). Moreover, ϕ can be chosen so that a, b = 0 by requiring ϕ(∞) = 0. We denote the measure on ℝ appearing in the representation (2) for ϕ by μ_{0}. Now we can consider a self-adjoint cyclic operator A_{0} and its cyclic vector φ such that μ_{0} is the spectral measure of A_{0} corresponding to φ (take for instance the operator of multiplication by x in L^{2}(μ_{0}) and put φ = 1). We denote by A_{λ} the corresponding family of rank-one perturbations and by μ_{λ} the spectral measures of these operators. We claim that the family A_{λ} posesses the desired properties (after the proper choice of ϵ_{n}).

After [14], we say that a point on the boundary of a complex domain is sectorially accessible if there exists an open triangle with a vertex at this point that lies completely within the domain. If f is a conformal map from ℂ_{+} to W we denote by Sect(f) the set of points x on ℝ at which f has nontangential limit f(x) ∈ ∂W that is sectorially accessible in W.

Now we want to show that if ϵ_{n} are chosen in a right way then almost all points from E are sectorially accessible in Γ.

Let O be an open set O = ∪I_{n} for some disjoint open intervals I_{n}. If |I_{n}| > 2ϵ we denote by I_{n}^{ϵ} the open interval with the same center of length |I_{n}| − 2ϵ. If |I_{n}| ≤ 2ϵ then I_{n}^{ϵ} stands for the empty set. After that we define O^{ϵ} = ∪I_{n}^{ϵ}. Note that |O \ O^{ϵ}| → |O| as ϵ → 0.

For each open set E_{n} choose ϵ_{n} so that |E_{n} \E_{n}^{ϵ_{n}}| < 1/2^{n}. It is not difficult to see that if a point x in E is not sectorially accessible from Γ then it must belong to E_{n} \ E_{n}^{ϵ_{n}} for arbitrarily large n’s. Hence x must belong to ∪_{n≥N}E_{n} \ E_{n}^{ϵ_{n}} for every N. But | ∪_{n≥N} E_{n} \ E_{n}^{ϵ_{n}}| ≤ 1/2^{N−1}. Therefore there is at most a zero set of such unaccessible points x.

To see that all μ_{λ} have absolutely continuous parts, notice that all points on the boundary of Γ that are not on the real line are sectorially accessible. Therefore, by The McMillan Sector Theorem, see [14], preimage of such points (the set where ϕ takes such nontangential boundary values) is of positive measure. Hence, there is a set of positive measure where the nontangential boundary values of Cμ_{0} have positive imaginary parts. This means that μ_{0}, and therefore all μ_{λ}, have nontrivial absolutely continuous parts.

Now let us show that μ_{λ} have non-trivial singular parts for a. e. λ ∈ B. The McMillan Twist Theorem [14] implies that ϕ has a nontangential derivative at a. e. point of Sect(ϕ). Together with the Sector Theorem this means that, since a. e. point in E is sectorially accessible, for a. e. y ∈ E there exists x ∈ ℝ such that the nontangential limit ϕ(x) is equal to y and ϕ has an angular derivative at x. It is well known (see for instance [16]) that this implies that μ_{−1/y} has a point mass at x. Recall that E = −1/B.

To fininsh this part of the proof we will show that μ_{λ} may have non-trivial singular parts only for a set of measure zero of λ ∉ B. Each open set E_{n} is a union of disjoint open intervals. Let us denote by S the intersection of all open sets E_{n} plus the endpoints of all such open intervals for all sets E_{n}. Then S \ E is a zero set. Since ϕ must tend to −1/λ nontangentially (μ_{λ})_{s}-a.e., it is enough to show that all the real nontangential limits of ϕ must belong to S. This certainly follows from general facts of the theory of conformal maps, but instead of referring the reader to those results we decided to include the following elementary argument. Let y ∉ S and suppose that ϕ(x + ih) → y as h → 0. Suppose that y ∉ E_{n} = ∪I_{k}. There must exist δ > 0 such that |ϕ(x + ih) − y| < ϵ_{n}/2 for h < δ. This means that for all such h the point ϕ(x + ih) stays inside of the same rectangle Q_{k}. But since y ∉ E_{n} and it is not an endpoint of the interval I_{k} (the base of Q_{k}), ϕ(x + ih) can never reach y staying inside Q_{k}.

Part II.

To make all parts of all μ_{λ} constructed in Part I dense we can apply Lemma 2 below together with the result of [2]. In [2] we constructed the family of spectral measures ν_{λ} such that each measure has absolutely continuous and singular parts, both densely supported in ℝ. Let us define the measure η_{0} as

η_{0} = ∫_{ℝ} ν_{λ}dμ_{0}(λ).

Once again we can choose a self-adjoint operator A_{0} and its cyclic vector φ so that η_{0} is the corresponding spectral measure. Denote by A_{λ} the rank-one perturbations defined by (1) and by η_{λ} the spectral measures of A_{λ}. Then by Lemma 2

η_{λ} = ∫_{B} ν_{ξ}dμ_{λ}(ξ).

In this way all η_{λ} will have densely supported absolutely continuous parts, densely supported singular parts for a. e. λ ∈ B and no singular part for a. e. λ ∉ B.

We will use the notation Cγ for the Cauchy integral of the measure γ with a different kernel:

Cγ(z) = ∫ℝ \frac{dγ(x)}{x − z}.  (4)

Let us consider the Pick functions

f_{λ}(z) = \frac{z}{1 + λz} , λ ∈ ℝ.  (5)

It is well known (see [16]) that if γ is a spectral measure of A_{0} corresponding to ϕ and γ_{λ} are spectral measures of the perturbations A_{λ} then

Cγ_{λ}(z) = f_{λ}(Cγ(z)), λ ∈ ℝ.

In such cases we will say that {γ_{λ}}_{λ∈ℝ} is the family generated by γ. Notice that γ_{0} = γ

The following result, which is essentially Lemma 5.3 of [3], will be useful

Lemma 1 Let μ, ν and η be positive Borel measures on ℝ, {μ_{λ}} be the family generated by μ. Then

η(A) = _{ℝ}^{∫} μ_{λ}(A)dν(λ)  (6)

for every Borel set A if and only if

Cη(z) = Cν((−Cμ)^{−1}(z)).  (7)

Proof.

⇒) Lemma 5.3 of [3]

⇐) By the definition of μ_{λ}, (7) implies

_{ℝ}^{∫} \frac{dη(x)}{x − z} = _{ℝ}^{∫} dν(λ)_{ℝ}^{∫} \frac{dμ_{λ}(x)}{x − z}

Since the kernels (x − z)^{−1} and their conjugates span the space of continuous functions, we get (6)

q.e.d.

Lemma 2

Let μ and ν be positive Borel measures on ℝ, {μ_{λ}} and {ν_{λ}} be the families generated by these measures correspondingly. Define the measure η as

η(A) = _{ℝ}^{∫} μ_{ξ}(A)dν(ξ)  (8)

for every Borel set A. Let η_{λ} be the family generated by η. Then

η_{λ}(A) = _{ℝ}^{∫} μ_{ξ}(A)dν_{λ}(ξ)

for all λ ∈ ℝ.

Proof.

Definition (8) implies Cη = Cν(−C_{μ}^{−1}) by the last lemma. Therefore

f_{λ}(Cη(z)) = f_{λ}(Cν(−C_{μ}^{−1}(z)))

i.e.

Cη_{λ}(z) = Cν_{λ}(−C_{μ}^{−1}(z))

and applying Lemma 1 again we obtain

η_{λ}(A) = ∫ μ_{ξ}(A)dν_{λ}(ξ)

for every Borel set A, λ ∈ ℝ q.e.d.

2 An example of coexistence of spectra

As was announced in the introduction, we now show that the words ”almost every” can be ommitted in the statement of Theorem 1 when the set B is simple. The following observations will be useful in our construction.

Consider the family of Pick functions

f_{λ}(z) = \frac{z}{1 + λ z} λ ∈ ℝ

as in (5). Let L(z) be a Pick function such that 0 ≤ Im L(z) ≤ π. For any λ ∈ ℝ consider

M_{λ}(z) = f_{λ}(L(z))
N_{λ}(z) = f_{λ}(H(L(z)))

where

H(z) = exp(az) + c, a, c ∈ ℝ
a ∈ (0, 1].

Both M_{λ} and N_{λ} are Pick functions admitting representations similiar to (2). We shall denote by μ_{λ} and ν_{λ} the measures appearing in the representations for M_{λ} and N_{λ} respectively.

As usual, we use the symbol γ^{ac} to denote the absolutely continuous part of a measure γ and γ^{s} for the singular part.

Lemma 3 Let μ_{λ} and ν_{λ} be defined as above and put

λ̃(λ) = −H^{−1} (−\frac{1}{λ}) .

Then

μ_{λ}^{s} = λ̃′(λ) νs ; λ̃.

Proof. We follow closely the proof of Lemma 2 in [2]. We shall first show

that

lim_{y↓0} \frac{Im M_{λ}(x + i y)}{Im Nλ̃(x + i y)} = λ̃′(λ)

for μ_{λ}^{s}-a.e. x. We use the notation z = x + i y.

First note that if λ and λ̃ are different from zero we have

(\frac{λ̃}{λ})^{2} |\frac{H(L(z)) − (−\frac{1}{λ̃})}{L(z) − (−\frac{1}{λ})}|^{2} · \frac{Im L(z)}{Im H(L(z))} = \frac{Im M_{λ}(z)}{Im Nλ̃(z)}  (9)

It is well known that for μ_{λ}^{s}-a.e. x.

lim_{y↓0} M_{λ}(x + i y) = ∞

and therefore using (5) and recalling that M_{0}(z) = L(z) we get

lim_{y↓0} L(x + i y) = −\frac{1}{λ}

for μ_{λ}^{s}-a.e. x. Therefore using the definition of λ̃ we obtain for μ_{λ}^{s}-a.e. x.

lim_{y↓0} \frac{H(L(x + i y)) − (−\frac{1}{λ̃})}{L(x + i y) − (−\frac{1}{λ})} = lim_{y↓0} \frac{H(L(x + i y)) − H(−\frac{1}{λ̃})}{L(x + i y) − (−\frac{1}{λ})} =
H′(−\frac{1}{λ}) = a exp a(−\frac{1}{λ})  (10)

Now

lim_{y↓0} \frac{Im H(L(x + i y))}{Im L(x + i y)} = lim_{y↓0} \frac{Im exp(a L(x + i y))}{Im L(x + i y)} =
lim_{y↓0} \frac{Im exp(a L(x + i y))}{Im L(x + i y)} = a exp a(−\frac{1}{λ}) for μ_{λ}^{s} − a.e. x (11)

Since L(x + i 0) = −\frac{1}{λ} for such points x.

¿From (9, 10, 11) we get

lim_{y↓0} \frac{Im M_{λ}(x + i y)}{Im Nλ̃(x + i y)} = (\frac{λ̃}{λ})^{2} (a exp a(−\frac{1}{λ})) = λ̃′(λ)

Since the Poisson integral P_{μ_{λ}}(x + i y) tends to infinity at μ_{λ}^{s}-a.e. x the last equation implies

lim_{y↓0} \frac{Im M_{λ}(x + i y)}{Im Nλ̃(x + i y)} = \frac{Im Cμ_{λ}(x + i y) + C_{1}y}{Im Cνλ̃(x + i y) + C_{2}y} = lim_{y↓0} \frac{P_{μ_{λ}}(z)}{P_{νλ̃}(z)} = λ̃(λ)

¿From here the result easily follows as in Lemma 2 of [2].

Example 1 Now we are ready to show that there exists a self-adjoint operator A_{0} such that the operators A_{λ}, defined as in (1), have dense singular spectrum for λ ∈ ℝ\[0, 1] and dense purely absolutely continuous spectrum for λ ∈ [0, 1]

Let E ⊂ ℝ be such that for every interval J ⊂ ℝ we have

| J |>| E ∩ J |> 0

and such that |E| < ∞. To see examples of such sets we refer to [6].

Put

dμ = \frac{1}{2} u(x) dx

where

u(x) = \{ \begin{matrix}1 & if x ∈ E \\ 0 & if x ∉ E \end{matrix}

Let

Cμ(z) = _{ℝ}^{∫} \frac{dμ(x)}{x − z}

and define

F (z) = exp Cμ(z)

Since F is a Pick function it can be represented as in (2). To evaluate the constants a and b in (2) observe that Cμ(z) → 0 when z goes to ∞ along the imaginary axis. Hence

F (z) = 1 + Cν(z)

where Cν(z) = _{ℝ}^{∫} \frac{d ν(x)}{x−z} for some finite measure ν.

Once we have the measure ν we proceed to define the family of operators A_{λ} in the standard way: Let A_{0} be the operator of multiplication by x in L^{2}(ν) and

A_{λ} = A + λ < φ, · > φ

where φ(x) = 1 for every x ∈ ℝ.

We claim the family of operators A_{λ} has the desired properties. To see this observe that the measures corresponding to A_{λ} are the ones generated by the family of Pick functions (see [16]).

Cν_{λ}(z) = f_{λ}(Cν(z)) = f_{λ}(exp Cμ(z) − 1)  (12)

¿From the definition of μ we have

Cμ(z) = \frac{1}{2}Cu(z)

where

Cu(z) = _{ℝ}^{∫} \frac{u(x)dx}{λ − z}

We shall denote by u_{λ} the measure which appears in the integral representation of the Pick function Cu_{λ} = f_{λ}(Cu(z)).

According to Lemma 3 with H(z) = exp \frac{1}{2}z − 1 we know that

u_{λ}^{s}(B) = λ̃′(λ) νs ; λ̃(B)

where

λ̃(λ) = (1 − exp(−\frac{1}{2λ}))^{−1}

Therefore if

u_{λ}^{s} is densely supported for every λ ≠ 0  (13)

then

νs ; λ̃ is densely supported forλ̃ ∈ ℝ\[0, 1]

To obtain (13) consider the Pick function G = exp Cu(z). Since

u(x) = \frac{1}{π} arg(exp Cu(x + i0))

we obtain that G has real boundary values a. e. on ℝ and therefore the measure asociated with G is purely singular. From the stability of the absolutely continuous part it follows that the measures η_{λ} associated with the Pick functions f_{λ}(exp Cu(z)) are purely singular too. By the choice of E the measure η_{0} associated with G and therefore all the measures η_{λ} are densely supported.

Since by Lemma 3

u_{λ}^{s}(J) = λ̃′(λ) ηs ; λ̃(J)

where

λ̃(λ) = −(exp(−\frac{1}{λ}))^{−1}

we get (13). Hence ν_{λ}^{s} is densely supported for every λ ∈ ℝ\[0, 1] as required.

Now we shall see that if λ ∈ [0, 1] then the measure ν_{λ} is purely absolutely continuous.

Consider the following support of the singular part of the measure ν_{λ}

Σ_{λ} = {x|Cν_{λ}(x + i0) = ∞}

see [10] for example, which can be seen to be equal to the set

\{x|Cν_{0}(x + i0) = −\frac{1}{λ}\}

using the definition of the famiy f_{λ}.

From the definition (12) and the properties of the Poisson kernel it follows that

arg (Cν_{0}(x + i0)) + 1) = arg (exp Cμ(x + i0)) =
Im Cμ(x + i0) = P μ(x + i0) =

\frac{π}{2} lim_{ϵ↓0} \frac{| E ∩ (x − ϵ, x + ϵ) |}{2ϵ}.

Since for x ∈ Σ_{λ} Cν_{0}(x + i0) = −\frac{1}{λ} ∈ ℝ we should have

lim_{ϵ↓0} \frac{| E ∩ (x − ϵ, x + ϵ) |}{2ϵ} = 0 or 2

and from the inequality

\frac{| E ∩ (x − ϵ, x + ϵ) |}{2ϵ} ≤ \frac{2ϵ}{2ϵ} = 1

it follows

lim_{ϵ↓0} \frac{| E ∩ (x − ϵ, x + ϵ) |}{2ϵ} = 0.

Therefore −\frac{1}{λ} = exp(\frac{1}{2}Re Cμ) − 1 which implies λ ∈ ℝ\[0, 1]. Thus if λ ∈ [0, 1] we have νs ; λ̃ = 0.

Remark.

In the case of a family of ergodic operators H_{ω} we can not have an analogous situation to the one described above taking instead of λ the parameter ω. This follows from a theorem of Kunz and Souillard see [1, p.169], which tells us that there exist sets ∑_{s} and ∑_{ac} such that

σ_{s}(H_{ω}) = _{∑s}
σ_{ac}(H_{ω}) = _{∑ac}

Coexistence of spectra in rank-one perturbation problems. 16

for a.e. ω.

Acknowledgment This work was partially supported by projects IN102998 PAPIIT-

UNAM and 27487E CONACyT.

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