Schrödinger Operators with Purely Discrete Spectrum
We prove that −Δ+V has purely discrete spectrum if V ≥ 0 and, for all M, |{x | V (x)<M}| < ∞ and various extensions.
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irk-123456789-56992010-02-03T12:01:04Z Schrödinger Operators with Purely Discrete Spectrum Simon, B. We prove that −Δ+V has purely discrete spectrum if V ≥ 0 and, for all M, |{x | V (x)<M}| < ∞ and various extensions. 2009 Article Schrödinger operators with purely discrete spectrum / B. Simon // Methods of Functional Analysis and Topology. — 2009. — Т. 15, № 1. — С. 61-66. — Бібліогр.: 25 назв. — англ. 1029-3531 http://dspace.nbuv.gov.ua/handle/123456789/5699 en Інститут математики НАН України |
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We prove that −Δ+V has purely discrete spectrum if V ≥ 0 and, for all M, |{x | V (x)<M}| < ∞ and various extensions. |
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Simon, B. Schrödinger Operators with Purely Discrete Spectrum |
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Simon, B. |
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Simon, B. |
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Schrödinger Operators with Purely Discrete Spectrum |
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Schrödinger Operators with Purely Discrete Spectrum |
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Schrödinger Operators with Purely Discrete Spectrum |
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Schrödinger Operators with Purely Discrete Spectrum |
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Schrödinger Operators with Purely Discrete Spectrum |
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schrödinger operators with purely discrete spectrum |
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Інститут математики НАН України |
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2009 |
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Schrödinger operators with purely discrete spectrum / B. Simon // Methods of Functional Analysis and Topology. — 2009. — Т. 15, № 1. — С. 61-66. — Бібліогр.: 25 назв. — англ. |
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2025-07-02T08:46:46Z |
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Methods of Functional Analysis and Topology
Vol. 15 (2009), no. 1, pp. 61–66
SCHRÖDINGER OPERATORS WITH PURELY DISCRETE
SPECTRUM
BARRY SIMON
Dedicated to A. Ya. Povzner
Abstract. We prove that −∆ + V has purely discrete spectrum if V ≥ 0 and, for
all M , |{x | V (x) < M}| < ∞ and various extensions.
1. Introduction
Our main goal in this note is to explore one aspect of the study of Schrödinger operators
(1.1) H = −∆ + V
which we will suppose have V ’s which are nonnegative and in L1
loc(R
ν), in which case (see,
e.g., Simon [15]) H can be defined as a form sum. We are interested here in criteria under
which H has purely discrete spectrum, that is, σess(H) is empty. This is well known to
be equivalent to proving (H +1)−1 or e−sH for any (and so all) s > 0 is compact (see [9,
Thm. XIII.16]). One of the most celebrated elementary results on Schrödinger operators
is that this is true if
(1.2) lim
|x|→∞
V (x) =∞.
But (1.2) is not necessary. Simple examples where (1.2) fails but H still has compact
resolvent were noted first by Rellich [10]—one of the most celebrated examples is in
ν = 2, x = (x1, x2), and
(1.3) V (x1, x2) = x2
1x
2
2
where (1.2) fails in a neighborhood of the axes. For proof of this and discussions of
eigenvalue asymptotics, see [11, 16, 17, 20, 21].
There are known necessary and sufficient conditions on V for discrete spectrum in
terms of capacities of certain sets (see, e.g., Maz’ya [6]), but the criteria are not always
so easy to check. Thus, I was struck by the following simple and elegant theorem:
Theorem 1. Define
(1.4) ΩM (V ) = {x | 0 ≤ V (x) < M}.
If (with | · | Lebesgue measure)
(1.5) |ΩM (V )| <∞
for all M, then H has purely discrete spectrum.
I learned of this result from Wang–Wu [25], but there is much related work. I found
an elementary proof of Theorem 1 and decided to write it up as a suitable tribute and
2000 Mathematics Subject Classification. 47B07, 35Q40, 47N50.
Key words and phrases. Compact resolvent, Schrödinger operators.
Supported in part by NSF grant DMS-0652919 and by Grant No. 2006483 from the United States–
Israel Binational Science Foundation (BSF), Jerusalem, Israel.
61
62 BARRY SIMON
appreciation of A. Ya. Povzner, whose work on continuum eigenfunction expansions for
Schrödinger operators in scattering situation [7] was seminal and inspired me as a gra-
duate student forty years ago!
The proof has a natural abstraction:
Theorem 2. Let µ be a measure on a locally compact space, X with L2(X, dµ) separable.
Let L0 be a selfadjoint operator on L2(X, dµ) so that its semigroup is ultracontractive
([1]): For some s > 0, e−sL0 maps L2 to L∞(X, dµ). Suppose V is a nonnegative
multiplication operator so that
(1.6) µ({x | 0 ≤ V (x) < M}) <∞
for all M. Then L = L0 + V has purely discrete spectrum.
Remark. By L0 + V , we mean the operator obtained by applying the monotone conver-
gence theorem for forms (see, e.g., [13, 14]) to L0 + min(V (x), k) as k →∞.
The reader may have noticed that (1.3) does not obey Theorem 1 (but, e.g.,
V (x1, x2) = x2
1x
4
2 + x4
1x
2
2
does). But out proof can be modified to a result that does include (1.3). Given a set Ω
in R
ν , define for any x and any ℓ > 0,
(1.7) ωℓ
x(Ω) = |Ω ∩ {y | |y − x| ≤ ℓ}|.
For example, for (1.3), for x ∈ ΩM ,
(1.8) ωℓ
x(ΩM ) ≤
Cℓ
|x|+ 1
.
We will say a set Ω is r-polynomially thin if
(1.9)
∫
x∈Ω
ωℓ
x(Ω)r dνx <∞
for all ℓ. For the example in (1.3), ΩM is r-polynomially thin for any M and any r > 0.
We’ll prove
Theorem 3. Let V be a nonnegative potential so that for any M, there is an r > 0 so
that ΩM is r-polynomially thin. Then H has purely discrete spectrum.
As mentioned, this covers the example in (1.3). It is not hard to see that if P (x) is
any polynomial in x1, . . . , xν so that for no v ∈ R
ν is �v · �∇P ≡ 0 (i.e., P isn’t a function
of fewer than ν linear variables), then V (x) = P (x)2 obeys the hypotheses of Theorem 3.
In Section 2, we’ll present a simple compactness criterion on which all theorems rely.
In Section 3, we’ll prove Theorems 1 and 2. In Section 4, we’ll prove Theorem 3.
It is a pleasure to thank Peter Stollmann for useful correspondence and Ehud de Shalit
for the hospitality of Hebrew University where some of the work presented here was done.
2. Segal’s Lemma
Segal [12] proved the following result, sometimes called Segal’s lemma:
Proposition 2.1. For A, B positive selfadjoint operators,
(2.1) ‖e−(A+B)‖ ≤ ‖e−Ae−B‖.
Remarks. 1. A + B can always be defined as a closed quadratic form on Q(A) ∩Q(B).
That defines e−(A+B) on Q(A) ∩Q(B) and we set it to 0 on the orthogonal complement.
Since the Trotter product formula is known in this generality (see Kato [5]), (2.1) holds
in that generality.
SCHRÖDINGER OPERATORS WITH PURELY DISCRETE SPECTRUM 63
2. Since ‖C∗C‖ = ‖C‖2, ‖e−A/2e−B/2‖2 = ‖e−B/2e−Ae−B/2‖, and since
‖e−(A+B)/2‖2 = ‖e−(A+B)‖,
(2.1) is equivalent to
(2.2) ‖e−A+B‖ ≤ ‖e−B/2e−Ae−B/2‖
which is the way Segal [12] stated it.
3. Somewhat earlier, Golden [4] and Thompson [22] proved
(2.3) Tr(e−(A+B)) ≤ Tr(e−Ae−B)
and Thompson [23] later extended this to any symmetrically normed operator ideal.
Proof. There are many; see, for example, Simon [18, 19]. Here is the simplest, due to
Deift [2, 3]: If σ is the spectrum of an operator
(2.4) σ(CD) \ {0} = σ(DC) \ {0}
so with σr the spectral radius,
(2.5) σr(CD) = σr(DC) ≤ ‖DC‖.
If CD is selfadjoint, σr(CD) = ‖CD‖, so
(2.6) CD selfadjoint⇒ ‖CD‖ ≤ ‖DC‖.
Thus,
(2.7) ‖e−A/2e−B/2‖2 = ‖e−B/2e−Ae−B/2‖ ≤ ‖e−Ae−B‖.
By induction,
(2.8) ‖(e−A/2n
e−B/2n
)2
n
‖ ≤ ‖e−A/2n
e−B/2n
‖2n ≤ ‖e−Ae−B‖.
Take n→∞ and use the Trotter product formula to get (2.1). �
In [18], I noted that this implies for any symmetrically normed trace ideal, IΦ, that
(2.9) e−A/2e−Be−A/2 ∈ IΦ ⇒ e−(A+B) ∈ IΦ.
I explicitly excluded the case IΦ = I∞ (the compact operators) because the argument
there doesn’t show that, but it is true—and the key to this paper!
Since
C ∈ I∞ ⇔ C∗C ∈ I∞
and e−(A+B) ∈ I∞ if and only if e−
1
2
(A+B) ∈ I∞, it doesn’t matter if we use the
symmetric form (2.2) or the following asymmetric form which is more convenient in
applications.
Theorem 2.2. Let I∞ be the ideal of compact operators on some Hilbert space, H. Let
A, B be nonnegative selfadjoint operators. Then
(2.10) e−Ae−B ∈ I∞ ⇒ e−(A+B) ∈ I∞.
Proof. For any bounded operator, C, define µn(C) by
(2.11) µn(C) = min
ψ1...ψn−1
sup
‖ϕ‖=1
ϕ⊥ψ1,...,ψn−1
‖Cϕ‖.
By the min-max principle (see [9, Sect. XIII.1]),
(2.12) lim
n→∞
µn(C) = sup(σess(|C|))
and µn(C) are the singular values if C ∈ I∞. In particular,
(2.13) C ∈ I∞ ⇔ lim
n→∞
µn(C) = 0.
64 BARRY SIMON
Let ∧ℓ(H) be the antisymmetric tensor product (see [8, Sects. II.4, VIII.10], [9,
Sect. XIII.17], and [18, Sect. 1.5]). As usual (see [18, eqn. (1.14)]),
(2.14) ‖∧m(C)‖ =
m
∏
j=1
µj(C).
Since µ1 ≥ µ2 ≥ · · · ≥ 0, we have
(2.15) lim
n→∞
µn(C) = lim
n→∞
(µ1(C) . . . µn(C))1/n.
(2.13)–(2.15) imply
(2.16) C ∈ I∞ ⇔ lim
n→∞
‖∧n(C)‖1/n = 0.
As usual, there is a selfadjoint operator, d ∧n (A) on ∧n(H) so
(2.17) ∧n(e−tA) = e−t d∧n(A)
so Segal’s lemma implies that
(2.18) ‖∧n(e−(A+B))‖ ≤ ‖∧n(e−A) ∧n (e−B)‖ = ‖∧n(e−Ae−B)‖.
Thus,
(2.19) lim
n→∞
‖∧n(e−(A+B))‖1/n ≤ lim
n→∞
‖∧n(e−Ae−B)‖1/n.
By (2.16), we obtain (2.10). �
3. Proofs of Theorems 1 and 2
Proof of Theorem 1. By Theorem 2.2, we need only show C = e∆e−V is compact. Write
(3.1) C = Cm + Dm
where
(3.2) Cm = CχΩm
, Dm = CχΩc
m
with χS the operator of multiplication by the characteristic function of a set S ⊂ R
ν .
‖e−V χΩc
m
‖∞ ≤ e−m
and ‖e∆‖ = 1, so
(3.3) ‖Dm‖ ≤ e−m
and thus,
(3.4) lim
m→∞
‖C − Cm‖ = 0.
If we show each Cm is compact, we are done. We know e∆ has integral kernel f(x−y)
with f a Gaussian, so in L2. Clearly, since V is positive, Cm has an integral kernel
Cm(x, y) dominated by
(3.5) |Cm(x, y)| ≤ f(x− y)χΩm
(y).
Thus,
∫
|Cm(x, y)|2 dνxdνy ≤ ‖f‖2L2(Rν)‖χΩm
‖L2(Rν) <∞
since |Ωm| <∞. Thus, Cm is Hilbert–Schmidt, so compact. �
Proof of Theorem 2. We can follow the proof of Theorem 1. It suffices to prove that
e−sL0e−sV is compact, and so, that e−sL0χΩm
is Hilbert–Schmidt.
That e−sL0 maps L2 to L∞ implies, according to the Dunford–Pettis theorem (see
[24, Thm. 46.1]), that there is, for each x ∈ X, a function fx( · ) ∈ L2(X, dµ) with
(3.6) (e−sL0g)(x) = 〈fx, g〉
SCHRÖDINGER OPERATORS WITH PURELY DISCRETE SPECTRUM 65
and
(3.7) sup
x
‖fx‖L2 = ‖e−sL0‖L2→L∞ ≡ C <∞.
Thus, e−sL0 has an integral kernel K(x, y) with
(3.8) sup
x
∫
|K(x, y)|2 dµ(y) = C <∞
(for K(x, y) = fx(y)). But e−sL0 is selfadjoint, so its kernel is complex symmetric, so
(3.9) sup
y
∫
|K(x, y)|2 dµ(x) = C <∞.
Thus,
(3.10)
∫
|K(x, y)χΩm
(y)|2 dµ(x)dµ(y) ≤ Cµ(Ωm) <∞
and e−sL0χΩm
is Hilbert–Schmidt. �
4. Proof of Theorem 3
As with the proof of Theorem 1, it suffices to prove that for each M, e∆χΩM
is compact.
e∆ is convolution with an L1 function, f . Let QR be the characteristic function of
{x | |x| < R}. Let FR be convolution with fQR. Then
(4.1) ‖e∆ − FR‖ ≤ ‖f(1−QR)‖1 → 0
as R→∞, so
(4.2) ‖e∆χΩM
− FRχΩM
‖ → 0
and it suffices to prove for each R, M,
(4.3) CM,R = FRχΩM
is compact. Clearly, this works if we show for some k, (C∗
M,RCM,R)k is Hilbert–Schmidt.
Let D be the operator with integral kernel
(4.4) D(x, y) = χΩM
(x)Q2R(x− y)χΩM
(y).
Since f is bounded, it is easy to see that
(4.5) (C∗
M,RCM,R)(x, y) ≤ cD(x, y)
for some constant c, so it suffices to show Dk is Hilbert–Schmidt.
Dk has integral kernel
(4.6) Dk(x, y) =
∫
D(x, x1)D(x1, x2) . . . D(xk−1, y) dx1 . . . dxk−1.
Fix y. This integral is zero unless |x−x1| < 2R, . . . , |xk−1−y| < 2R, so, in particular,
unless |x − y| ≤ 2kR. Moreover, the integrand can certainly be restricted to the regions
|xj − y| ≤ 2kR. Thus,
Dk(x, y) ≤ Q2kR(x − y)
(
∫
|xj−y|≤2kR
k−1
∏
j=1
χΩM
(xj) dx1 . . . dxk−1
)
χΩm
(y)(4.7)
= Q2kR(x − y)(ω2kR
y (ΩM )k−1)χΩM
(y)(4.8)
by the definition of ωℓ
x in (1.7).
Thus,
∫
|Dk(x, y)|2 dνxdνy ≤ C(kR)ν
∫
x∈Ω
[ω2kR
x (ΩM )]2k−2 dνx
so if 2k − 2 > r and (1.9) holds, Dk is Hilbert–Schmidt. �
66 BARRY SIMON
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Mathematics 253-37, California Institute of Technology, Pasadena, CA 91125, USA
E-mail address: bsimon@caltech.edu
Received 17/10/2008
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