On the spectrum of leaky surfaces with a potential bias

0.19 MB·

by  Pavel Exner

On the spectrum of leaky surfaces with a potential bias
Pavel Exner
To my friend Helge Holden on the occasion of his 60th birthday.
Abstract. We discuss operators of the type H = −∆+V(x)−αδ(x−Σ) with an
7
attractiveinteraction,α>0,inL2(R3),whereΣisaninfinitesurface,asymptotically
1
0 planar and smooth outside a compact, dividing the space into two regions, of which
2 one is supposed to be convex, and V is a potential bias being a positive constant
n V0 in one of the regions and zero in the other. We find the essential spectrum and
ask about the existence of the discrete one with a particular attention to the critical
a
J case, V0 = α2. We show that σdisc(H) is then empty if the bias is supported in the
3 ‘exterior’ region, while in the opposite case isolated eigenvaluesmay exist.
2
2010 Mathematics Subject Classification. Primary 81Q10; Secondary35J10
]
h Keywords. Schrödingeroperators, δ-interaction, potential bias, spectral properties
p
-
h
t 1. Introduction
a
m
An anniversary is usually an opportunity to look back at the achievements of
[
the jubilee. In Helge’s case the picture is impressive as he contributed signifi-

1 cantly to several different areas of mathematical physics. Nevertheless, one of
v
his works made a much larger impact than any other, namely the monograph
8
[AGHH] first published in 1988. It is a collective work but Helge’s hand is
8
2 unmistakably present in the exposition, and I add that it makes me proud to
6 be a part of the second edition of this book.
0
This also motivates me to choose a problem from this area as a topic of
.
1 this paper. I am going to discuss operators of the type
0
7
H = ∆+V(x) αδ(x Σ), α >0, (1)
1 − − −
:
v in L2(R3), where the δ-potential is supported by an infinite surface Σ dividing
i
X the space into two regions, of which one is supposed to be convex, and V
is a potential bias being a positive constant in one of the regions and zero
r
a in the other. The question to be addressed concerns spectral properties of
such operators, in particular, how they depend on the geometry of Σ. We
will observe similarities with recent results obtained in the two-dimensional
analogue of the present problem [EV16], especially a peculiar asymmetry. On
theotherhand,however,itisnotlikelytohavetheseresultsextendedtohigher
dimensions, cf. a remark at the end of Section 5.
2 STATEMENT OF THE PROBLEM AND THE RESULTS 2
2. Statement of the problem and the results
Let us begin with formulating the problem described above in proper terms.
2.1. Assumptions. As we shall see, the geometry of Σ will be decisive for
spectralpropertiesoftheoperator(1). Wefocusourattentiononthefollowing
class of surfaces:
(a) Σ is topologicallyequivalentto a plane dividingR3 into tworegions such
thatoneofthemisconvex. Thetrivialcaseoftwohalfspacesisexcluded.
(b) Σmaycontainatmost finitefamilies = C offiniteC1 curves, which
j
C { }
are either closed or have regular ends, and = P of points such that
j
P { }
outsidetheset thesurfaceisC2 smoothadmittingaparametrization
C∪P
with a uniformly elliptic metric tensor.
Todistinguishthetworegionsweshallrefertotheconvexoneas‘interior’and
denoteitΩ ,theotherwillbe‘exterior’,denotedasΩ . Thecurvefiniteness
int ext
in assumption (b) refers to the Hausdorff distance, i.e. to the metric inherited
fromtheambientthree-dimensionalEuclideanspace. Theassumptionimplies,
inparticular,thatΣisC2 smoothoutsideacompact. Notealsothatthecurves
indicatingthenon-smoothpartsofΣmayingeneraltouchorcross; theregular
ends mean that they can be prolonged locally without losing the C1 property.
Since Σis byassumptiontopologicallyequivalentto a plane,wecan use an
atlas consisting of a single chart, in otherwords, a map Σ: R2 R3 provided
→
we accept the licence that the fundamentalforms and quantities derived from
them may not exist at the points of , nevertheless, geodesic distances
C ∪P
remain well defined across these singularities. Furthermore, by assumption
(b)theprincipalcurvaturesk ,k ofΣarewelldefinedoutsideacompact. We
1 2
will suppose that
(c) Σ is asymptotically planar, that is, the principal curvature vanish as the
geodesic distance from a fixed point tends to infinity.
Equivalently one can require that both the Gauss and mean curvatures given
by K = k k and M = 1(k +k ), respectively, vanish asymptotically. We
1 2 2 1 2
also assume that
(d) there is a c>0 such that Σ(s) Σ(t) cs t holds for any s,t R2.
| − |≥ | − | ∈
Thisensures,inparticular,thattherearenocuspsatthepointsof where
C∪P
Σ is not smooth; in view of assumption (a) such a constant mustsatisfy c <1.
Given a bounded potential V, one can demonstrate in the same way as
in [BEKŠ94, Sec. 4] that under the stated assumptions the quadratic form
q = q defined by
α,Σ,V
q[ψ]:= ψ 2+(ψ,Vψ) α ψ(Σ(s))2g1/2(s)ds ds , (2)
1 2
k∇ k − ZR2| |
2 STATEMENT OF THE PROBLEM AND THE RESULTS 3
wheres= (s ,s )arethecoordinatesusedtoparametrizeΣandg =det(g )is
1 2 ij
the appropriate squared Jacobian defined by means of the metric tensor (g ),
ij
with the domain H1(R3), is closed and below bounded. Thus it is associated
with a unique self-adjoint operator which we identify with H = H of
α,Σ,V
(1) above. In fact such a claim is valid for a much wider class of potentials,
however, we focus here our attention on a particular case. By hypothesis (a)
above the surface Σ splits R3 into two regions, and we assume that
(e) V(x) =V >0 in one of these regions and V(x)= 0 in the other.
0
2.2. An auxiliary problem. In the trivial case we have excluded the prob-
lem is solved easily by separation of variables. It is useful to look at the
transverse part which we will needin thefollowing. It is given bytheoperator
d2
h= αδ(x)+V(x), (3)
−dx2 −
where V(x) = V for x > 0 and V(x) = 0 otherwise, associated with the
0
quadratic form φ φ′ 2 αφ(0)2+(φ,Vφ) defined on H1(R). Properties
7→ k k − | |
of this operator are easily found; we adopt without proof from [EV16] the
following simple results.
Lemma 2.1. (i) σ (h) =[0, ).
ess
∞
(ii) The operator h has no eigenvalues for V α2.
0
≥
(iii) The operator h has a unique eigenvalue µ= α2−V0 2 for V <α2.
−(cid:16) 2α (cid:17) 0
(iv) If V =α2 the equation hψ =0 has a bounded weak solution ψ / L2(R).
0
∈
(v) For V >0 and any ϕ C2(R ) L2(R ) we have
0 + +
∈
T
∞
(ϕ′ 2+V ϕ2)(x)dx V ϕ(0)2.
0 0
Z | | | | ≥ | |
0 p
Relationsbetweenthecouplingconstantandthepotentialbiaswillplayan
important role in the following. For the sake of brevity, we shall call the case
(iv) of the lemma critical, and similarly we shall use the terms subcritical for
case (iii) and supercritical for the situation where V >α2.
0
2.3. The results. Letuslookfirstattheessentialspectrum. Asusualinthe
Schrödingeroperatortheoryitisdeterminedbythebehavioroftheinteraction
at large distances. In view of assumption (c) we expect that asymptotically
the situation approaches the trivial case with separated variables mentioned
above, and indeed, we have the following result.
Theorem 2.2. σ (H)=[µ, ) holds under the assumptions (a)–(e), where
ess
∞
µ:= 1α−2(α2 V )2 for V <α2 and µ:=0 otherwise.
−4 − 0 0
3 PROOF OF THEOREM 2.2 4
The question about the existence of the discrete spectrum is more involved
and the potential bias makes the answer distinctively asymmetric. In par-
ticular, a critical or supercritical potential supported in the exterior region
prevents negative eigenvalues from existence.
Theorem 2.3. Under the stated assumptions, suppose that V(x)=0 holds in
Ω and V(x) =V α2 in Ω , then σ(H)=σ (H)=[0, ).
int 0 ext ess
≥ ∞
On the other hand, the operator (1) may have isolated eigenvalues in the
(sub)critical regime as we are going to illustrate on examples. In Section 5 we
discuss the case of a conical surface and show that the discrete spectrum of
H is nonempty provided V is small enough. The most interesting, though, is
0
the critical case, V = α2 with the bias in the interior region. In Section 6 we
0
discuss another example, this time with Σ being a ‘rooftop’ surface, and show
that for suitable values of parameters we have here σ (H)= .
disc
6 ∅
3. Proof of Theorem 2.2
Theargumentsplitsintotwoparts. Firstweshalldemonstratetheimplication
ν σ (H) if ν µ. (4)
ess
∈ ≥
To this goal, one has to find for any fixed ν µ and any ε > 0 an infinite-
≥
dimensional subspace Dom(H) such that (H ν)ψ < ε holds for any
L ⊂ k − k
ψ . We denote ζ := ν µ 0 and choose a pair of functions of unit L2
no∈rmL, f C2(R2) with the−pro≥perty that ( ∆ ζ)f < 1ε, and g C2(R)
∈ 0 − − 4 ∈ 0
suchthat g =1 and (h µ)g < 1ε. Su(cid:13)chfunctions(cid:13)canalwaysbefoundin
k k k − k 4 (cid:13) (cid:13)
view of the fact that the essential spectrum of the two-dimensional Laplacian
is [0, ) and of the properties of the operator h stated in Lemma 2.1.
∞
In the next step we choose a sequence of surface points a = Σ(s(j)) such
j
that s(j) as j . With each of them we associate the Cartesian
| | → ∞ → ∞
system of coordinates (x(j),y(j)) where x(j) are the Cartesian coordinates in
the tangential plane to Σ at the point a and y(j) is the distance from this
j
tangential plane. By assumption (b), the points a can be always chosen in
j
suchawaythatthiscoordinatechoicemakessense. Thisallowsustodefinethe
functions ψ (x(j),y(j)) = f(x(j))g(y(j)). By construction, each of them has a
j
compactsupportofthediameterindependentofj,henceinviewofassumption
(d) one can pick them so that suppψ Σ is simply connected. Using then a
j
∩
straightforwardtelescopicestimateincombinationwiththerequirement f =
k k
g = 1 we get the inequality
k k
(H ν)ψ ( ∆x ζ)f + (h µ)g +V ψ + (δ δ )ψ , (5)
k − jk≤(cid:13) − − (cid:13) k − k 0k j|Ajk k Σj− Σ jk
(cid:13) (cid:13)
whereΣ istheta(cid:13)ngentialplan(cid:13)eata and isthepartofthefunctionsupport
j j j
A
squeezed between Σ and Σ ; the last term is understood as the L2-norm over
j
3 PROOF OF THEOREM 2.2 5
the two surface segments contained in the border of . The first two terms
j
A
on the right-hand side of (5) are by construction bound by 1ε. Furthermore,
2
in view of the assumptions (a)–(c) in combination with the smoothness of the
functions f,g, which are the same for all the ψ , the other two terms tend to
j
zeroasj ,hence (H ν)ψ <εholdsforallj largeenough. Inaddition,
j
→∞ k − k
one can always choose the points aj in such a way that suppψj suppψj′ =
∩ ∅
holds for j =j′, which means that Weyl’s criterion hypothesis is satisfied.
6
Tocompletetheproofofthetheorem,wehavetodemonstratetheopposite
implication, in other words, to check the validity of the relation
σ (H) ( ,µ)= , (6)
ess
∩ −∞ ∅
which is equivalent to infσ (H) µ. While in the first part of the proof we
ess
≥
have extended to the present case the argument used in the two-dimensional
situation, now we choose a different approach because the localization esti-
mates employed in [EV16] become more involved here. We will need an aux-
iliary result which is a sort of modification of Proposition 2.5 in [EY02].
Lemma 3.1. Let h denote the operator (3) acting on the interval ( d,d)
N
−
with Neumann boundary conditions at the endpoints, associated with the form
φ φ′ 2 αφ(0)2 +(φ,Vφ) defined on H1( d,d). If V α2, there are
0
7→ k k − | | − ≤
positive c ,d such that for all d > d we have infσ(h ) µ c d−1. If, on
0 0 0 N 0
≥ −
the other hand, V > α2 holds, then h 0 for all d large enough.
0 N
≥
Proof. We observe that the ground-state eigenfunction of h corresponding
N
the eigenvalue µ <0 is of the form
d
ψ(x) =c χ coshκ (x d)+c χ coshκ (x+d),
1 (0,d) 1 2 (−d,0) 1
−
where κ := √ µ and κ := √V µ . Since the function has to be con-
1 d 2 0 d
− −
tinuous at x = 0 and satisfy ψ′(0+) ψ′(0 ) = αψ(0), we get the spectral
− − −
condition
κ tanhκ d+κ tanhκ d=α. (7)
1 1 2 2
Asa functionof µ , theleft-handside is increasing from√V tanh√V d, be-
d 0 0
−
having asymptotically as 2√ µ+ (( µ)−1/2), the equation (7) has a unique
− O −
solution for any fixed d > 0 provided V α2. Since the left-hand side
0
≤
is monotonous also with respect to d we have µ < µ where µ = µ of
d ∞ ∞
Lemma 2.1(iii).
To get a lower bound we have to estimate the left-hand side of (7) from
below. We will do that using the rough bound tanhx >1 2e−2x >1 x−1.
− −
Writing the solution of the appropriate estimating condition in the form µ˜ =
d
µ δ we find after a short computation that µ˜ =µ 2d−1+ (d−2) holds as
d
− − O
d which together with the inequality µ >µ˜ yields the result.
d d
→∞
If V > α2 no solution exists for a sufficiently small d. The condition
0
(7) can be then modified replacing one or both hyperbolic tangents by the
3 PROOF OF THEOREM 2.2 6
trigonometric one, however, we will not need it; it is enough to note that the
lowest eigenvalue of h — which certainly exists as h as a Sturm-Liouville
N N
operator on a finite interval has a purely discrete spectrum — is positive for d
large enough.
Consider now the neighborhood Ω := x R3 : dist(x,Σ) < d of the
d
{ ∈ }
surface. Furthermore, fix a point x Σ and divide Σ into two parts, Σ
0 R
∈
consisting of the points the geodesic distance from x is larger than R and
0
Σc = Σ Σ . Note that by assumptions (a) and (b) outside a compact Σ
R \ R
has a well defined normal and the points of Ω can be written as x +n u
d Σ xΣ
with u < d, where x is the point satisfying dist(x,Σ) = dist(x,x ); for
Σ Σ
| |
d small enough this part of Ω does not intersect itself, i.e. the point x is
d Σ
unique. Consequently, for R sufficiently large and d sufficiently small we may
define Ω := x +n u : x Σ , u < d and Ωc := Ω Ω . We
d,R { Σ xΣ Σ ∈ R | | } d,R d \ d,R
also introduce the Ωc := R3 Ω consisting of two connected components, so
together we have R3d=Ω \ Ωdc Ωc.
d,R∩ d,R∩ d
Using these notions we employ a bracketing argument. Changing the do-
main of H by additional Neumann conditions imposed at the boundaries of
the three domains we obtain a lower bound to our operator,
H ≥HΩd,R ⊕HΩcd,R ⊕HΩcd.
For the proof of (6) only the first part on right-hand side is relevant, because
HΩc correspondstoa compactregionanditsessentialspectrum isthusvoid,
d,R
and infσess(HΩc)=0 holds obviously. To analyze the first part, which we for
d
brevity denote as H we employ a geometric argument similar to that used
d,R
in [EK03], the difference being the constant potential V to one side of Σ .
0 R
The‘piercedlayer’Ω canberegardedasasubmanifoldinR3 equippedwith
d,R
the metric tensor
(G ) 0
G = µν , G =(δσ uhσ)(δρ uhρ)g , (8)
ij (cid:18) 0 1 (cid:19) µν µ − µ σ − σ ρν
referringtothecurvilinearcoordinates(s ,s ,u),whereg isthemetrictensor
1 2 ρν
of Σ and hσ is the corresponding Weingarten tensor; we conventionally use
R µ
the Greeknotation for the range (1,2) of the indices and the Latin for (1,2,3)
and we employ the Einstein summation convention. In particular, the volume
element of Ω is given by dΩ:=G1/2d2sdu with
d
G:= det(G =g[(1 uk )(1 uk )]2 =g(1 2Mu+Ku2)2; (9)
ij 1 2
− − −
for brevity we use the shorthand ξ(s,u) 1 2M(s)u+ K(s)u2. Next we
≡ −
introduce
̺ := ( max k , k )−1.
R { ΣR k 1k∞ k 2k∞}
3 PROOF OF THEOREM 2.2 7
By assumption (c) we have ̺ as R , and as self-intersections of
R
→ ∞ → ∞
Ω are avoided as long d < ̺ ; we see that the layer halfwidth d can be in
d,R R
fact chosen arbitrarily big provided R is sufficiently large. At the same time,
the transverse component of the Jacobian satifies the inequalities C (d,R)
−
ξ(s,u) C (d,R), where C (d,R) := (1 d̺−1)2, hence for a fixed d an≤d
≤ + ± ± R
R large the Jacobian is essentially given by the surface part; recall that the
metric tensor g is assumed to be unifomly elliptic, c δ g c δ
µν µν µν + µν
− ≤ ≤
with positive c .
±
Now we use these geometric notions to asses the spectral threshold of
H . Passing tothecurvilinearcoordinates(s,u),wewrite thecorresponding
d,R
quadratic form as
(∂ ψ,Gij∂ ψ) +(ψ,Vψ) α ψ(s,0)2 dΣ
i j G G
− Z | |
|s|>R
definedonH (Ωflat,dΩ),where(, ) meansthescalarproductinL2(Ωflat,dΩ),
1 d,R · · G d,R
the symbol dΣ stands for g1/2(s)ds and Ωflat := q : s > R, u <d . Using
d,R { | | | | }
the diagonal form (8) of the metric tensor together with (10) and the above
mentioned bound to the factor ξ(s,u) we find
(ψ,H ψ) ∂ ψ 2+V ψ 2 dΩ α ψ(s,0)2 dΣ
d,R G u
≥ZΩfld,aRt (cid:0)| | | | (cid:1) − Z|s|>R| |
ξ− ∂ ψ 2+V ψ 2 dΣdu α ψ(s,0)2 dΣ,
≥ R ZΩfld,aRt (cid:0)| u | | | (cid:1) − Z|s|>R| |
where ξ− := inf ξ(s,u). Introducing similarly ξ+ := sup ξ(s,u) and
R Ωd,R R Ωd,R
using Lemma 3.1 with the coupling constant α = α , we get
R −
ξ
R
ξ−
(ψ,H ψ) R µ c d−1 ψ 2
d,R G ≥ ξ+ − 0 k kG
R (cid:0) (cid:1)
provided d is large enough. It is clear from the above discussion that to any
ε > 0 one can choose R and d sufficiently large to get c d−1 < 1ε, and at the
0 2
−
same time, ̺ > d and ξR < ε . Consequently, infσ (H) > µ ε which
R ξ+ 2µ ess −
R
concludes the proof.
Remarks 3.2. (a)Theproofdidnotemploythepartofassumption(a)speaking
about convexity of one of the regions to which the surface divides the space
and, in fact, neither the fact that Σ is simply connected.
(b) The bound in Lemma 3.1 is a rough one but it suffices for the present
purpose;inrealitytheerrorcausedbytheNeumannboundaryisexponentially
small as d , similarly as in [EY02]. Note that we use such an estimate
→ ∞
in a different way than in the said work: there we made the error small by
choosing a large α while here the coupling constant is fixed but we choose a
large d which we are allowed to do being far enough in the asymptotic region.
4 PROOF OF THEOREM 2.3 8
4. Proof of Theorem 2.3
In view of Theorem 2.2 it is sufficient to check that q[ψ] 0 holds for any
ψ C2(R3). The contributionfrom Ω to the quadraticfor≥mis non-negative
∈ 0 int
and may be neglected; this yields the estimate q[ψ] q [ψ], where
ext
≥
q [ψ]:= ψ 2+V ψ 2 (x)dx α ψ(Σ(s))2g1/2(s)d2s. (10)
ext 0
ZΩext(cid:0)|∇ | | | (cid:1) − ZR2| |
To estimate the quantity from below, we note that by assumption (b) there is
a family of open connected subsets Σ , l = 1,...,N, of Σ which are mutually
l
disjoint and such that Σ = Σ and Σ is C2 smooth for any l. It may
∪l l |Σl
happen that N = 1 if Σ ( ) is connected, on the other hand, N 2
\ C ∪P ≥
has to hold, for instance, when one of the curves of the family is closed,
C
or more generally, if contains a loop. The number of the Σ ’s can be made
l
C
larger if we divide a smooth part of the surface by an additional boundary,
but by assumption (b) the partition can be always chosen to have a finite
number of elements, and moreover, only one of the Σ ’s is not precompact in
l
Σ. Next we associate the sets Ω := x +n u : x Σ , u < 0 Ω
l { Σ xΣ Σ ∈ l } ⊂ ext
with the Σ ’s, where the negativesign refers to the factthat the normalvector
l
points conventionally into the interior domain. Since the latter is convex by
assumption (a), no two halflines x Σ , u < 0 emerging from different
Σ l
{ ∈ }
points of Σ can intersect, and consequently, the sets Ω are mutually disjoint,
l
if = the closure Ω may be a proper subset of the exterior domain.
l l
C ∪P 6 ∅ ∪
This yields
q [ψ] ψ 2+V ψ 2 (x)dx α ψ(Σ(s))2g1/2(s)d2s
ext 0
≥ Xl ZΩl(cid:0)|∇ | | | (cid:1) − ZR2| |
and passing in the first integral to the curvilinear coordinates in analogy with
the previous proof we get
0
q [ψ] ((∂ ψ)Gij(∂ ψ)+V ψ 2)G1/2 (s,u)d2sdu
ext i j 0
≥ Z Z | |
Xl Ml −∞(cid:0) (cid:1)
α ψ(Σ(s))2g1/2(s)d2s,
− ZR2| |
where M is the pull-back of the surface component Σ by the map Σ. Ne-
l l
glecting the non-negative term (∂ ψ)Gµν(∂ ψ) and using (10) we arrive at
µ ν
the estimate
0
q [ψ] (∂ ψ 2+V ψ 2)(s,u)g1/2(s)(1 uk (s))
ext u 0 1
≥ Z Z | | | | −
Xl Ml −∞
(1 uk (s))d2sdu α ψ(Σ(s))2g1/2(s)d2s.
2
× − − ZR2| |
5 EXAMPLE OF A CONICAL SURFACE 9
However,1 uk (s) 1holdsforµ=1,2andu<0becauseboththeprincipal
µ
− ≥
curvaturesarenon-negativeinviewoftheconvexityassumption. Furthermore,
thedifferencebetween M andR2 isazeromeasureset,hencewefinallyfind
l l
∪
0
q[ψ] (∂ ψ 2+V ψ 2)(s,u)du αψ(s,0)2 g1/2(s)d2s,
u 0
≥ZR2(cid:26)Z−∞ | | | | − | | (cid:27)
where, with the abuse of notation, we have employed the symbol ψ(s,0) for
ψ(Σ(s)), but α √V holds by assumption, and consequently, the expression
0
≤
in the curly brackets is positive by Lemma 2.1(v). This concludes the proof.
5. Example of a conical surface
Consider now the the situation where Σ = is a circular conical surface of
θ
C
an opening angle 2θ (0,π), in other words
∈
= (x,y,z) R3 : z =cotθ x2+y2 .
θ
C ∈
(cid:8) p (cid:9)
This surface satisfies the assumptions (a)–(e) with consisting of a single
P
point,thetipofthecone,henceifthepotentialbiasissupportedintheexterior
of this cone, the spectrum of the corresponding operator is by Theorems 2.2
and 2.3 purely essential, σ(H)=[µ, ). Let us look nowwhat happensin the
∞
opposite case when the bias is in the interior.
In view of the symmetry it is useful to employ the cylindrical coordinates
relative to the axis of . Since our potential is independent of the azimuthal
θ
C
angle, the operator H now commutes with the corresponding component of
the angular momentum operator, i∂ , and allows thus for a partial-wave
ϕ
−
decomposition, H = H(m). Writing the wave function in the standard
m∈Z
way through its reducLed components,
ω (r,z)
ψ(r,ϕ,z)= m eimϕ,
√2πr
mX∈Z
we can rewrite the quadratic form (2) as the sum q[ψ] = q [ψ], where
m∈Z m
P
4m2 1
q [ψ]:= ω 2 + − ω (r,z)2drdz (11)
m k∇ mkL2(R2+) ZR2 4r2 | m |
+
+ZR2 V(r,z)|ωm(r,z)|2drdz−α2kωm|ΓθkL2(R2+),
+
where R2 is the halfplane (r,z) : r > 0, z R and Γ is the halfline
+ { ∈ } θ
z = r cotθ. We note first that it is sufficient to focus on the component with
vanishing angular momentum in the partial-wave decomposition.
5 EXAMPLE OF A CONICAL SURFACE 10
Proposition 5.1. Let V(x) = V χ (x) with V > 0, then q [ψ] µ holds
for any nonzero m Z and all ψ0 ΩHin1t(R3). 0 m ≥
∈ ∈
Proof. Ifm=0,thesecondtermontheright-handsideof(11)isnon-negative
6
and one estimate the form from below by neglecting it. Following the paper
[BEL14] where the case V = 0 is treated, we introduce in the halfplane R2
0 +
anotherorthogonalcoordinates, s measured along Γ and t in the perpendicu-
θ
lar direction; the axes of the (t,s) system are rotated with respect to those of
(r,z) around the point r =z =0 by the angle θ. In these coordinates we have
∞ ∞ ∞
q [ψ] ω 2 +V ds ω (s,t)2dt α ω (s,0)2ds,
m ≥ k∇ mkL2(R2+) 0Z0 Z−stanθ| m | − Z0 | m |
wherewiththeabuseofnotationwewriteω (s,t)forthefunctionvalueinthe
m
rotated coordinates. If ψ H1(R3) and m= 0, the reduced wave functionω
m
belongs to H1(R2). By ω˜∈ we denote its e6xtension to the whole plane by the
+ m
zerovalueintheotherhalfplane, (s,t): s R,t< stanθ ,whichnaturally
belongs to H1(R2). Such function{form a su∈bspace i−n H1(R2}), however,hence
we have
inf q [ψ]= inf ω˜ 2 +V ω˜ (s,t)2dsdt
ωm∈H1(R2+) m ωm∈H1(R2+)(cid:26)k∇ mkL2(R2+) 0ZR2| m |
α ω˜ (s,0)2ds
m
− ZR| | (cid:27)
inf ̺ 2 +V ̺(s,t)2dsdt α ̺(s,0)2ds .
≥̺∈H1(R2)(cid:26)k∇ kL2(R2+) 0ZR2| | − ZR| | (cid:27)
Noting finally that form on the right-hand side of the last inequality is asso-
ciated with the self-adjoint operator in L2(R2) which has separated variables,
∂2 I +I h, where h is the operator (3) with the variable x replaced by
− s ⊗ t s⊗
t, we obtain the desired claim from Lemma 2.1.
−
On the other hand, the component with zero momentum can give rise to a
nontrivialdiscretespectrum,atleastaslongthepotentialbiasisweakenough.
Proposition 5.2. To any integer N there is a number v (0,α2) such that
N
∈
#σ (H) N holds for 0 V <v .
disc 0 N
≥ ≤
Proof. The potential bias we consider is a bounded perturbation, therefore
H : V 0 is a type (A) holomorphic family in the sense of [Ka].
{ α,Cθ,V 0 ≥ }
This means, in particular, that the eigenvalues, if they exist, are continuous
functions of V . The same is by Theorem 2.2 true for the essential spectrum
0
threshold. Since the bias-free operator H has by [BEL14] an infinite
α,Cθ,0
number of isolated eigenvalues accumulating at 1α2, the continuity implies
−4
the result.