The uncertainty principle lemma under gravity and the discrete spectrum of Schrödinger operators

0.25 MB·English

by  Kazuo Akutagawa

THE UNCERTAINTY PRINCIPLE LEMMA UNDER GRAVITY
AND
THE DISCRETE SPECTRUM OF SCHRO¨DINGER OPERATORS
9
0
0
KAZUOAKUTAGAWA∗ ANDHIRONORIKUMURA†
2

n
a
J
3 Abstract. TheuncertaintyprinciplelemmafortheLaplacian∆onRnshows
1 the borderline-behavior of a potential V for the following question : whether
theSchro¨dingeroperator−∆+V hasafiniteorinfinitenumberofthediscrete
] spectrum. In this paper, we will give a generalization of this lemma on Rn
G to that onlargeclasses of complete noncompact manifolds. ReplacingRn by
D somespecificclassesofcomplete noncompact manifolds,includinghyperbolic
spaces,wealsoestablishsomecriterionsfortheabove-type question.
.
h
t
a
m
1. Introduction and Main Results
[

2 Let us start with the following crucial result:
v
TheoremRSK(Reed-Simon[14],Kirsch-Simon[9]). Let ∆+V betheSchro¨dinger
63 operator with a potential V C0(Rn) on L2(Rn), where−n 3. Assume that
∈ ≥
6 ( ∆+V)C∞(Rn) is essentially self-adjoint and that σess( ∆+V)=[0, ), where
4 σ− ( ∆+|Vc) denotes the essential spectrum of ∆+V.− ∞
ess
. − −
2 (i) Assume that, there exists R0 >0 such that V satisfies
1
(n 2)2
8 V(x) − for r := x R .
0 ≥− 4r2 | |≥ 0
: Then, the set σ ( ∆+V) of the discrete spectrum is finite.
v disc
−
i (ii) Assume that, there exist δ >0 and R1 >0 such that V satisfies
X
(n 2)2
r V(x) (1+δ) − for r R .
a ≤− 4r2 ≥ 1
Then, σ ( ∆+V) is infinite.
disc
−
Here, we recall the following fundamental two facts: First, σ ( ∆) = [0, ).
ess
Second, if V C0(Rn) and V(x) 0 as x , then ( ∆+V−)C∞(Rn) is∞es-
∈ → | | → ∞ − | c
sentially self-adjoint and σ ( ∆+V) = [0, ) (cf. [6]). Hence, the assumptions
ess
− ∞
for the potential V in Theorem RSK are reasonable if one has an interest in only
the asymptotic behavior of V near infinity for the question whether ∆+V has
−
a finite or infinite number of the discrete spectrum. With these understandings,
Date:January8,2009.
∗ supported in part by the Grants-in-Aid for Scientific Research (C), Japan Society for the
PromotionofScience, No.18540098.
† supported inpart bythe Grants-in-AidforScientific Research(C), Japan Society forthe Pro-
motionofScience, No.18540212.
1
2 Theuncertaintyprinciplelemmaundergravity
TheoremRSKgivesa completeanswerto this question. Inorderbothto knowthe
borderline-behavior of V and to prove the finiteness of σ ( ∆+V), the Uncer-
disc
−
tainty Principle Lemma below is crucial, whichis heavilyrelated to the Heisenberg
Uncertainty Principle (cf. [13, 6]).
Uncertainty Principle Lemma. When n 3, the following inequality holds :
≥
(n 2)2 u2
(1) u2dx − dx for u C∞(Rn),
ZRn|∇ | ≥ 4 ZRn r2 ∈ c
where u denotes the gradient of u.
∇
Inthispaper,wewillstudythefinitenessandinfinitenessofthediscretespectrum
of a Schr¨odinger operator under gravity, that is, a Schr¨odinger operator ∆ +V
g
−
on a complete noncompact n-manifold (M,g). Here, ∆ denotes the Laplace-
g
−
Beltrami operator with respect to g on C∞(M) and V C0(M). Throughout
∈
this paper, without particular mention, we always assume that V C0(M) and
∈
that ( ∆g+V)C∞(M) is essentially self-adjoint for the sake of simplicity. For our
− | c
purpose, we first prove the following uncertainty principle lemma under gravity,
that is, the one on a complete noncompact manifold with ends of a specific type.
Theorem 1.1 (Uncertainty Principle Lemma under Gravity). Let (M,g)
be a complete noncompact Riemannian n-manifold, where n 2. Assume that one
≥
of ends of M, denoted by E, has a compact connected C∞ boundary W :=∂E such
that the outward normal exponential map exp : +(W) E is a diffeomorphism
W N →
(see Fig. 1), where
+(W):= v TM v is outward normal to W .
W
N ∈ |
Assume also that the mea(cid:8)n curvature(cid:12)(cid:12)HW of W with respect to (cid:9)the inward unit
normal vector is positive. Take a positive constant R>0 satisfying
1
H on W,
W
≥ R
and set
ρ(x):=dist (x,W), r(x):=ρ(x)+R for x E.
g
∈
Then, for all u C∞(M), we have
∈ c
(2) u2dv
g
|∇ |
ZE
1 1 1 1 1 1
+ (∆ r)2 dr2 Ric ( r, r) u2dv + ∆ r u2dσ
≥ 4r2 4 g − 2|∇ | − 2 g ∇ ∇ g 2 g − R g
ZEn o ZW (cid:16) (cid:17)
1 1 1 1
+ (∆ r)2 dr2 Ric ( r, r) u2dv ,
≥ 4r2 4 g − 2|∇ | − 2 g ∇ ∇ g
ZEn o
where Ric , dv and dσ denote respectively the Ricci curvature, the Riemannian
g g g
volume measure of g and the (n 1)-dimensional Riemannian volume measure of
−
(W,g ). In particular, if (M,g) has a pole p M (cf. [5]), then
W 0
| ∈
1 1 1 1
(3) u2dv + (∆ r)2 dr2 Ric ( r, r) u2dv ,
|∇ | g ≥ 4r2 4 g − 2|∇ | − 2 g ∇ ∇ g
ZM ZMn o
where r(x):=dist (x,p ) for x M.
g 0
∈
KazuoAkutagawaandHironoriKumura 3
E
W =∂E
Figure 1
Remark 1.2. (i) Forx E, ( dr)(x )and(∆ r)(x )arerespectivelythe sec-
0 0 g 0
∈ ∇
ond fundamental form and the mean curvature of the level hypersurfacer−1(x )=
0
x E r(x)=r(x ) at x (with respect to the inward unit normal vector).
0 0
{(ii)∈ Let| (M,g) be th}e Euclidean n-space Rn = (Rn,g ) (resp. the hyperbolic n-
0
spaceHn( κ)=(Hn( κ),g )ofconstantnegativecurvature κ). ForeachR>0,
κ
we denote−B (0) the g−eodesic open ball of radius R centered−at the origin0 of Rn
R
(resp. Hn( κ) ) and
−
r(x):=ρ(x)+R=dist (x,∂B (0))+R =dist (x,0)
g R g
forx E :=Rn BR(0)(resp.E :=Hn( κ) BR(0)(cid:0)). Then,theter(cid:1)mappearing
∈ − − −
in the boundary integral of (2) can be described as
1 n−2 if (M,g)=Rn,
(4) ∆ r = R
(cid:16) g −R(cid:17)(cid:12)∂BR(0) ( (n−1) √κ coth(√κR)− R1 if (M,g)=Hn(−κ),
(cid:12)
and hence, when(cid:12) n 2, this term is non-negative for all R > 0 in the both cases.
≥
More generally, let (M,g) be a complete n-manifold with the following type end:
([0, ) N,dr2+f(r)2 g ) with f′(r)>0 for r R >0,
N 0
∞ × · ≥
where(N,g )isaclosedRiemannian(n 1)-manifoldandR >0issomepositive
N 0
−
constant. The condition f′(r) > 0 (r R ) implies that the boundary ∂E of
0 L
each E := [L, ) N has the positiv≥e mean curvature H = (n−1)f′(L) > 0
L ∞ × ∂EL f(L)
provided that L R . Then, if one choose R > 0 appropriately corresponding to
0
≥
E , the term ∆ r 1 is non-negative. However, for a general end E and
L g − R ∂EL
a sequence L(cid:16) with 0 <(cid:17)(cid:12)L < L < < L < , the non-negativity of
{ i} (cid:12) 1 2 ··· i ··· ր ∞
∆ r 1 doesnot(cid:12)necessarilyrequireanexpansionofg onE. Forinstance,
(cid:16) g −R(cid:17)(cid:12)∂ELi
(cid:12)
(cid:12)
4 Theuncertaintyprinciplelemmaundergravity
suppose that (M,g) has the following periodic end:
[0, ) N,dr2+(2+sinr)2g .
N
∞ ×
Then, for E2mπ =[2mπ,(cid:0)∞)× N (m≥ (n−11)π,m∈Z) a(cid:1)nd R:=2mπ, we obtain
1 n 1 1
∆ r = − 0.
g
− R ∂E2mπ 2 − 2mπ ≥
(iii) When (M,g)=(cid:16)(Rn,g ), w(cid:17)e(cid:12)(cid:12) have
0 (cid:12)
n 1 n 1
∆r = − , dr2 = − , Ric ( r, r)=0.
r |∇ | r2 g0 ∇ ∇
Hence, the inequality (3) includes the inequality (1) in the original uncertainty
principle lemma on Rn.
Applying Theorem1.1tothe hyperbolicn-spaceHn( κ),we obtainthe explicit
−
result below.
Theorem 1.3 (Uncertainty Principle Lemma on Hn( κ)). When n 2,
for u C∞(Hn( κ)) and R>0, the following holds − ≥
∈ c −
(5) u2dv
ZHn(−κ)−BR(0)|∇ | gκ
(n 1)2κ 1 (n 1)(n 3)κ
− + + − − u2dv .
≥ZHn(−κ)−BR(0)n 4 4r2 4 sinh2(√κr) o gκ
In particular,
(n 1)2κ 1 (n 1)(n 3)κ
(6) u2dv − + + − − u2dv .
ZHn(−κ)|∇ | gκ ≥ZHn(−κ)n 4 4r2 4 sinh2(√κr) o gκ
Here, r(x):=dist (x,0).
gκ
Remark 1.4. Since
sinh(√κr)
lim =1,
κց0 √κr
by letting κ 0 in (6), we recover the inequality (1) in the original uncertainty
principle lemցma on Rn.
Using this lemma, we also obtain the following explicit criterion.
Theorem 1.5. Let ∆ +V be the Schro¨dinger operator with a potential V on
L2(Hn( κ)), where n− 2g.κAssume that σ ( ∆ +V)=[(n−1)2κ, ).
− ≥ ess − gκ 4 ∞
(i) Assume that, there exist R >0 such that V satisfies
0
1 (n 1)(n 3)κ
V(x) + − − for r R .
≥− 4r2 4 sinh2(√κr) ≥ 0
(cid:16) (cid:17)
Then, σ ( ∆ +V) is finite.
disc − gκ
(ii) Assume that, there exist δ >0 and R >0 such that V satisfies
1
1
V(x) (1+δ) for r R .
≤− 4r2 ≥ 1
Then, σ ( ∆ +V) is infinite.
disc − gκ
We regards Theorem RSK and Theorem 1.5 as the model criterions. Using also
Theorem 1.1 in a sense of approximation, we have the following criterions.
KazuoAkutagawaandHironoriKumura 5
Theorem 1.6. Let (M,g) be a complete noncompact n-manifold with n 3 and
≥
∆ +V the Schro¨dinger operator with a potential V. Take a point p M, and
g 0
− ∈
set r(p):=dist (p,p ) for p M. Assume that, there exist some positive constants
g 0
∈
L,L′,K >0 and a small positive constant τ (0<τ <1) such that
(1.1) Lr−(2+τ), Ric L′r−(3+τ) for all large r,
g g
|R |≤ |∇ |≤
(1.2) Vol(B (p )) Ktn for all t>0,
t 0
≥
(1.3) σ ( ∆ +V)=[0, ),
ess g
− ∞
where denotes the Riemannian curvature tensor of g.
g
R
(i) Assume that, there exist δ >0 and R >0 such that V satisfies
0 0
(n 2)2
V(x) (1 δ ) − for r R .
≥− − 0 4r2 ≥ 0
Then, σ ( ∆ +V) is finite.
disc g
−
(ii) Assume that, there exist δ >0 and R >0 such that V satisfies
1 1
(n 2)2
V(x) (1+δ ) − for r R .
≤− 1 4r2 ≥ 1
Then, σ ( ∆ +V) is infinite.
disc g
−
Remark 1.7. ByTheorem(1.1)andRemark(1.8)in[2],(M,g)isasymptotically
locallyEuclidean(abbreviatedtoALE)oforderτ withfinitelymanyends. Namely,
there exists a relatively compact open set such that each component (i.e., each
O
end) of M has coordinates at infinity x = (x1, ,xn); that is, there exist
R > 0 and −a fiOnite subgroup Γ O(n) acting freely o·n··Rn B (0) such that the
R
endis diffeomorphicto Rn B⊂(0) Γ and,with respectt−o x=(x1, ,xn), the
R
− ···
metric g satisfies the following on the end:
(cid:0) (cid:1)(cid:14)
g =δ +O(x−τ), ∂ g =O(x−1−τ), ∂ ∂ g =O(x−2−τ).
ij ij k ij k ℓ ij
| | | | | |
In this case, one can easily check that σ ( ∆ ) = [0, ) for the ALE manifold
ess g
− ∞
(M,g). From the argument in Proof of Theorem 1.6-(ii) later, one can also easily
see that, if the potential V satisfies the decay condition in (ii) only on one end,
then σ ( ∆ +V) is infinite.
disc g
−
Theorem 1.8. Let (M,g) be an asymptotically hyperbolic n-manifold of class C2
with n 2, and ∆ +V the Schro¨dinger operator with a potential V. Assume
g
≥ −
that
(n 1)2
σ ( ∆ +V)=[ − , ).
ess g
− 4 ∞
Take a point p M, and set r(x):=dist (x,p ) for x M.
0 g 0
∈ ∈
(i) Assume that, there exist δ >0 and R >0 such that V satisfies
0 0
1
V(x) (1 δ ) for r R .
≥− − 0 4r2 ≥ 0
Then, σ ( ∆ +V) is finite.
disc g
−
(ii) Assume that, there exist δ >0 and R >0 such that V satisfies
1 1
1
V(x) (1+δ ) for r R .
≤− 1 4r2 ≥ 1
Then, σ ( ∆ +V) is infinite.
disc g
−
6 Theuncertaintyprinciplelemmaundergravity
Remark 1.9. Let M be a compact C∞ n-manifold with boundary (possibly
disconnected)andM itsinterior. Then,ametricgonM issaidtobeasymptotically
hyperbolic of class C2 if g satisfies the following: There exists a defining function
λ C∞(M) of ∂M such that the conformally rescaled metric g := λ2g has a C2-
∈
extention on M and that dλ2 =1 on ∂M (cf. [12, 4, 11]). R. Mazzeo [12] proved
| |g
that such (M,g) is complete and has sectional curvatures uniformly approaching
1 near ∂M. Moreover,σ ( ∆ )=[(n−1)2, ) (cf. [1, 10]).
− ess − g 4 ∞
In the next section, we first introduce Hardy’s inequality. We then prove The-
orem 1.1. Theorem 1.1, that is, the uncertainty principle lemma under gravity
arises from essentially this inequality. Using this theorem, we also prove the other
main results mentioned above, including Theorem RSK with a simple proof. In
Section 3, we give one more application of Thorem 1.1 and an interesting example
of a noncompact manifold on which this theorem holds.
Acknowledgements. The authors would like to thank Hideo Tamura, Hiroshi
Isozaki, Rafe Mazzeo and Richard Schoen for helpful discussions.
2. Hardy’s Inequality and Proof of Main Results
We first recall the following Hardy’s inequality [7, Theorem 327], which is an
essential source of the inequalities in (1) and (2).
Hardy’s Inequality. For any f C1 [0, ) and R>0, the following holds.
∈ c ∞
(7) ∞ f′(t)2dt (cid:0)∞ f(t)(cid:1)2dt f(R)2.
| | ≥ 4t2 − 2R
ZR ZR
Proof of Theorem 1.1. Let y =(y2, ,yn) be local coordinateson an opensubset
···
U of W. Under the identification E = +(W) by the outwardnormalexponential
∼N
map, then the metric g can be described as
g(r,y)=dr2+gW(r,y)dyαdyβ on [R, ) U (2 α,β n),
αβ ∞ × ≤ ≤
where (ρ,y) are the Fermi coordinates on [0, ) U and r := ρ+R. For each
∞ ×
u C∞(M), we have on [R, ) y (y U)
∈ c ∞ ×{ } ∈
∞
(8) ∂ u2 gWdr
r
| |
ZR
∞ 1p ∞ 1 1 1
= ∂ ( gW2u)2dr+ (∆ r)2 dr2 Ric ( r, r) u2 gWdr
r g g
| | 4 − 2|∇ | − 2 ∇ ∇
ZR p Z1R n o p
+ ∆ r u2 gW .
g
2 · · (R,y)
(cid:16) p (cid:17)(cid:12)
where gW := det(gW(r,y)). (cid:12)(cid:12)
αβ
Thepproof ofqthe identity (8) is the below: A direct computation shows that on
the interval [R, ) y
∞ ×{ }
∞ 1
(9) ∂ ( gW2u)2dr
r
| |
ZR
= 1 ∞ p∂r gW 2u2dr+ 1 ∞ ∂ gW ∂ (u2)dr+ ∞ gW ∂ u2dr.
r r r
4ZR (cid:0) pgW (cid:1) 2ZR ZR | |
(cid:0) p (cid:1) p
p
KazuoAkutagawaandHironoriKumura 7
Using integrationbyparts,we cancalculate the secondtermofthe righthand side
of the above as
(10) 1 ∞ ∂ gW ∂ (u2)dr = 1 ∂ gW u2 r=∞ 1 ∞ ∂2 gW u2dr
2 r r 2 r r=R − 2 r
ZR ZR
(cid:0) p (cid:1) 1 (cid:2)(cid:0)∞p (cid:1) (cid:3) 1 (cid:0) p (cid:1)
= ∂2 gW u2dr ∂ gW u2 .
Since ∆ r = ∂r√gW and ∂ (−∆2rZ)R=(cid:0) rpdr2 +(cid:1) Ric −( 2r,(cid:16)(cid:0)rr)p, we (cid:1)now(cid:17)(cid:12)(cid:12)(cid:12)g(Ret,y)the
g √gW − r g |∇ | g ∇ ∇
following two identities
∂2 gW = dr2+Ric ( r, r) gW +(∆ r)2 gW,
r − |∇ | g ∇ ∇ g
∂p gW 2 (cid:8) (cid:9)p p
r =(∆ r)2 gW.
g
(cid:0) pgW (cid:1)
p
These two idenptities combined with (9) and (10) imply the identity (8).
Hardy’s inequality (7) implies the following
∞ 1 ∞ u2 1 u2
(11) ∂ gW2u 2dr gWdr gW .
Note also ZthRat,|inr(cid:0)tperms of(cid:1)t|he co≥orZdRina4ters2p(r,y) on−[R2,(cid:16)R) pU ((cid:17)(cid:12)(cid:12)([RR,y,) ) W ),
∞ × ⊂(cid:12) ∞ ×
the volume element dv can be expressed as
g
dv (r,y)=dr dσ (r,y)= gW(r,y) dr dy2 dyn.
g g
···
q
SubstitutingtheHardy’sinequality(11)intotheidentity(8)combinedwith u2
|∇ | ≥
∂ u2 and integrating its both sides over W (locally with respect to dy2 dyn),
r
| | ···
we then get the inequality (2).
Now, we assume that (M,g) has a pole p M. For each small ε > 0, set
0
∈
E := M B (p ) in the inequality (2), where B (p ) denotes the open geodesic
ε 0 ε 0
−
ball of radius ε centered at p . Letting ε 0 in the integrationover W =∂B (p )
0 ε 0
ց
of (2), we then have
1 1 n 2
∆ r u2dσ = − σ u(p )2 εn−2+O(εn−1) 0,
2 g − ε g 2 n−1· 0 · →
Z∂Bε(p0)(cid:16) (cid:17)
whereσ denotesthe(n 1)-dimensionalvolumeVol(Sn−1(1))oftheunit(n 1)-
n−1
sphere Sn−1(1) of Rn. Com−bining this with (2), we obtain the inequality (3).− (cid:3)
Proof of Theorem 1.3. On the space Hn( κ), we have
−
∆ r =(n 1)√κ coth(√κr), dr2 =(n 1)κ coth2(√κr),
gκ − |∇ | −
Ric ( r, r)= (n 1)κ.
gκ ∇ ∇ − −
Applying the inequalities (2) and (3) to the space Hn( κ) combined with these
identities and (4), we obtain the inequalities (5) and (6).− (cid:3)
For both the self-containedness and the later use on Proofs of Theorems 1.5,
1.6, 1.8, we give here a simple proof of Theorem RSK, particularly the finiteness
assertion (i) by using the inequality (2), not the original inequality (1). For the
proofofthe infiniteness assertion(ii), wealsogive a unifiedview ofthe mechanism
for constructing a nice test function on each rotationally symmetric Riemannian
8 Theuncertaintyprinciplelemmaundergravity
n-manifold (Rn,dr2+h(r)2·gSn−1(1)). Here, gSn−1(1) denotes the standard metric
of constant curvature 1 on Sn−1(1).
Proof of Theorem RSK. Assertion (i). Suppose that σ ( ∆+V) is infinite.
disc
−
Then, there exist a family λ ∞ of negative eigenvalues and the corresponding
{ i}i=1
eigenfunctions ϕ ∞ satisfying
{ i}i=1
λ <λ λ λ 0, ϕ ϕ dx=δ for all i,j 1.
1 2 3 k i j ij
≤ ≤···≤ ≤···ր ZRn ≥
Decompose Rn into the two pieces as
Rn =B (0) Rn B (0) .
R0 ∪ − R0
(cid:0) (cid:1)
We consider the Neumann eigenvalue problem for ∆+V on both B (0) and
− R0
Rn B (0). Arrangeall the eigenvaluesofB (0)andRn B (0)inincreasing
− R0 R0 − R0
order, with repetition according to multiplicity:
µ µ .
1 2
≤ ≤···
From Domain monotonicity of eigenvalues (vanishing Neumann data) (cf. [3]), we
have
(12) µ λ ( <0 ) for all i 1.
i i
≤ ≥
Now applyingthe inequality(2)to Rn B (0)combinedwithRemark1.2-(ii),
− R0
(iii), we then obtain
(n 2)2 u2
u2dx − dx for u C∞(Rn).
ZRn−BR0(0)|∇ | ≥ 4 ZRn−BR0(0) r2 ∈ c
Therefore, the assumption for the potential V implies
u2+Vu2 dx 0 for u C∞(Rn),
ZRn−BR0(0) |∇ | (cid:17) ≥ ∈ c
(cid:0)
and hence the Schr¨odinger operator ∆ + V on Rn B (0) (with Neumann
− − R0
boundary condition) has no negative eigenvalue. On the other hand, since B (0)
R0
is compact, ∆+V on B (0) (with Neumann boundary condition) has only a
− R0
finite number of negative eigenvalues. These facts combined with (12) contradict
that σ ( ∆+V) is infinite.
disc
−
Assertion (ii). First, we prove the following.
Lemma2.1. LetAbeaSchro¨dingeroperator d2 +V(t)withitsdomainC∞(0, )
−dt2 c ∞
and V C0(0, ). Assume that A has a self-adjoint extension A on L2((0, );dt)
∈ ∞ ∞
and that σ (A)=[0, ). Assume also that, there exist δ >0 and R>0 such that
ess
∞
V satisfies b
b 1 e e
V(t) (1+δ) for t R.
≤− 4t2 ≥
Then, σ (A) is infinite. e e
disc
b
KazuoAkutagawaandHironoriKumura 9
Proof ofLemma2.1. ForafixedR R,wedefineacut-offfunctionχ(t)asfollows:
≥
0 if t [0,R],
e ∈
 R1(t−R) if t∈[R,2R],
χ(t):= 1 1 (t 2kR) iiff tt∈[[k2RR,,k2kRR],],
−kR − ∈
0 if t [2kR, ),
where k is a large positive constant defined later. S∈et ϕ(t)∞:= χ(t)t21 for t > 0.
Then, the direct computation shows that
1 1
ϕ′(t)2+V(t)ϕ(t)2 = χ′(t)2t+ V(t)+ ϕ(t)2+ (χ(t)2)′.
| | | | 4t2 2
(cid:16) (cid:17)
Integrating the both sides over (0, ), we have
∞
∞ ∞ ∞ 1
ϕ′ 2+Vϕ2 dt= χ′(t)2tdt+ V + ϕ2dt
{| | } | | 4t2
Z0 Z0 Z0 (cid:16) (cid:17)
∞ δ ∞ χ2
χ′(t)2tdt dt
≤ | | − 4 t
Z0 Z0
1 2R 1e 2kR δ kR χ2
tdt+ tdt dt
≤ R2 (kR)2 − 4 t
ZR ZkR Z2R
e
δ k
= 3 log .
− 4 2
Hence, there exists a large positiveecons(cid:16)tan(cid:17)t k =k (δ,n) such that
0 0
∞
ϕ′ 2+Vϕ2 dt<0 if k k .
0
{| | } ≥
Z0
Successively choosing R and k, we then get a sequence ϕ ∞ of functions in
{ i}i=1
W1,2((0, )) with compact support such that
∞
supp ϕ supp ϕ = if i=j,
i j
∩ ∅ 6
∞
ϕ′ 2+Vϕ2 dt<0 for all i 1.
{| i| i} ≥
Z0
Therefore, the min-max principle implies that σ (A) is infinite. (cid:3)
disc
Next, we consider a rotationally symmetric Riemabnnian n-manifold
(13) (M,g):=(Rn,dr2+h(r)2·gSn−1(1)).
The restriction ( ∆ ) of ∆ to the space of radial functions on M is given
g radial g
− | −
by
d2 h′(r) d
( ∆ ) = (n 1) on L2((0, );h(r)n−1dr).
− g |radial −dr2 − − h(r) dr ∞
Now we define a unitary operator U as
U :L2((0, );h(r)n−1dr) ψ hn−21ψ L2((0, );dr).
∞ ∋ 7−→ ∈ ∞
Then, the self-adjoint operator ( ∆ ) on L2((0, );h(r)n−1dr) is unitary
g radial
− | ∞
equivalenttotheself-adjointoperatorL :=U ( ∆ ) U−1onL2((0, );dr).
g g radial
◦ − | ◦ ∞
Remark also that the multiplication operator V : ψ V ψ is conserved under
7→ ·
10 Theuncertaintyprinciplelemmaundergravity
this unitary transformation, that is, U V U−1 = V. For u Dom(L ), we can
g
◦ ◦ ∈
calculate L u explicitly as
g
Lgu= hn−21 ( ∆g)radial (h−n−21u)
◦ − |
(cid:16) d2u (n 1)(n(cid:17) 3) h′ 2 n 1h′′
= + − − + − u.
−dr2 4 h 2 h
n (cid:16) (cid:17) o
With these understandings, we will now return the construction of a desired
test function on Rn. Note that, since the upper bound for the potential V(x) on
r = x R is given by the radial function (1 + δ)(n−2)2, hence we may
{assume| t|h≥at V1}is also a radial function on Rn. So−, we will c4orn2struct our desired
test function as a radial function. Since the Euclidian n-space Rn is given by (13)
with h(r):=r, the potential term of L is
g
(n 1)(n 3) h′ 2 n 1h′′ (n 1)(n 3)
− − + − = − − ,
4 h 2 h 4r2
(cid:16) (cid:17)
andhence the potentialofthe operatorU ( ∆ ) +V(r) U−1 is givenby
g radial
◦ − | ◦
(cid:16) (cid:17)
(n 1)(n 3)
− − +V(r).
4r2
If we set δ := (n 2)2δ > 0 in Lemma 2.1, the condition for the potential of the
−
operator U ( ∆ ) +V(r) U−1
g radial
e ◦ − | ◦
(cid:16) (cid:17)
(n 1)(n 3) 1
− − +V(r) (1+δ) for r R
4r2 ≤− 4r2 ≥ 1
is equivalent to that for the potential V(r)
e
(n 2)2
V(r) (1+δ) − for r R .
≤− 4r2 ≥ 1
1
For ϕ(r)=χ(r)r2 defined in Proof of Lemma 2.1, its unitary transformation is
U−1ϕ(r)=h−n−21ϕ(r)=r−n−21χ(r)r12 =χ(r)r−n−22.
Setting φ(x) := U−1ϕ(r(x)) = χ(r(x))r(x)−n−22 for x Rn, we then get, by
∈
Lemma 2.1,
∞ (n 1)(n 3)
φ2+Vφ2 dx=σ ϕ′ 2+ V + − − ϕ2 dt
ZRn{|∇ | } n−1Z0 n| | (cid:16) 4t2 (cid:17) o
(n 2)2 k
σ 3 − δlog <0 if k k .
n−1 0
≤ − 4 2 ≥
For each i 1, by setting also a nfunction φ (x) := U(cid:16)−1(cid:17)ϕo(r(x)) in W1,2(Rn) with
i i
≥
compact support, then the sequence φ ∞ satisfies the following:
{ i}i=1
supp φ supp φ = if i=j,
i j
∩ ∅ 6
φ 2+Vφ2 dx<0 for all i 1.
ZRn{|∇ i| i} ≥
Here, ϕ is the function defined in Proof of Lemma 2.1. Therefore, the min-max
i
principle implies that σ ( ∆+V) is infinite. (cid:3)
disc
−
Proof of Theorem 1.5. Assertion (i). By using Theorem 1.3, the proof of the