Two-body problem on spaces of constant curvature

0.34 MB·English

by  A. V. Shchepetilov

Two-body problem on spaces of constant curvature
A.V. Shchepetilov∗, I.E. Stepanova†
Abstract
6
0 The two-body problem with a central interaction on simply connected constant
0 curvaturespacesofanarbitrary dimensionisconsidered. Theexplicit expressionfor
2 thequantumtwo-bodyHamiltonian viaaradialdifferentialoperator andgenerators
n of the isometry group is found. We construct a self-adjoint extension of this Hamil-
a tonian. Someitsexactspectralseriesarecalculatedforseveralpotentialinthespace
J S3.

7 Wedescribealsothereducedclassicalmechanicalsystemonahomogeneousspace
1 ofaLiegroupintermsofthecoadjointactionofthisgroup. Usingthisapproachthe
description of the reduced classical two-body problem on constant curvature spaces

4 is given.
v
5
1 PACS numbers: 03.65.Fd, 02.40.Vh, 02.40.Ky, 02.40.Yy.
0
Mathematical Subject Classification: 70F05, 43A85, 22E70, 57S25, 70G65.
1
0
5 1 Introduction
0
/
h The simply connected constant curvature spaces Sn and Hn posses isometry groups as
p wide as the isometry group for the Euclidean space En and have no selected points or
-
h directions [1]. The one-body classical and quantum problems in the central field on these
t spaces were studied in many papers, among which we indicate basic ones [2]-[13].
a
m IncontrasttotheEuclideancase,theconfigurationspacesSn Sn,Hn Hnofthetwo-
× ×
body problem on spaces Sn and Hn are not spaces of a constant curvature. Only space
:
v isometries that preserve an interaction potential enter in the symmetry group of such
i
X problem a priori. This group does not suffice to ensure the integrability of the two-body
problem. Atthesametime,no”hidden”symmetriesorotherintegrabilitytoolsareknown
r
a fornontrivialpotential. Numericalexperimentsforthereducedclassicaltwo-bodyproblem
in spaces Sn and Hn, n = 2,3 [14] show the soft chaos in this system for some natural
interactive potentials. Numerical experiments ([15]) and analytical results ([16]-[18]) for
the restricted classical two-body problem on S2 and H2 also prove its nonintegrability.
Theclassicalmechanicaltwo-bodyprobleminconstantcurvaturespaceswasfirstcon-
sideredin[19],wherethemethodoftheHamiltonianreductionofsystemswithsymmetries
[20] was used to exclude the motion of a system as a whole. The description of reduced
mechanical systems, their classification, and conditions for existence of a global dynamic
were obtained using explicit analytic coordinate calculations on a computer. In [22], an
analogousquantummechanicalsystemwasconsideredinthe two-dimensionalcase,i.e.on
the spaces S2 H2. There, the quantum mechanical two-body Hamiltonian was expressed
through isometry group generators and a radial differential operator. The structure of
this expression is similar to the structure of reduced Hamilton function. The idea arises
∗Department of Physics, Moscow State University, 119992 Moscow, Russia, e-mail address:
[email protected]
†Institute forPhysicsofEarth,RAS,bol.Gruzinskaja10,123995Moscow,Russia
1
2
to seek a general procedure for simultaneous simplification both classical and quantum
problems without performing cumbersome calculations. We present such a procedure in
the present paper. The obtained expression for the quantum two-body Hamiltonian is
useful for solving at least three problems.
First it enables us to prove that the two-body Hamiltonian with the proper domain
is self-adjoint. Secondly, using this expression, one can reduce the spectral problem for
the two-body Hamiltonian to a sequence of systems of ordinary differential equations
enumeratedbyirreduciblerepresentationsoftheisometrygroup. Inthecaseofthesphere
S3 we found all separate differential equations for spectral values, which can be solved in
an explicit form for some interaction potentials. Thus, although the two-body problem
on spaces Sn and Hn seems to be non-integrable in any sense, some its energy levels can
be explicitly found. Such a situation appears in so called quasi exactly solvable models
[23]-[25]. There aretwoessentialdifferences however. First,usually quasiexactly solvable
models are artificially constructed. Second exact energy levels are obtained, as a rule,
from one differential equation with the specially selected potential, but this equation has
also other unknown spectral values. Conversely, in the problem under consideration we
select a separate differential equation from systems of ordinary differential equations and
find all its spectral values.
Finally, from the obtained expression for the two-body Hamiltonian we derive the
Hamilton function of the reduced two-body classical mechanical system, using the de-
scription of a reduced classical mechanical system on a homogeneous space in terms of
coadjoint orbits of the corresponding Lie group, founded in section 6.
This paper is an essential revision of papers [26] and [27], made by the first author.
Some results in those papers were not properly grounded. This was later done in [28].
Proofs in the present paper are extended and in some technically difficult cases they
contain references to [28]. More serious revision concerns the theorem 5 of the present
paper. Because of a confusion between left and right shifts on a group, the statement in
[27] corresponding to this theorem contains some additional erroneous eigenvectors.
2 Notations
Consider the sphere Sn as the space Rn with the metric
∪{∞}
n n 2
g = 4R2 dx2 1+ x2 , (1)
s i i
!, !
i=1 i=1
X X
where x , i = 1,...,n are Cartesian coordinates in Rn and R is the curvature radius.
i
Let ρs(, ) denotes the distance between two points in Sn. The identity component of a
· ·
whole isometry group for Sn, acting from the left, is SO(n+1). The set
∂ ∂
Xs =x x , 16i<j 6n,
ij i∂x − j∂x
j i
(2)
n n
1 ∂ ∂
Ys = 1 x2 +x x , i=1,...,n,
i 2 − j∂x i j∂x
i j
j=1 j=1
X X
 
is a base of Killing vector fields on Sn. It corresponds to some base in the Lie algebra
so(n+1).
Consider the hyperbolic space Hn as the unit ball Dn Rn with the metric
⊂
n n 2 n
g = 4R2 dx2 1 x2 , x2 <1. (3)
h i!, − i! i
i=1 i=1 i=1
X X X
3
Denote the distance between two points in Hn as ρh(, ). Let O (1,n) be an identity
0
· ·
component of a whole isometry group for Hn, acting from the left. Its Lie algebra is
so(1,n). The set
∂ ∂
Xh =x x , 16i<j 6n,
ij i∂x − j∂x
j i
(4)
n n
1 ∂ ∂
Yh = 1+ x2 x x , i=1,...,n
i 2 j∂x − i j∂x
i j
j=1 j=1
X X
 
is a base of Killing vector fields on Hn.
3 Special forms of free Hamiltonians
Let Q = Sn Sn and Q = Hn Hn be the configuration spaces of the two-body
s h
× ×
problems in Sn and Hn. The corresponding Hamiltonians are defined as
1 1
H = +U(ρs,h) Hs,h+U(ρs,h), (5)
s,h −2m △1−2m △2 ≡ 0
1 2
where and b are Laplace-Beltrami operators in thebdirect factors of Sn Sn and
1 2
△ △ ×
Hn Hn,correspondingtothe firstandthesecondparticles,andU isacentralpotential.
×
Here and below the subscript ”s” corresponds to the spherical case and the subscript ”h”
corresponds to the hyperbolic case.
Accordingtothegeneralconceptofquantummechanics[29],adomainofthe operator
H mustbeapropereverywheredensesubspaceinthespace 2(Q ,dµ )ofallsquare
s,h s,h s,h
L
integrablefunctionsonQ . ThissubspaceischoseninsuchawaythattheoperatorH
s,h s,h
bbecomes self-adjoint; the corresponding measure dµ is the product of measures on the
s,h
direct factors of Sn Sn (Hn Hn), invariant w.r.t. the group SO(n+1) (O (1,n)).b
0
× ×
To express the total Hamiltonian H through the radial differential operator and
s,h
generators of the isometry group, it is suffices to find such an expression for the free
Hamiltonian. Recall (see, for exampleb, [30]) that the Laplace-Beltrami operator on
△
spaces Sn and Hn is respectively self-adjoint with domains
W2,2 := φ 2(Sn,dµ ) φ 2(Sn,dµ ) ,
s ∈L s |△ ∈L s
Wh2,2 :=(cid:8)φ∈L2(Hn,dµh)|△φ∈L2(Hn,dµh(cid:9)) .
Theactionoftheoperator (cid:8)isconsideredinthesenseofdistributio(cid:9)ns. Theoperator on
△ △
Sn is essentially self-adjoint on the space C∞(Sn) of smooth functions, and the operator
on Hn is essentially self-adjoint on the space of finite smooth functions C∞(Hn) [31].
△ 0
Hence the free Hamiltonian Hs,h is self-adjoint on the product
0
W :=W2,2 W2,2 (6)
b s,h s,h ⊗ s,h
of two copies of the space W2,2 corresponding respectively to the first and the second
s,h
particles.
Let Fs,h be submanifolds of the space Q , corresponding to the constant value r of
r s,h
the function tan(ρs/(2R)) for the space Q and the function tanh(ρh/(2R)) for the space
s
Q . The submanifolds Fs, Fs are diffeomorphic to Sn (the value r = corresponds to
h 0 ∞ ∞
two diametrically opposite points of the sphere Sn) and Fh is diffeomorphic to Hn. For
0
0<r < thesubmanifoldFs isahomogeneousRiemannianspaceofthegroupSO(n+1)
∞ r
with the stationary subgroup K = SO(n 1). For 0 < r < 1 the submanifold Fh is a
∼ − r
homogeneous Riemannian space of the group O (1,n) with the stationary subgroup K.
0
4
Up to a zero measure set it holds Q =R (SO(n+1)/K), where R =(0, ) and
s + +
× ∞
also Q =I (O (1,n)/K), where I =(0,1). The operators Hs,h are Laplace-Beltrami
h × 0 − 0
ones for the metric g = 2m g(1) +2m g(2) on Q , where the metrics g(1) and g(2)
s,h 1 s,h 2 s,h s,h s,h s,h
have either form (1) or (3) on different copies of the spaces Sn abnd Hn, corresponding to
particles 1 and 2. e
3.1 Two-particle Hamiltonian on Sn Sn
×
Giventhepointx F onecanidentifythesubmanifoldF withthefactorspaceSO(n+
0 r r
∈
1)/SO(n 1) by the formula x = gKx , where gK is the left coset of the element g in
0
−
the group SO(n+1) w.r.t. its subgroup K. Let (r,y ,...,y ) be local coordinates in
1 2n−1
some neighborhood W of the point x Q and then (y ,...,y ) are the coordinates
0 s 1 2n−1
∈
in the open subset W F of the submanifold F . Then the metric g in W becomes1
r r s
T 2n−1
g =g (r)dr2+ g (r,y ,...,y )dy dy .e
s rr ij 1 2n−1 i j
i,j=1
X
e
The secondterm in this formula is the restriction of a metric g from the submanifold F
f r
ontothedomainW F . UsingthestandardexpressionfortheLaplace-Beltramioperator
r
trough local coordinates, one gets
T
∂ ∂
=(g detg )−1/2 grrdetg + . (7)
△gs rr ij ∂r ij∂r △gf
(cid:18) (cid:19)
p
e
To expressthe operator onF throughthe generatorsofthe Lie groupSO(n+1)one
△gf r
can use the following construction from [33].
Let Γ be a Lie group and Γ be its subgroup. The group Γ acts from the left on the
0
homogeneous space Γ/Γ . Left-invariant differential operators on the space Γ/Γ can be
0 0
representedby left-invariantdifferentialoperatorsonthe groupΓ thatare simultaneously
invariant w.r.t. the right action of the group Γ . This representation is one to one up to
0
operators summands, vanishing while acting onto functions that are invariant w.r.t. right
Γ -shifts.
0
Indeed, functions on the factor space Γ/Γ are in one to one correspondence with
0
functions on the group Γ that are invariant w.r.t. right Γ -shifts. This correspondence is
0
definedbytheformulaλ: f f˜:=f π,whereπ : Γ Γ/Γ isthecanonicalprojection
0
→ ◦ →
andf is a functiononthe factorspaceΓ/Γ . LetD be a differentialoperatoronΓ thatis
0
invariant w.r.t. left Γ-shifts and right Γ -shifts. Let f be a smooth function on the factor
0
space Γ/Γ . Then the formulaD f =Df˜defines the correspondenceD D , where D
0 u u u
→
is a differential operator on the space Γ/Γ , invariant w.r.t. left Γ-shifts.
0
Let e ,...,e be a base ofgthe Lie algebra of the group Γ, N := dimΓ and let L
1 N γ
and R denote respectively the left and right shifts by the element γ Γ. The algebra
γ
of left invariantdifferential operatorson the group Γ is generatedover R∈by left invariant
vector fields el,...,el , where el(γ)=dL (e ), γ Γ, i=1,...,N [33].
1 N i γ i ∈
Let now Γ = SO(n+1), Γ = K = SO(n 1), er(γ) = dR (e ), i = 1,...,N, N =
0 ∼ − i γ i
(n+1)(n+2)/2, x =(r ,0,...,0,r ,0,...,0) Sn Sn, where
0 1 2
∈ ×
n−1 n−1
m| {z } | {z } m
2 1
r =tan arctanr , r = tan arctanr . (8)
1 2
m +m − m +m
(cid:18) 1 2 (cid:19) (cid:18) 1 2 (cid:19)
1We suppose that this metric does not contain summands proportional to drdyi. It is valid for the
parametrization (8)below. Moredetailed consideration for amoregeneral parametrization andan arbi-
trarytwo-pointhomogeneousspacesM canbefoundin[28]. Theconnectionofthetwo-bodyHamiltonian
withthealgebraofinvariantdifferentialoperatorsontheunitspherebundleoverM wasdescribedin[32].
5
The set of Killing vector fields Xs,Ys, i,j = 1,...,n on the space Sn Sn, corre-
ij i ×
sponding to (2), coincides (up to a permutation) with the set
d N
e˜r(γx )= exp(τe )γx , x =x (r), 0<r< , (9)
i 0 dτ i 0 0 0 ∞
(cid:26) (cid:12)τ=0 (cid:27)i=1
(cid:12)
(cid:12)
under the proper choice of(cid:12)the basis e1,...,eN. Let f be a second order differential
△
operator on the group Γ such that ( ) = . Then it is left invariant and can be
△f u △gf
expressed in the form2
N N
= cijel el + ciel ,
△f|γ i◦ j γ i γ
i,j=1 i=1
X (cid:12) X (cid:12)
(cid:12) (cid:12)
where cij,ci are constant on the submanifold F . Let e be the unit element of the group
r
Γ. Obviously, er = el , i=1,...,N and
i|e i e
(cid:12)
(cid:12) N N
= cijer er + cier . (10)
△f|e i ◦ j e i e
i,j=1 i=1
X (cid:12) X (cid:12)
(cid:12) (cid:12)
It yields
N N
= cij e˜r e˜r + cie˜r =: (2) + (1) .
△gf x0 i ◦ j x0 i x0 △gf x0 △gf x0
One can find(cid:12)(cid:12)coefficiX,ije=n1ts cij in th(cid:12)(cid:12)e follXio=w1ing w(cid:12)(cid:12)ay. Consid(cid:12)(cid:12)(cid:12)er the ord(cid:12)(cid:12)(cid:12)ered set of vec-
tors Y1s|x0,...,Yns|x0,X1s2|x0,...,X1sn|x0 as a base in the linear space Tx0Fr. Let
Y1,...,Yn,X2,...,Xn be the dual basis. Then
(cid:8) (cid:9)
(cid:8) (cid:9)
n n
g = Y1 α Yi+β Xi + α Yi Yj +β Xi Xj +γ Yi Xj
f|x0  ⊗ i i ij ⊗ ij ⊗ ij ⊗ 
i=2 j=2
X (cid:0) (cid:1) X(cid:0) (cid:1)
+aY1 Y1, 
⊗
where
a= g(Ys,Ys) =2R2(m +m ),
1 1 |x0 1 2
α = g(Ys,Ys) =0,
i 1 i |x0
e
β = g(Ys,Xs) =0, i=2,...,n,
i 1 1i |x0
e
e
2 m 1 r2 2
α = g Ys,Ys =2R2 k − k δ ,
ij i j x0 k=1 (1(cid:0)+rk2)2(cid:1) ij
(cid:0) (cid:1)(cid:12) X
e (cid:12) 2 m r2
β = g Xs,Xs =8R2 k k δ , (11)
ij 1i 1j x0 (1+r2)2 ij
k=1 k
(cid:0) (cid:1)(cid:12) X
e (cid:12) 2 m r (1 r2)
γ = g Ys,Xs =4R2 k k − k δ , i,j =2,...,n.
ij i 1j x0 (1+r2)2 ij
k=1 k
(cid:0) (cid:1)(cid:12) X
2Hereweconsiderevector fieldsa(cid:12)sdifferentialoperators ofthefirstorder.
6
Therefore one gets
n
1 1
(2) = (Ys)2 + A (Xs )2+C (Ys)2 B Xs ,Ys , (12)
△gf x0 a 1 x0 2 s 1k s k − s{ 1k k} x0
(cid:12) (cid:12) kX=2h i
(cid:12) (cid:12)
(cid:12) (cid:12)
where , denotes the anticommutator and functions A ,B ,C have the form
s s s
{· ·}
m (1 r2)2(1+r2)2+m (1+r2)2(1 r2)2
A = 1 − 1 2 2 1 − 2 ,
s 4R2m m (r r )2(1+r r )2
1 2 1 2 1 2
−
m r (1 r2)(1+r2)2+m r (1 r2)(1+r2)2
B = 1 1 − 1 2 2 2 − 2 1 ,
s 2R2m m (r r )2(1+r r )2
1 2 1 2 1 2
−
m r2(1+r2)2+m r2(1+r2)2
C = 1 1 2 2 2 1 .
s R2m m (r r )2(1+r r )2
1 2 1 2 1 2
−
These functions can be expressed also through the coordinate r:
1 (1+r2)2 1 r4 1+r2
A (r)= + − cosζ+ (m m )sinζ ,
s R2 8mr2 8mr2 4m m r 1− 2
(cid:18) 1 2 (cid:19)
1 m m 1 r4
B (r)= 2− 1(1+r2)cosζ+ − sinζ ,
s −4R2 m m r 2mr2
(cid:18) 1 2 (cid:19)
1 (1+r2)2 1 r4 1+r2
C (r)= − cosζ (m m )sinζ ,
s R2 8mr2 − 8mr2 − 4m m r 1− 2
(cid:18) 1 2 (cid:19)
m m m m
1 2 1 2
ζ :=2 − arctanr, m:= .
m +m m +m
1 2 1 2
Theoperators and (2) (see(12))areinvariantw.r.t.reflectionsT : x
△gf x0 △gf x0 k k →
−xk,xj → xj,j 6= k, j(cid:12)(cid:12)= 1,...,n, k(cid:12)(cid:12) = 2,...,n of the sphere Sn. Therefore the operator
(1) is alsoinvariantw.r.t.these(cid:12)reflections. Reflections T changessignsofthe vector
△gf x0 k
fields(cid:12)(cid:12)(cid:12)X1sk|x0,Yks|x0, k =2,...,n,thereforetheoperator △(g1f) x0 isproportionalto Y1s|x0.
The more accurate analysis (see [28]) shows that the operator(cid:12)(cid:12)(cid:12)△(g1f) vanishes.
If we denote by Ys,l,Xs,l,Ys,l left invariant vector fields on the group SO(n + 1),
1 k k
corresponding to vectors Ys ,Xs ,Ys , k =2,...,n, we get
1|x0 1k|x0 k|x0
1 1 1
= D2+ A D + C D +B D , (13)
△f a 0 2 s 2 2 s 1 s 3
where operators D ,D ,D and D have the form
0 1 2 3
n n n
2 2 1
D =Ys,l, D = Ys,l , D = Xs,l , D = Xs,l,Ys,l .
0 1 1 k 2 k 3 −2 k k
kX=2(cid:16) (cid:17) Xk=2(cid:16) (cid:17) Xk=2n o
By direct calculations in the universal enveloping algebra U(so(n+1)), one can get
(see [32]) the following commutator relations for the operators D ,...,D
0 3
[D ,D ]= 2D , [D ,D ]=2D , [D ,D ]=D D , [D ,D ]= 2 D ,D , (14)
0 1 3 0 2 3 0 3 1 2 1 2 0 3
− − − { }
(n 1)(n 3) (n 1)(n 3)
[D ,D ]= D ,D + − − D , [D ,D ]= D ,D − − D .
1 3 0 1 0 2 3 0 2 0
−{ } 2 { }− 2
Thus we found the operator up to summands, annulled by functions that are
f
△
invariant w.r.t. right Γ -shifts.
0
7
We are to find now the first term in expression (7). At the point x one has
0
∂ m 1+r2 ∂ m 1+r2 ∂
= 2 1 1 2
∂r m +m 1+r2∂r − m +m 1+r2∂r
1 2 1 1 2 2
and therefore
∂ ∂ 8R2m m
1 2
g =g , = . (15)
rr ∂r ∂r (m +m )(1+r2)2
(cid:18) (cid:19) 1 2
Due to formulas (11) one gets
e
(1+r2)n ∂ rn−1 ∂
= + ,
△gs 8mR2rn−1∂r (1+r2)n−2∂r △gf
(cid:18) (cid:19)
where the first term is ethe radial part of the one particle Hamiltonian with the mass m.
Direct calculation at the point x gives for the measure dµ , corresponding to the
0 s
metric g in the space Q , the following expression
s s
rn−1
e dµs|x0 = (1+r2)ndr∧Y1∧···∧Yn∧X2∧···∧Xn
up to a constant factor.
The measure dµ is left invariant w.r.t. the group SO(n+1) and therefore it can be
s
represented in the form dµ = dν dµ , where dν = rn−1dr/(1+r2)n is a measure on
s s f s
R =(0, ), the same as for the on⊗e particlecase,anddµ is a measure onSO(n+1)/K
+ f
∞
left invariant w.r.t. the group SO(n+1).
Each Lie group admits unique (up to a constant factor) left-invariant and right-
invariant measures (Haar measures [34]). For the groups under consideration SO(n+1)
and O (1,n) such measures are two-side invariant. Hence there exist a unique two-side
0
invariant measure dη on the group SO(n+1) such that the integral of any integrable
s
function f on the space SO(n+1)/K w.r.t. the measure dµ equals the integral of the
f
function f˜on the group SO(n+1) w.r.t. the measure dη .
s
Definition 1. For a subgroup Γ of a Lie group Γ denote by 2(Γ,Γ ,dη) the space of
0 0
L
square-integrable functions on the group Γ (w.r.t. the measure dη on Γ) that are right
invariant w.r.t. Γ -shifts.
0
Theorem 1. The free two-particle Hamiltonian on the sphere Sn can be considered as the
self-adjoint differential operator (on the manifold Q =R SO(n+1))
s +
×
(1+r2)n ∂ rn−1 ∂
Hs = e , (16)
0 −8mR2rn−1∂r (1+r2)n−2∂r −△f
(cid:18) (cid:19)
with the domain b
D :=D(1) D(2) := 2(R ,dν ) 2(SO(n+1),K,dη ),
s s ⊗ s ⊂Hs L + s ⊗L s
where
D(1) := φ 2(R ,dν ) (1)φ 2(R ,dν ) ,
s ∈L + s |△s ∈L + s
D(2) :=nφ 2(SO(n+1),K,dη ) φ 2(oSO(n+1),K,dη ) ,
s ∈L s |△f ∈L s
(1+r2)n ∂ rn−1 ∂
(1) :=(cid:8) , (cid:9)
△s − rn−1 ∂r (1+r2)n−2∂r
(cid:18) (cid:19)
thesubgroupK isisomorphic tothegroupSO(n 1),anddη isaunique(uptoaconstant
s
−
factor) two-side invariant measure on the group SO(n+1). It means that there exists an
isometry of the initial space of functions 2(Q ,dµ ) onto the space that generates the
s s s
L H
isomorphism of Hamiltonians. The space D is everywhere dense in .
s s
H
8
Proof. Expression(7) representsthe HamiltonianHs via coordinates,correspondingwith
0
the representation of the space Q as the direct product R SO(n+1)/SO(n 1) up
s +
× −
to the zero measure set Fs Fs , which is inessenbtial when studying functions that are
0 ∪ ∞
integrable over this measure. Therefore
2(Q ,dµ )= 2(R ,dν ) 2(SO(n+1)/SO(n 1),dµ ).
s s + s f
L L ⊗L −
The isometry λ:f f˜of spaces
→
2(SO(n+1)/SO(n 1),dµ ) and 2(SO(n+1),SO(n 1),dη )
f s
L − L −
generatesthe isometryid λofspaces 2(R ,dν ) 2(SO(n+1)/SO(n 1),dµ )and
+ s f
⊗ L ⊗L −
. Calculations above imply that the isometry id λ transforms operator (7) into oper-
s
H ⊗
ator (16) and the space W into the space D .
s s
Remark 1. In the case n = 2 this result can be obtained by treating a basis in the Lie
algebra so(3) as a moving frame on the submanifold F [22]. For n>2 this is impossible
r
sincetheSO(n+1)-action on F is notfreeandtheprojection ofleft-invariant vectorfields
r
from SO(n+1) onto SO(n+1)/SO(n 1) is not well defined. By lifting the Hamiltonian
−
onto the symmetry group one can express it through group generators.
3.2 Two-particle Hamiltonian on the space Hn Hn
×
Theformalsubstitutionx ix ,r ir,R iR,j =1,...,n,(hereiistheimaginary
j j
→− →− →
unit) transforms objects on the sphere Sn into objects on the hyperbolic space Hn (see
also [19]). Therefore from results of the previous section one can obtain the two-particle
Hamiltonian on the space Hn Hn:
×
(1 r2)n ∂ rn−1 ∂ 1 1 1
Hh = − D¯2 A D¯ C D¯ B D¯ , (17)
0 −8mR2rn−1∂r (1 r2)n−2∂r − a 0− 2 h 2− 2 h 1− h 3
(cid:18) − (cid:19)
wherbe 0<r <1,
n n n
2 2 1
D¯ =Yh,l, D = Ys,l , D¯ = Xh,l , D¯ = Xh,l,Yh,l ,
0 1 1 k 2 k 3 −2 k k
Xk=2(cid:16) (cid:17) kX=2(cid:16) (cid:17) Xk=2n o
vectorfields Xh,l,Yh,l relateto vectorfields (4)inthe samewayasvectorfields Xs,l,Ys,l
k k k k
relate to vector fields (2), and
1 (1 r2)2 1 r4 1 r2
A (r)= − + − coshζ − (m m )sinhζ ,
h R2 8mr2 8mr2 − 4m m r 1− 2
(cid:18) 1 2 (cid:19)
1 (m m ) 1 r4
B (r)= 2− 1 (1 r2)coshζ+ − sinhζ ,
h −4R2 m m r − 2mr2
(cid:18) 1 2 (cid:19)
1 (1 r2)2 1 r4 1 r2
C (r)= − + − coshζ − (m m )sinhζ ,
h −R2 − 8mr2 8mr2 − 4m m r 1− 2
(cid:18) 1 2 (cid:19)
m m
1 2
ζ :=2 − arctanhr.
m +m
1 2
Commutator relations for operators D¯ ,D¯ ,D¯ ,D¯ have the form (see [32])
0 1 2 3
[D¯ ,D¯ ]=2D¯ , [D¯ ,D¯ ]=2D¯ , [D¯ ,D¯ ]=D¯ +D¯ , [D¯ ,D¯ ]= 2 D¯ ,D¯ , (18)
0 1 3 0 2 3 0 3 1 2 1 2 0 3
− { }
(n 1)(n 3) (n 1)(n 3)
[D¯ ,D¯ ]= D¯ ,D¯ − − D¯ , [D¯ ,D¯ ]= D¯ ,D¯ − − D¯ .
1 3 0 1 0 2 3 0 2 0
−{ }− 2 { }− 2
9
Theorem 2. The free two-particle Hamiltonian on the space Hn can be considered as the
self-adjoint differential operator (17) on the manifold Q =I O (1,n) with the domain
h 0
×
D :=D(1) D(2) := 2(R ,dν ) e 2(O (1,n),K,dη ),
h h ⊗ h ⊂Hh L + h ⊗L 0 h
where K =SO(n 1),
−
D(1) := φ 2(R ,dν ) (1)φ 2(R ,dν ) ,
h ∈L + h |△h ∈L + h
D(2) :=nφ 2(O (1,n),K,dη ) φ 2(Oo(1,n),K,dη ) ,
h ∈L 0 h |△h ∈L 0 h
(1 r2)n ∂ rn−1 ∂ rn−1dr
(1) :=(cid:8) − , dν = , (cid:9)
△h − rn−1 ∂r (1 r2)n−2∂r h (1 r2)n
(cid:18) − (cid:19) −
1 1 1
:= D¯2 A D¯ C D¯ B D¯ ,
△h −a 0− 2 h 2− 2 h 1− h 3
and dη is a unique (up to a constant factor) two-side invariant measure on the group
h
O (1,n).
0
The proof is analogous to the proof of theorem 1.
4 Self-adjointness of two-particle Hamiltonians
InEuclideanspacethe self-adjointnessofmany-particleHamiltonianswithpairwiseinter-
acting particles is usually proved using the Galilei invariance, which has no analog in the
spaces Sn and Hn [35]. The self-adjointness of one-particle Hamiltonians with singular
potential unbounded frombelow can be provedusing the perturbationtheory for the cor-
responding quadratic forms. The key point of the proof is the following estimate, called
the ”uncertainty principle” in [35]:
(Uψ,ψ)6 ψ 2,
k∇ k
where (, ) is the scalar product in the space 2(R3), is the gradient operator, and U
· · L ∇
is a potential. To derive this estimate for spaces Sn and Hn one needs some modification
of the proof w.r.t. the Euclidean case. Instead of tending to full generality, we are mostly
interested in physically significant potentials.
Fromtheself-adjointnessofthefreetwo-particleHamiltonianwithdomain(6)weshall
prove the self-adjointness of the two-particle Hamiltonian with an interaction using the
perturbation theory for the quadratic forms.
Let(, )bethescalarproductonfibersofthecotangentbundleT∗Q ,generatedbythe
s
· ·
metric g˜ , and be respectively the corresponding norm and the gradient operator.
s
k·k ∇
Let also f,ψ C∞(Q ) be real functions3 such that f is constant on submanifolds Fs,
∈ s r
i.e. f =f(r). Then it holds
(fψ) ψ f 2 ψ2 f 2 2ψ( f, (fψ))
ψ 2 = ∇ ∇ > k∇ k ∇ ∇
k∇ k f − f f2 − f2
(cid:13) (cid:13)
=(cid:13)(cid:13)(cid:13)ψ2fg˜2srr(f′)2− 2fψ2(cid:13)(cid:13)(cid:13)g˜srrf′∂∂r (fψ)
Equation(15)impliesthatg˜rr =(1+r2)2/(8R2m). IntegratingoverQ withthemeasure
s
3Allfunctionalspacesareassumedtocomprisecomplex-valuedfunctions.
10
dµ , one gets
s
1 (f′)2ψ2 2ψf′ ∂
ψ 2dµ > (fψ) (1+r2)2dµ
k∇ k s 8R2m f2 − f2 ∂r s
Z Z (cid:20) (cid:21)
Qs Qs
(19)
∞
1 (f′)2ψ2 1 ψf′rn−1 ∂
= (1+r2)2dµ (fψ)drdµ .
8R2m f2 s− 4R2m f2(1+r2)n−2∂r f
Z Z Z
Qs Frs 0
We now want to find a function f such that for every smooth function ψ, the function
ψf′rn−1 ∂
(fψ)
f2(1+r2)n−2∂r
has the form ∂ φ(r)ψ2 . Then the last integral in (19) can vanish identically. Solving
∂r
the system of equations
(cid:0) (cid:1)
(f′)2rn−1 f′rn−1
=φ′, =2φ,
f2(1+r2)n−2 f(1+r2)n−2
one gets
(1+r2)n−2 −1 (n 2)rn−2/4, n>3
φ(r)= 4 dr , φ(r) − , r 0,
− rn−1 ∼ (4 lnr)−1, n=2 →
(cid:20) Z (cid:21) (cid:26) | |
(n 2)r2−n/4, n>3
φ(r) − − , r , (20)
∼ (4 lnr)−1, n=2 →∞
(cid:26) − | |
f′ 2 (n 2)2/(4r2), n>3
f (r)∼ 4−r2ln2r −1, n=2 , r →0, r →∞.
(cid:18) (cid:19) (cid:26)
(cid:0) (cid:1)
It is easy to verify that for any choice of the integration constant the function φ(r) is
discontinues at some point, which does not occur in the Euclidean case for n > 3. Let
ω := x Q r(x)<δ and ω′ := x Q r(x)>δ−1 . Choose δ > 0 in such a way
δ { ∈ s| } δ ∈ s|
that the function φ(r) is continues in the domain (ω ω′ ) (Fs Fs ). On the space
(cid:8) 2δ∪(cid:9)2δ \ 0 ∪ ∞
Q (Fs Fs ) the continuity domains of the functions f′/f and φ(r) coincide; moreover
s\ 0 ∪ ∞
both functions are nonzero for r =0, . We set
6 ∞
r−2+r2, n>3
u (r)= ,
n r−2+r2 /ln2r, n=2
(cid:26)
and choose the constant κ>0 such t(cid:0)hat the in(cid:1)equality
(1+r2)2(f′)2
κu (r)6
n 8R2mf2
holds. Then due to (19) and (20) one gets the following inequality
κ u (r)ψ 2dµ 6 ψ 2dµ (21)
n s s
| | k∇ k
Z Z
Qs Qs
forthefunctionψ C∞(Q )withsupp ψ ω ω′. Writinginequality(21)separatelyfor
∈ s ⊂ δ∪ δ
therealandimaginarypartsofacomplex-valuedfunction,weobtain(21)foranarbitrary
function ψ C∞(Q ) with suppψ ω ω′.
∈ s ⊂ δ∪ δ