ANZOR KHELASHVILI,TEIMURAZ NADAREISHVILI
(TSU, HEPI)
PHYSICS OF SINGULAR POTENTIALS
I. Statement of problem (Scrhrodinger equation.Discrete Spectrum)
⎛ ∂ 1⎞
Pr = −i ⎜ + ⎟
From the demand, that H and
⎝ ∂r r ⎠ operators are
Hermitian it follows, that [1.D.Blockincev, 2.V.Pauli, 3.A.Messia]
limrR=u(0) =0
(1.1)
r→0
For regular potentials
limr2V =0
(1.2)
R = ar l + br − ( l +1)
(1.3)
r→0
r →0
Second term in (1.3) doesn’t obey (1.1) and is neglected usually
Singular potentials
(1.4)
lim r 2V → ±∞
r →0
Transition potentials
lim r 2V → ±V0 (V0 > 0)
r →0
(1.5)
Theorem. For attractive transition potentials Schrodinger equation
except standard solutions, may also have additional solutions. Proof:
l (l + 1)
(1.6)
u′′ + 2m [ E − V ( r ) ] u −
u = 0; u = Rr
2
r
At small r from (1.6) we obtain
(1.7)
u = ast r1/ 2+ P + aadd r1/ 2− P = ust + uadd
r →0
Where
P=
( l + 1/ 2 )
2
− 2mV0
For
1
(1.8)
0<P<1/2
(1.9)
Both standard and additional solutions satisfy (1.1) condition (For
P>1/2 only standard solutions stay!)
From (1.8) and (1.9) we obtain condition of existence of additional
states
(1.10)
l (l + 1) < 2mV0
uadd satisfy requirement, that [4.-L.Schiff. Quantum mechanics.]
integral from particle coordinate probability density is finite!
Remark:
In paragraph 35- “Falling on the center” [5-L.Landau, E.Lifchitz.
u
is considered at small r
Quantum mechanics]. behavior of R =
r
(1.11)
R = Ar −1/ 2+ P + Br1/ 2− P
r →0
In (1.11) for P<1/2 both terms are singular (second term is more
singular!) and in [6- R.Newton monograph] author notice: “If P<1/2,
then the second solution is irregular in sense, that it is dominant above
first solution”. So R.Newton come very close to additional state
problem, but doesn’t mention that they exist! In [5] potential is made
regular by cutting off it at some small r0 and the limit r0 → 0 is taken,
which selects less singular solution and so additional solutions are
neglected! But if we multiple (1.11) relation on r we get (1.7) relation,
where we have, no singularity in the 0<P<1/2 region and as mentioned
above standard and additional solutions are “equal in rights” members
of (1.7) relation!
I I. Introduction of self-adjoint extension parameter
In [7-K.Case.Phys.Rev.80,797(1950); 8-K.Meetz.Nuovo Cimento 34,
690(1964); 19-A.Perelomov, V.Popov.TMF.vol 4 (1970)] was shown,
that for attractive regular and transition potentials it isn’t enough to
2
know potential and is necessary to introduce one arbitrary constant,
which is equivalent to give boundary condition at the origin. Indeed,
when
2
2mV0 > ( l + 1/ 2 )
(2.1)
P is complex from (1.8), both ust and uadd solutions have same behavior
V
at the origin and for example for limV = − 02 at small distances one
r →0
r
have [5, 7]
u ≈ A r cos
(
2mV0 − ( l + 1/ 2 ) ln r + B
2
)
(2.2)
B is arbitrary constant. On the Mathematical language it means, that H
is Hermitian (symmetric), but isn’t Self-adjoint operator and it is
necessary to introduce 1 parameter to make H Self-adjoint
![M.Reed,B.Simon:vol 2]. As was shown in [7] if B is fixed constant,
then all eigensolutions form a complete orthonormal set, and Eeigenvalues are real! But in this case we have“falling” on the center
and energy isn’t bounded from below!
In the region
2
2mV0 < ( l + 1/ 2 )
(2.3)
based on the above mentioned paragraph of [5] , uadd solutions is
neglected. We notice above, that in the 0<P<1/2 region it is necessary
to preserve uadd !
Then for arbitrary E1 and E2 levels ortogonality condition is
∞
m( E2 − E1 ) ∫ u2u1dr = 2 P {a1st a2add − a2st a1add }
0
And for ortogonality right side of (2.4) is zero
3
(2.4)
a1st
a2st
(2.5)
= add
add
a1
a2
So, we get it is necessary to introduce self-adjoint τ extension
parameter
a
τ = add
(2.6)
ast
From (1.7) and (2.6) we have:
a) aadd = 0(τ = 0) .We retain only standard levels
b) ast = 0(τ = ±∞) .We retain only additional levels
c) When τ ≠ ±∞,0 ,then both levels exist at the same time!
For some unknown reasons the Nature choose only standard levels
yet! We think,
that other cases are also possible!
I I.I Model of Valent electron
V0 α
(3.1)
− ; V0 ,α > 0
r2 r
This potential “naturally” appears for coulomb potential in the KleinGordon equation. Following [12-W.Krolikowski; Bulletin De L’
academics
polonaise
Vol
XVII.83(1979);13A.A.Khelashvili,T.P.Nadareishvili, Bulletin of
Georgian
Acad.Sci:Vol 164.no1(2001)] we obtain general solution for (3.1)
potential
u = C1ρ 1/ 2+ P e− ρ / 2 F (1/ 2 + P − λ ,1 + 2 P; ρ ) + C2e1/ 2− P e− ρ / 2 F (1/ 2 − P − λ ,1 − 2 P; ρ ) (3.2)
Where P is given again by (1.8) and
2mα
(3.3)
;E<0
ρ = −8mE ⋅ r ; λ =
−8mE
we get transcendental equation for E
Γ(1/ 2 − λ − P)
τ
Γ(1 − 2 P )
(3.4)
=−
P
Γ(1/ 2 − λ + P)
Γ
(1
+
2
P
)
( −8mE )
V =−
4
where SAE parameter τ is
C
1
τ= 2
C1 ( − mE ) P
(3.5)
E depends on τ parameter. For τ = 0 and τ = ±∞ we obtain standard
and additional levels analitically
mα 2
mα 2
(3.6)
=−
Est ,add = −
2
2
2
2 [1/ 2 + nr ± P ]
2 ⎡1/ 2 + nr ± ( l + 1/ 2 ) − 2mV0 ⎤
⎢⎣
⎥⎦
nr = 0,1, 2
Remark: For V0 < 0 in (3.1), we get Kratzer Molecular potential and we
obtain for standard levels well known formula, but in this case isn’t
fulfilled (1.10) condition and so we have no additional levels for
Kratzer potential .
For alkaline metal atoms (Li,Na,K,Rb,Cs) is used (3.1) potential [14S.Frish .Optical specra of atoms;15 –M.Eliashevich.Atomic and
molecular spectroscopy].Spectra of this atoms is similar hydrogen
atom spectra
1
En′ = − R 2
(3.7)
n′
Where R is Rydberg constant and n′ is effective principal number
(3.8)
n′ = nr + l ′ + 1
And l ′ is defined from
(3.9)
l ′(l ′ + 1) = l (l + 1) − 8mV0
For l′ is taken only + sign in front of root P [12,13]
l ′ = −1/ 2 + P = −1/ 2 +
( l + 1/ 2 )
2
− 2mV0
(3.10)
So up to now wasn’t considered additional levels (- sign in front of
root). Then in [13] the root is expand is expand for small V0
5
Est = − R
1
(n + ∆ )
st
l
2
; n = nr + l + 1
(3.11)
Where ∆ lst is Rydberg correction (quantum defect)
2mV0
∆ lst = −
2l + 1
(3.12)
For Eadd we can’t take small V0 ,because l (l + 1) < 2mV0 So for Est at
V0 → 0 one get hydrogen atom spectra; Eadd exist only for “strong”
values of V0 !
So it is expectable, that in the Model of Valent electron, beside the
well known Est levels, may also exist Eadd and (3.4) transcendental
equation levels.
Remark: Our formalism works everywhere, where (3.1) potentials
works: for excited (Rydberg) atoms, for alkaline isoelectronic ions and
etc.
I V . Singular (Spiked) Oscillator model
V
(4.1)
V = − 02 + gr 2 ; V0 , g > 0
r
Use: Calogero model, Fractional statistics and anyons, Quantum Hall
effect, Spin chains, Two dimensional QCD.
Γ( −1/ 4 2m / g E + 1/ 2 − P / 2)
Γ(1 − P )
τ
(4.2)
=−
P/2
Γ
+
(1
P
)
Γ( −1/ 4 2m / g E + 1/ 2 + P / 2)
( 2mg )
For τ = 0 and τ = ±∞ we get standard and additional levels
Est ,add = 2 g / 2m {2nr + 1 ± P} ; nr = 0,1, 2...
(4.3)
Remark: For V = −V0 / r 2 + W (r ) potential (where W is regular
potential) we can define generally quantum defect by
6
∆ lst = P − ( l + 1/ 2 ) as a deviation from W(r), because when V0 = 0 ,
then P=l+1/2 and ∆ lst = 0 .
V. Scattering Problems (Continuous Spectrum).
V
(5.1)
V = − 02 ;V0 > 0
r
This interaction is realized in nature- physical applications: .
1).Gharge interacting with a point dipole [14H.Camblong...Phys.Rev.
Lett. 87, 220402 (2001)]
2).Interaction of a neutral, but polarizable atom with a charged wire
[15-J.Denschlag; Phys.Rev.Lett.81.737. (1998)
3)Aaronov-Bom effect [16 –J.Audretsh…J.Phys.A28,2359 (1995)].
4).Black holes [Gupta,Shabad…]
U k (r ) = kr { A( k ) J P (kr ) + B( k ) J − P (kr )} ; k 2 = 2mE ; E > 0
(5.2)
In (5.2) for 0<P<1/2 r J − P (kr ) is regular at the origin and we keep it!
a).Introduction of SAE parameter
∞
u
(5.3)
I = ∫ r 2 Rk∗′ (r ) Rk ( r ) dr = 2πδ ( k ′ − k ) ; Rk = k
r
0
We use integrals from [17-J.Audretsch.J.Phys.A34,235 (2001)] for
Bessels functions and get
P
−P
⎫⎪
2sin π P ⎧⎪⎛ k ⎞ ∗
k⎞
⎛
∗
′
′
I = AA + BB + ( B A + A B ) cos π P δ ( k ′ − k ) +
B
k
A
k
A
k
B
k
−
(
)
(
)
(
)
(
)
⎨
⎬
⎜ ⎟
⎜ ′⎟
π ( k 2 − k ′2 ) ⎪⎩⎝ k ′ ⎠
⎝k ⎠
⎭⎪
{
(5.4)
∗
∗
∗
∗
}
B∗ (k ′)
B(k ) −2 P
−2 P
′
=
k
k =τP
(
)
∗
A (k ′)
A(k )
(5.5)
(5.5) is analog of (2.5) for continuous spectrum.
From (5.4) and (5.5) we get
7
AA∗ ⎡⎣τ P2 k 4 P + 2τ P k 2 P cos π P + 1⎤⎦ = 2π
(5.6)
Based on the methodology of [17] and [18 - S.Alliluev.JETP.Vol
61,p15 (1970)] articles,where is considered
R
I = lim ∫ uk∗′ ( r ) uk ( r ) =
R →∞
0
1
k ′2 − k 2
R
⎡ ∗ duk
∗ duk ′ ⎤
−
u
u
′
k
k
⎢⎣ dr
dr ⎥⎦ 0
(5.7)
One can show, that τ parameter is introduced from the lower limit of
the (5.7) integral as it was for the bound states and (5.6) is introduced
from the upper limit of this integral (For bound states wave function
decrease at large distances and we no analog of (5.6) relation).
b).Phase Shifts Calculation.
uk ( r ) = kr { A(k ) + B(k ) N P ( kr )}
(5.8) in (5.8) Second term is regular at the origin for 0<P<1/2 and we
keep it!
C
u
lim R = sin ( kr − lπ / 2 + δ l ) ; R =
r →∞
r
r
(5.9)
δ P = [l + 1/ 2 − P ]π / 2 − arctgB / A
(5.10)
or using (5.5) definition of SA parameter we get:
(5.11)
δ P = [l + 1/ 2 − P ]π / 2 − arctg (τ P k 2 P )
In the literature is known only first term [17.A.M.Perelomov;
V.S.Popov.TMF. Vol 4.No1(1970)]
τ P = 0;δ Pst = [l + 1/ 2 − P ]π / 2
a).B=0;
b).A=0;
τ P = ±∞;δ Padd = δ lst ± π / 2
(5.12)
(5.13)
+ sign in (5.13) is excluded, comparing it with asymptotic expression
8
2
sin ( kr − Pπ / 2 − π / 4 )
π kr
Remarks: 1. From (5.11) we see that δ P is depended on the energy
N P ( kr ) ≈
(k
2
= 2mE ) for
τ P ≠ 0, ∞
,so
scale
invariance
is
violated!
V0
attractive potential, for which δ P > 0 .As
2
r
one see from (5.11) may be δ P < 0 or we get repulsive potential! So τ P
parameter
may
change
the
NATURE
of
potential!
We have two possibilities: a).From the Physical motivation restrict
τ P parameter (Don’t change attractive potential by repulsive!) or as
one see from (5.11) demand
[l + 1/ 2 − P ]π / 2 − arctg (τ P k 2 P ) > 0
2. We considered V = −
b).Agree, that τ P > 0 can change potential nature!
2
dσ
1 ∞
= f (θ ) ; f (θ ) =
∑ (2l + 1) ( Sl − 1) Pl ( cosθ )
2ik l =0
dΩ
Sl = e 2iδl
4π ∞
σ = 2 ∑ (2l + 1)sin 2 δ l
k l =0
1 ∞
f (θ ) =
∑ fl Pl (cosθ )
2ik l =0
1
1 2iδl
fl =
e −1
( Sl − 1) =
2ik
2ik
2
σ l = 4π ( 2l + 1) fl
(
)
(5.14)
(5.15)
(5.16)
(5.17)
(5.18)
(5.19)
Remarks: 1. l (l + 1) < 2mV0 (1.10).So in (5.14 – 5.19) one need
SAE for l ,which satisfy (1.10).(l=0 always satisfy it!).
V
2. Total cross section σ is infinite for V = − 02 in usual quantum
r
9
mechanics (τ = 0 ) .We show, that for A=0 (τ=+∞) and small k σ is
again infinite, but in general case, when
δ P = [l + 1/ 2 − P ]π / 2 − arctg (τ P k 2 P ) − σ can become finite! This
problem needs more careful investigation!
VI.Scattering length a
In [17] is calculated scattering length a for the following potential
in the l=0 state
V
(6.1)
V ( r ) = − 02 θ ( R − r )
r
We now obtain more general formula using SAE.When r<R wave
function is
(6.2)
⎧ Ar1/ 2+ P + Br1/ 2− P ;2mV0 < 1/ 4; P = 1/ 4 − 2mV0
⎪
χ =⎨
⎪r1/ 2 sin (ν ln r + γ 0 ) ;2mV0 > 1/ 4;ν = 2mV0 − 1/ 4
⎩
γ 0 is SAE parameter, when one have “falling” on the center and is
known in the literature [5,17]
For r>R
χ0 = C ( r − a )
“Sewing” condition at r=R gives
(1 − 2 P ) AR P + BR − P (1 + 2 P )
a = −R
;2mV0 < 1/ 4
−P
P
(1 + 2 P ) AR + BR (1 − 2 P )
(6.3)
(6.4)
1 − 2ν ctg (ν ln R − γ 0 )
;2mV0 > 1/ 4
(6.5)
1 + 2ν ctg (ν ln R − γ 0 )
When B=0 we get [17] article formula
1 − 2P
(6.6)
a = −R
1 + 2P
As P<1/2, a<0 and it corresponds to attractive potential, but from
(6.4) a have no definite sign- we can’t say one have attractive or
a = −R
10
repulsive interaction!
SAE now we define
τ = −A/ B
a τ + b1
a= 1
a2τ + b2
(6.7)
(6.8)
a1 = − (1 − 2 P ) R1+ P ; b1 = (1 + 2 P ) R1− P
a2 = (1 + 2 P ) R ; b2 = − (1 − 2 P ) R
P
In the region
τ1 < τ < τ 2
where
τ1 =
−P
(6.9)
(6.10)
1 − 2P 1
1 + 2P 1
τ
;
=
2
1 + 2P R P
1 − 2P R2P
V0
θ ( R − r ) is
2
r
attractive potential and τ parameter from (6.10) region can change
it NATURE! Again one have two alternatives : a). From the
physical motivation exclude (6.10) region. b). Agree that τ can
change interaction nature!
Remarks: 1). We expand (6.4) and (6.5) near 2mV0 = 1/ 4 and get
relation between τ and γ 0
a>0 and we have repulsive interaction! V ( r ) = −
2ctgγ 0 =
τ +1
τ −1
(6.11)
2). σ tot = 4π a 2 (τ ) = σ (τ ) depends on !σ > 0 demand restrict τ!
VII.Scattering effective radios ( 2mV0 < 1/ 4 )
∞
r0 = 2 ∫ ⎡⎣u02 ( r ) − χ 02 ( r ) ⎤⎦dr
0
where
11
(7.1)
u0 = C ( r − a )
(7.2)
⎧⎪ Ar1/ 2+ P + Br 1/ 2− P ; r < R
(7.3)
χ0 = ⎨
⎪⎩C ( r − a ) ; r > R
2 1− P
⎧ τ2
⎫⎪
R( )
3
2
3
2⎪
2 P+2
−
+ τ R2 ⎬
r0 = 2 / 3C ( R − a ) + a − D ⎨
R
2 (1 − P )
⎪⎭
⎪⎩ 2 ( p + 1)
{
}
C ( R − a)
D = 1/ 2− P
(7.4)
R
− τ R1/ 2+ P
r0 > 0 demand restrict τ!
VIII.Modification of Rutherford formula
For Model of valence electron (3.1) potential for scattering case
we get
u = C1ρ 1/ 2+ P e − ρ / 2 F (1/ 2 + P − λ ,1 + 2 P; ρ ) + C2 ρ 1/ 2− P e F (1/ 2 − P − λ ,1 − 2 P; ρ )
(8.1)
Where P is given again by (1.8) and
mα
mα
(8.2)
= −iη ; E > 0; k = 2mE ;η =
;
ρ = 2ikr ; λ = −i
k
k
SAE parameter is
C
−2 P
(8.3)
τ = 2 ( 2ik )
C1
st
⎤
(8.4)
lim u = sin ⎡⎣ kr + η ln 2kr − ( P − 1/ 2 ) π / 2 − δ cou
l + δP ⎦
−ρ /2
r →∞
Where
st
δ coul
= arg Γ (1/ 2 + P − λ )
δ P = arctg Ψ; Ψ = τ P ( 2k )
2P
(8.5)
Γ (1 − 2 P ) Γ (1/ 2 + P − λ )
Γ (1 + 2 P ) Γ (1/ 2 − P − λ )
st
δ = δ P − δ coul
+ [l + 1/ 2 − P ]
12
(8.6)
(8.7)
Γ (1/ 2 + P + λ ) 2iarctg Ψ
e
(8.9)
Γ (1/ 2 − P − λ )
1 + iΨ
(8.10)
S = S Pst
1 − iΨ
iψ = 1 is pole!
And we again obtain (3.4) transendental equation for bound states.
(1.10)
l (l + 1) < 2mV0
SP = e
1 ⎧⎪ ⎣
f (θ ) =
⎨
2ik ⎪
⎩
⎡ −1/ 2 + 1/ 4 + 2 mV0 ⎤
⎦
∑
l =0
iπ [l +1 / 2 − P ]
( 2l + 1) Pl ( cos θ ) ⎡⎢ S Pst
⎣
∞
1 + iΨ ⎤ ⎫⎪
+
⎬
∑ (2l + 1) Pl (cos θ ) ⎡⎣ S Pst − 1⎤⎦
1 − iΨ ⎥⎦ ⎪ ⎡−1/ 2+ 1/ 4+ 2 mV0 ⎤
⎦
⎭ ⎣
(8.11)
(8.12)
f (θ ) = f SAE + fVE
When V0 → ∞ ,in the (8.9) leading term is f SAE and for small V0 is
fVE
dσ
2
2
2
= f (θ ) = fVE + f SAE + 2 Re fV∗E f SAE
(8.13)
dΩ
2mV0
2
P = ( l + 1/ 2 ) − 2mV0 ≈ ( 2l + 1) −
(8.14)
2l + 1
We keep in this case in f SAE only l=0 term and get
⎡ 2 (π V0 )2 m3 2
⎤
dσ
1
= 2 2 4
sin θ − 2π mV0 2m / E cos 2(σ 1/ 2 − σ 0 ⎥ +
⎢1 +
d Ω 4η k sin θ / 2 ⎢⎣
E
⎥⎦
0
0
sin 2 δ SAE
2 sin δ SAE
0
− η 3 2 [cos (η ln sin 2 θ / 2 − 2σ 0 − 2π mV0 − δ SAE
)−
2
k
k si n θ
0
(8.13)
π mV0 2m / E sin θ cos (η ln sin 2 θ / 2 − 2π mV0 − δ SAE
− 2σ −1/ 2 )
where the first term is usual Rutherford formula modified for VE
model last two terms are caused by SAE procedure and are similar
to the short range interactions. So SAE can again play a role in
potential nature! This formalism can be used also for, π
π+scattering, where is used Klein-Gordon equation.
13
VIII. Concluding remarks. Summary
1. Our main result: We show, that for lim r 2V = −V0 (V0 > 0)
r →0
potentials in the region ( l + 1/ 2 ) > 2mV0 (no “falling onto center!)
it is necessary to keep second additional solution in the 0<P<1/2
interval (We have our variant of Landau mentioned paragraph!)
and it is also necessary to introduce self-adjoint extension τ
parameter. in both bound states and scattering problems.
2. Physical quantities E , a, r0 ,σ depend on τ parameter and by this
reason physical picture is different then in usual quantum
mechanics! (As was mentioned above SAE can change nature of
V
potential, δ p became energy dependent for V = − 02 potential and
r
so on.
We have three possibilities:
1).It should be found another strong requirement in the quantum
mechanic mathematical formalism, which “destroys” additional
states!
2) If it isn’t possible, try to ‘’struggle’ against τ parameter by
physical demands: r0 > 0,σ > 0 ,don’t change physical nature of
interaction and so on.
3).Admit SAE existence and find new levels, and so on. And now
it stay open the following question: Why the NATURE “select”
only standard states (τ = 0) ?!
2
14