Quantum mechanical calculation of spectral statistics of a modified Kepler problem.

Abstract

For a modified Kepler problem, we reexamine jumps in the saturation spectral rigidity and large oscillations of the level number variance with near zero minima. Earlier discrepancy between the periodic orbit theory and numerical calculation is cleared by a quantum mechanical calculation. A new class of radial periodic orbits is included establishing a complete correspondence between the periodic orbit theory and the quantum mechanical approach. We show that the diagonal approximation for the level density in the periodic orbit theory already gives a good fit with the numerical calculation. Even greater accuracy is achieved by considering coherent interference between the classical periodic orbits term and the Balian-Bloch term. This procedure produces improved results for the hard-wall rectangular billiards as well.

Full text

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arXiv:1012.5828v1 [quant-ph] 28 Dec 2010
Quantum mechanical calculation of spectral statistics of modified Kepler problem
Tao Ma, Rostislav Serota
Department of Physics, University of Cincinnati, Cincinnati, OH 45221
(Dated: Dec. 28, 2010)
Discrepancy between periodic orbit theory and numerical calculation of a modified Kepler problem
is cleared by a quantum mechanical calculation. The diagonal approximation already gives a good fit
for the numerical calculation. A better result yet is gained by considering the coherent interference
between the classical periodic orbits and the Balian-Bloch term. This approach produces improved
results for the rectangular billiards as well.
PACS numbers:
I. INTRODUCTION
In a companion paper, [1] we investigated level correla-
tions in a modified Kepler (MK) problem and discovered
two important effects: quantum-Hall-like jumps in the
saturation level rigidity as a function of running energy
and persistent oscillations of the level number variance
as a function of the interval width. Unlike rectangular
billiards (RB), [2] where superposition of a number of
incommensurate harmonics led to a finite lower limit of
the variance, in the modified Kepler problem the level
variance can achieve near zero values.
In other words, variance can decrease with the increase
of the interval width due to the oscillatory behavior and,
moreover, at certain values of the interval width the num-
ber of levels is nearly identical for each of the members of
the statistical ensemble. This is a highly unusual circum-
stance from a statistical viewpoint and is due to specifics
of the strong semiclassical spectral correlations in classi-
cally integrable systems.
Previously, we used a periodic orbit theory [1] to in-
vestigate level correlations. However, an important class
of periodic obits - radial oscillations for a fixed angle -
were omitted. Here we utilized a quantum mechanical
approach that confirms a corrected periodic orbits ap-
proach. Both are in complete agreement now and are in
excellent agreement with the numerical simulation.
The key ingredient of our numerical work is the use
of parametric averaging (for instance, the aspect ratio
of equal area rectangles for RB; for MK see discussion
in appendix). This allows us to do statistical sampling
for a fixed running energy in the semiclassical spectrum
and should be contrasted to the traditional technique of
spectral averaging in which the key effects discussed in
this work are washed out.
II.LEVEL CORRELATION FUNCTION AND
LEVEL NUMBER VARIANCE
A. average spectral staircase
The eigen-energy of Modified Kepler problem E =
2pω + l2, where ω =√2β [1]. The average cumulative
number of levels is
?N (E)? =E3/2
3ω
+1
2(E1/2+E
2ω), (1)
which is the scaling relation we use in the numerical cal-
culation.But in the theoretical calculation, only the
first term is considered. We define the scaled energy
ǫ = ?N (E)? ≈ E3/2/3ω.
The average spectral staircase of rectangular billiard
(RB) is given in Eq. (5) of Ref. [2].
B.variance
1.level fluctuation
The level fluctuation [2]
δρ = ρ − ?ρ? = δρ(2)+ δρ(1),(2)
where δρ(1)is the Balian-Bloch term explained in ap-
pendix and δρ(2)is given by
δρ(2)(ǫ) =
1
(3ωǫ)1/3(ω
Mr)1/2×
?
?
Mθ≤
Mr
γ(cir)
2cos
?
πM2
r(3ωǫ)2/3+ M2
Mrω
θω2
−π
4
?
+
?
Mr>0
cos
?πMr(3ωǫ)2/3
1
πMθ
ω
−π
4
??
+
?
Mθ>0
1
(3ωǫ)1/3
sin(2πMθ(3ωǫ)1/3),
(3)
where the second term is due to radial periodic orbits
and the third term due to circular periodic orbits. The
level density fluctuation equation is the starting point to

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2
calculate spectral statistics. For RB,
δρ(2)= (1/4ǫπ3)1/4
?
M1,M2=1
4
cos(2π
?
(˜M1
˜
M1
2+˜M2
2?
2)1/4
4ǫ
π−π
4)
2+˜M2
+
?
M1=1
2
cos(2π˜
M1
?
1/2
4ǫ
π−π
4)
˜
M1
+
?
M2=1
2
cos(2π˜
M2
?
1/2
4ǫ
π−π
4)
˜
M2
.
(4)
where ˜M1= M1α1/4and ˜M2= M2α−1/4.
2. level correlation function
The correlation function of levels is defined as
K(ǫ;E) = ?δρ(ǫ − E/2)δρ(ǫ + E/2)?, where the ensem-
ble average ?·? is defined as parametric averaging over an
ensemble of ω.
Using diagonal approximation,
K(ǫ;E) ≈
?
Mr=1
??
Mr
γ(cir)
?
+1
4
?
2ω
Mr(3ωǫ)2/3cos
?2πMrE
(3ωǫ)1/3
?
,
(5)
where γ(cir)= (2β
(61) of Ref. [1].
Comparing Eq. 5 with Eq. (62) of [1], we have an
extra term due to radial periodic orbits, which is the
single summation term in Eq. 42 of [2].
3ǫ)1/3, which is the same γ(cir)in Eq.
3.variance of MK
Thevariance
?ǫ+E/2
?
is calculated fromΣ(ǫ,E)=
?ǫ+E/2
ǫ−E/2
ǫ−E/2K(ǫ1,ǫ2)dǫ1dǫ2.
Σ(ǫ,E) ≈
Mr=1
??
Mr
γ(cir)
?
+1
4
?
2ω
π2M3
r
sin2
?πMrE
(3ωǫ)1/3
?
(6)
.
Compared with the Eq. 70 of [1], there is the extra
contribution from radial periodic orbits.
The new variance equation is
Σ(ǫ,E) ≈
?
?πMrE
Mr=1


?
Mr
γ(cir)
?
+1
8+18
?
1 +
?
2Mr
ω
?2

2ω
π2M3
r
sin2
(3ωǫ)1/3
?
.
(7)
4.numerical simulation of MK
The ensemble averaging is averaging over an ensemble
of β from a normal distribution centered at β0= 3×106
and with a width β0/20. For RB, the normal distribution
is centered at 1 with a width 0.2.
0100020003000 40005000
E0
50
100
150
200
?
FIG. 1: We use the same values of β = 3×106, and ǫ = 2×105
same as in Fig. 2 of Ref. [1]. The red line is the numerical
result. The black line the old theoretical result from Eq. 6.
The green line is the new theoretical result from Eq. 7.
C.variance of RB
In the old theory, the variance is [2]
Σ(ǫ;E;α) =
?
ǫ
π5∆
?
?
M1,M2=1
4sin2[E
?
M1
π∆
ǫ(˜
2+˜
M1
2+˜M2
2)]
(˜
M2
2)3/2
+
?
M1=1
sin2[E
?
˜
M1
π∆
ǫ
3
˜
M1
2]
+
?
M2=1
sin2[E
?
˜
M2
π∆
ǫ
3
˜
M2
2]
?
.
(8)
In the new theory, the variance is
Σ(ǫ;E;α)
=
?
ǫ
π5∆
?
?
?
M1,M2=1
4sin2[π∆
(˜
ǫ(˜
M1
M1
2+˜
2)3/2
M2
2)]
2+˜M2
+
?
M1=1
1
2
1 +
?
1 −
?
˜
M1(π
ǫ)1/4
?2?
?2?
sin2[π∆
ǫ
˜
M1
3
2]
˜
M1
+
?
M2=1
1
2
?
1 +
?
1 −
?
˜
M2(π
ǫ)1/4
sin2[π∆
ǫ
˜
M2
3
2]
˜
M2
?
(9)
.
Unlike the modified Kepler problem, the average of an
ensemble of α does not converge to the average value of

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α. Hence the correct variance should be given by
?
Σ(ǫ;E;α)ρ(α)dα,(10)
which does not converge to Σ(ǫ;E;αaverage).
0 2000 400060008000
E30
40
50
60
70
80
90
?
0 2500 5000
40
60
80
FIG. 2: The red line is the numerical result. The blue line is
the theoretical result without ensemble averaging over α. The
green line is theoretical fit using Eq. 9 with ensemble averag-
ing. The black line is theoretical fit using Eq. 8 with ensemble
averaging. We see that the ensemble averaging fits well in all
the range, while it fits well only for E ≤ 3000 without ensem-
ble averaging. The variance with coherent interference is a
little better than without.
III.LEVEL RIGIDITY
A. rigidity as function of energy interval width
0 2000 4000 6000
E0
20
40
60
?3
FIG. 3: Rigidity of MK as a function of the energy interval
width. ǫ = 2 × 105.
0 2500 5000
E0
10
20
30
?3
FIG. 4: Rigidity of RB as a function of the energy interval
width. ǫ = 105.
B.rigidity of MK
The saturation level rigidity is calculated by
∆3,MK(ǫ,E) =
2
E4
?E
0
dx(ǫ3− 2xǫ2+ x3)Σ(ǫ,x). (11)
∆3(ǫ,E) =
?
Mr=1
??
Mr
γ(cir)
?
+1
4
?
ω
2π2M3
r
. (12)
Compared with the Eq. 69 of [1], the extra contribu-
tion to from radial orbits
1
4
?
Mr>0
ω
2π2M3
r
=
ω
2π2ζ(3) = 37.29. (13)
1. long term trend of rigidity
For Mr= 1 and (3ǫ
2β)1/3≫ 1,
∆∞
3≈ const +(2β)1/6
12
(3ǫ)1/3.(14)
But for β = 3 × 106, this requires ǫ ≫ 3 × 106.
2.discrepancy of rigidity
In Fig. 5, there is discrepancy between numerical and
theoretical results of rigidity. We explain it in this sec-
tion.
The new rigidity is that
∆3(ǫ,E) =
?
Mr=1


?
Mr
γ(cir)
?
+1
8+18
?
1 +
?
2Mr
ω
?2

ω
2π2M3
r
.
(15)

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4
C.rigidity of RB
In the old theory, the rigidity is
∆3(ǫ) =
1
π5/2ǫ1/2
?
?
?
M1,M2>0
1
(˜M1
1/4
˜
M2
2+˜M2
2)3/2
+
?
M1>0
1/4
˜
M1
3+
M2>0
3].
(16)
In the new theory, the rigidity is
∆3(ǫ) =
1
π5/2ǫ1/2[
?
?
?
M1>0,M2>0
1/8
˜
M1
1/8
˜
M2
1
(˜M1
1/8
˜
M1
1/8
˜
M2
2+˜M2
2)3/2
+
?
?
M1>0
3+
M1>0
3(1 −
?
?
˜
M1(π
ǫ)1/4)2]
+
M2>0
3+
M2>0
3(1 −
˜
M2(π
ǫ)1/4)2].
(17)
0 100000 200000300000 400000500000
Ε
40
50
60
70
80
?3
?
FIG. 5: We use the same values of β as in Ref. [1]. The red
line is the numerical result. The black line is the theoretical
result from Eq. 12. The green line is the theoretical result
from Eq. 15.
IV.CONCLUSIONS
We presented a complete quantum mechanical deriva-
tion of the level correlation function of a modified Kepler
problem. With the inclusion of a class of periodic orbits
omitted in [1] (as well as number of other corrections), it
is in full agreement with the periodic orbit theory.
The account of long-range level correlations leads to
striking properties of the level number variance and sat-
uration level rigidity, as given by eqs. (7) and (15). The
latter are the main results of this work and are in excel-
lent agreement with the numerical calculation, as shown
in Figs. 1 and 5.
0 20000400006000080000 100000
Ε
0
5
10
15
20
25
30
?3
?
FIG. 6: The rigidity of RB. The red line is the numerical
result. The black line the old theoretical result from Eq. 16.
The green line the new theoretical result from Eq. 17. The
inset shows the rigidity change against energy width and ǫ =
105.
Appendix A: Derivation of correlation function,
variance and rigidity
1.fluctuation of the level density
The level density in the variable k has two parts
ρ(k) = ρ(2)(k) + ρ(1)(k), where k = (l,p), ρ(2)(k) is
a delta function at lattice points in the first quadrant
(“1Q”) and half delta function at lattice points on the
half-axes, ρ(1)(k) is half delta function at lattice points
on the half-axes. The level density in the variable ǫ is [2]
ρ(ǫ) ≈
?
1Q
d2kδ(ǫ − φ(k))ρ(k),(A1)
where φ(k) = (2pω+l2)3/2/3ω. To use the Poisson sum-
mation formula, we make ρ(k) fully periodic by sym-
metrizing the integral to the whole plane (“WP”). Then
ρ(ǫ) ≈1
4
?
WP
d2kδ(ǫ − φ′(k))ρ(k), (A2)
where φ′(k) = (2|p|ω + l2)3/2/3ω.
We consider ρ(2)(ǫ) first.
summation formula, ρ(2)(ǫ) becomes a summation over
Fourier component I(2)
w with w = 2π(Mθ,Mr) with inte-
ger Mθand Mr:
After applying Poisson
ρ(2)(ǫ) =
?
Mθ,Mr
I(2)
w(ǫ). (A3)
A Fourier component of ρ(2)(ǫ) looks like
I(2)
w(ǫ) =1
4
?
Lǫ,WP
dk?
1
|∇kφ′(k)|eiw·k,(A4)
where Lǫis the constant energy surface. For the case of
Mθ= Mr= 0,
I(2)
2π(0,0)(ǫ) =
?
1Q
dk?
1
|∇kφ|= 1(A5)

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which is the average level density. The w = 0 component
gives the average level density and other components give
the fluctuation δρ(ǫ). The fluctuation of level density is
given by
δρ(2)(ǫ) =


?
Mθ,Mr?=0
+
?
Mθ=0,Mr?=0
+
?
Mθ?=0,Mr=0

I(2)
w(ǫ).
(A6)
a.case of Mθ,Mr > 0
For every w = 2π(Mθ,Mr) with Mθ,Mr> 0, there are
four symmetric ws: 2π(±Mθ,±Mr). We redefine I(2)
as the sum of these four terms:
w (ǫ)
I(2)
w(ǫ) =
?
WP
dk?
1
|∇kφ′|cos(2πMθl)cos(2πMrp) (A7)
where Mθ,Mr> 0. Also notice that the integral is sym-
metric between positive and negative l or p. Using this
symmetry, we return to the first quadrant.
I(2)
w(ǫ) = 4
?
1Q
?2ω
r(3ωǫ)2/3+ πM2
Mrω
?πM2
dk?
1
|∇kφ|cos(2πMθl)cos(2πMrp)
=
1
(3ωǫ)1/3
?πM2
Mr×
?
cos
θω2
?
?
(C−(ǫ) + C+(ǫ))+
sin
r(3ωǫ)2/3+ πM2
Mrω
θω2
(S−(ǫ) + S+(ǫ))
?
(A8)
,
where C/S±(ǫ) = C/S(
C(x) and S(x) are the Fresnel integrals defined asC
?x
which is true for large β.
The terms C−(ǫ) and S−(ǫ) are important because
they are almost step functions and they are the underly-
ing origin of quantum jumps. If
?
2
Mrω(Mr(3ωǫ)1/3± Mθω)) and
S(x) =
0dt
function. We treat C+(ǫ) = S+(ǫ) =
cos
sin(π
2t2).The Fresnel integrals are almost step
1
2, if
Mθω
√Mrω≫ 1,
?
2
Mrω(Mr(3ωǫ)1/3− Mθω) < 1,(A9)
both C−(ǫ) and S−(ǫ) are −1
S−(ǫ) + S+(ǫ) are 0, which makes the oscillation terms
?πM2
not important.
To simplify the problem, we approximate C−(ǫ) +
C+(ǫ) and S−(ǫ) + S+(ǫ) using the step function.
?
2, both C−(ǫ) + C+(ǫ) and
cos
sin
r(3ωǫ)2/3+ πn2ω2
Mrω
?
(A10)
C/S−(ǫ) + C/S+(ǫ) ≈
1
0
ifMθ≤
otherwise.
Mr
γ(cir),
(A11)
Then I(2)
2π(Mθ,Mr)(ǫ) can be approximated as
I(2)
2π(Mθ,Mr)(ǫ) ≈
?
Mθ≤
Mr
γ(cir)
2
(3ωǫ)1/3
?ω
Mr
cos
?πM2
r(3ωǫ)2/3+ πM2
Mrω
θω2
−π
4
?
.
(A12)
a.Quantum jumps
The quantum jumps happen
when
Mθ=
Mr
γ(cir).(A13)
For β = 3×106, when Mθ= 1 and Mr= 2, ǫ = 2.5×105.
When Mθ= 1 and Mr= 1, another big jump will happen
at ǫ = 2×106, which is not shown in Fig. 1 of [1]. When
Mθ= 1 and Mr= 3, ǫ = 0.74 × 105. This jump is also
in Fig. 5.
b. Quantum and classical correspondence
Mθcorrespond to winding numbers of the classical orbits
in the angular direction and radial direction respectively.
The orbit periods are always multiples of Tmin =
When Mθ= 1 and Mr= 1 and ǫ = 2 × 106, the system
has reached its minimal period.
Mr and
2π
ω.
b.case of Mθ= 0, Mr > 0
There are two symmetric terms in Eq. A7.
I(2)
2π(0,Mr)(ǫ)
?
= 2
1Q
dk?
1
|∇kφ|cos(2πMrp)
?ω
≈
1
(3ωǫ)1/3
Mr
cos(πMr(3ωǫ)2/3
ω
−π
4).
(A14)
Classically, I(2)
bits. The rigidity before jumps comes from radial peri-
odic orbits.
2π(0,Mr)(ǫ) corresponds to radial periodic or-
c.case of Mθ > 0 and Mr = 0
There are two symmetric terms in Eq. A7.
I(2)
2π(Mθ,0)(ǫ) = 2
?
1Q
1
dk?
1
|∇kφ|cos(2πMθl)
1
πMθ
=
(3ωǫ)1/3
sin(2πMθ(3ωǫ)1/3).
(A15)
Classically, I(2)
periodic orbits After collecting Eq. A12, A14, A15, we
get Eq. 3.
2π(Mθ,0)(ǫ) corresponds to isolated angular