Slides

Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Numerical solution of the three body scattering
problem
Leonid Rudenko
Department of Computational Physics, St Petersburg State University
3 April, 2009
Numerical
solution of
the three
body
scattering
problem
Outline
Leonid
Rudenko
• Introduction
• The Three-body scattering problem
• Faddeev Equation
• Solution expansion
• Malfliet-Tjon potential
• Numerical calculations: eigenvalues, eigenfunctions,
geometrical connections
• Conclusions
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Introduction
Numerical
solution of
the three
body
scattering
problem
Introduction
Leonid
Rudenko
• Expanding of the tree body wave function
• Basis of hyperspherical adiabatic harmonics
• The standard projection procedure leads to effective
equations for the expansion coefficients
• How can be solved?
• Have to find out properties of the basis functions
Numerical
solution of
the three
body
scattering
problem
Introduction
Leonid
Rudenko
• Geometrical connections Aij = hφi , ∂ρ φj i
• Develop a formalism to study different models
- scattering of a particle off a two body bound state
- scattering of a particle off a resonant state of a pair of
particles
• Investigate eigenvalues λk , eigenfunctions φk , and
geometrical connections Aij of the operator
h(ρ) = −ρ−2 ∂θ + V (ρ cos θ).
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Faddeev Equation in {ρ, θ}
Coordinates
Numerical
solution of
the three
body
scattering
problem
Faddeev equation in {ρ, θ}
coordinates
Leonid
Rudenko
The Faddeev s-wave equation in polar coordinates
µ
¶
µ
¶
1
1 2
2
−∂ρ − 2 U(ρ, θ) + − 2 ∂θ + V (ρ cos θ) U(ρ, θ) −
4ρ
ρ
√ Z θ (θ)
+
3
d θ0 U(ρ, θ0 )
−EU(ρ, θ) = −V (ρ cos θ)
4 θ− (θ)
where θ+ (θ) = π/2 − |π/6 − θ| and θ− (θ) = |π/3 − θ|
Numerical
solution of
the three
body
scattering
problem
Faddeev equation in {ρ, θ}
coordinates
Leonid
Rudenko
The boundary conditions should be imposed for the scattering
solution:
• Regularity of the solution
U(0, θ) = U(ρ, 0) = U(ρ, π/2) = 0
• Asymptotics as ρ → ∞
U(ρ, θ) ∼ ϕ(x)ρ1/2 sin(qy )/q +
√
+a(q)ϕ(x)ρ1/2 exp(iqy ) + A(θ, E ) exp(i E ρ),
where x = ρ cos θ.
Numerical
solution of
the three
body
scattering
problem
Faddeev equation in {ρ, θ}
coordinates
Leonid
Rudenko
• The bound-state wave function of the two body target
system ϕ(x) obeys the equation
{−∂x2 + V (x)}ϕ(x) = ²ϕ(x)
• It is assumed that the discrete spectrum of the Hamiltonian
−∂x2 + V (x) consists of one negative eigenvalue ² at most
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Basis functions
Numerical
solution of
the three
body
scattering
problem
Expansion basis
Leonid
Rudenko
• As basis we choose the eigenfunctions set of the operator
h(ρ)
¡ −2 2
¢
−ρ ∂θ + V (ρ cos θ) φk (θ|ρ) = λk (ρ)φk (θ|ρ)
• ρ is an external parameter for h(ρ)
• EV and EF of h(ρ) inherit parametrical dependence on ρ
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Spectral properties
• The main spectral properties of the operator h(ρ) as
ρ → ∞ and of the two body Hamiltonian −∂x2 + V (x)
λ = lim λ0 (ρ) = ²,
ρ→∞
φ0 (θ|ρ) =
√
ρ ϕ(ρ cos θ)(1 + O(ρ−µ ))
• Excited states φk (θ|ρ), k ≥ 1, when ρ → ∞ the
asymptotic behavior is given by the formulas
µ
λk (ρ) ∼
2k
ρ
¶2
,
2
φk (θ|ρ) ∼ √ sin(2kθ)
π
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Perturbation Theory
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Perturbation theory - introduction
• If the parameter ρ is large the support of the potential
V (ρ cos θ) is small on the interval [0, π/2]
• The estimates for EV and EF of the operator h(ρ) can be
obtained by the perturbation theory considering the
potential V (ρ cos θ) as a small perturbation
λk
∼ λ0k + λ1k ,
φk (θ|ρ) ∼ φ0k (θ|ρ) + φ1k (θ|ρ)
• The leading terms λ0k and φ0k for k ≥ 1 are given by
λ0k = (2k)2 /ρ2 ,
2
φ0k (θ|ρ) = √ sin(2kθ)
π
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Correction terms
• The first order correction terms λ1k and φ1k are defined in
terms of the matrix elements of the potential V (ρ cos θ)
Z
Vnk (ρ) =
hφ0n | V | φ0k i
π/2
=
0
d θ φ0n (θ|ρ)V (ρ cos θ)φ0k (θ|ρ)
• As the model situation, the function V (x) is taken in the
form of Yukawa potential
V (x) = C
exp(−µx)
x
where C and µ are some parameters.
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Calculation of corrections
• As we set ρ → ∞, we can approximately calculate integral
Vnk (ρ) ∼ (−1)n+k
16nkC [1 − exp(−µR)(1 + µR)] 1
.
π
µ2
ρ3
• The first order correction term λ1k is
λ1k = Vkk ∼ ρ−3
• The second order correction term λ2k is
λ2k =
X
|Vnk |2
(0)
n6=k
(0)
λk − λn
λ2k ∼ ρ−4
,
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Calculation of corrections
• For eigenfunctions the first correction term is given by
φ1k =
X
Vnk
(0)
n6=k λk
−
(0)
λn
φ0n ∼ ρ−1
• From equation
¶
µ
1 2
− 2 ∂θ + V (ρ cos θ) φk (θ|ρ) = λk (ρ)φk (θ|ρ)
ρ
one can obtain formula for Aij
Aij =
hφi | ∂V
∂ρ | φj i
λi − λj
Numerical
solution of
the three
body
scattering
problem
Aij
Leonid
Rudenko
• Rough calculation of Aij leads to dependence
Aij ∼ ρ −2 , i, j 6= 0
Aij ∼ ρ −5/2 , i = 0 or j = 0
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Solution Expansion
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Solution expansion
• The operator h(ρ) is Hermitian on [0, π/2] interval and its
eigenfunction set {φk (θ|ρ)}∞
0 is complete.
X
φk (θ|ρ)Fk (ρ)
U(ρ, θ) = φ0 (θ|ρ)F0 (ρ) +
k≥1
• Substituting this expansion in Faddev equation and
projecting on basis functions lead to the following system
of equations for Fk (ρ), k = 0, 1, ...
µ
¶
1
−∂ρ2 − 2 + λk (ρ) − E Fk (ρ) =
4ρ
µ
¶
∞
X
∂Fi (ρ)
2Aki
+ Bki (ρ)Fi (ρ) − Wki (ρ)Fi (ρ)
∂ρ
i=0
Numerical
solution of
the three
body
scattering
problem
Solution expansion
Leonid
Rudenko
• The nonadiabatical matrix elements Aki (ρ), Bki (ρ) and
potential coupling matrix Wki (ρ) are given by integrals
Z
Aki (ρ) = hφk | ∂ρ φi i =
0
Z
Bki (ρ) =
hφk | ∂ρ2 φi i
π/2
=
0
π/2
d θφ∗k (θ|ρ)
d θφ∗k (θ|ρ)
∂φi (θ|ρ)
,
∂ρ
∂ 2 φi (θ|ρ)
,
∂ρ 2
√ Z π/2
Z θ+ (θ)
3
Wki (ρ) =
d θφ∗k (θ|ρ)V (ρ cos θ)
d θ0 φi (θ0 |ρ)
4 0
θ− (θ)
Numerical
solution of
the three
body
scattering
problem
Solution expansion
Leonid
Rudenko
• Presence of first derivative Fi0 (ρ) in the right part of the
expression
• Matrix form of equation
µ
¶
1
2
−∂ρ − 2 + Λ(ρ) − E F (ρ) = 2AF 0 (ρ)+(B −W )F (ρ)
4ρ
Numerical
solution of
the three
body
scattering
problem
Solution expansion
Leonid
Rudenko
• Transform equation in form
F 00 (ρ) + PF 0 (ρ) + QF (ρ) = 0
³
´
where P = 2A and Q = 4ρ12 − Λ(ρ) + E + B − W are
two-dimensional infinite matrices
• Introduce
F = UG
and elements of G are new unknown quantities
Numerical
solution of
the three
body
scattering
problem
Solution expansion
Leonid
Rudenko
¡
¢
¡
¢
UG 00 + 2U 0 + PU G 0 + U 00 + PU 0 + QU G = 0
• Equation for matrix U
1
U 0 = − PU = −AU
2
• As ρ → ∞
Z
U(ρ) = I +
∞
AUd ρ
ρ
Numerical
solution of
the three
body
scattering
problem
Solution expansion
Leonid
Rudenko
• Form without first derivative
µ
¶
1 0 1 2
UG + − P − P + Q UG = 0
2
4
00
• Substitute expressions for P and Q
1
−
4ρ2
−Λ(ρ) + E + B − W )UG = 0
G 00 + U −1 (−A0 (ρ) − A2 (ρ) +
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Solution expansion
• As ρ → ∞ asymptotic behavior U(ρ) ∼ I
µ
¶
1
−∂ρ2 − 2 I + Λ(ρ) − E G =
4ρ
¡
¢
= −A0 (ρ) − A2 (ρ) + B(ρ) − W (ρ) G
• The right hand side terms vanish for large ρ faster than
ρ−2 . Hence this terms can be neglected.
µ
¶
1
−∂ρ2 − 2 + λas
(ρ)
−
E
Gk (ρ) = 0
k
4ρ
2
2
as
here λas
0 (ρ) = ² and λk (ρ) = (2k) /ρ , k ≥ 1
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Solution expansion
• Solutions to these equations with appropriate asymptotics
can be expressed in terms of the Bessel (Y , J) functions
and the Hankel (H (1) ) functions of the first kind as
following
r
πqρ Y0 (qρ) − J0 (qρ)
√
G0 (ρ) ∼
+
2
2q
r
πqρ (1)
a0 (q)
H (qρ)e iπ/4
2 0
s √
π E ρ (1) √
Gk (ρ) ∼ ak (E )
H2k ( E ρ)e i(π/4+kπ)
2
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Malfliet-Tjon Potential
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Malfliet-Tjon potential
• A potential, which describes the modeling nucleon-nucleon
interaction acting in the triplet state with the spin 3/2
• Form of the Yukawa terms superposition
V (x) = V1
exp(−µ1 x)
exp(−µ2 x)
+ V2
x
x
with parameters listed in the table
Coefficient
V1
V2
µ1
µ2
Value
−626, 885MeV
1438, 72MeV
1, 55 Fm−1
3, 11 Fm−1
Numerical
solution of
the three
body
scattering
problem
Malfliet-Tjon potential
Leonid
Rudenko
3
V(x)
V(x), Fm^(-2)
2
1
0
-1
-2
0
1
2
x, Fm
3
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Unit system
• Proton-neutron interaction
¶
µ 2
~ 2
− ∂x + V (x) ψ(x) = E ψ(x),
2µ
• Reduced mass
mp mn
m2
m
'
= , mp ' mn = m
mp + mn
2m
2
³
m
m ´
−∂x2 + 2 V (x) − 2 E ψ(x) = 0
~
~
• [V ] = [E ] = MeV
µ=
Numerical
solution of
the three
body
scattering
problem
Unit system
Leonid
Rudenko
•
~2
' 41, 47 Fm2 ∗ MeV
m
• Obtain equation
³
´
e (x) − E
e ψ(x) = 0
−∂x2 + V
e ] = [E
e ] = Fm−2
where [V
• Set x = ρ cos θ in polar coordinates and ρ is parameter
Numerical
solution of
the three
body
scattering
problem
ρ=1
Leonid
Rudenko
V(rho*cos(Theta)), rho=1
30
25
V(rho*cos(Theta))
20
15
10
5
0
-5
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
ρ = 10
Leonid
Rudenko
V(rho*cos(Theta)), rho=10
3
V(rho*cos(Theta))
2
1
0
-1
-2
1,2
1,4
Theta
1,6
Numerical
solution of
the three
body
scattering
problem
ρ = 100
Leonid
Rudenko
V(rho*cos(Theta)), rho=100
4
V(rho*cos(Theta))
3
2
1
0
-1
-2
1,52
1,53
1,54
1,55
Theta
1,56
1,57
1,58
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Numerical Calculations
Numerical
solution of
the three
body
scattering
problem
Numerical calculations
Leonid
Rudenko
• Finite-difference approach
−φk (θ − h) + 2φk (θ) − φk (θ + h)
+
ρ 2h 2
+V (ρ cos θ)φk (θ) = λk φk (θ)
where h = θi+1 − θi , i = 1, ..., N and N ∼ 100000
• Boundary conditions
φk (0) = φk (π/2) = 0
Numerical
solution of
the three
body
scattering
problem
Software
Leonid
Rudenko
• CLAPACK
• Intel Math Kernel Library 10.0.1 for Unix
• ’DSTEVX’: computes selected eigenvalues and, optionally,
eigenvectors of a real symmetric tridiagonal matrix.
Eigenvalues and eigenvectors can be selected by specifying
either a range of values or a range of indices.
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Eigenvalues
Numerical
solution of
the three
body
scattering
problem
Asymptotics
Leonid
Rudenko
• k =0
lim λ0 (ρ) = ²
ρ→∞
• k ≥1
µ
λk (ρ) ∼
2k
ρ
¶2
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
λ0 (ρ)
0,04
lambda0
0,02
0,00
-0,02
-0,04
-0,06
-0,08
0
10
20
30
rho
40
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
λ10 (ρ)
100
lambda10
asymptotic
80
60
40
20
0
0
2
4
6
8
10
rho
12
14
16
18
20
Numerical
solution of
the three
body
scattering
problem
Eigenvalues
Leonid
Rudenko
lambda0
lambda1
0,30
lambda2
lambda3
0,25
lambda4
lambda5
0,20
Eigenvalues
0,15
0,10
0,05
0,00
-0,05
-0,10
0
5
10
15
20
rho
25
30
35
40
Numerical
solution of
the three
body
scattering
problem
Deviation from asymptotics
Leonid
Rudenko
• ρ ∼ 10 – deviation is big and varies from 10% up to 90%.
For small k the absolute deviation is bigger than for big k.
• ρ ∼ 100 – deviation is about 5 − 7%
• ρ = 700 – deviation is less than 1%
• ρ ∼ 1000 – deviation is about 0, 7%
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Eigenfunctions
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 1
1,4
phi0(Theta), rho=1
1,2
1,0
0,8
0,6
0,4
0,2
0,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 3
1,4
phi0(Theta), rho=3
1,2
1,0
0,8
0,6
0,4
0,2
0,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φ0 (θ), ρ = 5
Leonid
Rudenko
phi0(Theta), rho=5
1,4
1,2
1,0
0,8
0,6
0,4
0,2
0,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φ0 (θ), ρ = 10
Leonid
Rudenko
phi0(Theta), rho=10
1,8
1,6
1,4
1,2
1,0
0,8
0,6
0,4
0,2
0,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φ0 (θ), ρ = 30
Leonid
Rudenko
phi0(Theta), rho=30
3,0
2,5
2,0
1,5
1,0
0,5
0,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 1
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 3
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 5
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 10
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 30
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 100
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 300
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 500
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
φ0 (θ), ρ = 1000
18
16
14
12
10
8
6
4
2
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Step-by-step
18
17
16
rho=1
15
rho=3
14
rho=5
13
12
rho=10
11
rho=30
10
9
rho=100
8
rho=300
7
rho=500
6
5
rho=1000
4
3
2
1
0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Step-by-step
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
rho=1
rho=3
rho=5
rho=10
rho=30
rho=100
rho=300
rho=500
rho=1000
3
2
1
0
1,48
1,50
1,52
1,54
Theta
1,56
1,58
Numerical
solution of
the three
body
scattering
problem
φ0 (θ): conclusions
Leonid
Rudenko
• As ρ → ∞ eigenfunction φ0 transforms into δ-function
• Computational problems: decreasing of spacing, time of
computation, size of data, useless zero-area
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
phi_k(Theta), rho=1
k=1
k=2
k=3
1,0
k=4
k=5
0,5
0,0
-0,5
-1,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
k=1
phi_k(Theta), rho=3
k=2
k=3
k=4
1,0
k=5
0,5
0,0
-0,5
-1,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
k=1
phi_k(Theta), rho=5
k=2
k=3
k=4
1,0
k=5
0,5
0,0
-0,5
-1,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
k=1
phi_k(Theta), rho=10
k=2
k=3
1,0
k=4
k=5
0,5
0,0
-0,5
-1,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
k=1
phi_k(Theta), rho=30
k=2
1,2
k=3
1,0
k=4
0,8
k=5
0,6
0,4
0,2
0,0
-0,2
-0,4
-0,6
-0,8
-1,0
-1,2
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
k=1
phi_k(Theta), rho=100
k=2
1,2
k=3
1,0
k=4
0,8
k=5
0,6
0,4
0,2
0,0
-0,2
-0,4
-0,6
-0,8
-1,0
-1,2
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
k=1
phi_k(Theta), rho=300
k=2
1,2
k=3
1,0
k=4
0,8
k=5
0,6
0,4
0,2
0,0
-0,2
-0,4
-0,6
-0,8
-1,0
-1,2
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
k=1
phi_k(Theta), rho=500
k=2
1,2
k=3
1,0
k=4
k=5
0,8
0,6
0,4
0,2
0,0
-0,2
-0,4
-0,6
-0,8
-1,0
-1,2
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
φk (θ), k ≥ 1
Leonid
Rudenko
phi_k(Theta), rho=1000
k=1
1,2
k=2
1,0
k=3
0,8
k=4
k=5
0,6
0,4
0,2
0,0
-0,2
-0,4
-0,6
-0,8
-1,0
-1,2
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Step-by-step φ1 (θ)
rho=1
Leonid
Rudenko
rho=3
phi1(Theta)
rho=5
rho=10
1,0
rho=30
rho=100
rho=300
0,5
rho=500
rho=1000
0,0
-0,5
-1,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Orthogonality and deviations
Leonid
Rudenko
• Check orthogonality on each step
• When does eigenfunction reach its asymptotic behavior?
max |φk,ρ (θ) − sin(2kθ)|
θ
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
ρ = 500
k
1
2
3
4
5
6
7
...
14
15
16
17
18
19
20
max deviation
0, 025055987697
0, 050095937266
0, 075103861927
0, 100064122846
0, 124961227783
0, 149780111430
0, 174506106355
...
0, 343853748940
0, 367366621881
0, 390677267607
0, 413776421157
0, 436655434712
0, 459306336123
0, 481722165969
Numerical
solution of
the three
body
scattering
problem
Deviation in case k = 20, ρ = 500
Leonid
Rudenko
| phi_20 - sin(2k*Theta) |, k=20
0,5
0,4
0,3
0,2
0,1
0,0
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
ρ = 1000
k
1
2
3
4
5
6
7
...
14
15
16
17
18
19
20
max deviation
0, 012548181991
0, 025094300382
0, 037636297454
0, 050172115034
0, 062699700358
0, 075217084552
0, 087722341482
...
0, 174758410342
0, 187098541399
0, 199410537020
0, 211691972534
0, 223941583831
0, 236157876824
0, 248338419864
Numerical
solution of
the three
body
scattering
problem
Deviation in case k = 20, ρ = 1000
Leonid
Rudenko
| phi_k - sin(2k*Theta) |, k=20
0,25
0,20
0,15
0,10
0,05
0,00
0,0
0,2
0,4
0,6
0,8
Theta
1,0
1,2
1,4
1,6
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Geometrical connections
Numerical
solution of
the three
body
scattering
problem
Geometrical connections
Leonid
Rudenko
• Off-diagonal matrix of geometrical connections A with
elements
Z
Aki (ρ) =
hφk | φ0i i
=
0
π/2
d θ φ∗k (θ|ρ)
∂φi (θ|ρ)
∂ρ
• Determines behavior of right part of equation
µ
¶
1
2
−∂ρ − 2 I + Λ(ρ) − E G =
4ρ
¡
¢
= −A0 (ρ) − A2 (ρ) + B(ρ) − W (ρ) G
Numerical
solution of
the three
body
scattering
problem
Geometrical connections
Leonid
Rudenko
•
Aki =
• Aki = −Aik and Akk = 0
hφk | ∂V
∂ρ | φi i
λk − λi
Numerical
solution of
the three
body
scattering
problem
Aki , k = 0 or i = 0
Leonid
Rudenko
1,0
num. calc.
O(rho^(-2,5))
0,8
O(rho^(-2))
A01
0,6
0,4
0,2
0,0
500
1000
rho
1500
2000
Numerical
solution of
the three
body
scattering
problem
Aki , k 6= 0 and i 6= 0
Leonid
Rudenko
3
num. calc.
O(rho^(-2))
O(rho^(-1,5))
A12
2
1
0
500
1000
rho
1500
2000
Numerical
solution of
the three
body
scattering
problem
Aki , k 6= 0 and i 6= 0
Leonid
Rudenko
0,0
-0,2
-0,4
-0,6
num. calc.
A13
-0,8
O(rho^(-2))
O(rho^(-1,5))
-1,0
-1,2
-1,4
-1,6
-1,8
-2,0
500
1000
rho
1500
2000
Numerical
solution of
the three
body
scattering
problem
The last note about Aki
Leonid
Rudenko
• The behavior of geometrical connections Aki is regular (no
singularity or special points) in case of Malfliet-Tjon
potential.
Numerical
solution of
the three
body
scattering
problem
Leonid
Rudenko
Thank you!