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decay and nuclear structure
F. Nowacki1
XIII Nuclear Physics Workshop
Kazimierz Dolny
September 27-October 1, 2006
1 Strasbourg-Madrid
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Shell-Model collaboration; OF :(
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Outline
I decay
I
Shell model
I 2 calculations
I 0 calculations
I
summary
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decay
p
n
e
W
e
2
e
W
e
p
n
authorized by SM
! 76 Cd
!124Se
!136 Te
!130
Ba
! 116 Xe
!82 Sn
! Kr
!96100 Ru
! 150Mo
! 48 Sm
! Ti
Pd
Ge
Sn
136
Xe
130
Te
116
Cd
82
Se
100 Mo
96
Zr
150
Nd
48
Ca
76
124
110
Q (keV)
2013
2040
2288
2479
2533
2802
2995
3034
3350
3667
4271
Ab.(
232
Th = 100)
12
8
6
9
34
7
9
10
3
6
0.2
1
0.8
Arbitrary scale
Transition
110
0.6
0.4
0.2
0
0
1
Energy spectrum of emitted electrons
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decay
p
n
e
W
0
=
W
e
p
n
violation of leptonic number conservation
! 76 Cd
!124Se
!136 Te
!130
Ba
! 116 Xe
!82 Sn
! Kr
!96100 Ru
! 150Mo
! 48 Sm
! Ti
Pd
Ge
Sn
136
Xe
130
Te
116
Cd
82
Se
100 Mo
96
Zr
150
Nd
48
Ca
76
124
110
Q (keV)
2013
2040
2288
2479
2533
2802
2995
3034
3350
3667
4271
Ab.(
232
Th = 100)
12
8
6
9
34
7
9
10
3
6
0.2
1
0.8
Arbitrary scale
Transition
110
0.6
0.4
0.2
0
0
1
Energy spectrum of emitted electrons
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Shell Model Problem
I
I
Define a valence space
Derive an effective interaction
H = E ! Heff eff = E eff
CORE
I
Build and diagonalize the
Hamiltonian matrix.
In principle, all the spectroscopic properties are described
simultaneously (Rotational band AND decay half-life).
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Shell Model Problem
I
I
Define a valence space
Derive an effective interaction
H = E ! Heff eff = E eff
CORE
I
Build and diagonalize the
Hamiltonian matrix.
In principle, all the spectroscopic properties are described
simultaneously (Rotational band AND decay half-life).
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decay
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Structure Function
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Quenching
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ap
Shell Model Problem
I
I
Define a valence space
Derive an effective interaction
H = E ! Heff eff = E eff
CORE
I
Build and diagonalize the
Hamiltonian matrix.
In principle, all the spectroscopic properties are described
simultaneously (Rotational band AND decay half-life).
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decay
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Structure Function
Gamow-Teller SF
Quenching
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Shell Model Problem
I
I
Define a valence space
Derive an effective interaction
H = E ! Heff eff = E eff
CORE
I
Build and diagonalize the
Hamiltonian matrix.
In principle, all the spectroscopic properties are described
simultaneously (Rotational band AND decay half-life).
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Shell Model problem
I
I
I
Hilbert space
hamiltonian:
E
transition operator:
H = hjOji
G matrix:
I Valence space
I Heff :eff : = E eff :
I heff :jOeff :jeff : i
M. Hjorth-Jensen, T.T.S. Kuo and E. Osnes,
Realistic effective interactions for nuclear systems,
Physics Reports 261 (1995) 125-270
need some phenomenology from experimental data:
I energies of states of (semi) magic nuclei
effective charge
I systematics of B(E2) transitions
GT transitions
quenching factor
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Shell Model problem
I
I
I
Hilbert space
hamiltonian:
E
transition operator:
H = hjOji
G matrix:
I Valence space
I Heff :eff : = E eff :
I heff :jOeff :jeff : i
M. Hjorth-Jensen, T.T.S. Kuo and E. Osnes,
Realistic effective interactions for nuclear systems,
Physics Reports 261 (1995) 125-270
need some phenomenology from experimental data:
I energies of states of (semi) magic nuclei
effective charge
I systematics of B(E2) transitions
GT transitions
quenching factor
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decay
Specificity of ( )0 :
NO EXPERIMENTAL DATA !!!
prediction for m very difficult
easier for m (A)/m (A’)
What is the best isotope to observe ( )0 decay ?
What is the influence of the structure of the nucleus on ( )0 matrix
elements ?
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decay
Specificity of ( )0 :
NO EXPERIMENTAL DATA !!!
prediction for m very difficult
easier for m (A)/m (A’)
What is the best isotope to observe ( )0 decay ?
What is the influence of the structure of the nucleus on ( )0 matrix
elements ?
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decay
Specificity of ( )0 :
NO EXPERIMENTAL DATA !!!
prediction for m very difficult
easier for m (A)/m (A’)
What is the best isotope to observe ( )0 decay ?
What is the influence of the structure of the nucleus on ( )0 matrix
elements ?
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Two neutrinos mode
The theoretical expression of the half-life of the 2 mode can be
written as:
2 2
[T12=2 ℄ 1 = G2 jMGT
j;
with
2
MGT
=
X h0 jj~t jj1 ih1 jj~t jj0 i
+
f
m
I
I
I
I
+
m
+
m
Em + E0
+
i
G2 contains the phase space factors and the axial coupling
constant gA
summation over intermediate states
to quench or not to quench ? (eff : )
does a good 2 ME guarantee a good 0 ME ?
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Two neutrinos mode
The theoretical expression of the half-life of the 2 mode can be
written as:
2 2
[T12=2 ℄ 1 = G2 jMGT
j;
with
2
MGT
=
X h0 jj~t jj1 ih1 jj~t jj0 i
+
f
m
I
I
I
I
+
m
+
m
Em + E0
+
i
G2 contains the phase space factors and the axial coupling
constant gA
summation over intermediate states
to quench or not to quench ? (eff : )
does a good 2 ME guarantee a good 0 ME ?
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Two neutrinos mode
The theoretical expression of the half-life of the 2 mode can be
written as:
2 2
[T12=2 ℄ 1 = G2 jMGT
j;
with
2
MGT
=
X h0 jj~t jj1 ih1 jj~t jj0 i
+
f
m
I
I
I
I
+
m
+
m
Em + E0
+
i
G2 contains the phase space factors and the axial coupling
constant gA
summation over intermediate states
to quench or not to quench ? (eff : )
does a good 2 ME guarantee a good 0 ME ?
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Two neutrinos mode
The theoretical expression of the half-life of the 2 mode can be
written as:
2 2
[T12=2 ℄ 1 = G2 jMGT
j;
with
2
MGT
=
X h0 jj~t jj1 ih1 jj~t jj0 i
+
f
m
I
I
I
I
+
m
+
m
Em + E0
+
i
G2 contains the phase space factors and the axial coupling
constant gA
summation over intermediate states
to quench or not to quench ? (eff : )
does a good 2 ME guarantee a good 0 ME ?
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Lanczos Algorithm
In the first step we write:
Hj1i = E j1i + E j2i
11
12
h jHj1i hHi, the mean value of H.
there E11 is just 1
In the second step:
Hj2i = E j1i + E j2i + E j3i
21
22
H
23
h jHj2i
The hermiticity of
implies E21 = E12 , E22 is just 2
and E23 is obtained by normalization :
ji H
ji
E23 3 = (
ji
E22 ) 2 - E21 1
At rank N, the following relations hold:
HjNi
ENN
jN
1
;
ENN =
=
ENN
1
1 =
EN
1N
j
i + E jNi + E
i H
and ENN+1 N + 1 = (
NN
NN+1
jN + 1i
hNjHjNi
j i
ENN ) N - ENN
1
jN
i
1
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Lanczos Algorithm
In the first step we write:
Hj1i = E j1i + E j2i
11
12
h jHj1i hHi, the mean value of H.
there E11 is just 1
In the second step:
Hj2i = E j1i + E j2i + E j3i
21
22
H
23
h jHj2i
The hermiticity of
implies E21 = E12 , E22 is just 2
and E23 is obtained by normalization :
ji H
ji
E23 3 = (
ji
E22 ) 2 - E21 1
At rank N, the following relations hold:
HjNi
ENN
jN
1
;
ENN =
=
ENN
1
1 =
EN
1N
j
i + E jNi + E
i H
and ENN+1 N + 1 = (
NN
NN+1
jN + 1i
hNjHjNi
j i
ENN ) N - ENN
1
jN
i
1
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Lanczos Algorithm
In the first step we write:
Hj1i = E j1i + E j2i
11
12
h jHj1i hHi, the mean value of H.
there E11 is just 1
In the second step:
Hj2i = E j1i + E j2i + E j3i
21
22
H
23
h jHj2i
The hermiticity of
implies E21 = E12 , E22 is just 2
and E23 is obtained by normalization :
ji H
ji
E23 3 = (
ji
E22 ) 2 - E21 1
At rank N, the following relations hold:
HjNi
ENN
jN
1
;
ENN =
=
ENN
1
1 =
EN
1N
j
i + E jNi + E
i H
and ENN+1 N + 1 = (
NN
NN+1
jN + 1i
hNjHjNi
j i
ENN ) N - ENN
1
jN
i
1
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Lanczos Algorithm
In the first step we write:
Hj1i = E j1i + E j2i
11
12
h jHj1i hHi, the mean value of H.
there E11 is just 1
In the second step:
Hj2i = E j1i + E j2i + E j3i
21
22
H
23
h jHj2i
The hermiticity of
implies E21 = E12 , E22 is just 2
and E23 is obtained by normalization :
ji H
ji
E23 3 = (
ji
E22 ) 2 - E21 1
At rank N, the following relations hold:
HjNi
ENN
jN
1
;
ENN =
=
ENN
1
1 =
EN
1N
j
i + E jNi + E
i H
and ENN+1 N + 1 = (
NN
NN+1
jN + 1i
hNjHjNi
j i
ENN ) N - ENN
1
jN
i
1
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Lanczos Algorithm
It is explicit that we have built a tridiagonal matrix
hI jHjJ i = hJ jHjI i =
0E
B
E
B
B
0
B
B
0
B
B
B
11
12
E12
E22
E32
0
0
E23
E33
E43
0
0
E34
E44
0 if jI
0
0
0
E45
Jj > 1
0
0
0
0
1
CC
CC
CC
CC
A
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Lanczos Strength Function
Let be an operator acting on some initial state jini i, we obtain the
state jini i whose norm is the sum rule of the operator in the
initial state:
S = jj
jini ijj =
ph j
j i
ini
2
ini
Depending on the nature of the operator ,
the state jini i belongs to the same nucleus (if is a e.m transition
~j , , ...)
operator) or to another (Gamow-Teller, nucleon transfer, ayj =a
If the operator does not commute with H, jini i is not necessarily
an eigenvector of the system,
BUT it can be developped in energy eigenstates:
P
P
jini i = S (Ei )jEi i and hini j
2 jini i = S 2 (Ei )
i
i
where S 2 (Ei ) is the strength function (or structure function)
(S (Ei ) = hEi j
jini i)
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Lanczos Structure Function
If we carry on the Lanczos procedure
using ji = jini i as initial pivot.
then H is again diagonalized to obtain the eigenvalues jEi i
U is the unitary matrix that diagonalizes H and gives the expression
of the eigenvectors in terms of the Lanczos vectors:
U=
ji
j2i
j3i
:
:
jN i
S (Ei )
0 jE jE i jE i
BB
B
1
2
3
:::
jEN i 1
CC
CA
= U (1; i )
How good is the Strength function obtained at iteration N compared
to the exact one?
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Lanczos Structure Function
Any distribution can be characterized by the moments of the
distribution.
E
= h
jH j
i =
X
Ei jhEi j
jini ij2
X
= h
j(H E ) j
i = (Ei
i
mn
)n jhEi j
E
n
jini ij2
Gaussian distribution characterized by two
moments (E ; 2 m2 )
p1 exp (E2E2 )2
g E
2
( )=
=
(
)
g(E)
i
2σ
E
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Lanczos Structure Function
Lanczos method provides a natural way of determining the basis ji.
j i = phj
ij
i .
Initial vector 1
j i = (H E11 )j1i
j i = (H E22 )j2i E12 j1i
ENN +1 jN + 1i = (H ENN )jNi
EN 1N jN 1i
E12 2
E23 3
Each Lanczos iteration gives information about two new moments of the distribution.
= h1jH j1i = E
= h
j(H E11 )2 j
i = m2
m3
+ E
E22 =
:::
E11
2
E12
where
2
E23
ENN
= hNjH jNi
;
ENN +1
= EN +1N
=
m2
m4
m2
m32
m22
m2
Diagonalizing Lanczos matrix after N iterations gives an approximation to
the distribution with the same lowest 2N moments.
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Evolution of Strength Distribution
GT Strength on 48 Sc
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Evolution of Strength Distribution
GT Strength on 48 Sc
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Evolution of Strength Distribution
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Evolution of Strength Distribution
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Evolution of Strength Distribution
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Evolution of Strength Distribution
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Evolution of Strength Distribution
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Evolution of Strength Distribution
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Evolution of Strength Distribution
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Ca(p,n)48 Sc Strength Function
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Ca(p,n)48 Sc Strength Function
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Quenching of GT strength in the pf -shell
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Quenching of GT operator
If we write
X
j^i i = j0~!i + n jn~!i;
6
X
n=0
j^f i = 0 j0~!i +
6
n0 jn~!i
n=0
then
0
1
X
h^f k T k ^i i = 0 T + 0 T A
2
2
0
6
n n
n
;
n=0
I n 6= 0 contributions negligible
I 0
projection of the physical wavefunction in the 0~!
space is Q
2
transition quenched by Q 2
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structure function
2
MGT
=
Xh
m
0+
f
jj~t jj1m ih1m jj~t jj
+
+
0+
i
i
Em + E0
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structure function
2
MGT
=
Xh
m
0+
f
jj~t jj1m ih1m jj~t jj
+
+
0+
i
i
Em + E0
Calculation in three steps:
I
calculate the final and initial states
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structure function
2
MGT
=
Xh
m
0+
f
jj~t jj1m ih1m jj~t jj
+
+
0+
i
i
Em + E0
Calculation in three steps:
I
calculate the final and initial states
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structure function
2
MGT
=
Xh
m
0+
f
jj~t jj1m ih1m jj~t jj
+
+
0+
i
i
Em + E0
Calculation in three steps:
I
I
calculate the final and initial states
generate the doorway states ~ t j0+
t+ j0+f i
i i and ~
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structure function
2
MGT
=
Xh
m
0+
f
jj~t jj1m ih1m jj~t jj
+
+
0+
i
i
Em + E0
Calculation in three steps:
I
I
I
calculate the final and initial states
generate the doorway states ~ t j0+
t+ j0+f i
i i and ~
take a doorway and use of Lanczos Strength Function method:
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structure function
2
MGT
=
Xh
m
0+
f
jj~t jj1m ih1m jj~t jj
+
+
0+
i
i
Em + E0
Calculation in three steps:
I
I
I
calculate the final and initial states
generate the doorway states ~ t j0+
t+ j0+f i
i i and ~
take a doorway and use of Lanczos Strength Function method:
at iteration N, N 1+ states in the intermediate nucleus, with excitation energies Em
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structure function
2
MGT
=
Xh
m
0+
f
jj~t jj1m ih1m jj~t jj
+
+
0+
i
i
Em + E0
Calculation in three steps:
I
I
I
calculate the final and initial states
generate the doorway states ~ t j0+
t+ j0+f i
i i and ~
take a doorway and use of Lanczos Strength Function method:
at iteration N, N 1+ states in the intermediate nucleus, with excitation energies Em
overlap with the other doorway, enter energy denominators and add
up the N contributions
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2 half-lifes
0.02
48
Ca
0.01
0
-0.01
-0.02
0
5
10
15
20
Energy [MeV]
2 strength function in 48 Ca, 130 Te and 136 Xe
Parent nuclei
T12=2 (g :s:) th.
T12=2 (g :s:) exp
48 Ca
3:7E19
4:2E19
76 Ge
1:15E21
1:4E21
82 Se
3:4E19
8:3E19
130 Te
4E20
2:7E21
136 Xe
6E20
> 8:1E20
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2 half-lifes
0.02
130
Te
0.01
0
-0.01
-0.02
0
5
10
15
20
Energy [MeV]
2 strength function in 48 Ca, 130 Te and 136 Xe
Parent nuclei
T12=2 (g :s:) th.
T12=2 (g :s:) exp
48 Ca
3:7E19
4:2E19
76 Ge
1:15E21
1:4E21
82 Se
3:4E19
8:3E19
130 Te
4E20
2:7E21
136 Xe
6E20
> 8:1E20
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2 half-lifes
0.01
136
Xe
0.005
0
-0.005
-0.01
0
5
10
15
20
Energy [MeV]
2 strength function in 48 Ca, 130 Te and 136 Xe
Parent nuclei
T12=2 (g :s:) th.
T12=2 (g :s:) exp
48 Ca
3:7E19
4:2E19
76 Ge
1:15E21
1:4E21
82 Se
3:4E19
8:3E19
130 Te
4E20
2:7E21
136 Xe
6E20
> 8:1E20
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Neutrinoless mode:
Exchange of a light neutrino, only left-handed currents
d
u
WL
e−
e−
d
WL
u
The theoretical expression of the half-life of the 0 mode can be
written as:
[T10=2 (0+ ! 0+ )℄ 1 = G0 jM 0 j2 hm i2
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Neutrinoless mode:
CLOSURE APPROXIMATION then
hf jjO K jji i
( )
with
OK =
( )
X
;K
Wijkl
h
(ayi ayj ) (a~k a~l )
i
K
ijkl
We are left with a “standard” nuclear structure problem
M(0 )
= MGT0
(
)
(0 )
gV2
MF
gA2
P h( : )t t j0 i
;
P
h0 j ; ht t j0 i
= h0f j
+
gV2
gA2
+
f
n
nm
nm
n
m
m
n
m
+
i
+
i
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Neutrinoless mode:
CLOSURE APPROXIMATION then
hf jjO K jji i
( )
with
OK =
( )
X
;K
Wijkl
h
(ayi ayj ) (a~k a~l )
i
K
ijkl
two-body operator
We are left with a “standard” nuclear structure problem
M(0 )
= MGT0
(
)
(0 )
gV2
MF
gA2
P h( : )t t j0 i
;
P
h0 j ; ht t j0 i
= h0f j
+
gV2
gA2
+
f
n
nm
nm
n
m
m
n
m
+
i
+
i
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Neutrinoless mode:
CLOSURE APPROXIMATION then
hf jjO K jji i
( )
with
OK =
( )
X
;K
Wijkl
h
(ayi ayj ) (a~k a~l )
i
K
ijkl
two-body operator
We are left with a “standard” nuclear structure problem
M(0 )
= MGT0
(
)
(0 )
gV2
MF
gA2
P h( : )t t j0 i
;
P
h0 j ; ht t j0 i
= h0f j
+
gV2
gA2
+
f
n
nm
nm
n
m
m
n
m
+
i
+
i
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Update of 0 results
hm i for T
48
1
2
= 1025 y.
MGT
0
1-F
Ca
Ge
82
Se
96
Zr
100
Mo
110
Pd
116
Cd
124
Sn
128
Te
130
Te
136
Xe
0.85
0.90
0.42
0.67
2.35
2.35
1.14
1.10
1.10
0.67
0.24
0.45
1.92
0.35
0.41
2.52
2.59
2.11
2.36
2.13
1.77
1.16
1.19
1.13
1.13
1.13
1.13
150
hopeless
for
SM !
76
Nd
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Dependance on the effective interaction
The results depend only weakly on the effective interactions provided
they are compatible with the spectroscopy of the region.
For the lower pf shell we have three interactions that work properly,
KB3, FPD6 and GXPF1. Their predictions for the 2 and the
neutrinoless modes are quite close to each other
MGT (2 )
MGT (0 )
KB3
FPD6
GXPF1
0.083
0.667
0.104
0.726
0.107
0.621
Similarly, in the r 3g and r 4h spaces, the variations among the
predictions of spectroscopically tested interactions is small (10-20%)
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Influence of deformation
Changing adequately the effective interaction we can increase or decrease the
deformation of parent, grand-daughter or both, and so gauge its effect on the decays.
A mismatch of deformation can reduce the matrix elements by factors 2-3. In fact
the fictitious decay Ti-Cr, using the same energetics that in Ca-Ti, has matrix elements
more than twice larger. If we increase the deformation in both Ti and Cr nothing
happens. On the contrary, if we reduce the deformation of Ti, the matrix elements are
severely quenched. The effect of deformation is therefore quite important and cannot
be overlooked
48 Ca
MGT (2 )
MGT (0 )
!
48 Ti
0.083
0.667
48 Ti
!
48 Cr
0.213
1.298
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Influence of the spin-orbit partner
Similarly, we can increase artificially the excitation energy of the spin-orbit partner of
the intruder orbit. Surprisingly enough, this affects very little the values of the matrix
elements, particularly in the neutrinoless case. Even removing the spin-orbit partner
completely produces minor changes
48 Ca
MGT (2 )
MGT (0 )
!
48 Ti
48 Ti
!
48 Cr
0.083
0.667
0.213
1.298
Without spin-orbit partner
48 Ca
MGT (2 )
MGT (0 )
!
48 Ti
0.049
0.518
48 Ti
!
48 Cr
0.274
1.386
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decay
6
7
5
6
82
Se
4
Te
4
3
3
2
2
1
MGT
MGT
130
5
0
-1
1
0
-1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
-7
+
0
+
0-
1
+
1-
2
+
2-
3
+
3-
4
+
4-
5
+
5-
6
+
6-
7
+
7-
8
+
8-
9
0+
+
0-
1
+
1-
2
+
2-
3
+
3-
4
+
4-
5
+
5-
6
+
6-
7
+
7-
8
+
8-
9
+
9-
10
10-
+
11
J components (particle-particle representation)
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Summary I
I
I
I
Large scale shell model calculations with high quality effective
interactions are available or will be in the immediate future for all
but one of the neutrinoless double beta emitters
The theoretical spread of the values of the nuclear matrix
elements entering in the lifetime calculations is greatly reduced if
the ingredients of each calculation are examined critically and
only those fulfilling a set of quality criteria are retained
A concerted effort of benchmarking between LSSM and QRPA
practitioners would be of utmost importance to increase the
reliability and precision of the nuclear structure input for the
double beta decay processes
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Summary II
I
I
The most favorable case appears when both nuclei (emitter and
daugther) are spherical (superfluid)
not far from semi-magic:
82
Se, 116 Cd, 124 Sn, 130 Te, 136 Xe
With the present work, 116 Cd is the best case BUT ...
I uncertainties with the interaction
I what happens when we enlarge the valence space ?
I region of 96 Zr/100 Mo remains to be studied more carefully
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matrix elements
=
h0 k
X
0GT
=
h0 k
X 0
h (n :m )tn
!GT
=
h0 k
X
T
=
h0 k
P
=
h0
R
=
1 s
(g
6
(0 )
MGT
Gamow-Teller SF
+
f
+
f
+
f
+
f
+
f
n ;m
n ;m
n ;m
k0 i;
h(n :m )tn tm
F
+
i
tm
h! (n :m )tn tm
k0 i=M
+
i
k0 i=M
+
i
0F
GT ;
(0 )
(0 )
GT
h0 k
+
=
f
=
F!
;
X
n;m
h0 k
+
f
=
X 0
h tn
n;m
h0 k
+
f
htn tm
X
n;m
k0 i
+
+
2rn;m
n ;m
1
2
h k
g s1 ) 0+
f
2
+
k0 i
+
i
n;m
hR (n :m )tn tm
i
k0 i gg
+
i
V
A
(0 )
=MGT
;
g 2
i
h! tn tm
V
gA
(0 )
=MGT
;
g 2
V
gA
(0 )
=MGT
;
)
(
+
+
X
(
V
gA
k0 i
tm
X 0
1
n :m ℄tn tm k0i i=MGT0 ;
h [(n :^rn;m )(m :^rn;m )
3
n ;m
X ki h0 r n;m [(n m ):(^rn;m ^r n;m )℄tn tm k0 i gV =M 0 ;
+
g 2
i
gA
)
GT
(0 )
=MGT
:
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matrix elements
hi)
0
h (r ; hi)
h! (r ; hi)
h (r ; hi)
h (r ;
R
=
=
e
e
=
h0 (x )
=
(x )
=
d
dx
hi
hi
hir );
h him R h (hir );
him ( 2 R him R h) + 4R
R0
( me r );
r
h + me R0 h0 (
=
Mi
d
(x ) ;
dx
2
2
=
0 0
0
2
e
r
0
[sin(x )Cint (x )
cos(x )Sint (x )℄;
[sin(x )Cint (x ) +
cos(x )Sint (x )℄:
Mp
2
0
Æ ( r );
Sint (x ) and Cint (x ) being the integral sinus and cosinus functions,
Sint (x ) =
Z1 sin( )
x
d ;
Cint (x ) =
Z1 cos( )
x
d
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