Isolde Shell Model Course for Non Practitioners:Lecture 5: Lanczos Strength Functions
E. Caurier, G. Martínez-Pinedo,F. Nowacki, A. Poves and K. Sieja
Isolde Shell Model Course for Non PractitionersISOLDE/CERN, October 14th-18th, 2013
Slater determinant representation
We represent a Slater determinant by a machine word, where each stateis a bit (0 empty 1 occupied)
1/2 3/2 1/2- 1/2 -1/2 -3/2 1/2 3/2 1/2- 1/2 -1/2 -3/2
0 0 1 1 1 1 1 1 1 10 0
12 11 10 9 8 7 6 5 4 3 2 1
Mn
i=
Mp
0p1/2 0p3/2 0p1/2 0p3/2
≡ a†10a†9a†8a†7 b†4b†3b†2b†1|0〉
After diagonalization the eigenstates are a linear combination of Slaterdeterminants (basis states)
|Ψα〉 =∑
i
cαi |i〉
Computation of transition operators
We want to compute the transition matrix elements for a one-bodyoperator, O, between many body wave functions |Ψi〉 and |Ψ f 〉 obtainedfrom a shell-model diagonalization:
〈Ψ f |O|Ψi〉
One-body operator:
O =
A∑k=1
Ok ⇒ O =∑α,β
〈α|O|β〉a†αaβ
We need to compute:the value of our one body operator between single particle wavefunctions 〈α|O|β〉the one body density matrix elements 〈Ψ f |a†αaβ|Ψi〉
Computation of transition operators
For the evaluation of 〈α|O|β〉 one needs to know the single particlewave functions, usually harmonic oscillator.
〈α|O|β〉 =
∫d3rφ∗α(r)O(r, s, t)φβ(r)
For the one body density matrix elements (same procedure as forthe hamiltonian):
a†5a2|001011〉 = |011001〉
We can use this procedure to compute:
Electric transitions:∑
k ekrLk Yk
LMβ decay:
Fermi decay:∑
k tk±
Gamow-Teller decay:∑
k σk tk±
Fermi Transitions
B(F) =1
2Ji + 1
∑Mi,M f
|〈J f M f ; T f Tz f |
A∑k=1
tk±|JiMi; TiTzi〉|
2
B(F) = [Ti(Ti + 1) − Tzi(Tzi ± 1)]δJi,J f δTi,T f δTz f ,Tzi±1
Energetics:
EIAS = Qβ + sign(Tzi)[EC(Z + 1) − EC(Z) − (mn − mH)]∆EC = EC(Z + 1) − EC(Z) = 1.4136(1)Z/A1/3 − 0.91338(11) MeV
Selection rule:∆J = 0 ∆T = 0 πi = π f
Sum rule (sum over all the final states, S =∑
f |〈 f |O|i〉|2 = 〈i|O†O|i〉):
S (F) = S −(F) − S +(F) = 2Tzi = (N − Z)
For a neutron S −(F) = 1, S +(F) = 0.
Gamow-Teller Transitions
B(GT ) =g2
A
2Ji + 1
∑Mi,M f ,q
|〈J f M f ; T f Tz f |
A∑k=1
σkq tk±|JiMi; TiTzi〉|
2
σ±1 = ∓1√
2(σ1 ± iσ2), σ0 = σ3
B(GT ) =g2
A
2Ji + 1|〈J f ; T f Tz f ||
A∑k=1
σk tk±||Ji; TiTzi〉|
2
gA = −1.2701 ± 0.0025
Selection rule:
∆J = 0, 1 (no Ji = 0→ J f = 0) ∆T = 0, 1 πi = π f
Ikeda sum rule:
S (GT ) = S −(GT ) − S +(GT ) = 3g2A(N − Z)
For a neutron, S −(GT ) = 3g2A, S +(GT ) = 0.
Decay rate
β decay half-life calculation
Determine initial state |Ψi〉.
Determine all posible final states |Ψ f 〉.
Compute 〈Ψ f |O|Ψi〉
λ f =ln 2t1/2
=ln 2K
f (Q f )[B f (F) + B f (GT )]
Determine total decay rate:
λ =ln 2T1/2
=∑
f
λ f
Example of beta-decay: ft-value
64 30Zn00+
991.542+
1799.412+ 1910.300+
2609.450+3005.722+ 3186.801+ 3261.971 3321.8(1) 3365.961+ 3425.161+ 3795.261+
4454.21+ 4608.4(1) 4712.4(1) ~0.017
4712
<0.17
2913
0.15 210
3~0.
030460
90.1
0 3616.5
0.74 445
4.30.0
43346
2.40.2
8 2654.4
0.17 254
4.21.2
2 3795.1
0.64 280
3.71.5
9 1995.8
0.11 118
6.04.0
6 3425.0
50.5
9 2433.7
1.05 162
5.70
0.18 151
5.2419
.513.
1 3365.8
67.0
0 2374.3
52.0
8 1566.5
01.8
7 1455.8
1.25 756
.520.0
78332
20.1
3 1411.3
0.12 326
12.0
9 2270.3
90.3
0 1462
<0.13
1352
0.15 318
79.2
4 2195.2
11.8 138
7.34
5.57 127
6.54 D
3005.5
E2201
4.3M1
(+E2)
1206.2
M1+E
2261
0 E01.6
5 1617.8
E2809
1910 E0
8.08 918
.76E2
0.216
110.9
3.59 179
9.61 E2
13.65
807.86
E2+M
1
43991
.52E2
stable
1.80 ps
2.0 ps 0.95 ns
0.20 ps
0.057 ps 0.042 ps 0.4 ps 0.023 ps 0.031 ps
3.2 fs
64 31Ga ≈
33.7% 6.5
<0.6% >7.5
<0.25% >7.5
26.8% 5.22.64% 6.10.21% 7.225.3% 5.15.9% 5.73.59% 5.6
1.24% 5.5~0.13% ~6.3~0.3% ~5.8
0+ 02.630 m
QEC=7165
1521
f t1/2 =K
B(F) + B(GT ), K = 6147.0 ± 2.4 s
f (Q) phase space function.
B(F) Fermi matrix element.
B(GT ) Gamow-Teller matrix element.
Summary
5
10
15
0
E (
MeV
)
(Z−1,A)
+GT
0
5
10
15
20
E (
MeV
)
(Z−1,A)
(Z,A)
GT−
F
Electron capture
Beta decay
In neutron rich nuclei GT+ strength represent a small part of Ikeda sumrule (S − = 3(N − Z) + S + so S − S +). Its value is rather sensitive tomany body correlations.
For neutron rich nuclei GT− constitutes most of the Ikeda sum rule. Mostof the strength is located in a resonance with a width of several MeV andat energy: EGT − EIAS = 7.0 − 28.9(N − Z)/A MeV.
Fermi transitions only contribute to the β− direction. All the strength(N − Z) is located at the IAS state at an energy with respect of the parentstate: QIAS = ∆EC − (mn − mp)
Strength Function
Let Ω be an operator acting on some initial state |Ψi〉, we obtain thestate |Ω〉 = Ω|Ψi〉 whose norm is the sum rule of the operator Ω in theinitial state:
S = 〈Ω|Ω〉 = 〈Ψi|Ω†Ω|Ψi〉
Depending on the nature of the operator Ω, the state |Ω〉 belongs to thesame nucleus (if Ω is a e.m. transition operator) or to another(Gamow-Teller, nucleon transfer: a†, a, double-beta, . . . )
If the operator Ω does not commute with H, |Ω〉 is not an eigenvector ofthe system, but it can be developped in energy eigenstates, (|Eα〉):
|Ω〉 =∑α
〈Eα|Ω|Ψi〉|Eα〉, and S = 〈Ψi|Ω†Ω|Ψi〉 =
∑α
|〈Eα|Ω|Ψi〉|2
We can denote S (Eα) = |〈Eα|Ω|Ψi〉|2, the strength function (or
structure function).
Lanczos Strength Function
If we carry on the Lanczos procedure using |Σ〉 = Ω|Ψi〉/√
S as initialpivot.then H is again diagonalized to obtain the eigenvalues |Ei〉
U is the unitary matrix that diagonalizes H after N Lanczos iterations andgives the expression of the eigenvectors in terms of the Lanczos vectors:
U =
|E1〉 |E2〉 |E3〉 ... |EN〉
|Σ〉
|2〉|3〉::|N〉
S (Eα) = S |U(1, α)|2 = S |〈Eα|Σ〉|
2 = |〈Eα|Ω|Ψi〉|2
How good is the Strength function obtained after N iterations comparedwith the exact one?
Moments of a distribution
Any distribution can be characterized by the moments of thedistribution.
E = 〈Σ|H|Σ〉 =1S
∑α
Eα|〈Eα|Ω|Ψ〉|2
mn = 〈Σ|(H − E)n|Σ〉 =1S
∑α
(Eα − E)n|〈Eα|Ω|Ψ〉|2
Gaussian distribution characterized by twomoments (E, σ2 = m2)
g(E) =1
σ√
2πexp−
(E − E)2
2σ2
E
g(E
)
2σ
Moments of a distribution
In general we only need a finite number of momenta. We can define abasis of |α〉 states.
mn = 〈Ω|(H − E)n|Ω〉 =
N∑α
(Eα − E)n|〈α|Ω|Ψ〉|2 (∀n ≤ M)
Eα ≈ 〈α|H|α〉
With N states we can reproduce 2N moments of the distribution.
Lanczos Strengh Function
Lanczos method provides a natural way of determining the basis |α〉.
Initial vector |1〉 =|Ω〉√〈Ω|Ω〉
.
E12|2〉 = (H − E11)|1〉E23|3〉 = (H − E22)|2〉 − E12|1〉. . .ENN+1|N + 1〉 = (H − ENN)|N〉
−EN−1N |N − 1〉where
ENN = 〈N|H|N〉, ENN+1 = EN+1N
Each Lanczos iteration givesinformation about two new momentsof the distribution.
E11 = 〈1|H|1〉 = E
E212 = 〈1|(H − E11)2|1〉 = m2
E22 =m3
m2+ E
E223 =
m4
m2−
m23
m22
− m2
Diagonalizing Lanczos matrix after N iterations gives an approximation tothe distribution with the same lowest 2N − 1 moments.
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution Strength Distribution
GT− Strength on 48Ca
Evolution of Strength Distribution
Evolution of Strength Distribution
GT− Strength on 48Ca
Evolution of Strength Distribution
GT− Strength on 48Ca
48Ca(p,n)48Sc Strength Function
Quenching GTmatrix elements from beta decay
Gamow-Teller matrix elements measured on beta decay of A = 41–52nuclei:
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
R(GT) Th.
R(G
T)
Exp
.
0.770.74
0.0 0.2 0.4 0.6 0.8 1.00.0
0.2
0.4
0.6
0.8
1.0
T(GT) Th.T
(GT
) E
xp.
0.770.74
GMP et al, PRC 53, R2602 (1996)
Comparison with measured GT+ strength
Nucleus Uncorrelated Correlated Expt.(IPM) Unquenched Q = 0.74
48Ti 4.16 1.21 0.66 1.19 ± 20 (0.427 ± 0.108)51V 5.15 2.42 1.33 1.2 ± 0.154Fe 10.19 5.98 3.27 3.3 ± 0.555Mn 7.96 3.64 1.99 1.7 ± 0.256Fe 9.44 4.38 2.40 2.8 ± 0.358Ni 11.9 7.24 3.97 3.8 ± 0.459Co 8.52 3.98 2.18 1.9 ± 0.162Ni 7.83 3.65 2.00 2.5 ± 0.120Ne 5.0 0.55 0.30 0.35 ± 0.2
Where is the remaining strength?
If we write:|i〉 = α|0~ω〉 +
∑n,0
βn|n~ω〉,
| f 〉 = α′|0~ω〉 +∑n,0
β′n|n~ω〉
then
〈 f |O|i〉2 =
αα′〈0~ω|O|0~ω〉 +∑n,0
βnβ′n〈n~ω|O|n~ω〉
2
Projection of the physical wave function in the 0~ω space is Q ≈ α2
(α ≈ α′)
transition quenched by Q2.
Quenching Spin Operators (M1)
(P. von Neumann-Cosel et al.)
Where is the remaining strength?
no-core shell-model calculations for 12C indicate that the strength maybe located at very high excitation energies.
0 50 100 150 200E (MeV)
0
0.02
0.04
0.06
0.08
GT
Str
engt
h
0
0.02
0.04
0.06
0.08
GT
Str
engt
h
4~ω
6~ω
/10
/10
67% in 0~ω states
78% in 0~ω states
0 50 100 150 200E (MeV)
0
0.1
0.2B
(M1)
(µ N2 )
0
0.1
0.2
0.3
B(M
1) (
µ N2 )
4~ω
6~ω
/10
/10
64% in 0~ω states
75% in 0~ω states
