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Themes and challenges of Modern Science
•Complexity out of simplicity -- Microscopic
How the world, with all its apparent complexity and diversity can be constructed out of a few elementary building blocks and their interactions
•Simplicity out of complexity – Macroscopic
How the world of complex systems can display such remarkable regularity and simplicity
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If we have more than one quadrupole phonon (boson) what total angular momenta do we have?
Use m-scheme for BOSONS (no Pauli Principle)
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Deformed Nuclei• What is different about non-spherical nuclei?
• They can ROTATE !!!
• They can also vibrate– For axially symmetric deformed nuclei there are two low
lying vibrational modes called and
• So, levels of deformed nuclei consist of the ground state, and vibrational states, with rotational sequences of states (rotational bands) built on top of them.
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0+
2+
4+
6+
8+
E(I) ( ħ2/2I )J(J+1)
R4/2= 3.33
Rotational Motion in nuclei
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0+2+4+
6+
8+
Rotational states
Vibrational excitations
Rotational states built on (superposed on)
vibrational modes
Ground or equilibrium
state
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Systematics and collectivity of the lowest vibrational
modes in deformed nuclei
Notice the smooth systematics and low
energies for the mode near mid-shell,
compared with the erratic behavior and
higher energies of the mode. This points to
lower collectivity of the vibration.
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IBA – A Review and Practical Tutorial
Drastic simplification of
shell model
Valence nucleons
Only certain configurations Simple Hamiltonian – interactions
“Boson” model because it treats nucleons in pairs
2 fermions boson
F. Iachello and A. Arima
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Shell Model Configurations
Fermion configurations
Boson configurations
(by considering only configurations of pairs of fermions
with J = 0 or 2.)
The IBA
Roughly, gazillions !!Need to simplify
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Modeling a NucleusModeling a Nucleus
154Sm 3 x 1014 2+ states
Why the IBA is the best thing since jackets
Shell model
Need to truncate
IBA assumptions1. Only valence nucleons2. Fermions → bosons
J = 0 (s bosons)
J = 2 (d bosons)
IBA: 26 2+ states
Is it conceivable that these 26 basis states are correctly chosen to account for the
properties of the low lying collective
states?
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Why s, d bosons?
Lowest state of all e-e First excited state
in non-magic s nuclei is 0+ d e-e nuclei almost always 2+
- fct gives 0+ ground state - fct gives 2+ next above 0+
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Why the IBA ?????• Why a model with such a drastic simplification ???
• Answer: Because it works !!!!!• By far the most successful general nuclear collective
model for nuclei ever developed. Wide variety of collective structures.
• Extremely parameter-economic
Deep relation with Group Theory !!! Dynamical symmetries, group chains, quantum numbers
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Dynamical Symmetries
• Shell Model - (Microscopic)
• Geometric – (Macroscopic)
• Third approach — “Algebraic”
Phonon-like model with microscopic basis explicit from the start.
Group Theoretical
Sph.Def.
Shell Mod. Geom. Mod.
IBA
Collectivity
Mic
rosc
opic
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IBA ModelsIBA Models
IBA – 1 No distinction of p, n
IBA – 2 Explicitly write p, n parts
IBA – 3, 4 Take isospin into account p-n pairs
IBFM Int. Bos. Fermion Model for Odd A nuclei
H = He – e(core) + Hs.p. + Hint
IBFFM Odd – odd nuclei
[ (f, p) bosons for = - states ]
Parameters !!!: IBA-1: ~2 Others: 4 to ~ 20 !!!
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• F. Iachello and A. Arima, The Interacting Boson Model (Cambridge University Press, Cambridge, England, 1987).
• F. Iachello and P. Van Isacker, The Interacting Boson-Fermion Model (Cambridge University Press, Cambridge, England, 2005)
• R.F. Casten and D.D. Warner, Rev. Mod. Phys. 60 (1988) 389.
• R.F. Casten, Nuclear Structure from a Simple Perspective, 2nd Edition (Oxford Univ. Press, Oxford, UK, 2000), Chapter 6 (the basis for most of these lectures).
• D. Bonatsos, Interacting boson models of nuclear structure, (Clarendon Press, Oxford, England, 1989)
• Many articles in the literature
Background, ReferencesBackground, References
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Note key point:Note key point:
Bosons in IBA are pairs of fermions in valence shell
Number of bosons for a given nucleus is a fixed number
1549262 Sm
N = 6 5 = N NB =
11
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Me et al.
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An early conference on the IBA(Featuring young Alison)
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Review of phonon creation and destruction operatorsReview of phonon creation and destruction operators
is a b-phonon number operator.
For the IBA a boson is the same as a phonon – think of it as a collective excitation with ang. mom. zero (s) or 2 (d).
What is a creation operator? Why useful?
A) Bookkeeping – makes calculations very simple.
B) “Ignorance operator”: We don’t know the structure of a phonon but, for many predictions, we don’t need to know its microscopic basis.
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Brief, simple, trip into the Group Theory of the IBA
DON’T BE SCARED
You do not need to understand all the details but try to get the idea of the
relation of groups to degeneracies of levels and quantum numbers
A more intuitive (we will see soon) name for this application of Group Theory is
“Spectrum Generating Algebras”
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To understand the relation, consider operators that create, destroy s and d bosons
s†, s, d†, d operators Ang. Mom. 2
d† , d = 2, 1, 0, -1, -2
Hamiltonian is written in terms of s, d operators
Since boson number is conserved for a given nucleus, H can only contain “bilinear” terms: 36 of them.
s†s, s†d, d†s, d†d
Gr. Theor. classification
of Hamiltonian
IBAIBA has a deep relation to Group theory
Note on ” ~ “ ‘s: I often forget them
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Concepts of group theory First, some fancy words with simple meanings: Generators, Casimirs, Representations, conserved quantum numbers, degeneracy splitting
Generators of a group: Set of operators , Oi that close on commutation. [ Oi , Oj ] = Oi Oj - Oj Oi = Ok i.e., their commutator gives back 0 or a member of the set
For IBA, the 36 operators s†s, d†s, s†d, d†d are generators of the group U(6).
Generators: define and conserve some quantum number.
Ex.: 36 Ops of IBA all conserve total boson number = ns + nd N = s†s + d† d
Casimir: Operator that commutes with all the generators of a group. Therefore, its eigenstates have a specific value of the q.# of that group. The energies are defined solely in terms of that q. #. N is Casimir of U(6).
Representations of a group: The set of degenerate states with that value of the q. #.
A Hamiltonian written solely in terms of Casimirs can be solved analytically
ex: † † † † † †, d s d sd s s s n n d ss s s sd s n n
† † †
† †
†
†
1 1, 1
1 1 1, 1
1 1, 1
s d s d s
s d s
s d s d s
d s s s d s
d s d s
d s
d sn n n s sd s n n
n s s d s n n
n s s n n n n
n n n n n n
n n n n
d s n n
or: † † †,d s s s d s
e.g: † † †
† †
,N s d N s d s dN
Ns d s dN
† † 0Ns d Ns d
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Sub-groups:
Subsets of generators that commute among themselves.
e.g: d†d 25 generators—span U(5)
They conserve nd (# d bosons)
Set of states with same nd are the representations of the group [ U(5)]
Summary to here:
Generators: commute, define a q. #, conserve that q. #
Casimir Ops: commute with a set of generators
Conserve that quantum #
A Hamiltonian that can be written in terms of Casimir Operators is then diagonal for states with that quantum #
Eigenvalues can then be written ANALYTICALLY as a function of that quantum #
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Simple example of dynamical symmetries, group chain, degeneracies
[H, J 2 ] = [H, J Z ] = 0 J, M constants of motion
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Let’s illustrate group chains and degeneracy-breaking.
Consider a Hamiltonian that is a function ONLY of: s†s + d†d
That is: H = a(s†s + d†d) = a (ns + nd ) = aN
In H, the energies depend ONLY on the total number of bosons, that is, on the total number of valence nucleons.
ALL the states with a given N are degenerate. That is, since a given nucleus has a given number of bosons, if H were the total Hamiltonian, then all the levels of the nucleus would be degenerate. This is not very realistic (!!!) and suggests that we
should add more terms to the Hamiltonian. I use this example though to illustrate the idea of successive steps of degeneracy breaking being related to different groups
and the quantum numbers they conserve.
The states with given N are a “representation” of the group U(6) with the quantum number N. U(6) has OTHER representations, corresponding to OTHER values of N,
but THOSE states are in DIFFERENT NUCLEI (numbers of valence nucleons).
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H’ = H + b d†d = aN + b nd
Now, add a term to this Hamiltonian:
Now the energies depend not only on N but also on nd
States of a given nd are now degenerate. They are “representations” of the group U(5). States with
different nd are not degenerate
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N
N + 1
N + 2
nd
1 2
0
a
2a
E
0 0
b
2b
H’ = aN + b d†d = a N + b nd
U(6) U(5)
H’ = aN + b d†d
Etc. with further term
s
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Concept of a Dynamical Symmetry
N
OK, here’s the key point -- get this if nothing else:
Spectrum generating algebra !!
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OK, here’s what you need to remember from the Group Theory
• Group Chain: U(6) U(5) O(5) O(3)
• A dynamical symmetry corresponds to a certain structure/shape of a nucleus and its characteristic excitations. The IBA has three dynamical symmetries: U(5), SU(3), and O(6).
• Each term in a group chain representing a dynamical symmetry gives the next level of degeneracy breaking.
• Each term introduces a new quantum number that describes what is different about the levels.
• These quantum numbers then appear in the expression for the energies, in selection rules for transitions, and in the magnitudes of transition rates.
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Group Structure of the IBAGroup Structure of the IBA
s boson :
d boson :
U(5)vibrator
SU(3)rotor
O(6)γ-soft
1
5U(6)
Sph.
Def.
Magical group
theory stuff
happens here
Symmetry Triangle of
the IBA
(everything we do from here
on will be discussed in the
context of this triangle. Stop
me now if you do not
understand up to here)
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Most general IBA Hamiltonian in terms with up to four boson operators (given N)
IBA Hamiltonian
Mixes d and s components of the wave functions
d+
d
Counts the number of d bosons out of N bosons, total. The
rest are s-bosons: with Es = 0 since we deal only with
excitation energies.
Excitation energies depend ONLY on the number of d-bosons.
E(0) = 0, E(1) = ε , E(2) = 2 ε.
Conserves the number of d bosons. Gives terms in the
Hamiltonian where the energies of configurations of 2 d bosons
depend on their total combined angular momentum. Allows for
anharmonicities in the phonon multiplets.d
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U(5)
Spherical, vibrational nuclei
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What J’s? M-scheme
Look familiar? Same as quadrupole vibrator.
6+, 4+, 3+, 2+, 0+
4+, 2+, 0+
2+
0+
3
2
1
0
nd
Simplest Possible IBA Hamiltonian – given by energies of the bosons with NO interactions
† †
d d s s
d s
H n n
d d s s
Excitation energies so, set s = 0, and drop subscript d on d
dH n
What is spectrum? Equally spaced levels defined by number of d bosons
= E of d bosons + E of s bosons
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E (nd, , J) = α nd + β nd (nd + 4) + 2γ ( +4)
+ 2δJ (J + 1)
Degeneracy breaking of multiplets [Cj terms]
But Ψ’s remain unchanged since H [ U(5)]) cannot change (mix) nd
But, U(5) is a rich symmetry that allows anharmonicities
Harmonic U(5): β, γ, δ = 0; E ~ nd
U(5) U(5)
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EE2 Transitions in the IBA2 Transitions in the IBAKey to most tests
Very sensitive to structure
E2 Operator: Creates or destroys an s or d boson or recouples two d bosons. Must conserve N
T = e Q = e[s† + d†s + χ (d† )(2)]d d
Specifies relative strength of this term
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E2 transitions in U(5)
• χ = 0 so
• T = e[s† + d†s]
• Can create or destroy a single d boson, that is a single phonon.
d
6+, 4+, 3+, 2+, 0+
4+, 2+, 0+
2+
0+
3
2
1
0
nd
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0+
2+
6+. . .
8+. . .
Vibrator (H.O.)
E(I) = n ( 0 )
R4/2= 2.0
Vibrator
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IBA HamiltonianComplicated and not
really necessary to use all these terms and all
6 parameters
Simpler form with just two parameters – RE-GROUP TERMS ABOVE
H = ε nd - Q Q Q = e[s† + d†s + χ (d† )(2)]d d
Competition: ε nd term gives vibrator.
Q Q term gives deformed nuclei.
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Relation of IBA Hamiltonian to Group Structure
We will see later that this same Hamiltonian allows us to calculate the properties of a nucleus ANYWHERE in the triangle simply by
choosing appropriate values of the parameters
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SU(3)
Deformed nuclei
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M
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Examples of energies of different representations of SU(3), that is, different (, )
E = A [ 2 + 2 + + 3 ( + ) ] + BJ (J + 1) f (, )
N = 16
E ( ) = A [ f (28, 2) – f (32, 0) ] = A [ (28)2 + (2)2 + 2(28) + 3(28 + 2) ]
– [ (32)2 + 0 + 0 + 3(32 + 0) ]
= A [ 934 - 1120 ] = -186 A, A < 0
So if E ( ) ~ 1 MeV, then A ~ 5-6 keV More generally,
E ( ) = A [ (2N)2 + 3 (2N) ] – [ (2N - 4)2 +4+(2N - 4)(2) +3 (2N – 4 + 2)] = A 4N2 + 6N – [ 4N2 – 16N + 16 + 4 + 4 N – 8 + 6N – 12 + 6)]
= A (12N – 6)
~ 2N - 1
(28, 2)
(32, 0) 10
20
20
20
20
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Typical SU(3) SchemeTypical SU(3) Scheme
SU(3) O(3)
K bands in (, ) : K = 0, 2, 4, - - - -
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Signatures of SU(3)Signatures of SU(3)
E = E
B ( g ) 0
Z 0
B ( g ) B ( g )
E ( -vib ) (2N - 1)
1/6
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Signatures of SU(3)Signatures of SU(3)
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Totally typical example
Similar in many ways to SU(3). But note that the two excited excitations are not degenerate as they should be in SU(3). While SU(3) describes an axially symmetric rotor,
not all rotors are described by SU(3) – see later discussion
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E2 Transitions in SU(3)
Q = (s†d + d†s) + (- 7
2) (d†d )(2)
Q is also in H and is a Casimir operator of SU(3), so conserves , .
B(E2; J + 2 →J)yrast
B(E2; +12 → +
10 )
N2 for large N
Typ. Of many IBA predictions → Geometric Model as N → ∞ Δ (, ) = 0
2 2 13
2 2 34 2 3 2 5B
J JN J N J
J Je
SU(3)
1 1
1 1
( 2;4 2 ) 10 (2 2)(2 5)
( 2;2 0 ) 7 (2 )(2 3)
B E N N
B E N N
E2: Δ (, ) = 0
2 2 3
5BN N
e
Alaga rule Finite N correction
“β”→ γ Collective B(E2)s
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Example of finite boson number effects in the IBA
B(E2: 2 0): U(5) ~ N; SU(3) ~ N(2N + 3) ~ N2
B(E2)
~N
N2
N
Mid-shell
H = ε nd - Q Q and keep the parameters constant.What do you predict for this B(E2) value??
!!!
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O(6)
Axially asymmetric nuclei(gamma-soft)
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Transition Rates
T(E2) = eBQ Q = (s†d + d†s)
B(E2; J + 2 → J) = 2B
J+ 1
24
J Je N - N2 2 J + 5
B(E2; 21 → 01) ~ 2B
+ 4e
5N N
N2
Consider E2 selection rules Δσ = 0 0+(σ = N - 2) – No allowed decays! Δτ = 1 0+( σ = N, τ = 3) – decays to 22 , not 12
O(6) E2 Δσ = 0 Δτ = 1
Note: Uses χ = o
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196Pt: Best (first) O(6) nucleus -soft
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Xe – Ba O(6) - like
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Classifying Structure -- The Symmetry TriangleClassifying Structure -- The Symmetry Triangle
Most nuclei do not exhibit the idealized symmetries but rather lie in transitional regions. Mapping the triangle.
Sph.
Deformed
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The IBA: convenient model that spans the entire The IBA: convenient model that spans the entire triangle of colllective structurestriangle of colllective structures
H = ε nd - Q Q Parameters: , (within Q)/ε
Sph. Driving Def. Driving
H = c [
ζ ( 1 – ζ ) nd 4NB Qχ ·Qχ - ]
Competition: : 0 to infinity /ε
Span triangle with and
Parameters already known for many nuclei
c is an overall scale factor giving the overall energy scale. Normally, it is fit to the first 2+ state.
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