IBA – An Introduction and Overview

Basic Ideas, underpinnings, Group Theory, basic predictions

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

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

Why do we need to bother with such a model?

Remember 3 x 1014 ?We simply MUST simplify the problem.

As it turns out, the IBA is:a)The most successful macroscopic modelb)The only collective model in which it is even possible in practice to calculate many observables

Shell Model ConfigurationsFermion

configurations

Boson configurations

(by considering only configurations of pairs of fermions

with J = 0 or 2.)

0+ s-boson

2+ d-boson

• Valence nucleons only

• s, d bosons – creation and destruction operators

H = Hs + Hd + Hinteractions

Number of bosons fixed in a given nucleus: N = ns + nd

= ½ # of val. protons + ½ # val. neutrons

valenceIBM Assume fermions couple in pairs to bosons of spins 0+ and 2+

s boson is like a Cooper paird boson is like a generalized pair. Create ang. mom. with d bosons

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+

The IBA – an audacious, awesome leapThe IBA – an audacious, awesome leap

154Sm 3 x 1014 2+ states

Or, why the IBA is the best thing since tortellini Magnus

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

could possibly be correctly chosen to

account for the properties of the low

lying collective states?

Why the IBA ?????• Why a model with such a drastic simplification –

Oversimplification ???

• Answer: Because it works !!!!!• By far the most successful general nuclear collective

model for nuclei ever developed• Extremely parameter-economic

Deep relation with Group Theory !!! Dynamical symmetries, group chains, quantum numbers

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 !!!

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

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.

Most general IBA Hamiltonian in terms with up to four boson operators (given, fixed N)

IBA Hamiltonian

AARRGGHHH !!!We will greatly simplify this soon but it is useful to look at its

structure

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

† †

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 nWhat is spectrum? Equally spaced levels defined by number of d bosons

Most general IBA Hamiltonian in terms with up to four boson operators (given N)IBA Hamiltonian

These terms CHANGE the numbers of s and d bosons:

MIX basis states of the model

Crucial for structure

Ba Ce Nd

Sm

Gd

Dy

Er

Yb

11

12

13

14

15

16

17

84 86 88 90 92 94 96

Crucial for masses

Most general IBA Hamiltonian in terms with up to four boson operators (given N)IBA Hamiltonian

Complicated 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.

Note: 3 parameters. BUT: H’ = aH have identical wave functions, q.#s, sel. rules, trans. rates. Only the energy SCALE differs.

STRUCTURE – 2 parameters. MASSES – need scale

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”

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

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

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 #

Simple example of dynamical symmetries, group chain, degeneracies

[H, J 2 ] = [H, J Z ] = 0 J, M constants of motion

Let’s ilustrate 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).

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

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

Concept of a Dynamical Symmetry

N

OK, here’s the key point -- get this if nothing else:

Spectrum generating algebra !!

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.

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)

Dynamical Symmetries – The structural benchmarks

• U(5) Vibrator – spherical nucleus that can oscillate in shape

• SU(3) Axial Rotor – can rotate and vibrate

• O(6) Axially asymmetric rotor ( “gamma-soft”) – squashed deformed rotor

Dynamical SymmetriesDynamical Symmetries

I. U(6) U(5) O(5) O(3) U(5) N nd n J

II. U(6) SU(3) O(3) SU(3) N ( , ) K J

III. U(6) O(6) O(5) O(3) O(6) N σ J

Vibrator

Rotor

Gamma-soft rotor

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.

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

U(5)

Spherical, vibrational nuclei

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

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

U(5) H = εnd + anharmonic terms nd = # d bosons in wave function ε = energy of d boson 4 3 2 1 0 E Quantum Numbers

N = ns + nd = Total Boson No.

nd = # d bosons

n = # of pairs of d bosons coupled to J = 0+

nΔ = # of triplets of d bosons coupled to J = 0+

(400) (420)

(300)

(410) (401) (400) (410) (400) (400) 8+ 0+

6+

2+ 2+ 4+ 4+ 5+ 6+

0+ 2+ 3+ 4+

0+ 2+ 4+

0+

2+

(301) (310) (300) (300)

(210) (200) (200)

(000)

(100)

U(5)

(nd n nΔ)

Harmonic Spectrum

(d† d†)L ss |0 >

(d† d† d†)L sss |0 >

(d)

(d)

(d)

(d)

(d)

d†s | 0 >

Important as a benchmark of structure, but also since the

U(5) states serve as a convenient set of

basis states for the IBA

U(5)Multiplets

Which nuclei are U(5)?• No way to tell a priori (until better microscopic understanding of IBA

is available).

• More generally, phenomenological models like the IBA predict nothing on their own. They can predict relations among observables for a given choice of Hamiltonian parameters but they don’t tell us which parameter values apply to a given nucleus. They don’t tell us which nuclei have which symmetry, or perhaps none at all. They need to be “fed”.

• The nuclei provide their own food: but the IBA is not gluttonous – a couple of observables allow us to pinpoint structure.

• Let the nuclei tell us what they are doing !!!!• Don’t force an interpretation on them

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

χ is generally fit as a parameter but has characteristic values in each dynamical symmetry

E2 electromagnetic transition rates in the IBA

Finite, fixed number of bosons has a huge effect compared ot the geometrical model

Note: TWO factors in B(E2). In geometrical model B(E2) values are proportional to the number of phonons in the

initial state.

In IBA, operator needs to conserve total boson number so gamma ray transitions proceed by operators of the form:s†d. Gives TWO square roots that compete.

B(E2: J J-2)

Yrast (gsb) states

J

2 4 6 8 10 12

6

5

4

3

2

1

Geom. Vibrator

IBA, U(5), N=6

Finite Boson Number Effects: B(E2) Values

Slope = 1.51

2+ 0+

Classifying Structure -- The Symmetry TriangleClassifying Structure -- The Symmetry Triangle

Have considered vibrators (spherical nuclei). What

about deformed nuclei ??

Sph.

Def.

SU(3)

Deformed nuclei(but only a special subset)

M

( or M, which is not exactly the same as K)

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

Typical SU(3) SchemeTypical SU(3) Scheme

SU(3) O(3)

K bands in (, ) : K = 0, 2, 4, - - - -

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

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

Another 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??

!!!

Signatures of SU(3)Signatures of SU(3)

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

O(6)

Axially asymmetric nuclei(gamma-soft)

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

196Pt: Best (first) O(6) nucleus -soft

Xe – Ba O(6) - like

Classifying Structure -- The Symmetry TriangleClassifying Structure -- The Symmetry Triangle

Most nuclei do not exhibit the idealized symmetries but rather lie in transitional regions.

Trajectories of structural evolution. Mapping the triangle.