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Review

- Short range force, Pauli Principle Shell structure, magic numbers, concept of valence nucleons
- Residual interactions favoring of 0+ coupling: 0+ ground states for all even-even nuclei
- Concept of seniority lowest states have low seniority, huge simplification (n-body calculations often reduce to 2-body !)
- constant g factors, constant energies in singly magic or near magic nuclei, parabolic B(E2) systematics, change in sign of quadrupole moments (prolate-oblate shapes) across a shell

Effects of monopole interactions

Between 40Zr and 50Sn protons fill 1g9/2 orbit.

Large spatial overlap with neutron 1g7/2 orbit.

1g7/2 orbit more tightly bound. Lower energy

How does structure evolve?

- Benchmarks
- Magic nuclei – spherical, stiff
- Nuclei with only one kind of valence nucleon: like the 2-particle case
- Nuclei with both valence protons and neutrons: mixing of configurations, complex wave functions with components from every configuration (~10big).
- Is there another way? YES !!!
- Macroscopic perspective. Many-body approach, collective coordinates, modes

Microscopic origins of collectivity

correlations, configuration mixing and deformation: Residual interactions

Crucial for structure

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 b gand g
- 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.

Benchmark

700

333

100

0

Rotor J(J + 1)

Amplifies structural differences

Centrifugal stretching

Deviations

6+690

4+330

2+100

0+0

JE (keV)

Identify additional

degrees of freedom

The value of paradigms

?

Interpretation without rotor paradigm

Exp.

6+

4+

2+

0+

Rotational states built on (superposed on) vibrational modes

Vibrational excitations

Rotational states

Ground or equilibrium state

Systematics and collectivity of the lowest vibrational modes in deformed nuclei

Notice that the the b mode is at higher energies (~ 1.5 times the g vibration near mid-shell)* and fluctuates more. This points to lower collectivity of the b vibration.

* Remember for later !

How can we understand collective behavior

- Do microscopic calculations, in the Shell Model or its modern versions, such as with density functional theory or Monte Carlo methods. These approaches are making amazing progress in the last few years. Nevertheless, they often do not give an intuitive feeling for the structure calculated.
- Collective models, which focus not on the particles but the structure and symmetries of the many-body, macroscopic system itself. They are not predictive in the same way as microscopic calculations but they can reveal coherent behavior more clearly in many cases.
- We will illustrate collective models with the IBA, historically, by far the most successful and parameter-efficient collective model.

IBA – A Review and Practical Tutorial

F. Iachello and A. Arima

- Drastic simplification of
- shell model
- Valence nucleons
- Only certain configurations
- Simple Hamiltonian – interactions
- “Boson” model because it treats nucleons in pairs
- 2 fermions boson

Shell Model Configurations

Fermion configurations

Roughly, gazillions !!

Need to simplify

The IBA

Boson configurations

(by considering only configurations of pairs of fermions with J = 0 or 2.)

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+

Modeling a Nucleus

3 x 1014

2+ states

2. Fermions → bosons

J = 0 (s bosons)

J = 2 (d bosons)

IBA: 26 2+ states

Why the IBA is the best thing since jackets

154Sm

Shell model

Need to truncate

IBA assumptions

Is it conceivable that these 26 basis states are correctly chosen to account for the properties of the low lying collective states?

1. Only valence nucleons

Bosons in IBA are pairs of fermions in valence shell

Number of bosons for a given nucleus is a fixednumber

N=6 5 =NNB= 11

Basically the IBA is a Hamiltonian written in terms of s and d bosons and their interactions. It is written in terms of boson creation and destruction operators. Let’s briefly review their key properties.

Review of phonon creation and destruction operators

- What is a creation operator? Why useful?
- 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.

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. 0 (s) or 2 (d).

IBAhas a deep relation to Group theory

That relation is based on the 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.

Gr. Theor. classification of Hamiltonian

Group is called

U(6)

s†s, s†d, d†s, d†d

Brief, simple, trip into the Group Theory of the IBA

Next 8 slides give an introduction to the Group Theory relevant to the IBA. If the discussion of these is too difficult or too fast, don’t worry, you will be able to understand the rest anyway. Just take a nap for 5 minutes. In any case, you will have these slides on the web and can look at them later in more detail if you want.

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 name for this application of Group Theory is

“Spectrum Generating Algebras”

e.g:

or:

Concepts of group theory

First, some fancy words with simple meanings:Generators, Casimirs, Representations, conserved quantum numbers, degeneracy splitting

Generatorsof a group: Set of operators , Oi that close on commutation.

[ Oi , Oj] = OiOj - OjOi = 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†

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

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, J2] = [H, JZ] = 0 J, M constants of motion

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).

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 + 2

2a

N + 1

a

2

2b

Etc. with further terms

1

b

N

0

0

0

nd

E

U(6) U(5)

H’ = aN + b d†d

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 IBA

U(5)

vibrator

U(6)

SU(3)

rotor

O(6)

γ-soft

1

s boson :

5

d boson :

Magical group theory stuff happens here

Symmetry Triangle of the IBA

Def.

Sph.

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+d

d

Mixes d and s components of the wave functions

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

Simplest Possible IBA Hamiltonian –

given by energies of the bosons with NO interactions

3

2

1

0

nd

6+, 4+, 3+, 2+, 0+

4+, 2+, 0+

2+

0+

= E of d bosons + E of s bosons

Excitation energies so, set s = 0, and drop subscript d on d

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

What J’s? M-scheme

Look familiar? Same as quadrupole vibrator.

U(5) also includes anharmonic spectra

Key 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)]

Specifies relative strength of this term

2

1

0

nd

6+, 4+, 3+, 2+, 0+

4+, 2+, 0+

2+

0+

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.

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)]

Competition:ε nd term gives vibrator.

Q Qterm gives deformed nuclei.

This is the form we will use from here on

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

- Characteristic signatures:
- Degenerate bands within a group
- Vanishing B(E2) values between groups
- Allowed transitions
- between bands within a group
- Where? N~ 1-4, Yb, Hf

SU(3) O(3)

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

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

B(E2)

N2

~N

N

Example of finite boson number effects in the IBAB(E2: 2 0): U(5) ~ N; SU(3) ~ N(2N + 3) ~ N2

H =ε nd - Q Q and keep the parameters constant.

What do you predict for this B(E2) value??

!!!

Deformed

Classifying Structure -- The Symmetry Triangle

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

Mapping the Entire Triangle

H =ε nd - Q Q

Parameters: , c (within Q)

/ε

2 parameters

2-D surface

/ε

168-Er very simple 1-parameter calculation

c

H =ε nd - Q Q

ε = 0

- H =- Q Q
- is just scale factor

So, only parameter is c

/ε

for the SU(3) – O(6) leg

2

H = - κ Q • Q

κ is just energy scale factor

Ψ’s, B(E2)’s independent of κ

Results depend only on χ

[ and, of course, vary with NB]

Can plot any observable as a set of

contours vs. NBand χ.

Mapping the entire triangleTechnique of orthogonal crossing contours (OCC)

H has two parameters. A given observable can only specify one of them. What does this imply? An observable gives a contour of constant values within the triangle

= 2.9

R4/2

At the basic level : 2 observables (to map any point in the symmetry triangle)

Preferably with perpendicular trajectories in the triangle

2.7

2.9

2.5

3.1

3.3

2.2

A simple way to pinpoint structure. What do we need?

Simplest Observable: R4/2

Only provides a locus of structure

We have a problem

What we have:

What we need:

Lots of

Just one

+2.0

+2.9

+1.4

+0.4

+0.1

-1

-0.1

-0.4

-2.0

-3.0

Fortunately:

Mapping Structure with Simple Observables – Technique of Orthogonal Crossing Contours

γ - soft

Vibrator

Rotor

Burcu Cakirli et al.

Beta decay exp. + IBA calcs.

Complementarity of macroscopic and microscopic approaches. Why do certain nuclei exhibit specific symmetries? Why these evolutionary trajectories?

What will happen far from stability in regions of proton-neutron asymmetry and/or weak binding?

Collective models and masses, binding energies, or separation energies

Crucial for structure

Crucial for masses

Two-neutron separation energies

BindingEnergies

S2n = A + BN + S2n (Coll.)

Normal behavior: ~ linear segments with drops after closed shells

Discontinuities at first order phase transitions

Use any collective model to calculate the collective contributions toS2n.

Which 0+ level is collective and which is a 2-quasi-particle state?

Evolution of level energies in rare earth nuclei

But note:

McCutchan et al

Do collective model fits, assuming one or the other 0+ state, at 1222 or 1422 keV, is the collective one. Look at calculated contributions to separation energies. What would we expect?

Collective contributions to masses can vary significantly for small parameter changes in collective models, especially for large boson numbers where the collective binding can be quite large.

S2n(Coll.) for alternate fits to Er with N = 100

S2n(Coll.) for two calcs.

B.E.(z,c)

IBA

B.E (MeV)

Gd – Garcia Ramos et al, 2001

Masses: a new opportunity – complementary observable to spectroscopic data in pinning down structure, especially in nuclei with large numbers of valence nucleons. Strategies for best doing that are still being worked out. Particularly important far off stability where data will be sparse.

Cakirli et al, 2009

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