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
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
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
Microscopic origins of collectivity
correlations, configuration mixing and deformation: Residual interactions
Crucial for structure
J = 2, one phonon vibration
More than one phonon? What angular momenta? M-scheme for bosons
E(I) (ħ2/2I )J(J+1)
Rotor J(J + 1)
Amplifies structural differences
degrees of freedom
The value of paradigms
Interpretation without rotor paradigm
Rotational states built on (superposed on) vibrational modes
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 !
IBA – A Review and Practical Tutorial
F. Iachello and A. Arima
Roughly, gazillions !!
Need to simplify
(by considering only configurations of pairs of fermions with J = 0 or 2.)
Why s, d
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+
3 x 1014
2. Fermions → bosons
J = 0 (s bosons)
J = 2 (d bosons)
IBA: 26 2+ states
Why the IBA is the best thing since jackets
Need to truncate
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
Note key point:
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
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
s†s, s†d, d†s, d†d
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”
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).
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
H’ = aN + b d†d = a N + b nd
N + 2
N + 1
Etc. with further terms
H’ = aN + b d†d
OK, here’s the key point :
Spectrum generating algebra !!
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.
s boson :
d boson :
Magical group theory stuff happens here
Symmetry Triangle of the IBA
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.
Mixes d and s components of the wave functions
Most general IBA Hamiltonian in terms with up to four boson operators (given N)
U(5)Spherical, vibrational nuclei
Simplest Possible IBA Hamiltonian –
given by energies of the bosons with NO interactions
6+, 4+, 3+, 2+, 0+
4+, 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
E2 Transitions in the IBA
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
6+, 4+, 3+, 2+, 0+
4+, 2+, 0+
E(I) = n (0 )
8+. . .
6+. . .
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
Typical SU(3) Scheme
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
B(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??
O(6)Axially asymmetric nuclei(gamma-soft)
Note: Uses χ = o
196Pt: Best (first) O(6) nucleus
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)
168-Er very simple 1-parameter calculation
H =ε nd - Q Q
ε = 0
So, only parameter is c
“Universal” IBA Calculations
for the SU(3) – O(6) leg
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.
Universal O(6) – SU(3) Contour Plots
H = -κ Q • Q
χ = 0 O(6)
χ = - 1.32 SU(3)
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
At the basic level : 2 observables (to map any point in the symmetry triangle)
Preferably with perpendicular trajectories in the triangle
A simple way to pinpoint structure. What do we need?
Simplest Observable: R4/2
Only provides a locus of structure
What we have:
What we need:
Mapping Structure with Simple Observables – Technique of Orthogonal Crossing Contours
γ - soft
Burcu Cakirli et al.
Beta decay exp. + IBA calcs.
Evolution of Structure
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
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
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.
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