IBA Lecture part 2

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+dd+d Counts the number of d bosons out of N bosons, total. The rest are s-bosons: with E s = 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

U(5) Spherical, vibrational nuclei

What J’s? M-scheme Look familiar? Same as quadrupole vibrator. U(5) also includes anharmonic spectra 6 +, 4 +, 3 +, 2 +, , 2 +, n d Simplest Possible IBA Hamiltonian – given by energies of the bosons with NO interactions Excitation energies so, set  s = 0, and drop subscript d on  d What is spectrum? Equally spaced levels defined by number of d bosons = E of d bosons + E of s bosons

More than one phonon? What angular momenta? M-scheme for bosons

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

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

E2 transitions in U(5) χ = 0 That is: T = e[s† + d†s] Why? So that it can create or destroy a single d boson, that is, a single phonon. 6 +, 4 +, 3 +, 2 +, , 2 +, n d

Vibrator (H.O.) E(I) = n (   0 ) R 4/2 = 2.0 Vibrator

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

Input $diag eps = 0.20, kappa = 0.00, chi =-0.00, nphmax = 6, iai = 0, iam = 6, neig = 3, mult=.t.,ell=0.0,pair=0.0,oct=0.0,ippm=1,print=.t. $ $em E2SD=1.0, E2DD=-0.00 $ SLCT Output L P = 0+ Basis vectors |NR> = |ND,NB,NC,LD,NF,L P> | 1> = | 0, 0, 0, 0, 0, 0+> | 2> = | 2, 1, 0, 0, 0, 0+> | 3> = | 3, 0, 1, 0, 0, 0+> | 4> = | 4, 2, 0, 0, 0, 0+> | 5> = | 5, 1, 1, 0, 0, 0+> | 6> = | 6, 0, 2, 0, 0, 0+> | 7> = | 6, 3, 0, 0, 0, 0+> Energies Eigenvectors 1: : : : : : : L P = 1+ No states L P = 2+ Energies L P = 3+ Energies L P = 4+ Energies L P = 5+ Energies L P = 6+ Energies Transitions: 2+ -> 0+ (BE2) 2+,1 -> 0+,1: ,1 -> 0+,2: ,1 -> 0+,3: ,2 -> 0+,1: ,2 -> 0+,2: ,2 -> 0+,3: ,3 -> 0+,1: ,3 -> 0+,2: ,3 -> 0+,3: and 0+ -> 2+ (BE2) 0+,1 -> 2+,1: ,2 -> 2+,1: ,3 -> 2+,1: ,1 -> 2+,2: ,2 -> 2+,2: ,3 -> 2+,2: ,1 -> 2+,3: ,2 -> 2+,3: ,3 -> 2+,3: Transitions: 4+ -> 2+ (BE2) 4+,1 -> 2+,1: ,1 -> 2+,2: ,1 -> 2+,3: ,2 -> 2+,1: ,2 -> 2+,2: ,2 -> 2+,3: ,3 -> 2+,1: ,3 -> 2+,2: ,3 -> 2+,3: U(5) Basis Energies Pert. Wave Fcts.

Deformed nuclei Use the same Hamiltonian but re- write it in more convenient and physically intuitive form

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 n d -  Q  Q Q = e[s † + d † s + χ (d † ) (2) ] Competition: ε n d term gives vibrator.  Q  Q term 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 = -1.32

In a region of increasing  softness, can simulate simply by decreasing |  | towards zero

SU(3) Deformed nuclei

Any calculation deviating from U(5) gives wave functions where n d is no longer a good quantum number. If the wave function is expressed in a U(5) – vibrator – basis, then it contains a mixture of terms. Understanding these admixtures is crucial to understanding IBA calculations Wave functions in SU(3): Consider non-diagonal effects of the QQ term in H on n d components in the wave functions Q operator: Q = (s † d + d † s) + (d † d ) (2) QQ = [ { (s † d + d † s) + χ(d † d ) (2) } x { (s † d + d † s) + (d † d ) (2) } ] ~ s † d s † d + s † d d † s + s † d d † d ….  n d = …. 2, 0, 1

  M

Typical SU(3) Scheme SU(3)  O(3) K bands in (,  ) : K = 0, 2, 4,  Characteristic signatures: Degenerate bands within a group Vanishing B(E2) values between groups Allowed transitions between bands within a group Where? N~ 104, Yb, Hf

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

Data on E2 transitions in deformed nuclei Intraband --- STRONG, typ. ~ 200 W.u. in heavy nuclei Interband --- Collective but much weaker, typ W.u. Alaga Rules for Branching ratios Brief excursion to discuss E2 transitions and the Alaga rules in the geometric model

0  g  b  | (b +   g )> =  g  g > = 1 g g

B(E2) branching ratios between rotational levels belonging a pair of intrinsic states

= 1.43 x [ … ]

Ratio of yrast B(E2) values: compare to U(5) and SU(3) predictions

Signatures of SU(3)

E  = E  B (   g )  0 Z   0 B (   g ) B (   g ) E (  -vib )  (2N - 1)  1/6

Example of finite boson number effects in the IBA B(E2: 2  0): U(5) ~ N; SU(3) ~ N(2N + 3) ~ N 2 B(E2) ~N N2N2 N Mid-shell H = ε n d -  Q  Q and keep the parameters constant. What do you predict for this B(E2) value?? !!!

******************** Input file contents ******************** $diag eps = 0.00, kappa = 0.02, chi = , nphmax = 6, iai = 0, iam = 6, neig = 5, mult=.t.,ell=0.0,pair=0.0,oct=0.0,ippm=1,print=.t. $ $em E2SD=1.0, E2DD= $ ************************************************************* L P = 0+ Basis vectors |NR> = |ND,NB,NC,LD,NF,L P> | 1> = | 0, 0, 0, 0, 0, 0+> | 2> = | 2, 1, 0, 0, 0, 0+> | 3> = | 3, 0, 1, 0, 0, 0+> | 4> = | 4, 2, 0, 0, 0, 0+> | 5> = | 5, 1, 1, 0, 0, 0+> | 6> = | 6, 0, 2, 0, 0, 0+> | 7> = | 6, 3, 0, 0, 0, 0+> Energies Eigenvectors 1: : : : : : : L P = 1+ No states L P = 2+ Energies L P = 3+ Energies L P = 4+ Energies L P = 5+ Energies L P = 6+ Energies Binding energy = , eps-eff = SU(3) Wave fcts. in U(5) basis

O(6) Axially asymmetric nuclei (gamma-soft)

Note: Uses χ = 0

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

Input $diag eps = 0.0, kappa = 0.02, chi =-0.0, nphmax = 6, iai = 0, iam = 6, neig = 5, mult=.t.,ell=0.0,pair=0.0,oct=0.0,ippm=1,print=.t. $ $em E2SD=1.0, E2DD=-0.00 $ Output L P = 0+ Basis vectors |NR> = |ND,NB,NC,LD,NF,L P> | 1> = | 0, 0, 0, 0, 0, 0+> | 2> = | 2, 1, 0, 0, 0, 0+> | 3> = | 3, 0, 1, 0, 0, 0+> | 4> = | 4, 2, 0, 0, 0, 0+> | 5> = | 5, 1, 1, 0, 0, 0+> | 6> = | 6, 0, 2, 0, 0, 0+> | 7> = | 6, 3, 0, 0, 0, 0+> Energies Eigenvectors 1: : : : : : : L P = 1+ No states L P = 2+ Energies L P = 3+ Energies L P = 4+ Energies L P = 5+ Energies L P = 6+ Energies Binding energy = , eps-eff = O(6) Basis Pert. Wave Fcts. Energies

Thus far, we have only dealt with nuclei corresponding to one of the three dynamical symmetries. Probably <1% of nuclei do that. So, how do we treat the others? That is, how do we calculate with the IBA AWAY from the vertices of the symmetry triangle? A couple of interesting examples first, then a general approach -- - The technique of Orthogonal Crossing Contours (OCC) More General IBA calculations 

CQF along the O(6) – SU(3) leg H = -κ Q Q Only a single parameter,  H = ε n d -  Q  Q Two parameters ε /  and 

 H = ε n d -  Q  Q ε = 0  /ε 168-Er very simple 1-parameter calculation H = -  Q  Q  is just scale factor So, only parameter is 

1

“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 N B ] Can plot any observable as a set of contours vs. N B and χ. 2

Universal O(6) – SU(3) Contour Plots H = -κ Q Q χ = 0 O(6) χ = SU(3) SU(3)

Systematics and collectivity of the lowest vibrational modes in deformed nuclei Notice that the the  mode is at higher energies (~ 1.5 times the  vibration near mid-shell)* and fluctuates more. This points to lower collectivity of the  vibration. * Remember for later ! Max. Coll. Increasing  softness

Os isotopes from A = 186 to 192: Structure varies from a moderately gamma soft rotor to close to the O(6) gamma- independent limit. Describe simply with: H = -κ Q Q  : 0  small as A decreases 3

Backups

Crucial for structureCrucial for masses Collective models and masses, binding energies, or separation energies

Mapping the Triangle with a minimum of data -- exploiting an Ising-type Model –The IBA Mapping the Triangle with a minimum of data -- exploiting an Ising-type Model –The IBA Competition between spherical-driving (pairing – like nucleon) and deformation-driving (esp. p-n) interactions H = aH sph + bH def Structure ~ a/b Sph. Def.

O(6) SU(3)U(5) 