IBA – An Introduction and Overview Basic Ideas, underpinnings, Group Theory, basic predictions • Shell Model Sph. Def. - (Microscopic) • Geometric – (Macroscopic) • Third approach — “Algebraic” Dynamical Symmetries Group Theoretical Phonon-like model with microscopic basis explicit from the start. IBA Shell Mod. Geom. Mod. 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 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 model b) The only collective model in which it is even possible in practice to calculate many observables Shell Model Configurations Fermion configurations Boson configurations (by considering only configurations of pairs of fermions with J = 0 or 2.) Assume valence fermions couple in pairs to bosons of spins 0+ and 2+ IBM 0+ s-boson 2+ d-boson s boson is like a Cooper pair d boson is like a generalized pair. Create ang. mom. with d bosons • 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 Why s, d bosons? s Lowest state of all e-e nuclei is 0+ - fct gives 0+ ground state d First excited state in non-magic e-e nuclei almost always 2+ - fct gives 2+ next above 0+ The IBA – an audacious, awesome leap Or, why the IBA is the best thing since tortellini Magnus 154Sm Shell model Need to truncate IBA assumptions 1. Only valence nucleons 2. Fermions → bosons J = 0 (s bosons) J = 2 (d bosons) 3 x 1014 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? IBA: 26 2+ 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 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: Bosons in IBA are pairs of fermions in valence shell Number of bosons for a given nucleus is a fixed number 154 62 Sm 92 N = 6 5 = N NB = 11 Review of phonon creation and destruction operators 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. 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). IBA Hamiltonian Most general IBA Hamiltonian in terms with up to four boson operators (given, fixed N) AARRGGHHH !!! We will greatly simplify this soon but it is useful to look at its structure Simplest Possible IBA Hamiltonian H d nd s ns d d † d † s s s Excitation energies so, set s = 0, and drop subscript d on d H nd What is spectrum? Equally spaced levels defined by number of d bosons 3 6+, 4+, 3+, 2+, 0+ 2 4+, 2+, 0+ 1 2+ 0 0+ nd What J’s? M-scheme Look familiar? Same as quadrupole vibrator. IBA Hamiltonian Most general IBA Hamiltonian in terms with up to four boson operators (given N) These terms CHANGE the numbers of s and d bosons: MIX basis states of the model Crucial Crucialfor forstructure masses 17 Yb 16 Er 15 Dy 14 Gd 13 12 Sm Ba Ce Nd 11 84 86 88 90 92 94 96 IBA Hamiltonian Most general IBA Hamiltonian in terms with up to four boson operators (given N) 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 Competition: ε nd QQ Q = e[s†d + d†s + χ (d† d )(2)] term gives vibrator. 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” IBA has a deep relation to Group theory 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 Note on ” ~ “ ‘s: I often forget them Gr. Theor. classification of Hamiltonian 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). † † some † quantum † Generators conserve ex: d † s, s:†define s n d nand d ss s s sd s n d n number. s s † Ex.: 36 Ops of IBA all conserve boson †number d † sntotal s n d n s s sd s n d n s N = s†s + d†d = ns + nd n s s †s d †s n d n s † † † all N , s d N s d s † with e.g: Operator Casimir: that commutes the generators of adN group. n s s s n d 1 n s n d 1, n s 1 Therefore, its † The energies are defined eigenstates have a specific value of the q.#† of that group. Ns d s n d 1 n s n s n s 1 n d 1, ndN s 1 solely in terms of that q. #. N is Casimir † U(6). Nsof d Ns † d 0 n d 1 n s n d 1, n s 1 d † s n dThe n s set of degenerate states with that value of the q. #. Representations of a group: d † s, s † s d † s Hamiltonian written solely in terms of Casimirs can be solved analytically or: A 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: That is: s†s + d†d 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 2a N + 2 a H’ = aN + b d†d = a N + b nd N+1 b 2 1 0 0 2b N 0 nd E U(6) H’ = aN U(5) + b d† d OK, here’s the key point -- get this if nothing else: Concept of a Dynamical Symmetry N 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) s boson : 1 d boson : 5 vibrator U(6) SU(3) rotor Magical group theory stuff happens here O(6) γ-soft Symmetry Triangle of the IBA Def. Sph. (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 Symmetries Vibrator Rotor Gamma-soft rotor I. U(6) N II. U(6) SU(3) N III. U(6) N U(5) nd O(5) SU(3) O(3) n J U(5) O(3) ( , ) KJ O(6) σ O(5) O(3) J O(6) 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 Competition: ε nd QQ Q = e[s†d + d†s + χ (d† d )(2)] term gives vibrator. 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 IBA Hamiltonian Counts the number of d bosons out of N bosons, total. Conserves the of with d bosons. Gives in the The rest arenumber s-bosons: Es = 0 since we terms deal only Hamiltonian wherewith theexcitation energiesenergies. of configurations of 2 d + bosons depend on their total combined angular ExcitationAllows energies ONLY on the in number of dd d momentum. fordepend anharmonicities the phonon multiplets. bosons. E(0) = 0, E(1) = ε , E(2) = 2 ε. 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 H d n d s n s = E of d bosons + E of s bosons d d † d † s s s Excitation energies so, set s = 0, and drop subscript d on d H nd What is spectrum? Equally spaced levels defined by number of d bosons 3 6+, 4+, 3+, 2+, 0+ 2 4+, 2+, 0+ 1 2+ 0 0+ nd What J’s? M-scheme Look familiar? Same as quadrupole vibrator. U(5) H=ε n d + anharmonic terms nd = # d bosons in wave function ε = energy of d boson U(5) Multiplets 4 8+ 3 6+ 2 4+ 1 2+ 0 0+ E (400) (300) (200) (100) (000) 6+ 4+ 2+ (400) (300) 3+ (200) d (400) 5+ 0+ (300) 4+ 2+ (210) (400) 4+ (310) 0+ † (410) (301) † L (d d ) † s|0> U(5) (nd n nΔ) Harmonic Spectrum Quantum Numbers 2+ (401) † † (410) 2+ † L 0+ (420) sss |0 > (d d d ) ss |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 N = ns + nd = Total Boson No. nd = # d bosons n = # of pairs of d bosons coupled to J = 0 (d) (d) + + (d) (d) nΔ = # of triplets of d bosons coupled to J = 0 (d) 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 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 E2 electromagnetic transition rates in the IBA T = e Q = e[s†d + d†s + χ (d†d )(2)] Specifies relative strength of this term χ is generally fit as a parameter but has characteristic values in each dynamical symmetry 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. Finite Boson Number Effects: B(E2) Values 6 5 SlopeGeom. = 1.51Vibrator B(E2: J J-2) 4 Yrast (gsb) states 3 IBA, U(5), N=6 2 1 2 4 6 2+ J 8 0+ 10 12 Classifying Structure -- The Symmetry Triangle Def. Sph. Have considered vibrators (spherical nuclei). What about deformed nuclei ?? 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 (, ) 0 2 0 1 (28, 2) N = 16 (32, 0) E ( 0 2 ) = A [ f (28, 2) – f (32, 0) ] =A 2 [ (28) + (2)2 + 2(28) + 3(28 + 2) ] – [ (32)2 + 0 + 0 + 3(32 + 0) ] = A [ 934 - 1120 ] = -186 A, A < 0 So if E ( 0 2 ) ~ 1 MeV, then A ~ 5-6 keV More generally, E ( 0 2) = 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 Typical SU(3) Scheme SU(3) K bands in (, ) : K = 0, 2, 4, - - - - O(3) 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) † † † (2) Q = (s d + d s) + (- 7 ) (d d ) 2 Q is also in H and is a Casimir operator of SU(3), so conserves , . E2: Δ , ) = 0 SU(3) ( J 2 J 1 3 2 B(E2; J + 2 →J)yrast e B 2N J 2N 3 J 4 2 J 3 2 J 5 + B(E2; 2 1+ → 0 1 ) e B2 N 2 N 3 5 2 N for large N B( E 2;4 1 2 1 ) 10 (2 N 2)(2 N 5) B( E 2;2 1 0 1 ) 7 (2 N )(2 N 3) Alaga rule Finite N correction Typ. Of many IBA predictions → Geometric Model as N → ∞ Δ , ) = 0 ( “β ” → γ C o l l e c t i v e B ( E 2 ) s Another example of finite boson number effects in the IBA 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?? B(E2) Mid-shell N2 ~N N !!! Signatures of SU(3) Signatures of SU(3) E =E B( g) Z 0 0 B( B( g) g) E( -vib ) 1/6 (2N - 1) O(6) Axially asymmetric nuclei (gamma-soft) Transition Rates Q = (s† d + d†s) T(E2) = eBQ O(6) E2 Note: Uses Δσ = 0 Δτ = 1 J B(E2; J + 2 → J) = e 2B N - N 2 B(E2; 21 → 01) ~ e 2B J + 1 J 2 4 2 J + 5 N N + 4 5 N2 Consider E2 selection rules Δσ = 0 0+(σ = N - 2) – No allowed decays! Δτ = 1 0+( σ = N, τ = 3) – decays to 2 2 , not 2 1 χ= o 196Pt: Best (first) O(6) nucleus -soft Xe – Ba O(6) - like Classifying 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.