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Symmetries Galore

Symmetries Galore. R. F. Casten Yale CERN, August, 2014. “Not all is lost inside the triangle” (A. Leviatan , Seville, March, 2014). Themes and challenges of Modern Science Complexity out of simplicity -- Microscopic

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Symmetries Galore

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  1. Symmetries Galore R. F. Casten Yale CERN, August, 2014 “Not all is lost inside the triangle” (A. Leviatan, Seville, March, 2014)

  2. Themes and challenges of Modern Science • Complexity out of simplicity -- Microscopic • How the world, with all its apparent complexity and diversity can be constructed out of a few elementary building blocks and their interactions • Simplicity out of complexity – Macroscopic • How the world of complex systems can display such remarkable regularity and simplicity What is the force that binds nuclei? Why do nuclei do what they do? What are the simple patterns that nuclei display and what is their origin ? We will look at a model that lives in both camps

  3. The macroscopic perspective (many-body quantal system) often exploits the idea of symmetries These describe the basic structure of the object. Geometrical symmetries describe the shape. Symmetry descriptions are usually analytic and parameter-free ( except for scale). (Scary group theory comment.) One model, the Interacting Boson Approximation (IBA) model, is expressed directly in terms of symmetries. Unfortunately, very few physical systems, especially in atomic nuclei, manifest a symmetry very well. We will see that this statement is now being greatly modified and symmetries may play a much larger role than heretofore.

  4. Simple Observables - Even-Even Nuclei 2+ 1000 Relative B(E2) values 4+ 330 100 2+ 0 0+ Jπ E (keV)

  5. 8+ 6+ 4+ 2+ 0+ Example of a geometrical symmetry – ellipsoidal, axially symmetric nuclei Nuclei that are non-spherical can rotate (1952)!Rotational energies follow the quantum symmetric top: E(J)Rot.= (ħ2/2I )J(J+1) E(0)Rot. 0, E(2)Rot. 6, E(4)Rot. 20 R4/2= 3.33

  6. Amplifies structural differences Centrifugal stretching Deviations 6+690 4+330 2+100 0+0 JE (keV) Identify additional degrees of freedom Value of paradigms Paradigm Benchmark 700 333 100 0 Rotor (ħ2/2I ) J(J + 1) ? Without rotor paradigm

  7. 8+ 6+ 4+ 2+ 0+ Rotational states built on (superposed on) vibrational modes So, identification of the characteristic predictions of a symmetry both tells us the basic nature of the system and can highlight specific deviations from the perfect symmetry which can point to additional degrees of freedom. Real life example Vibrational excitations Rotor E(I)  (ħ2/2I )I(I+1) R4/2= 3.33 Rotational states Ground or equilibrium state

  8. The IBA – A collective model built on a highly truncated shell model foundation(only configurations with pairs of nucleons coupled to states with angular momentum 0 (s bosons) or 2 (d bosons)Embodies the finite number of valence nucleons. Like the shell model but opposed to traditional collective models, the predictions depend on N and Z

  9. Shell Model Configurations Fermion configurations The IBA Boson configurations (by considering only configurations of pairs of fermions with J = 0 or 2.) s bosons and d bosons as the basic building blocks of the collective states Roughly, gazillions !! Need to simplify Huge truncation of the shell model

  10. Symmetries of the IBA U(5) vibrator Sph. SU(3) rotor O(6) γ-soft s and d bosons: 6-Dim. problem U(6) Magical group theory stuff happens here R4/2= 2.5 Three Dynamic symmetries, nuclear shapes What are these symmetries? Idealized structures whose predictions follow analytic formulas, with states labeled by good quantum numbers. Def. R4/2= 2.0 R4/2= 3.33 IBA Symmetry Triangle

  11. Proliferation of Symmetries O(6) PDS – SU(3) QDS PDS QDS AoR However, most nuclei do not exhibit the idealized symmetries: So, is their role just as benchmarks? Most nuclei lie inside the triangle where chaos and disorder are thought to reign. ENTER PDS,QDS: Recent work (Leviatan, Van Isacker, Alhassid, Bonatsos, Cejnar, Pietralla, Cakirli, rfc, ..): symmetries not limited to vertices: triangle permeated by symmetry elements: AoR, QDS, PDS

  12. PDS, QDS: what are these things? They are various situations in which some of the features of a Dyn.Sym. persist even though there is considerable symmetry-breaking. PDS: Some of the levels have the pure symmetry [such as SU(3)] and others are severely mixed QDS: Some of the degeneracies characteristic of a symmetry persist and some of the wave function correlations persist.

  13. . . . Typical SU(3) Scheme (for N valence nucleons) . (b,g) vibrations • Characteristic signatures: • Degenerate bands within a group • Vanishing B(E2) values between groups (l,m) SU(3) O(3) What do real nuclei look like – what are the data??

  14. (b,g) vibrations Totally typical example Similar in many ways to SU(3). But note that the excited (b,g) vibrational excitations are not degenerate as they should be in SU(3). Also there are collective B(E2) values from the g band to the ground band. Most deformed rotors are not SU(3).

  15. Clearly, SU(3) is severely broken Or so we thought

  16. B(E2) values in deformed rotor nuclei (typical values) b 12 Wu 2 Wu 300 Wu So, “clearly”, this violates the predictions of SU(3). So, realistic calculations have broken the symmetries (mixed their basis states). The degeneracies, quantum numbers, selection rules of the symmetry no longer apply. …. Or so we thought

  17. What is an SU(3)-PDS?A bit weird!! It is based on the IBA SU(3) symmetry BUT breaks it while retaining pure SU(3) symmetry ONLY the ground and gbands which preserve SU(3) exactly. All other states are severely mixed! Why would we need such a thing? We have excellent fits to the data with numerical IBA calculations that break SU(3). Why would we think g band preserves SU(3)?

  18. Partial Dynamical Symmetry (PDS) SU(3) (l,m) (l,m) (l,m) PDS: ONLY g and ground bands are pure SU(3). (l,m) So, expect PDS to predict vanishing B(E2) values between these bands as in SU(3). How can that possibly work since empirically these B(E2) values are collective !? BUT, g to ground B(E2)s CAN be finite in the PDS

  19. SU(3) PDS: B(E2) valuesg, ground states pure SU(3), others mixed. Generalized T(E2) T(E2)PDS = e [ A (s+d + d+s) + B (d+d)] = e [T(E2)SU(3) + C (s+d + d+s) ] First term gives zero from SU(3) selection rules. But not second. Hence, relative g to ground B(E2) values: M(E2: Jg – Jgr) eC < Jgr | (s+d + d+s) | Jg > -------------------- = ------------------------------------ eC cancels M(E2: Jg – J’ gr) eC < J’gr | (s+d + d+s)| Jg > Relative INTERband g to gr B(E2)’s are parameter-free! So, Leviatan:PDS works well for 168Er (Leviatan, PRL, 1996). Accidental or a new paradigm? Tried extensive test for 47 Rare earth nuclei. Key data will be relative g to ground B(E2) values.

  20. Illustrative results

  21. Transitional nuclei

  22. Lets look into these predictions and comparisons a little deeper. Compare to “Alaga Rules” – what you would get for a pure rotor for RATIOS of squares of transition matrix elements [B(E2) values] from one rotational band to another. 5:100:70 Alaga

  23. 168-Er: The Alaga rules, valence space, and collectivity PDS has pure g, gr band K, so why differ from Alaga? Ans: PDS (from IBA) is valence space model: predictions are Nval– dep. PDS differs from Alagasolely due to N-Depeffects that arise from the fermion underpinning (Pauli Principle). Valence collective models that agree with the PDS have minimal mixing. Predictions deviating from the PDS signal configuration mixing.

  24. Using the PDS to better understand collective model calculations (and collectivity in nuclei) • PDS B(E2: g – gr): the sole reason they differ from the Alaga rules is that they take into account the finite number of valence nucleons. • Therefore, IBA calculations (like WCD) that agree with the PDS differ from the Alaga rules purely because of finite valence nucleon number effects. Not heretofore recognized. • IBA – CQF deviates further from the Alaga rules, agrees better with the data (and has one fewer parameter). • IBA – CQF: The differences from the PDS are due to mixing • Can use the PDS to disentangle valence space from mixing!

  25. Now a highly selective O(6) PDS Again, as one enters the triangle expect vertex symmetries to be highly broken. Consider structures descending from O(6)

  26. (s,t)

  27. Calculate wave functions throughout triangle, expand in O(6) basis sgood along a line descending from O(6) Remnants of pure O(6) (for the ground state band only) persist along a line in the triangle extending into the region of well-deformed rotational nuclei

  28. Order, Chaos, and the Arc of regularity

  29. Order and chaos? What happens generally inside the triangle 6+, 4+, 3+, 2+, 0+ 4+, 2+, 0+ 2+ 0+ 3 2 1 0 nd Arc of Regularity is a narrow zone in the triangle with high degree of order amidst regions of chaotic behavior. What is going on along the AoR? Whelan, Alhassid, ca 1989

  30. Role of the arc as a “boundary” between two classes of structures AoR – E(0+2) ~ E(2+2) b < g b < g Not just a theoretical curiosity: 8 nuclei in the rare earth region have been found to lie along the arc

  31. All SU(3) degeneracies and all analytic ratios of 0+bandhead energies persist along the Arc. The symmetry underlying the Arc of Regularity is an SU(3)-based Quasi Dynamical Symmetry QDS

  32. Summary O(6) PDS – SU(3) QDS PDS QDS AoR Elements of structural symmetries abound throughout the triangle

  33. Principal Collaborators: R. BurcuCakirli D. Bonatsos Klaus Blaum Aaron Couture Thanks to Ami Leviatan, Piet Van Isacker, Michal Macek, Norbert Pietralla, C. Kremer for discussions.

  34. BACKUPS

  35. SU(3) (l,m) (l,m) (l,m) • States labelled by quantum numbers • Degenerate bands within irrep (l,m) (l,m) T(E2) = eQ = e[(s† d + d †s) - (d † d )(2)]

  36. Creation and destruction operators as “Ignorance operators” Example: Consider the case we have just discussed – the spherical vibrator. Why is the B(E2: 4 – 2) = 2 x B(E2: 2– 0) ?? Difficult to see with Shell Modelwave functions with 1000’s of components However, as we have seen, it is trivial using destruction operatorsWITHOUT EVER KNOWING ANYTHING ABOUT THE DETAILED STRUCTURE OF THESE VIBRATIONS !!!! These operators give the relationships between states. 4,2,0 2E 2 E 2 1 0 0

  37. Finite Valence Space effects on Collectivity PDS B(E2: g to Gr) Alaga Note: 5 N’s up, 5 N’s down: 1 with N Most dramatic, direct evidence for explicit valence space size in emerging collectivity R.F. Casten, D.D. Warner and A. Aprahamian , Phys. Rev. C 28, 894 (1983).

  38. ~ 0.03 ~ 0.1 x ~0.03 ~ 0.003

  39. Concepts of group theory 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

  40. 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).

  41. 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

  42. H’ = aN + b d†d = a N + b nd 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

  43. Concept of a Dynamical Symmetry • Each successive term: • Introduces a new sub-group • A new quantum number to label the states described by that group • Adds an eigenvalue term that is a function of the new quantum number, hence • Breaks a previous degeneracy N Spectrum generating algebra !!

  44. Now, WHY are spin increasing transitions so small in Alaga rules and so much larger in the data than in the PDS? 2g 2, 0 DR = 4 !! Jtot Jint , ~R 0, 4 4 ~ Forbidden 0, 2 2 0, 0 0 Jtot Jint , R Deviations from Alaga: g – gr. band mixing. E2 matrix element : ME(E2) = [ a < || >unpert + b < || > mix ] Can be larger or smaller than the unperturbed. BUT, if unperturbed is forbidden, then ME(E2) = [b < | > mix] > unpert (which is zero). So mixing always increases forbidden/weak transitions.

  45. 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).

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