Symmetrization problem in a quadrupole-octupole collective - PowerPoint PPT Presentation

Symmetrization problem in a quadrupole-octupole collective approaches Artur Dobrowolski , ... Institute of Physics, Dept. Math. Phys.,UMCS, Lublin, Poland Kazimierz Dolny, 2011 1 / 30

Table of contents Intrinsic frame Surface collective variables Intrinsic groups Uniqueness of quantum states 2 / 30

Collaboration z , Andrzej G´ o´ zd´ IF UMCS, Lublin, Poland Jerzy Dudek , IPHC/IReS, Strasbourg, France Agnieszka Szulerecka , IF UMCS, Lublin, Poland edrak , Aleksandra P¸ IF UMCS, Lublin, Poland Katarzyna Mazurek IFJ, Krak´ ow, Poland 3 / 30

Idea of intrinsic frame It is expected that the intrinsic frame: • excludes contributions from global translational motion; • introduces rotational degrees of fredom explicitely; • shows intrinsic symmetries of nuclei; • separates different kinds of intrinsic motions (sometimes), see, e.g. Bohr Hamiltonian in quadrupole a 0 , a 2 variables. 4 / 30

Surface collective variables The equation of nuclear surface in the laboratory frame is: � � � ( α lab λµ ) ⋆ Y λµ ( θ, φ ) R ( θ, φ ) = R 0 1 + λµ α lab λµ are spherical tensors in respect to SO(3) . The equation of nuclear surface has the same form after the transformation to the intrinsic frame: � α lab D λ µ ′ µ (Ω) α lab λµ → α λµ = λµ ′ µ ′ ( θ, φ ) → ( θ w , φ w ) , where new angles are measured in respect to the intrinsic frame α λµ are spherical tensors in respect to SO(3) . 5 / 30

Intrinsic groups G Jin-Quan Chen, Jialun Ping & Fan Wang: Group Representation Theory for Physicists, World Scientific, 2002. Def. For each element g of the group G , one can define a corresponding operator g in the group linear space L G as: gS = Sg , for all S ∈ L G . The group formed by the collection of the operators g is called the intrinsic group of G . IMPORTANT PROPERTY: [ G , G ] = 0 6 / 30

Intrinsic group G Action of the rotation intrinsic group ¯ g ∈ SO(3) . Transformations of coordinates: λµ ) ′ = ¯ ( α lab g α lab λµ = α lab λµ ( α λµ ) ′ = ¯ � D λ µ ′ µ ( g − 1 ) α λµ ′ g α λµ = µ ′ Ω ′ = ¯ g Ω = Ω g . Action in the space of functions of intrinsic variables: g − 1 Ω) ¯ g ψ ( α λµ , Ω) = ψ (¯ g α λµ , ¯ 7 / 30

Intrinsic vs laboratory frames One may introduce the intrinsic frame which is tightly associated with a nucleus. This frame always follows the nucleus. In order to describe the nuclear dynamics we need the Euler angles which determine mutual orientation of the intrinsic vs laboratory frame. The transition from the intrinsic to laboratory frame λµ = � λ � α lab ν = − λ D λ ∗ µν (Ω) α λν f k ( α λµ , Ω) = 0 , k = 1 , 2 , 3 (e.g. in a pure quadrupole case, ( λ = 2), the conditions f k ( α λµ , Ω) = 0 are given explicite as α 22 = α 2 − 2 ∈ R , α 21 = α 2 − 1 = 0 set the intrinsic frame allong the elipsoid principal axes) 8 / 30

Uniqueness of quantum states In practice, the transformation to intrinsic frame is not reversible because of insufficient number of required conditions, an important problem is to discover this ambiguity. One needs to construct the group of transformations which fulfill the condition h ∈ G s : h → (( α λν ) ′ , Ω ′ ) ( α λν , Ω) − which leave invariant the corersponding laboratory coordinates: α lab λµ (( α λν ) ′ , Ω ′ ) = α lab λµ ( α λν , Ω) If Ψ( α λµ , Ω) = Ψ( α lab λµ ) then Ψ(( α λµ ) ′ , Ω ′ ) = Ψ( α lab λµ ) 9 / 30

Uniqueness of quantum states If Ψ( α λµ , Ω) = Ψ( α lab λµ ) then Ψ(( α λµ ) ′ , Ω ′ ) = Ψ( α lab λµ ) CONTRADICTION; generally while working in the intrinsic frame, for most of square integrable functions Ψ( α λµ , Ω) � = Ψ(( α λµ ) ′ , Ω ′ ). The group G s is the SYMMETRIZATION GROUP. The symmetrization condition for states. For all ¯ h ∈ G s : ¯ h Ψ( α λµ , Ω) = Ψ( α λµ , Ω) 10 / 30

The symmetrization group Let us consider the standard quadrupole case of the collective variables α 20 , α 22 , Ω. This definition of intrinsic variables requires 3 conditions α 2 ± 1 = 0 and α 2 − 2 = α 22 ∈ R . These requirements give the following set of equations: D 2 0 , ± 1 ( g − 1 ) = 0 D 2 − 2 ± 1 ( g − 1 ) + D 2 2 ± 1 ( g − 1 ) = 0 D 2 02 ( g − 1 ) − D 2 0 , − 2 ( g − 1 ) = 0 D 2 − 2 , − 2 ( g − 1 ) + D 2 2 , − 2 ( g − 1 ) D 2 − 22 ( g − 1 ) + D 2 22 ( g − 1 ) . = 11 / 30

The symmetrization group The symmetrization group is: g ∈ O h Because the quadrupole variables are invariant in respect to the space inversion, the operation C i belongs to the symmetrization group. The generators of the group O h : R 1 ≡ ¯ R (0 , π, 0) : ( β, γ ) → ( β, γ ) R 2 ≡ ¯ R (0 , 0 , π/ 2) : ( β, γ ) → ( β, − γ ) R 3 ≡ ¯ R ( π/ 2 , π/ 2 , π/ 2) : ( β, γ ) → ( β, γ − π/ 3) 12 / 30

E.g. Bohr Hamiltonian H Bohr = ˆ ˆ H vib ;2 ( β, γ ) + ˆ H rot (Ω) + ˆ H vr ( β, γ, Ω) where � 1 � H vib ;2 = 1 ∂β β 4 ∂ ∂ 1 ∂γ sin(3 γ ) ∂ ∂ ˆ ∂γ + β 2 ∂β − + V ( β, γ ) β 2 sin(3 γ ) β 4 2 J 2 H rot = 1 ˆ � k . 2 J k k =1 , 2 , 3 J 2 1 ˆ � sin 2 ( γ − (2 π/ 3) k ) − ˆ k H vr = H rot . 8 β 4 k =1 , 2 , 3 13 / 30

E.g. Bohr Hamiltonian The vibrational hamiltonian has an octahedral symmetry: Sym ( ˆ H vib ;2 ) = O h . The generators of the group O h : R 1 ≡ ¯ R (0 , π, 0) : ( β, γ ) → ( β, γ ) R 2 ≡ ¯ R (0 , 0 , π/ 2) : ( β, γ ) → ( β, − γ ) R 3 ≡ ¯ R ( π/ 2 , π/ 2 , π/ 2) : ( β, γ ) → ( β, γ − π/ 3) H vib ;2 = ˆ ˆ H vib ;2 ( ∂/∂β, ∂/∂γ, β, cos(3 γ )) is a function of O h -invariants. The symmetrization group is here identical to the vibrational Hamiltonian symmetry group. 14 / 30

E.g. Bohr Hamiltonian The rotational hamiltonian has a dihedral symmetry: Sym ( ˆ H rot ) = D 2 h . The generators of the group D 2 h : ¯ J 2 k → J 2 C 2 y : ( β, γ ) → ( β, γ ) , k ¯ J 2 k → J 2 C 2 z : ( β, γ ) → ( β, γ ) , k H rot = ˆ ˆ H rot ( J x , J y , J z ) is a function of D 2 h -invariants. The coupling term has also a dihedral symmetry: Sym ( ˆ H vr ) = D 2 h . 15 / 30

E.g. Bohr Hamiltonian The symmetry group for eigensolutions of the Bohr hamiltonian (weak vib.-rot. coupling): ˆ ˆ ˆ H Bohr = H vib ;2 + H rot ↓ ↓ ↓ G H = O h × D 2 h ↓ ↓ Γ v Γ r In general, using the spectral theorem one can always decompose the Hamiltonian into a series of the subhamiltonians of given symmetry ˆ � ˆ � H = H G = E G | Ψ G �� Ψ G | G G 16 / 30

Basis of O h group built of rotational functions √ r J 2 J + 1 D J MK (Ω) ∗ MK (Ω) = 1 r (+) J ( r J MK + r J MK (Ω) = M − K ) , K ≥ 0 � 2(1 + δ K 0 ) 1 r ( − ) J ( r J MK − r J MK (Ω) = √ M − K ) , K > 0 2 R J M A B (Ω), A - irrep. of O h , B-irrep. of D 2 . A 1 A 1 K =0 (Ω) = r (+)0 R J =0 M =0 J = 0 , (Ω) 00  T 1 A 1 K =0 (Ω) = r (+)1 R J =1 M M 0 (Ω)   T 1 B 1 K =1 (Ω) = r ( − )1 J = 1 , R J =1 M M 1 (Ω) T 1 B 3 K =1 (Ω) = r (+)1 R J =1 M  M 1 (Ω)  17 / 30

The schematic model The quadrupole+octupole model Hamiltonian: H = ˆ ˆ H vib + ˆ H rot No vib-rot coupling terms ⇒ the eigenfunctions: Ψ Γ JM ν ( α, Ω) = φ Γ J ( α ) R JM ν (Ω) The reduced (in respect to J 3 = M ) matrix elements: � � Ψ Γ ′ J ′ ν ′ || Q lab � φ Γ ′ J ′ | Q λµ | φ Γ J �� R J ′ ν ′ || D λ⋆ λ || Ψ Γ J ν � = · µ || R J ν � µ The reduced B ( E λ ) probability: λ || Γ J ν �| 2 / (2 J + 1) B ( E λ ; (Γ J ν ) → (Γ ′ J ′ ν ′ )) = |� Γ ′ J ′ ν ′ || Q lab 18 / 30

Another group of symmetrization Another choice of intrinsic variables ( α 20 , α 21 , Ω): ( α λ − µ = ( − 1) µ α ∗ α 2 ± 2 = 0 and α 21 = − α 2 − 1 ∈ R , λµ ). It leads to the set of equations for allowed rotations and the symmetrization group: D 2 ± 20 ( g ) ∗ = 0 ± 2 , 1 ( g ) ∗ − D 2 D 2 ± 2 , − 1 ( g ) ∗ = 0 10 ( g ) ∗ + D 2 D 2 − 1 , 0 ( g ) ∗ = 0 11 ( g ) ∗ − D 2 − 1 − 1 ( g ) ∗ − D 2 D 2 1 , − 1 ( g ) ∗ D 2 − 11 ( g ) ∗ . = The symmetrization group is: g ∈ D 2 h 19 / 30

G s for quadrupole-octupole model The quadrupole+octupole model. The intrinsic variables ( α 20 , α 21 , { α 3 µ } , Ω). The intrinsic frame defined as: α 22 = α 2 , − 2 and α 2 , ± 1 = 0. The symmetrization group is: g ∈ O 20 / 30

Symmetrization for a quadrupole-octupole model Choice of collective variables: α 32 = α 3 − 2 , α 30 = α 3 ± 3 = α 3 ± 1 = 0 α 22 = α 2 , − 2 , α 2 , ± 1 = 0 . − 3 − 2 ( g ) ∗ + D 3 0 = D 3 − 32 ( g ) ∗ (1) 2 − 2 ( g ) ∗ + D 3 22 ( g ) ∗ = D 3 − 2 − 2 ( g ) ∗ + D 3 D 3 − 22 ( g ) ∗ (2) − 1 − 2 ( g ) ∗ + D 3 0 = D 3 − 12 ( g ) ∗ (3) 0 − 2 ( g ) ∗ + D 3 0 = D 3 02 ( g ) ∗ (4) 1 − 2 ( g ) ∗ + D 3 0 = D 3 12 ( g ) ∗ (5) 3 − 2 ( g ) ∗ + D 3 0 = D 3 32 ( g ) ∗ (6) The symmetrization group is: g ∈ D 4 21 / 30

The symmetrization problem The physical state space: K = { φ ( α, Ω) : g φ = φ, for all g ∈ G s } The collective hamiltonians ˆ H are generally defined in wider space K coll consisted of all functions, not only symmetrized. One needs to restrict ˆ H to the physical subspace ! 22 / 30