Generator Coordinate Method and Symmetries Andrzej G´ o´ zd´ z , ... Institute of Physics, Dept. Math. Phys.,UMCS, Lublin, Poland Kazimierz 2011 1 / 31
Collaboration Artur Dobrowolski , IF UMCS, Lublin, Poland Aleksandra P¸ edrak , IF UMCS, Lublin, Poland Agnieszka Szulerecka , IF UMCS, Lublin, Poland 2 / 31
GCM and GHW equation 1/2 Trial function � | Ψ � = d α f ( α ) | α � O α = set of generator variables, | α � = set of intrinsic generator functions and unknown f ( α )= weight functions. Variational principle δ � Ψ | ˆ H | Ψ � = 0 , where � Ψ | Ψ � = 1 . GHW equation H f ( α ) = E ˆ ˆ N f ( α ) , One needs to solve the integral equation. 3 / 31
GCM and GHW equation 2/2 Notation: The integral Hamilton operator � d α ′ H ( α, α ′ ) f ( α ′ ) , ˆ H f ( α ) ≡ O where the hamiltonian kernel H ( α, α ′ ) ≡ � α | ˆ H | α ′ � . The overlap operator � d α ′ N ( α, α ′ ) f ( α ′ ) , ˆ N f ( α ) ≡ O where the overlap kernel N ( α, α ′ ) ≡ � α | α ′ � . 4 / 31
GCM as a projection method 1/2 The eigenequation of ˆ N ˆ N w n ( α ) = λ n w n ( α ) . The set of w n ( α ) for λ n � = 0 is the basis in the space K w of the weight functions f . It allows to construct the basis in the corresponding many-body space K : The natural states 1 � | ψ n � = √ λ n d α w n ( α ) | α � . O This basis allows to construct the projection operator � P K coll = | ψ n �� ψ n | λ n > 0 5 / 31
GCM as a projection method 2/2 Standard way – solution of the GHW equation: • Solve the overlap operator equation, and find: λ n and w n ( α ). • Construct the natural states. • Compute 1 � � ψ k | ˆ d α d α ′ w k ( α ) ⋆ � α | ˆ H | α ′ � w l ( α ′ ) . √ λ k λ l H | ψ l � = O • Solve the eigenvalue problem � l H kl h l = Eh k . • Construct the weight function 1 � f ( α ) = √ λ n h n w n ( α ) n 6 / 31
Hamiltonian symmetries and GHW equation 1/3 Let G be a symmetry group of the Hamiltonian ˆ H : g − 1 = ˆ g ˆ ˆ H ˆ H Fundamental property ˆ ˆ ⇒ H φ n ( x ) = E n φ n ( x ) H (ˆ g φ n ( x )) = E n (ˆ g φ n ( x )) . Assume the GCM ansatz: � φ n ( x ) = d α f n ( α )Φ 0 ( α ; x ) , O where Φ 0 ( α ; x ) ≡ � x | α � . 7 / 31
Hamiltonian symmetries and GHW equation 2/3 One expects the same property for the GHW equation H ( g f n ( α )) = E n ˆ ˆ N ( g f n ( α )) . Transform the left hand side of the above condition: H gf n ( α ) = [ ˆ ˆ H , g ] f n ( α ) + g ˆ H f n ( α ) = [ ˆ H , g ] f n ( α ) + E n g ˆ N f n ( α ) = [ ˆ H , g ] f n ( α ) + E n [ g , ˆ N ] f n ( α ) + E n ˆ N gf n ( α ) It implies the following condition: Compatibility condition (CC) [ ˆ H , g ] f n ( α ) = E n [ ˆ N , g ] f n ( α ) 8 / 31
Hamiltonian symmetries and GHW equation 3/3 If Practical and sufficient condition (PSC) H g − 1 = ˆ N g − 1 = ˆ g ˆ g ˆ H and N the compatibility condition is always fuflfilled. Are the condition CC and PSC equivalent ? It is an open question. Symmetry in GCM A physical system decribed in the GCM formalism has the symmetry of the Hamiltonian ˆ H if the PSC, or more generally CC condition is fulfilled. 9 / 31
Symmetry group action 1/2 The symmetry group of the Hamiltonian ˆ H is defined in the many-body space K . One needs to find its realization in the space of weight functions K w . A natural symmetry group G action which relates both spaces K and K w : ˆ g | α � = | g α � , for all g ∈ G It implies (the integral should be G -invariant) � � d α f ( g − 1 α ) | α � g | Ψ � = d α f ( α ) | g α � = ˆ O O G action in the weight space gf ( α ) = f ( g − 1 α ) 10 / 31
Symmetry group action 2/2 Using this action: Invariant kernels g − 1 = ˆ g ˆ A ⇔ A ( g α, g α ′ ) = A ( α, α ′ ) . A ˆ ˆ The integral has to be G -invariant. Symmetry conserving GCM: For all g ∈ G N ( g α, g α ′ ) = N ( α, α ′ ) , H ( g α, g α ′ ) = H ( α, α ′ ) . and G compatible intrinsic generating function: g | β � , where ˆ g ′ | β, g � = either | β, g ′ g � or | β, gg ′ � . | β, g � = ˆ 11 / 31
An important example: an abelian group 1/2 Let the symetry group G be an abelian group and | α = ( α 1 , α 2 , . . . , α n ) � = ˆ g ( α = ( α 1 , α 2 , . . . , α n )) |−� Note that � d α ′ �−| ˆ N χ Γ ( α ) = ˆ g ( α ′ ) |−� χ Γ ( α ′ ) = g ( α ) † ˆ G � � d α ′ �−| ˆ d α ′ �−| ˆ g ( α − 1 α ′ ) |−� χ Γ ( α ′ ) = g ( α ′ ) |−� χ Γ ( αα ′ ) G G �� � d α ′ �−| ˆ ˆ N χ Γ ( α ) = g ( α ′ ) |−� χ Γ ( α ′ ) χ Γ ( α ) , G where χ Γ ( α ) are characters of the symmetry group G . 12 / 31
An important example: an abelian group 2/2 To this class belongs a series of problems: Axial symmetry and simplified angular momentum conservation n · ˆ � | α � = exp( − i α� J ) Rotations in the space of number of particles and the particle number conservation | α � = exp( − i φ ˆ N ) and many others. 13 / 31
Remark: GCM as ”restoration” of symmetries 1/2 w ν Γ κ ( g , α ) are eigenstates of ˆ Assume ˜ N and G required symmetry: G = Sym ( ˆ N ) and G � = Sym ( ˆ H ) ⊂ G . 1 � � | ν Γ κ � = √ λ ν Γ g | α � = dg d α ˜ w ν Γ κ ( g , α )ˆ G O � � w ν Γ κ ′ ( α ) P Γ = d α κκ ′ | α � O κ ′ where � w ν Γ κ ′ ( α )∆ Γ κκ ′ ( g ) ⋆ . w ν Γ κ ( g , α ) = dim (Γ) ˜ κ ′ 14 / 31
An important note: GCM as ”restoration” of symmetries 2/2 Set of generator functions � w ν Γ κ ′ ( α ) P Γ κκ ′ | α � , κ ′ where w ν Γ κ ′ ( α ) can be considered as the weight functions, allows to force (”restore”) the required symmetry G of the integral Hamiltonian ˆ H . 15 / 31
Symmetries in the intrinsic frame The most important symmetries are seen ONLY in the INTRINSIC FRAME of the NUCLEUS 16 / 31
Quantum rotations Figure: The rotated body probability spin orientations for the rotator wave functions ψ ∼ D 5 M 2 − D 5 M , − 2 (left) and ψ ∼ D 5 M 3 − D 5 M , − 3 (right) 17 / 31
Microscopic intrinsic frame 1/3 DEF. Intrinsic Frame (Biedernharn, Louck) � f k ( � x 1 + � a ,� x 2 + � a , . . . ,� x A + � a ) = � f k ( � x 1 ,� x 2 , . . . ,� x A ) � f k ( ˆ x 1 , ˆ x 2 , . . . , ˆ x A ) = ˆ R � R � R � R � f k ( � x 1 ,� x 2 , . . . ,� x A ) � R ki � = f k ( � x 1 ,� x 2 , . . . ,� x A ) k ( ∂� , ∂� , ∂� f i f i f i , ) �≡ � 0 , ∂ x n ;1 ∂ x n ;2 ∂ x n ;3 for any i and all n. 18 / 31
Microscopic intrinsic frame 2/3 A natural choice of the microscopic intrinsic frame: The following conditions relate the laboratory x n ; l and intrinsic ( y n ; l , Ω) coordinates: � x n ; l = R CM D kl (Ω − 1 ) y n ; k + l k Def. of rotational variables � l k = R (Ω − 1 ) � f k ( � x 1 , . . . ,� x A ) The center of mass condition 3 A � m n y n ; k = 0 n =1 19 / 31
Microscopic intrinsic frame 3/3 Principal axes frame: Def. of rotation variables Q ( lab ) � D ii ′ (Ω) Q i ′ j ′ ( y ) , D j ′ j (Ω − 1 ) ( x ) = ij i ′ j ′ A � Q ij ( y ) = m n y ni y nj . n =1 Principal axes rotating frame Q ij ( y ) = 0 for i � = j . 20 / 31
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 The groups G and G are antyisomorphic. 21 / 31
Hamiltonian and tranformations 1/3 Hamiltonian in the intrinsic frame H ( x ) → ¯ ˆ H ( y , Ω) = ¯ H ( y , ¯ J x , ¯ J y , ¯ J z ) Possible form: generalized rotor H ( y , ¯ ¯ J x , ¯ J y , ¯ J z ) = ¯ H 0 ( y ) + h 00 ( y ) ¯ J 2 ∞ λ � �� � h λ 0 ( y ) ˆ � h λµ ( y ) ˆ λµ ( y ) ˆ T λ � T λ µ + ( − 1) µ h ⋆ T λ + 0 + , − µ λ =1 µ =1 � λ 2 =2 ⊗ ¯ � λ 3 =3 ⊗ . . . ⊗ ¯ � λ n − 1 = n − 1 ⊗ ¯ � λ = n �� ˆ �� ¯ J ⊗ ¯ T λ µ = . . . J J J J , µ 22 / 31
Hamiltonian and tranformations 2/3 Laboratory rotations R ( ω ) ∈ SO(3) : ˆ ˆ R ( ω ) f ( y , Ω) = f ( y , ω − 1 Ω) Rotational invariance R ( ω )ˆ ˆ H ( x )ˆ R ( ω − 1 ) = ˆ H ( x ) It implies R ( ω )¯ ˆ H ( y , ¯ J x , ¯ J y , ¯ J z )ˆ R ( ω − 1 ) = ¯ H ( y , ¯ J x , ¯ J y , ¯ J z ) 23 / 31
Hamiltonian and tranformations 3/3 Important classes of transformations in the intrinsic frame: 1. Rotation intrinsic group R ( ω ) ∈ SO(3) : ¯ ¯ ω y , Ω ω − 1 ) R ( ω ) f ( y , Ω) = f (¯ It does not change laboratory variables x . 2. Intrinsic transformations of y only g − 1 y , Ω) ˆ g ∈ G vib : ˆ gf ( y , Ω) = f (ˆ 3. Intrinsic transformations of Ω only gf ( y , Ω) = f ( y , Ω g − 1 ) ˆ g ∈ G rot : ˆ 24 / 31
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