Dynamical Generation of Fermion Mixing Luca Smaldone in - PowerPoint PPT Presentation

Dynamical Generation of Fermion Mixing Luca Smaldone in collaboration with Massimo Blasone and Petr Jizba Department of Physics ”E.R.Caianiello” and INFN Sezione di Napoli, Gruppo Collegato di Salerno 29th Indian Summer School Prague, 5th September 2017 1

Contents 1. Neutrino Mixing in QFT 2. Patterns of Dynamical Symmetry Breaking 3. Conclusions and Perspectives 2

Motivations • Flavor mixing is a central ingredient in the Standard Model; • Mixing transformations for fields have been shown to be non–trivial since they induce a condensate structure in the vacuum 1 ; • This suggests the idea of dynamical generation of mixing in a similar way as it happens for the masses. 1 M.Blasone, G.Vitiello (1995). 3

Neutrino Mixing in QFT

Massive neutrinos Mixing transformations defines fields with definite mass ν 1 , ν 2 : ν e ( x ) = cos θ ν 1 ( x ) + sin θ ν 2 ( x ) ν µ ( x ) = − sin θ ν 1 ( x ) + cos θ ν 2 ( x ) with 2 m eµ tan 2 θ = m e − m µ These relations define neutrinos with definite mass. 4

Mass eigenstates Free fields with definite masses can be expanded as: � � � 1 e i k · x , − k ,i ( t ) β r † u r k ,i ( t ) α r k ,i + v r √ ν i ( x ) = i = 1 , 2 − k ,i V k ,r A mass-eigenstate neutrino is defined as: k ,i � = α r † | ν r k ,i | 0 � 1 2 , i = 1 , 2 where the mass vacuum satisfies: α r k ,i | 0 � 1 2 = β r k ,i | 0 � 1 2 , i = 1 , 2 5

Flavor Charges Flavor charges 2 : � � d 3 x ν † d 3 x ν † Q ν e ( t ) = e ( x ) ν e ( x ) , Q ν µ ( t ) = µ ( x ) ν µ ( x ) Total flavor charge: Q = Q ν e ( t ) + Q ν µ ( t ) Because of mixing, only this last is conserved. 2M.Blasone, P.Jizba, G.Vitiello (2001) 6

Standard Flavor eigenstates Flavor eigenstates, are usually taken as a simple combination of mass eigenstates: | ν r cos θ | ν r k , 1 � + sin θ | ν r k ,e � P = k , 2 � | ν r − sin θ | ν r k , 1 � + cos θ | ν r k ,µ � P = k , 2 � ACHTUNG! These are NOT eigenstates of the flavor charges: cos 4 θ + sin 4 θ + 2 | U k | sin 2 θ cos 2 θ < 1 P � ν r k ,e | : Q ν e (0) : | ν r k ,e � P = 2 (1 − | U k | ) sin 2 θ cos 2 θ > 0 P � ν r k ,e | : Q ν µ (0) : | ν r k ,e � P = with U k = u r † 2 , k u r 1 , k . 7

Mixing generator In a finite volume, mixing relations are rewritten G − 1 ν α θ ( t ) ν α e ( x ) = 1 ( x ) G θ ( t ) G − 1 ν α θ ( t ) ν α µ ( x ) = 2 ( x ) G θ ( t ) where mixing generator has been introduced � �� � � d 3 x ν † 1 ( x ) ν 2 ( x ) − ν † G θ ( t ) = exp θ 2 ( x ) ν 1 ( x ) 8

Decomposition of the mixing generator The mixing generator can be decomposed as 3 : G θ = B (Θ 1 , Θ 2 ) R ( θ ) B − 1 (Θ 1 , Θ 2 ) where    � � � � � �  α r † k , 2 + β r † α r † k , 1 + β r † � k , 1 α r − k , 1 β r e iψ k − k , 2 α r − k , 2 β r e − iψ k R ( θ ) ≡ exp  θ − k , 2 − k , 1  k ,r � � Θ k ,i ǫ r � �� α r k ,i β r − k ,i e − iφ k i − β r † − k ,i α r † k ,i e iφ k,i B i (Θ i ) ≡ exp , , i = 1 , 2 k ,r and B ( ϑ 1 , ϑ 2 ) ≡ B 1 ( ϑ 1 ) B 2 ( ϑ 2 ). 3M.Blasone, M.V.Gargiulo, G.Vitiello (2015) 9

B i (Θ k ,i ), i = 1 , 2 are Bogoliubov transformations which induces a mass shift and R ( θ ) is a rotation. Their action on the mass vacuum is: | � B − 1 (Θ 1 , Θ 2 ) | 0 � 1 , 2 0 � 1 , 2 ≡ � � � cos Θ k ,i + ǫ r sin Θ k ,i α r † k ,i β r † = | 0 � 1 , 2 − k ,i k ,r R − 1 ( θ ) | 0 � 1 , 2 = | 0 � 1 , 2 A rotation of fields is not a rotation at the level of creation and annihilation operators. 10

Flavor Vacuum The flavor vacuum is defined by 4 : | 0( t ) � e,µ ≡ G − 1 θ ( t ) | 0 � 1 , 2 In the infinite volume limit: (2 π )3 ln ( 1 − sin 2 θ | V k | 2 ) d 3 k 2 V � V →∞ 1 , 2 � 0 | 0( t ) � e,µ = lim lim = 0 V →∞ e where � k , 2 | 2 � = 0 | v r † | V k | 2 − k , 1 u s ≡ m 1 � = m 2 for r,s 4M.Blasone, G.Vitiello (1995). 11

Vacuum condensate k ,i α k ,i | 0 � e µ = sin 2 θ | V k | 2 , with • Condensation density: e µ � 0 | α r † i = 1 , 2. Same result for antiparticles. for k ≫ √ m 1 m 2 . • | V k | 2 ≃ ( m 2 − m 1 ) 2 4 k 2 12

Flavor eigenstates A neutrino flavor-state can be defined: k ,σ ( t ) � = α r † | ν r σ, k | 0( t ) � e µ , σ = e, µ At any time these are eigenstates of the flavor charges: Q σ ( t ) | ν r k ,σ ( t ) � = | ν r k ,σ ( t ) � Note that | ν r k ,σ ( t ) � � = | ν r k ,σ � P because � � B ( m 1 , m 2 ) , R − 1 ( θ ) � = 0 13

Bogoliubov vs Pontecorvo Bogoliubov and Pontecorvo do not commute! As a result, flavor vacuum gets a non-trivial term: � � | 0 � e,µ ≡ G − 1 B ( m 1 , m 2 ) , R − 1 ( θ ) | � θ | 0 � 1 , 2 = | 0 � 1 , 2 + 0 � 1 , 2 14

Flavor Vacuum and Condensate Structure The flavor vacuum is characterized by a condensate structure: (1 − sin 2 θ | V k | 2 ) − ǫ r sin θ cos θ | V k | ( α r † � � � k , 1 β r † − k , 2 + α r † k , 2 β r † | 0 � e,µ = − k , 1 ) k r + ǫ r sin 2 θ | V k || U k | ( α r † − k , 2 ) + sin 2 θ | V k | 2 α r † � k , 1 β r † − k , 1 − α r † k , 2 β r † k , 1 β r † − k , 2 α r † k , 2 β r † | 0 � 1 , 2 − k , 1 • SU(2) (Perelomov) coherent state. • Condensate structure on vacuum as in systems with SSB (e.g. superfluids, superconductors). • Exotic condensates: mixed pairs due to a non-diagonal Bogoliubov transformation. • Note that | 0 � e µ � = | a � 1 ⊗ | b � 2 ⇒ entanglement. 15

Why is vacuum structure so important? • Dark Energy contribution of the neutrino mixing 5 : � K Λ = 128 π 3 G d k k 2 ( ω k, 1 + ω k, 2 ) | V k | 2 0 If the cut-off is chosen of the order K ∼ √ m 1 m 2 , with m 1 = 7 × 10 − 3 eV and m 2 = 5 × 10 − 2 eV one gets: Λ ∼ 10 − 56 cm − 2 which is compatible with modern experimental upper bounds. 5M.Blasone, A.Capolupo,S.Capozziello, S.Carloni, G.Vitiello, (2005) 16

Patterns of Dynamical Symmetry Breaking

Chiral symmetry Consider the Lagrangian L = i ¯ ψγ µ ∂ µ ψ + U ( ψψ ) where � � ψ I ψ = . ψ II � � U ψψ is assumed to be invariant under chiral transformations U (2) V × U (2) A : g = ( g, g 5 ) , σαγ 5 g = e iω α σα 2 , g 5 = e iω α , α = 0 , 1 , 2 , 3 2 so the entire Lagrangian is invariant as well. 17

Chiral Symmetry Breaking The vector and axial Noether charges are: σ α J α = ψγ µ 2 ψ µ σ α J α = ψγ µ γ 5 2 ψ 5 µ If we add a diagonal mass term L M = − mψψ the conservation law of axial currents is explicitly broken: ∂ µ J α = 0 µ ∂ µ J α = iψ γ 5 m ψ 5 µ 18

Isospin Symmetry Breaking Adding a mass-shift term � � ∆ m 0 L ∆ m = − ψ ψ 0 − ∆ m the Isospin symmetry is broken to U (1) 0 V × U (1) 3 V (the subscript index indicates the generator) ∂ µ J 0 = 0 µ i ∂ µ J 1 = 2∆ mψ σ 1 ψ µ 1 ∂ µ J 2 = 2 ψ ∆ mσ 1 ψ µ ∂ µ J 3 = 0 µ 19

Total Flavor Charge Conservation Finally we add to the Lagrangian, an off-diagonal term � � 0 h L h = − ψ ψ 0 h The current evolution are now ∂ µ J 0 = 0 µ i ∂ µ J 1 = 2( m I − m II ) ψ σ 1 ψ µ − 1 ∂ µ J 2 = 2 ψ [2 hσ 3 − ( m I − m II ) σ 1 ] ψ µ ∂ µ J 3 = − ihψ σ 1 ψ µ Conservation of the total flavor charge Q 0 . 20

Dynamical Generation of Fermion Mixing Dynamical generation of mixing occurs if 6 → U (1) 0 U (2) V × U (2) A − V , at the ground state level. SSB is characterized by the existence of some (quasi)-local operators Φ i so that � Ω | [ Q α (0) , Φ i (0)] | Ω � = � Ω | ϕ α i | Ω � � = 0 , on some dressed vacuum. ϕ α i are called order parameters . We look at order parameters of the form ψ i ψ j ± ψ k ψ l with i, j, k, l = I , II. 6M. Blasone, P. Jizba, L. S., in preparation (2017). 21

Patterns of SSB Symmetry Group Order Parameter Broken Charges U (2) V × U (2) A - - Q α U (2) V � ψ I ψ I + ψ II ψ II � � = 0 5 U (1) 0 V × U (1) 3 Q α 5 ; Q 1 ; Q 2 � ψ I ψ I ± ψ II ψ II � � = 0 V � ψ I ψ I ± ψ II ψ II � � = 0 U (1) 0 Q α 5 ; Q 1 ; Q 2 ; Q 3 V � ψ I ψ II + ψ II ψ I � � = 0 22

Vacuum degeneracy (1) All charges are time independent: [ Q α , H ] = 0 [ Q α 5 , H ] = 0 , α = 0 , 1 , 2 , 3 Degenerate vacua: θ 5 � = e iθ α Q α + iθ 5 ,α Q α | � θ, � 5 | Ω � | Ω � is a fiducial vacuum. 23

Vacuum degenracy (2) Consider a fiducial vacuum | Ω( m ) � , where only � v = � Ω( m ) | ψ j ( x ) ψ j ( x ) | Ω( m ) � � = 0 j =I , II We find � � θ 5 0 | ψ j ( x ) γ 5 ψ j ( x ) | θ 5 0 � = i sin θ 5 0 v j =I , II � θ 5 3 | ψ II ( x ) γ 5 ψ II ( x ) − ψ I ( x ) γ 5 ψ I ( x ) | θ 5 3 � = i sin θ 5 3 v � θ 5 1 | ψ I ( x ) γ 5 ψ II ( x ) + ψ II ( x ) γ 5 ψ I ( x ) | θ 5 1 � = i sin θ 5 1 v � θ 5 2 | ψ I ( x ) γ 5 ψ II ( x ) − ψ II ( x ) γ 5 ψ I ( x ) | θ 5 2 � = sin θ 5 2 v in contrast with our hypothesis. 24