Baryon in the sextet gauge model Zoltan Fodor, Kieran Holland, - PowerPoint PPT Presentation

Baryon in the sextet gauge model Zoltan Fodor, Kieran Holland, Julius Kuti, Santanu Mondal, Daniel Nogradi, Chik Him Wong Lattice Higgs Collaboration ( L at HC )

Composite Higgs Model • Strongly coupled gauge theory. • Electroweak symmetry breaking is realized by spontaneous chiral symmetry breaking; and three Goldstone bosons are eaten up to give three massive gauge bosons ( W ± , Z 0 ). • Higgs is a composite particle of vacuum quantum numbers 0 ++

Possible dark matter candidates Electroweak singlet Goldstone boson → Possible for the models with more than three PNGBs. Weak singlet ⇒ can be light. N f = 2 pseudo-real SU(2) gauge model: Flavour symmetry group is SU(2 N f ) Most attractive channel breaks SU(4) − → Sp(4) ⇒ 5 Goldstone bosons. One of these can be a DM candidate (Lewis,Pica,Sannino in PRD 85, 014504 (2012)).

Dark Baryon Stable, electrically neutral, neutron like state, but has weak hypercharge. ⇒ Needed to be heavy enough from experimental constraint N f = 2 SU(3) gauge theory with fermions in sextet representation: Symmetry breaking: SU ( 2 ) × SU ( 2 ) − → SU ( 2 ) : 3 PNGBs → three massive electroweak gauge bosons.

Why N f = 2, SU(3) Sextet Gauge model? Minimal realization of composite Higgs mechanism. Walking behaviour L at HC : Phys. Lett. B 718, 657, PoS of LATTICE 2014; Kogut and Sinclair, Phys. Rev. D 84, 074504 (2011), Phys. Rev. D 81, 114507 (2010) Expectedly low S parameter S ∼ N ( N +1) . N f 2 2 Estimate from resonance spectrum shows it is not QCD like (T. Appelquist and F. Sannino, Phys. Rev. D 59 , 067702 (1999) [hep-ph/9806409] ) .

Symmetry braking: SU (2) R × SU (2) L − → SU (2) V Exactly three Goldstone modes − → eaten up to give the three massive gauge bosons. Can give a light composite scalar state with Higgs quantum numbers (0 ++ ) L at HC : PoS LATTICE 2013, 062 (2014); Z. Fodor, PoS of LATTICE 2014

Is neutral baryon stable? Charge assignments: → +2 u − 3 → − 1 d − 3 Sextet neutron udd : electrically neutral Lightest because of elctromagnetic correction Stable .....unlike QCD

Constructing nucleon operator in continuum Color singlet: 6 ⊗ 6 ⊗ 6 = 1 ⊕ 2 × 8 ⊕ 10 ⊕ 10 ⊕ 3 × 27 ⊕ 28 ⊕ 2 × 35 (1) → Only one singlet possible. Ψ ≡ (Ψ 0 , Ψ 1 ,..., Ψ 5 ) a vector in the sextet representation of SU(3) T ′ T ABC Ψ A Ψ B Ψ C aa ′ bb ′ cc ′ ψ aa ′ ψ bb ′ ψ cc ′ ≡ ε abc ε a ′ b ′ c ′ ψ aa ′ ψ bb ′ ψ cc ′ = (2) → ψ ′ aa ′ = U ab U a ′ b ′ ψ bb ′ ψ aa ′ − (3) Then ε abc ε a ′ b ′ c ′ ψ aa ′ ψ bb ′ ψ cc ′ → ε abc ε a ′ b ′ c ′ U ad U a ′ d ′ U be U b ′ e ′ U cf U c ′ f ′ ψ dd ′ ψ ee ′ ψ ff ′ − = ε def ε d ′ e ′ f ′ det U det U ψ dd ′ ψ ee ′ ψ ff ′ = ε def ε d ′ e ′ f ′ ψ dd ′ ψ ee ′ ψ ff ′ (4) since det U = 1.

→ Singlet → T ABC symmetric. Correct J PC ⇒ nucleon operator antisymmetric under exchange of spin indices. ⇒ Symmetric in flavour (Spin Statistics Theorem).

Flavour SU(2) irrep: 2 × 2 × 2 = 1 A +2 × 2 M +4 s (5) Thus sextet nucleon belongs to 2 M irrep. An example: Tritium isotope H 3 with pnn or the Helium isotope He 3 with ppn as baryon ground states. Color singlet contituents ⇒ spin-flavour structure will be similar as of sextet nucleon. This comes from a Slater determinant combining mixed representations of permutations.

sextet baryon in quark language In quark language our two fermions have SU(2) flavor symmetry and eight states can be formed: uuu, uud, udu, udd, duu, dud, ddu, ddd They are groupped into an isospin quadruplet and two isospin doublets. The quadruplet belongs to the symmetric rep. � 3 2 , 3 � � = uuu � 2 √ � 3 2 , 1 � � = ( uud + udu + duu ) / 3 � 2 √ � 3 2 , − 1 � � = ( ddu + dud + udd ) / 3 � 2 � 3 2 , − 3 � � = ddd � 2

We also have two doublets which have mixed symmetries: � 1 2 , 1 � � = − (2 ddu − udd − dud ) / sqrt 6 � 2 � 1 2 , − 1 � � = (2 uud − udu − duu ) / sqrt 6 � 2 where the mixed symmetry means symmetry under 1 → 2 and 2 → 1 and no definite symmetry under 1 → 3. The other doublet: � 1 2 , 1 � � = ( udd − dud ) / sqrt 2 � 2 � 1 2 , − 1 � � = ( udu − duu ) / sqrt 2 � 2 anti-symmetric under 1 → 2 and no symmetry under 1 → 3. From the combination of the two mixed reps it is possible to construct an anty-symmetric spin-flavor wave function.

Nucleon operator in lattice with staggered fermion A ( x ) [ u β j 5 ) ij d γ j B α i ( x ) = T ABC u α i B ( x ) ( C γ 5 ) βγ ( C ∗ γ ∗ C ( x )] Looking for a operator as local as possible. Staggered fields: u α i = 1 Γ α i 8 ∑ η χ u ( η ) η where η ≡ ( η 1 , η 2 , η 3 , η 4 ), Γ( η ) = γ η 1 1 γ η 2 2 γ η 3 3 γ η 4 4 . − 1 η C γ 5 Γ T η ) χ B u ( η ′ ) χ C 8 2 ∑ Tr ( C γ 5 Γ ′ Diquark ≡ [ ... ] = d ( η ) ηη ′ − 1 δ ηη ′ S ( η ) χ B u ( η ′ ) χ C 8 2 ∑ = d ( η ) ηη ′ − 1 8 2 ∑ S ( η ) χ B u ( η ) χ C = d ( η ) , S ( η ) is a sign factor η

Diquark populates 16 corner of the hypercube. Writing the third quark in staggered basis: 1 u ( η ′ ) ∑ B α i ( x ) = − T ABC 8 3 ∑ Γ α i η ′ χ A S ( η ) χ B u ( η ) χ C d ( η ) η ′ η To make the operator confined in a single time-slice an extra term has to be added to or subtracted from the diquark part. → Similar to the construction of single time-slice staggered meson operator. → corresponds to the parity partner. 1 u ( η ′ ) ∑ B α i ( x ) = − T ABC Γ α i η ′ χ A S ( η ) χ B u ( η ) χ C 8 3 ∑ d ( η ) η η ′ η ≡ ( η 1 , η 2 , η 3 ) , η ′ ≡ ( η ′ 1 , η ′ 2 η ′ 3 )

Hence B α i ( x ) is sum of 64 terms with proper sign. Local terms vanish individually when contracted with T ABC in color-space → Different from normal QCD where the color contraction tensor is ε abc , antisymmetric. The next simple type of terms is diquark sitting one of the 8 corners and the third quark in any other corner. We use operators of this type for our pilot calculation.

Operators used IV z VIII y x Table: Operator set a Label Operators IV xy χ u (1 , 1 , 0 , 0) χ u (0 , 0 , 0 , 0) χ d (0 , 0 , 0 , 0) IV yz χ u (0 , 1 , 1 , 0) χ u (0 , 0 , 0 , 0) χ d (0 , 0 , 0 , 0) χ u (1 , 0 , 1 , 0) χ u (0 , 0 , 0 , 0) χ d (0 , 0 , 0 , 0) IV zx χ u (1 , 1 , 1 , 0) χ u (0 , 0 , 0 , 0) χ d (0 , 0 , 0 , 0) VIII

IV z VIII y x Table: Operator set b Label Operators χ u (0 , 0 , 0 , 0) χ u (1 , 1 , 0 , 0) χ d (1 , 1 , 0 , 0) IV xy χ u (0 , 0 , 0 , 0) χ u (0 , 1 , 1 , 0) χ d (0 , 1 , 1 , 0) IV yz χ u (0 , 0 , 0 , 0) χ u (1 , 0 , 1 , 0) χ d (1 , 0 , 1 , 0) IV zx χ u (0 , 0 , 0 , 0) χ u (1 , 1 , 1 , 0) χ d (1 , 1 , 1 , 0) VIII

Small ensemble test with different operators 3 x64, β = 3.20, m = 0.007, t max = 20 Vol. = 32 Set a VIII IV_xy 0.65 IV_yz IV_zx 0.6 M N 0.55 0.5 0.45 11 12 13 14 15 Set b 0.65 0.6 M N 0.55 0.5 0.45 11 12 13 14 15 t min Figure: Comparing nucleon mass obtained by different operators.

Chiral extrapolation Preliminary 3 X64, β = 3.200 Vol. = 32 M N = a + b m, a=0.33(2), b=38(2) 2 /dof = 0.5 χ 0.6 0.5 M N 0.4 0.3 0.2 0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 m Figure: Chiral extrapolation

Hadron-Spectrum so far M = M 0 + C m M N 0.6 M a 1 M ρ 0.5 M a 0 M π 0.4 M 0.3 0.2 0.1 0 0 0.002 0.004 0.006 0.008 m Figure: Hadron masses versus quark mass

Direct Detection PHYSICAL REVIEW D 88, 014502 (2013) 10 4 10 2 10 0 Rate, event / (kg · day) 10 − 2 10 − 4 10 − 6 10 − 8 10 − 10 10 − 12 N f = 2 N f = 6 10 − 14 XENON100 [1207.5988], 95% CL exclusion 10 − 16 10 − 2 10 − 1 10 0 10 1 10 2 M χ = M B [TeV]

Conclusion and outlook • The value of nucleon mass in sextet gauge model, from our preliminary calculation is 0.33(2) in lattice unit, which is 3193 ± 167 GeV when converted to physical unit. • We also have ensembles on 40 3 × 80 and 48 3 × 96, and also at a finer lattice spacing 3 . 25, thus more systematic studies can be done. • Construction of operators with no mixing in taste space is needed for more precise calculation. • Calculations of magnetic moment and electric charge radius are needed to compare with direct detection experiments.