Axion predictions in Grand Unified Theories Planck 2018 Bonn, - PowerPoint PPT Presentation

Axion predictions in Grand Unified Theories Planck 2018 Bonn, 24/05/18 Anne Ernst based on JHEP 02(2018)103 , in collaboration with Andreas Ringwald and Carlos Tamarit

Motivation search for a well-motivated model solving fundamental problems of the Standard Model axion is a good candidate for Cold Dark Matter can we use GUT to constrain the axion mass? unification of gauge couplings one gauge group instead of 3 why SO(10)? simplest SU(5) models: disfavoured neutrinos massive: seesaw mechanism

Status models have been studied before SO (10) × U (1) P Q [Lazarides, Kim, Bajc et al, Babu et al, Altarelli et al, …] however, a few things were missing: a systematic identification of axion field and decay constant in the presence of gauge symmetries a systematic calculation of the couplings to other particles a direct calculation of associated domain wall number two-loop analysis of unification constraints including threshold corrections

GUT model building: non-SUSY SO(10) SO (10) 4 C 2 L 2 R 4 C 2 L 1 R 3 C 2 L 1 R 1 B − L 3 C 2 L 1 Y scale 3 , 2 , 0 , 1 3 , 2 , 1 � � � � 16 F (4 , 2 , 1) (4 , 2 , 0) := Q M Z 3 6 1 , 2 , − 1 � � (1 , 2 , 0 , − 1) := L M Z 2 (¯ � ¯ � ¯ � ¯ 4 , 1 , 1 3 , 1 , 1 2 , − 1 3 , 1 , 1 � � � 4 , 1 , 2) := d M Z 2 3 3 1 , 1 , 1 � � 2 , 1 (1 , 1 , 1) := e M Z � ¯ � ¯ � ¯ 4 , 1 , − 1 3 , 1 , − 1 2 , − 1 3 , 1 , − 2 � � � := u M Z 2 3 3 � 1 , 1 , − 1 � 2 , 1 (1 , 1 , 0) := N M BL • one generation of SM fermions + heavy right- handed neutrinos fits perfectly into one 16 representation of SO(10) • most general Yukawa coupling: � � L Y = 16 F Y 10 10 H + Y 120 120 H + Y 126 126 H 16 F + h . c .,

GUT model building: non-SUSY SO(10) SO (10) 4 C 2 L 2 R 4 C 2 L 1 R 3 C 2 L 1 R 1 B − L 3 C 2 L 1 Y scale 3 , 2 , 0 , 1 3 , 2 , 1 � � � � 16 F (4 , 2 , 1) (4 , 2 , 0) := Q M Z 3 6 1 , 2 , − 1 � � (1 , 2 , 0 , − 1) := L M Z 2 (¯ � ¯ � ¯ � ¯ 4 , 1 , 1 3 , 1 , 1 2 , − 1 3 , 1 , 1 � � � 4 , 1 , 2) := d M Z 2 3 3 1 , 1 , 1 � � 2 , 1 (1 , 1 , 1) := e M Z � ¯ � ¯ � ¯ 4 , 1 , − 1 3 , 1 , − 1 2 , − 1 3 , 1 , − 2 � � � := u M Z 2 3 3 � 1 , 1 , − 1 � 2 , 1 (1 , 1 , 0) := N M BL • one generation of SM fermions + heavy right- handed neutrinos fits perfectly into one 16 representation of SO(10) • most general Yukawa coupling: � � L Y = 16 F Y 10 10 H + Y 120 120 H + Y 126 126 H 16 F + h . c ., minimality

Role of PQ-symmetry in GUT model building complex representation necessary to 10 H reproduce realistic mass relations reduced predictivity in Yukawa sector ⇣ ⌘ Y 10 10 H + ˜ Y 10 10 ∗ L Y = 16 F H + Y 126 126 H 16 F + h . c .

Role of PQ-symmetry in GUT model building complex representation necessary to 10 H reproduce realistic mass relations reduced predictivity in Yukawa sector ⇣ ⌘ Y 10 10 H + ˜ Y 10 10 ∗ L Y = 16 F H + Y 126 126 H 16 F + h . c . solution: impose global U(1) symmetry! 16 F → 16 F e i α U (1) PQ : 10 H → 10 H e − 2 i α 126 H → 126 H e − 2 i α

Role of PQ-symmetry in GUT model building complex representation necessary to 10 H reproduce realistic mass relations reduced predictivity in Yukawa sector ⇣ ⌘ Y 10 10 H + ˜ Y 10 10 ∗ L Y = 16 F H + Y 126 126 H 16 F + h . c . solution: impose global U(1) symmetry! 16 F → 16 F e i α U (1) PQ : 10 H → 10 H e − 2 i α 126 H → 126 H e − 2 i α broken global symmetry Goldstone boson! Color-anomalous symmetry solution to Strong CP -problem

Peccei-Quinn solution U (1) PQ assume existence of anomalous global symmetry, spontaneously broken at a scale f A U (1) P Q : A → A A + ✏ ∈ [0 , 2 π ) Goldstone boson: f A f A f A anomaly induces SU (3) C − SU (3) C − U (1) P Q effective change in the Lagrangian : δ L = − g 2 A G aµ ν ˜ G a µ ν 32 π 2 f A can rewrite CP violating term as ✓ A ◆ L e ff = − g 2 G aµ ν ˜ + ¯ G a θ µ ν 32 π 2 f A | {z } θ

Peccei-Quinn solution non-perturbative effects introduce potential for A ! A 0 minimum at + ¯ θ = θ = 0 f A

Peccei-Quinn solution non-perturbative effects introduce potential for A ! A 0 minimum at + ¯ θ = θ = 0 f A Dynamical solution of strong CP problem ! Particle excitation of field A: the axion.

Peccei-Quinn solution non-perturbative effects introduce potential for A ! A 0 minimum at + ¯ θ = θ = 0 f A Dynamical solution of strong CP problem ! Particle excitation of field A: the axion. Axion properties are described by axion decay constant f A p m A ( T ) f A = χ ( T ) Q: Can we use GUT to constrain ? f A

Peccei-Quinn solution non-perturbative effects introduce potential for A ! A 0 minimum at + ¯ θ = θ = 0 f A Dynamical solution of strong CP problem ! Particle excitation of field A: the axion. Axion properties are described by axion decay constant f A p m A ( T ) f A = χ ( T ) A: It depends… Q: Can we use GUT to constrain ? f A

Higgs sector of SO(10): symmetry breaking chains SO (10) 4 C 2 L 2 R 4 C 2 L 1 R 3 C 2 L 1 R 1 B − L 3 C 2 L 1 Y 3 C 1 em scale (1 , 2 , 1 (1 , 2 , 1 (1 , 2 , 1 10 H (1 , 2 , 2) 2 ) 2 , 0) 2 ) (1 , 0) =: H d M Z (1 , 2 , − 1 (1 , 2 , − 1 (1 , 2 , − 1 2 ) 2 , 0) 2 ) (1 , 0) =: H u M Z 45 H (1 , 1 , 3) (1 , 1 , 0) := σ (1 , 1 , 0 , 0) (1 , 1 , 0) (1 , 0) M PQ 126 H (10 , 1 , 3) (10 , 1 , 1) (1 , 1 , 1 , − 2) (1 , 1 , 0) := ∆ R (1 , 0) M BL (15 , 2 , 1 (1 , 2 , 1 (1 , 2 , 1 (15 , 2 , 2) 2 ) 2 , 0) 2 ) (1 , 0) := Σ d M Z (15 , 2 , − 1 (1 , 2 , − 1 (1 , 2 , − 1 2 ) 2 , 0) 2 ) (1 , 0) := Σ u M Z 210 H (1 , 1 , 1) := φ (1 , 1 , 0) (1 , 1 , 0 , 0) (1 , 1 , 0) (1 , 0) M U cannot break SO(10) down to the Standard Model (as it 126 H leaves an SU(5) subgroup unbroken), we need at least one additional rep choosing , obtain the two-step symmetry breaking chain 210 H M U − 210 H M BL − 126 H M Z − 10 H SO (10) 4 C 2 L 2 R 3 C 2 L 1 Y 3 C 1 em − → − → − →

Higgs sector of SO(10): symmetry breaking chains SO (10) 4 C 2 L 2 R 4 C 2 L 1 R 3 C 2 L 1 R 1 B − L 3 C 2 L 1 Y 3 C 1 em scale (1 , 2 , 1 (1 , 2 , 1 (1 , 2 , 1 10 H (1 , 2 , 2) 2 ) 2 , 0) 2 ) (1 , 0) =: H d M Z (1 , 2 , − 1 (1 , 2 , − 1 (1 , 2 , − 1 2 ) 2 , 0) 2 ) (1 , 0) =: H u M Z 45 H (1 , 1 , 3) (1 , 1 , 0) := σ (1 , 1 , 0 , 0) (1 , 1 , 0) (1 , 0) M PQ 126 H (10 , 1 , 3) (10 , 1 , 1) (1 , 1 , 1 , − 2) (1 , 1 , 0) := ∆ R (1 , 0) M BL (15 , 2 , 1 (1 , 2 , 1 (1 , 2 , 1 (15 , 2 , 2) 2 ) 2 , 0) 2 ) (1 , 0) := Σ d M Z (15 , 2 , − 1 (1 , 2 , − 1 (1 , 2 , − 1 2 ) 2 , 0) 2 ) (1 , 0) := Σ u M Z 210 H (1 , 1 , 1) := φ (1 , 1 , 0) (1 , 1 , 0 , 0) (1 , 1 , 0) (1 , 0) M U cannot break SO(10) down to the Standard Model (as it 126 H leaves an SU(5) subgroup unbroken), we need at least one additional rep choosing , obtain the two-step symmetry breaking chain 210 H M U − 210 H M BL − 126 H M Z − 10 H SO (10) 4 C 2 L 2 R 3 C 2 L 1 Y 3 C 1 em − → − → − →

Physical PQ symmetry U (1) PQ : 16 F → 16 F e i α 10 H → 10 H e − 2 i α 126 H → 126 H e − 2 i α

Physical PQ symmetry linear combination of gauge and global symmetries U (1) PQ : 16 F → 16 F e i α axion is massless at the perturbative level 10 H → 10 H e − 2 i α 126 H → 126 H e − 2 i α orthogonal to all gauge symmetries, in particular B-L lower decay constant! f A ∼ M Z For an explicit construction of the physical axion, check out our paper! visible axion!

Lifting the axion from the electroweak scale Model 1 Model 2 Model 3 extend PQ symmetry extra scalar singlet extra scalar multiplet

M1: extended PQ symmetry U (1) PQ : 16 F → 16 F e i α 10 H → 10 H e − 2 i α 126 H → 126 H e − 2 i α

M1: extended PQ symmetry minimal extension: include U (1) PQ : in the PQ symmetry 16 F → 16 F e i α axion construction similar as before, but now obtain 10 H → 10 H e − 2 i α 126 H → 126 H e − 2 i α 210 H → 210 H e 4 i α

M1: extended PQ symmetry minimal extension: include U (1) PQ : in the PQ symmetry 16 F → 16 F e i α axion construction similar as before, but now obtain 10 H → 10 H e − 2 i α 126 H → 126 H e − 2 i α 3 ⇠ h 210 H i f A ⇠ v U 3 210 H → 210 H e 4 i α fixed by the requirement of gauge coupling unification!

M1: RGE running d α − 1 i ( µ ) = − a i b ij X 2 π − 8 π 2 α − 1 d ln µ j ( µ ) j 50 40 30 20 10 6 10 9 10 12 10 15 (no threshold corrections)

M1: Matching conditions matching conditions depend on the group structure and the contained particles size of threshold corrections depends on the masses of heavy scalars (more specifically, the deviation from the threshold scale) 1 Y ( M BL ) = 3 2 R ( M BL ) + 2 4 C ( M BL ) − λ 1 Y 00 − 1 00 − 1 α − 1 5 α 5 α 12 π 2 L ( M BL ) − λ 2 L 00 − 1 α − 1 2 L ( M BL ) = α G ( M U ) − λ 00 12 π 00 � 1 2 R ( M U ) = α � 1 2 R α 12 π 4 C ( M BL ) − λ 3 C 00 − 1 α − 1 3 C ( M BL ) = α G ( M U ) − λ 00 12 π 00 � 1 2 L ( M U ) = α � 1 2 L α 12 π G ( M U ) − λ 00 00 � 1 4 C ( M U ) = α � 1 4 C α 12 π