Finding Z ′ s Responsible for R K ( ∗ ) by Ben Allanach (University of Cambridge) • Simplified models that explain R K and R K ( ∗ ) • Can we directly discover the Z ′ s? • A less simplified model BCA, Gripaios, You, arXiv:1710.06363 ; BCA, Davighi, arXiv:1809.01158 ; BCA, Corbett, Dolan, You, arXiv:1810.02166 ; BCA, Davighi, Melville, arXiv:1812.04602 Ben Allanach (University of Cambridge) 1
During the 1990s We wanted to be the Grand Architects, searching for the string model to rule them all 2
During the 2010s We are happy with any beyond the Standard Model roof 3
Ben Allanach (University of Cambridge) 4
Simplified Models for c µ LL At tree-level, we have: Ben Allanach (University of Cambridge) 5
B s − ¯ B s Mixing s s ¯ Z ′ ¯ b b M Z ′ < g sb ¯ ∼ L 600 TeV Ben Allanach (University of Cambridge) 6
Simplified Z ′ Models 1 ıve model: only include couplings to ¯ s and µ + µ − Na¨ bs / b ¯ ( less model dependent ). + g µµ L min. g sb L Z ′ sγ ρ P L b + h.c. L Z ′ µγ ρ P L µ , � � ⊃ ρ ¯ ρ ¯ Z ′ which contributes to the O µ LL coefficient with L g µµ 4 πv 2 g sb c µ L ¯ LL = − , α EM V tb V ∗ M 2 ts Z ′ � 2 � 36 TeV L g µµ ⇒ g sb ✭ = ✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭✭ − 1 . 33 ± 0 . 34 − 0 . 93 ± 0 . 24 (clean) . L M Z ′ 1 BCA, Queiroz, Strumia, Sun arXiv:1511.07447 Ben Allanach (University of Cambridge) 7
Simplified Z ′ Models 2 � � L i λ ( Q ) L i λ ( L ) ij γ ρ Q ′ ij γ ρ L ′ Z ′ Q ′ L j + L ′ L Z ′ f = ρ , L j L V d L and PMNS U = V † After CKM mixing of V = V u † ν L V e L , � u L V Λ ( Q ) V † γ ρ u L + d L Λ ( Q ) γ ρ d L + L = � ν L U Λ ( L ) U † γ ρ ν L + e L Λ ( L ) γ ρ e L Z ′ ρ , where Λ ( Q ) ≡ V † Λ ( L ) ≡ V † d L λ ( Q ) V d L , e L λ ( L ) V e L . 2 BCA, Corbett, Dolan, You, arXiv:1810.02166 Ben Allanach (University of Cambridge) 8
Limiting Cases Mixed Up Model: all quark mixing is in left-handed ups 0 0 0 0 0 0 Λ ( Q ) = g bs Λ ( L ) = g µµ 0 0 1 , 0 1 0 , 0 1 0 0 0 0 Mixed Down Model: all quark mixing is in left-handed downs 0 0 0 0 0 0 Λ ( Q ) = g tt V † · Λ ( L ) = g µµ 0 0 0 · V, 0 1 0 , 0 0 1 0 0 0 Ben Allanach (University of Cambridge) 9
⇒ g bs = V ∗ ts V tb g tt ≈ − 0 . 04 g tt : the quark couplings are weaker than the leptonic ones Ben Allanach (University of Cambridge) 10
Ben Allanach (University of Cambridge) 11
Ben Allanach (University of Cambridge) 12
Third Family Hypercharge Model Add complex SM singlet scalar θ and gauged U (1) F : SU (3) × SU (2) L × U (1) Y × U (1) F � θ � ∼ Several TeV SU (3) × SU (2) L × U (1) Y � H � ∼ 246 GeV SU (3) × U (1) em • SM fermion content • anomaly cancellation • 0 F charges for first two generations Ben Allanach (University of Cambridge) 13
Unique Solution F Q ′ i = 0 F u R ′ i = 0 F d R ′ i = 0 F L ′ i = 0 F e R ′ i = 0 F H = − 1 / 2 F Q ′ 3 = 1 / 6 F u ′ R 3 = 2 / 3 F d ′ R 3 = − 1 / 3 F L ′ 3 = − 1 / 2 F e ′ R 3 = − 1 F θ � = 0 ′ L Ht ′ 3 L H c b ′ L H c τ ′ ′ R + Y b Q ′ L = Y t Q 3 R + Y τ L 3 R + H.c., • First two families massless at renormalisable level • Their masses and fermion mixings generated by small non-renormalisable operators This explains the hierarchical heaviness of the third family and small CKM angles Ben Allanach (University of Cambridge) 14
Z − Z ′ mixing angle � 2 � M Z g F sin α z ≈ ≪ 1 . g 2 + g ′ 2 M ′ � Z This gives small non-flavour universal couplings to the Z boson propotional to g F and: − sin θ w B µ + cos θ w W 3 � � Z µ = cos α z + sin α z X µ , µ Ben Allanach (University of Cambridge) 15
Example Case Take a simple limiting case: V u L = 1 ⇒ V d L = V , the CKM matrix. V u R = V d R = V e R = 1 for simplicity and the ease of passing bounds. 1 0 0 1 0 0 V d L = 0 cos θ sb − sin θ sb , V e L = 0 0 1 , 0 sin θ sb cos θ sb 0 1 0 V e R = 1 ⇒ V ν L = V e L U † , where U is the PMNS matrix. 16
Ben Allanach (University of Cambridge) 17
Conclusions The answers to the questions raised by R K ( ∗ ) may provide a direct experimental probe into the flavour problem. Ben Allanach (University of Cambridge) 18
R ( ∗ ) K in Standard Model R K = BR ( B → Kµ + µ − ) R K ∗ = BR ( B → K ∗ µ + µ − ) BR ( B → Ke + e − ) , BR ( B → K ∗ e + e − ) . These are rare decays (each BR ∼ O (10 − 7 ) ) because they are absent at tree level in SM. 19
R K ( ∗ ) Measurements LHCb results from 7 and 8 TeV: q 2 = m 2 ll . q 2 /GeV 2 LHCb 3 fb − 1 SM σ 0 . 745 +0 . 090 R K [1 , 6] 1 . 00 ± 0 . 01 2.6 − 0 . 074 0 . 66 +0 . 11 R K ∗ [0 . 045 , 1 . 1] 0 . 91 ± 0 . 03 2.2 − 0 . 07 0 . 69 +0 . 11 R K ∗ [1 . 1 , 6] 1 . 00 ± 0 . 01 2.5 − 0 . 07 2 . 0 R K ∗ 0 R K ∗ 0 1 . 0 1 . 5 0 . 8 0 . 6 1 . 0 LHCb BIP 0 . 4 CDHMV 0 . 5 LHCb EOS 0 . 2 BaBar flav.io LHCb LHCb JC Belle 0 . 0 0 . 0 0 1 2 3 4 5 6 0 5 10 15 20 q 2 [GeV 2 /c 4 ] q 2 [GeV 2 /c 4 ] 20
Statistics 3 c µ � χ 2 SM − χ 2 ¯ LL best clean − 1 . 33 ± 0 . 34 4.1 dirty − 1 . 33 ± 0 . 32 4.6 all − 1 . 33 ± 0 . 23 6.2 C µ c µ c µ � χ 2 SM − χ 2 9 = (¯ LL + ¯ LR ) / 2 best clean − 1 . 51 ± 0 . 46 3.9 dirty − 1 . 15 ± 0 . 17 5.5 all − 1 . 19 ± 0 . 15 6.7 Table 1: A fit to flavour anomalies for ‘clean’ ( R K , R K ∗ , B s → µµ ) and ‘dirty’ (100 others) observables 3 D’Amico, Nardecchia, Panci, Sannino, Strumia, Torre, Urbano 1704.05438 Ben Allanach (University of Cambridge) 21
c l Wilson Coefficients ¯ ij In SM, can form an EFT since m B ≪ M W : sγ µ P i b )(¯ O l ij = (¯ lγ µ P j l ) . c l ij � � � O l L eff ⊃ ij , Λ 2 l,ij l = e,µ,τ i = L,R j = L,R α � V tb V ∗ c l LL O l c l LR O l � = ¯ LL + ¯ ts LR 4 πv 2 l = e,µ,τ c l RL O l c l RR O l � +¯ RL + ¯ RR c l ij = (36 TeV / Λ) 2 c l ⇒ ¯ ij . c l ij ∼ ±O (1) all predicted by weak interactions in SM. Ben Allanach (University of Cambridge) 22
Which Ones Work? Options for a single BSM operator : c e • ¯ ij operators fine for R K ( ∗ ) but are disfavoured by global fits including other observables. c µ • ¯ LR disfavoured: predicts enhancement in both R K and R K ∗ c µ c µ RL disfavoured: they pull R K and R K ∗ in opposite • ¯ RR , ¯ directions . c µ LL = − 1 . 33 ± 0 . 34 fits well globally 4 . • ¯ 4 D’Amico et al, 1704.05438 . Ben Allanach (University of Cambridge) 23
Ben Allanach (University of Cambridge) 24
Widths: pick g bs to fit anomalies at each point. Ben Allanach (University of Cambridge) 25
Ben Allanach (University of Cambridge) 26
Z − X mixing Because F H = − 1 / 2 , Z − X mix: g ′ 2 − gg ′ g ′ g F − B µ N = v 2 M 2 − gg ′ g 2 − W 3 − gg F µ 4 g ′ g F − gg F g 2 F (1 + 4 F 2 θ r 2 ) − X µ • v ≈ 246 GeV is SM Higgs VEV • g F = U (1) F gauge coupling • r ≡ v F /v ≫ 1 , where v F = � θ � • F θ is F charge of θ field Ben Allanach (University of Cambridge) 27
� 1 6 u L Λ ( u L ) γ ρ u L + 1 6 d L Λ ( d L ) γ ρ d L − L Xψ = g F 1 2 n L Λ ( n L ) γ ρ n L − 1 2 e L Λ ( e L ) γ ρ e L + 2 3 u R Λ ( u R ) γ ρ u R − � 1 3 d R Λ ( d R ) γ ρ d R − e R Λ ( e R ) γ ρ e R Z ′ ρ , 0 0 0 V † Λ ( I ) ≡ I ξV I , ξ = 0 0 0 0 0 1 Z ′ couplings , I ∈ { u L , d L , e L , ν L , u R , d R , e R } Ben Allanach (University of Cambridge) 28
Important Z ′ Couplings 0 0 0 d L 1 sin 2 θ sb ′ 1 / g F 0 2 sin 2 θ sb Z s L + 6 d L cos 2 θ sb 1 0 2 sin 2 θ sb b L 0 0 0 e L − 1 ′ / Z 0 1 0 µ L 2 e L 0 0 0 τ L Put | sin θ sb | = | V ts | = 0 . 04 , so g µµ ≫ g bs , which helps us survive B s − B s constraint 29
12 Allanach and Davighi, 2018 10 B s -B s bar 8 R K (*) LEP LFU g F 6 4 2 R K (*) 0 0 2 4 6 8 10 M Z' /T eV Ben Allanach (University of Cambridge) 30
Example Case Predictions Mode BR Mode BR Mode BR b ¯ t ¯ ν ′ t 0.42 b 0.12 ν ¯ 0.08 µ + µ − τ + τ − ∼ O (10 − 4 ) 0.08 0.30 other f i f j LEP LFU � 2 � M Z M Z ′ g 2 ≤ 0 . 004 ⇒ g F ≤ 1.3 TeV . F M Z ′ It’s worth LHCb, BELLE II chasing BR ( B → K ( ∗ ) τ ± τ ∓ ) . 31
Ben Allanach (University of Cambridge) 32
Other conclusions • The answers to the questions raise by R K ( ∗ ) may provide a direct experimental probe into the flavour problem. • Focused on tree-level explanations of R K ( ∗ ) as they are usually harder to discover: Z ′ and leptoquarks. • News on R ( ∗ ) K expected in 2019 . At the current central value, Belle II can reach 5 σ by mid 2021. LHCb’s R K ∗ would be close to 5 5 σ by 2020. • R K ( ∗ ) ⇒ HL-LHC, HE-LHC and FCC-hh 5 Albrecht et al , 1709.10308 Ben Allanach (University of Cambridge) 33
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