Applications and Phenomenology QFT II - Weeks 3 & 4 1. Leptonic Decays of Hadrons: from π → 𝓂 ν to B → 𝓂 ν QFT in Hadron Decays. Decay Constants. Helicity Suppression in the SM. 2. On the Structure and Unitarity of the CKM Matrix The CKM Matrix. The GIM Mechanism. CP Violation. The Unitarity Triangle. 3. Introduction to the “Flavour Anomalies”: Semi-Leptonic Decays B → D (*) 𝓂 ν . The Spectator Model. Form Factors. Heavy Quark Symmetry. B → K (*) 𝓂 + 𝓂 - . FCNC. Aspects beyond tree level. Penguins. The OPE. 4. Introduction to Radiative Corrections: B → μ ν γ The (infrared) pole structure of gauge field theory amplitudes. Collinear and Infrared Safety. Peter Skands & Ulrik Egede Monash University — 2020
Recap of (applied) QFT ๏ Want to: • Start from assumed field content & Lagrangian (e.g., SM). • Compute scattering cross sections and decay rates. Total and differential ๏ • Compare to experimental measurements. ๏ Recipe in perturbative QFT : • Set up (relativistically normalised) in- and outgoing states . Interaction picture: plane-wave states (eigenstates of free theory) ๏ • Compute (Lorentz-invariant) transition amplitudes . QFT under the hood: Dyson’s Formula, Wick Contractions ๏ ➨ For practical calculations: Feynman rules & diagrams ๏ Sum over amplitudes, square, and keep terms to given perturbative order. ๏ • Integrate over the relevant (Lorentz-invariant) phase space (s). 2 Peter Skands Monash University
Recap: Decay Rates See, e.g., PDG review (pdg.lbl.gov) section 47: kinematics Partial decay rate ( a.k.a ., p 1 , m 1 Example: “partial width”) of particle of mass M into n bodies, in its CM: p 2 , m 2 P , M p 3 , m 3 Γ i → f = ∫ d Γ i → f = (2 π ) 4 2 M ∫ | ℳ | 2 d Φ n ( P ; p 1 , …, p n ) Lorentz-invariant Matrix Element Lorentz-invariant phase-space element : n n d 3 p i d Φ n ( P ; p 1 , . . . , p n ) = δ 4 ( P − � � p i ) (2 π ) 3 2 E i i =1 i =1 = d 4 p i with on-shell a.k.a. : dLIPS condition ( L.I. ) 3 Peter Skands Monash University
<latexit sha1_base64="xFO50REUgwcW6wg2AH8i8R+qT58=">ACAXicbZDLSsNAFIYn9VbrLepGcDNYhIpQEqnosujGZRV7gSaGyXTSDp1MwlyEurGV3HjQhG3voU738Zpm4VWDwz8f/nMHP+MGVUKsf5sgoLi0vLK8XV0tr6xuaWvb3TkokWmDRxwhLRCZEkjHLSVFQx0kFQXHISDscXk789j0Rkib8Vo1S4seoz2lEMVJGCuy9i5uKl9K7Y08lXqzNzXVg4Ciwy07VmRb8C24OZBXI7A/vV6CdUy4wgxJ2XWdVPkZEopiRsYlT0uSIjxEfdI1yFMpJ9NxjDQ6P0YJQIc7iCU/XnRIZiKUdxaDpjpAZy3puI/3ldraJzP6M81YpwPHso0gyqBE7igD0qCFZsZABhQc1fIR4gbAyoZVMCO78yn+hdVJ1a9XT61q5XsvjKIJ9cAqwAVnoA6uQAM0AQYP4Am8gFfr0Xq23qz3WvBymd2wa+yPr4BQ3iWGg=</latexit> <latexit sha1_base64="DxiVFPgwJV9bn9P3sFN0+zhqplM=">AB/nicbVDLSsNAFJ3UV62vqLhyEyxCRSiJVHRZdOyin1AE8NketMOnUzCzEQoeCvuHGhiFu/w51/47TNQlsPDHM4517uvSdIGJXKtr+NwtLyupacb20sbm1vWPu7rVknAoCTRKzWHQCLIFRDk1FYNOIgBHAYN2MLye+O1HEJLG/F6NEvAi3Oc0pAQrLfnmwdVdxU3ow6mrYgv0x1MfTnyzbFftKaxF4uSkjHI0fPL7cUkjYArwrCUXcdOlJdhoShMC65qYQEkyHuQ1dTjiOQXjZdf2wda6VnhbHQjytrqv7uyHAk5SgKdGWE1UDOexPxP6+bqvDSyhPUgWczAaFKbP0qZMsrB4VQBQbaYKJoHpXiwywETpxEo6BGf+5EXSOqs6ter5ba1cr+VxFNEhOkIV5KALVEc3qIGaiKAMPaNX9GY8GS/Gu/ExKy0Yec8+gPj8wdXvJRq</latexit> <latexit sha1_base64="AfWJwId7WSOTxVgr1+v4rpCDKC8=">AB+nicdVDJSgNBEO1xjXGb6NFLYxA8jT1qtoMQ9KDHCGaBJISaTkebdM8M3T1KiPkULx4U8eqXePNv7CyCij4oeLxXRVW9IBZcG0I+nLn5hcWl5dRKenVtfWPTzWzVdJQoyqo0EpFqBKCZ4CGrGm4Ea8SKgQwEqwf9s7Ffv2VK8yi8MoOYtSVch7zHKRgrdxMy0CT7CPD3DrHKSEjpslHimUSCGHiZf383mfWFIgpFjKYd8jE2TRDJWO+97qRjSRLDRUgNZNn8SmPQRlOBVslG4lmsVA+3DNmpaGIJluDyenj/CeVbq4FylbocET9fvEKTWAxnYTgnmRv/2xuJfXjMxvWJ7yM4MSyk0W9RGAT4XEOuMsVo0YMLAGquL0V0xtQI1NK21D+PoU/09qh5/5B1eHmfLp7M4UmgH7aJ95KMCKqMLVEFVRNEdekBP6Nm5dx6dF+d12jrnzGa20Q84b58nF5Ko</latexit> Recap: Decay Rates Partial decay rate ( a.k.a ., p 1 , m 1 Example: “partial width”) of particle of mass M into n bodies, in its CM: p 2 , m 2 P , M p 3 , m 3 Γ i → f = ∫ d Γ i → f = (2 π ) 4 2 M ∫ | ℳ | 2 d Φ n ( P ; p 1 , …, p n ) Total Width = sum over partial widths Γ i = ∑ Γ i → j Branching fractions = Γ j / Γ Example: π + decays (see, e.g., pdg.lbl.gov ) DECAY MODES DECAY MODES π π j i BR ( π + → µ + ν µ ) ] (99 . 98770 ± 0 . 00004) % Average Lifetime BR ( π + → e + ν e ) ) × 10 − 4 ] ( 1 . 230 ± 0 . 004 τ = 1 / Γ This agrees with the SM prediction. Our first application: weak leptonic decays of hadrons = ℏ / Γ if not using natural units 4 Peter Skands Monash University
Recap: Master Formula for 2-body decays ๏ In 2-body decays, the kinematics are fully constrained (up to an overall solid angle) m 1 M ๏ m 2 32 π 2 M 2 ∫ | ℳ fi | 2 d Ω VALID FOR ALL | p * | 2-BODY DECAYS Γ i → f = ⇒ Exercise problem E1a: derive this formula from the one on the previous page. with p * the 3-momentum of either of the decay products in the rest frame of M : p * = 1 [ M 2 − ( m 1 + m 2 ) 2 ][ M 2 − ( m 1 − m 2 ) 2 ] 2 M Question: why does it not matter Exercise problem E1b: derive this formula for p * which 3-momentum we use? 5 Peter Skands Monash University
<latexit sha1_base64="xhmb6MZ13pN73CfLqgNOuEvAq0=">ACEXicbVDNS8MwHE3n15xfVY9egkPoEcrEz0OvXic4D5grSXN0i0sTWuSCqPsX/Div+LFgyJevXnzvzHbetDNB4GX934/kveChFGpbPvbKCwtr6yuFdLG5tb2zvm7l5LxqnApIljFotOgCRhlJOmoqRTiIigJG2sHwauK3H4iQNOa3apQL0J9TkOKkdKSb1puQu9OrPsKdFUM3SjVl6QCj6EbIJG5PB37WrSGFd8s21V7CrhInJyUQY6Gb365vRinEeEKMyRl17ET5WVIKIoZGZfcVJIE4SHqk6mHEVEetk0RgeaUHw1jowxWcqr83MhRJOYoCPRkhNZDz3kT8z+umKrzwMsqTVBGOZw+FKYM6/KQe2KOCYMVGmiAsqP4rxAMkEFa6xJIuwZmPvEhap1WnVj27qZXrl3kdRXADoEFHAO6uAaNEATYPAInsEreDOejBfj3fiYjRaMfGcf/IHx+QMiQ5ta</latexit> Pion Decay ๏ Want to calculate M for: π − ( q ) → µ − ( p ) + ¯ ν µ ( k ) First problem: the SM Lagrangian p does not include a “pion” q How are we supposed to apply k Feynman rules without a π - μ - ν vertex? What is really going on? m π = 0.13 GeV q = (m π ,0,0,0) p m W = 80.4 GeV q (how) familiar is this? W propagator: σ ρ k − i ( g ρσ − q ρ q σ / M 2 ig ρσ W ) q 2 − M 2 M 2 It’s the weak force : W exchange W W between quark and lepton currents 6 Peter Skands Monash University
<latexit sha1_base64="xhmb6MZ13pN73CfLqgNOuEvAq0=">ACEXicbVDNS8MwHE3n15xfVY9egkPoEcrEz0OvXic4D5grSXN0i0sTWuSCqPsX/Div+LFgyJevXnzvzHbetDNB4GX934/kveChFGpbPvbKCwtr6yuFdLG5tb2zvm7l5LxqnApIljFotOgCRhlJOmoqRTiIigJG2sHwauK3H4iQNOa3apQL0J9TkOKkdKSb1puQu9OrPsKdFUM3SjVl6QCj6EbIJG5PB37WrSGFd8s21V7CrhInJyUQY6Gb365vRinEeEKMyRl17ET5WVIKIoZGZfcVJIE4SHqk6mHEVEetk0RgeaUHw1jowxWcqr83MhRJOYoCPRkhNZDz3kT8z+umKrzwMsqTVBGOZw+FKYM6/KQe2KOCYMVGmiAsqP4rxAMkEFa6xJIuwZmPvEhap1WnVj27qZXrl3kdRXADoEFHAO6uAaNEATYPAInsEreDOejBfj3fiYjRaMfGcf/IHx+QMiQ5ta</latexit> Application to Pion Decay π − ( q ) → µ − ( p ) + ¯ ν µ ( k ) ๏ Want to calculate M for: • What is really going on? m π = 0.13 GeV q = (m π ,0,0,0) p m W = 80.4 GeV q σ ig ρσ ρ k W propagator: M 2 W − i g w (how) familiar is this? u ( p ) γ σ (1 − γ 5 ) v ( k ) ¯ Lepton current: L σ ( p , k ) = 2 2 − i g w u γ ρ (1 − γ 5 ) u d Quark current: v ¯ ¯ Why not? 2 2 7 Peter Skands Monash University
The Quark Current ๏ The quark-antiquark pair • Bouncing around inside the pion ➜ not free plane-wave states. p q ig ρσ ℳ ( π → μ ¯ ν ) = Q ρ ( q ) L σ ( p , k ) M 2 σ ρ W k ๏ What do we know about the quark current? g w • Must be proportional to g w q ρ f ( q 2 ) ⟹ Q ρ ( q ) = 2 2 • Carries a 4-vector index, ρ q 2 = m π 2 = const. • Since the pion has spin 0 (no spin g w q ρ f π = vector), the only 4-vector is: q 2 2 f π : “Pion decay constant” 8 Peter Skands Monash University
<latexit sha1_base64="xhmb6MZ13pN73CfLqgNOuEvAq0=">ACEXicbVDNS8MwHE3n15xfVY9egkPoEcrEz0OvXic4D5grSXN0i0sTWuSCqPsX/Div+LFgyJevXnzvzHbetDNB4GX934/kveChFGpbPvbKCwtr6yuFdLG5tb2zvm7l5LxqnApIljFotOgCRhlJOmoqRTiIigJG2sHwauK3H4iQNOa3apQL0J9TkOKkdKSb1puQu9OrPsKdFUM3SjVl6QCj6EbIJG5PB37WrSGFd8s21V7CrhInJyUQY6Gb365vRinEeEKMyRl17ET5WVIKIoZGZfcVJIE4SHqk6mHEVEetk0RgeaUHw1jowxWcqr83MhRJOYoCPRkhNZDz3kT8z+umKrzwMsqTVBGOZw+FKYM6/KQe2KOCYMVGmiAsqP4rxAMkEFa6xJIuwZmPvEhap1WnVj27qZXrl3kdRXADoEFHAO6uAaNEATYPAInsEreDOejBfj3fiYjRaMfGcf/IHx+QMiQ5ta</latexit> M and the (spin-summed*) |M| 2 *: actually, initial state is spin 0 and final state only has a single non-zero helicity configuration π − ( q ) → µ − ( p ) + ¯ ν µ ( k ) ๏ So the matrix element for is: 2 g 2 w ℳ = G F G F = ( p ρ + k ρ ) f π [ ¯ u ( p ) γ ρ (1 − γ 5 ) v ( k ) ] 8 M 2 Exercise problem E2: fill in the details W 2 • Use the Dirac eqs. for the neutrino and muon: / / kv ( k ) = 0 u ( p )( p − m μ ) = 0 ¯ ➤ Only a term proportional to the muon mass survives ๏ ℳ = G F f π m μ ¯ u ( p )(1 − γ 5 ) v ( k ) 2 ⟹ | ℳ | 2 = G 2 μ Tr [ ( p + m μ )(1 − γ 5 ) k (1 + γ 5 ) ] F f 2 π m 2 / / 2 = 8( p ⋅ k ) (how) familiar is this? 9 Peter Skands Monash University
Putting it Together | ℳ | 2 = 4 G 2 F f 2 π m 2 μ ( p ⋅ k ) From previous slide: 32 π 2 M 2 ∫ | ℳ fi | 2 d Ω | p * | We also have the master Γ i → f = formula for 1 → 2 decays 2 ( 1 − π ) m 2 p * = m π μ with cf. your derivation of p * m 2 p = ( E μ , p * ) and q k = (p * ,- p * ) q = (m π ,0,0,0) ⟹ ( k ⋅ p ) = ( k ⋅ ( q − k )) = m π | p * | 10 Peter Skands Monash University
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