The Sakharov conditions for generating a matter-antimatter asymmetry are: (1) baryon number violation, (2) C and CP violation, (3) departure from thermal equilibrium. The Standard Model satisfies all three conditions in principle. Why is the SM insufficient to explain the observed baryon asymmetry?
ABecause baryon number is exactly conserved in the Standard Model
BBecause while the SM has all three ingredients (sphaleron processes violate B+L, the CKM phase provides CP violation, and the electroweak phase transition provides departure from equilibrium), the amount of CP violation is too small by ~10 orders of magnitude, and for m_H = 125 GeV the electroweak phase transition is a smooth crossover rather than the required first-order transition
CBecause the SM predicts equal amounts of matter and antimatter
DBecause the Sakharov conditions require physics at the Planck scale
Electroweak sphalerons (non-perturbative field configurations) violate B+L but conserve B-L, satisfying condition (1). The CKM phase provides CP violation, but its contribution to the baryon asymmetry is proportional to a Jarlskog-like invariant that is tiny (~10^{-20}), failing condition (2) quantitatively. For condition (3), a first-order electroweak phase transition would provide the necessary departure from equilibrium, but the Higgs mass of 125 GeV makes the transition a smooth crossover. Both conditions (2) and (3) must be enhanced by new physics.
Question 2 Short Answer
In thermal leptogenesis, the lightest right-handed neutrino N_1 decays to lepton-Higgs pairs (N_1 -> l H and N_1 -> l-bar H-bar). CP violation in the interference between tree and loop diagrams produces a lepton asymmetry. What is the Davidson-Ibarra bound?
Think about your answer, then reveal below.
Model answer: The Davidson-Ibarra bound is an upper limit on the CP asymmetry epsilon_1 in the decay of N_1: |epsilon_1| <= (3 M_1 / (16 pi v^2)) * (m_3 - m_1), where M_1 is the mass of N_1, v = 246 GeV is the Higgs vev, and m_3, m_1 are the heaviest and lightest light neutrino masses. For hierarchical light neutrinos with m_3 ~ sqrt(Delta m^2_{atm}) ~ 0.05 eV, successful leptogenesis requires M_1 > ~10^9 GeV. This implies that the right-handed neutrinos must be very heavy, and the mechanism cannot be directly tested at colliders. However, the connection between leptogenesis and neutrino masses (through the seesaw mechanism) makes it a testable framework: the neutrino mass matrix parameters constrain the leptogenesis parameter space.
The Davidson-Ibarra bound is one of the most important results in leptogenesis because it sets a minimum scale for the mechanism. Alternative scenarios (resonant leptogenesis, where two right-handed neutrinos are nearly degenerate) can evade this bound and allow leptogenesis at lower scales, potentially within reach of collider experiments.
Question 3 Multiple Choice
Electroweak sphalerons convert a lepton asymmetry into a baryon asymmetry. This conversion is described by the relation B = (28/79) * (B - L) in the SM (or B = (8/23) * (B - L) in the MSSM). Why is B - L the relevant quantity?
ABecause B - L is easier to measure than B or L separately
BBecause electroweak sphalerons violate B + L but conserve B - L — any primordial B + L asymmetry is washed out by sphalerons in thermal equilibrium, while the B - L asymmetry is preserved; leptogenesis generates a nonzero B - L (specifically L != 0 with initial B = 0), which sphalerons then redistribute into both B and L, creating the observed baryon asymmetry
CBecause B and L are not individually defined in the Standard Model
DBecause B - L is quantized while B + L is not
Sphalerons are non-perturbative electroweak processes that change B and L by multiples of 3 (one unit per generation) while keeping B - L fixed. At temperatures above ~100 GeV, sphalerons are in thermal equilibrium and efficiently erase any B + L asymmetry. Leptogenesis produces L != 0 (and B = 0), giving B - L = -L. Sphalerons then convert this to B = (28/79) * (B - L) and L = -(51/79) * (B - L), producing the observed baryon asymmetry. This is why B - L must be nonzero for any baryogenesis mechanism that operates above the electroweak scale.