PHYSICS OF NEUTRINOS. Concha Gonzalez-Garcia. Nufact07 Summer Institute, July (ICREA-University of Barcelona & YITP-Stony Brook)

Transcription

1 PHYSICS OF MASSIVE NEUTRINOS (ICREA-University of Barcelona & YITP-Stony Brook) Nufact07 Summer Institute, July 2007

2 Plan of Lectures I. Standard Neutrino Properties and Mass Terms (Beyond Standard) II. Effects of ν Mass and Neutrino Oscillations (Vacuum) III. Neutrino Oscillations in Matter IV. The Emerging Picture and Some Lessons

3 Plan of Lecture I Standard Neutrino Properties and Mass Terms (Beyond Standard) Historical Introduction The Standard Model of Massless Neutrinos Mass-related Neutrino Properties: Helicity versus Chirality, Majorana versus Dirac Neutrino Mass Terms Beyond the SM: Dirac, Majorana, the See-Saw Mechanism...

4 Discovery of ν s At end of 1800 s radioactivity was discovered and three types identified: α, β, γ β : an electron comes out of the radioactive nucleus.

5 Discovery of ν s At end of 1800 s radioactivity was discovered and three types identified: α, β, γ β : an electron comes out of the radioactive nucleus. Energy conservation e should have had a fixed energy (A, Z) (A, Z + 1) + e E e = M(A, Z + 1) M(A, Z)

6 Discovery of ν s At end of 1800 s radioactivity was discovered and three types identified: α, β, γ β : an electron comes out of the radioactive nucleus. Energy conservation e should have had a fixed energy (A, Z) (A, Z + 1) + e E e = M(A, Z + 1) M(A, Z) But 1914 James Chadwick showed that the electron energy spectrum is continuous

7 Discovery of ν s At end of 1800 s radioactivity was discovered and three types identified: α, β, γ β : an electron comes out of the radioactive nucleus. Energy conservation e should have had a fixed energy (A, Z) (A, Z + 1) + e E e = M(A, Z + 1) M(A, Z) But 1914 James Chadwick showed that the electron energy spectrum is continuous Do we throw away the energy conservation?

8 Discovery of ν s The idea of the neutrino came in 1930, when W. Pauli tried a desperate saving operation of the energy conservation principle. In his letter addressed to the Liebe Radioaktive Damen und Herren (Dear Radioactive Ladies and Gentlemen), the participants of a meeting in Tubingen. He put forward the hypothesis that a new particle exists as constituent of nuclei, the neutron ν, able to explain the continuous spectrum of nuclear beta decay

9 Discovery of ν s The idea of the neutrino came in 1930, when W. Pauli tried a desperate saving operation of the energy conservation principle. In his letter addressed to the Liebe Radioaktive Damen und Herren (Dear Radioactive Ladies and Gentlemen), the participants of a meeting in Tubingen. He put forward the hypothesis that a new particle exists as constituent of nuclei, the neutron ν, able to explain the continuous spectrum of nuclear beta decay (A, Z) (A, Z+1)+e +ν

10 Discovery of ν s The idea of the neutrino came in 1930, when W. Pauli tried a desperate saving operation of the energy conservation principle. In his letter addressed to the Liebe Radioaktive Damen und Herren (Dear Radioactive Ladies and Gentlemen), the participants of a meeting in Tubingen. He put forward the hypothesis that a new particle exists as constituent of nuclei, the neutron ν, able to explain the continuous spectrum of nuclear beta decay (A, Z) (A, Z+1)+e +ν The ν is light (in Pauli s words: the mass of the ν should be of the same order as the e mass ), neutral and has spin 1/2

11 Discovery of ν s The idea of the neutrino came in 1930, when W. Pauli tried a desperate saving operation of the energy conservation principle. In his letter addressed to the Liebe Radioaktive Damen und Herren (Dear Radioactive Ladies and Gentlemen), the participants of a meeting in Tubingen. He put forward the hypothesis that a new particle exists as constituent of nuclei, the neutron ν, able to explain the continuous spectrum of nuclear beta decay (A, Z) (A, Z+1)+e +ν The ν is light (in Pauli s words: the mass of the ν should be of the same order as the e mass ), neutral and has spin 1/2 In order to distinguish them from heavy neutrons, Fermi proposed to name them neutrinos.

12 Discovery of ν s The idea of the neutrino came in 1930, when W. Pauli tried a desperate saving operation of the energy conservation principle. In his letter addressed to the Liebe Radioaktive Damen und Herren (Dear Radioactive Ladies and Gentlemen), the participants of a meeting in Tubingen. He put forward the hypothesis that a new particle exists as constituent of nuclei, the neutron ν, able to explain the continuous spectrum of nuclear beta decay (A, Z) (A, Z+1)+e +ν The ν is light (in Pauli s words: the mass of the ν should be of the same order as the e mass ), neutral and has spin 1/2 In order to distinguish them from heavy neutrons, Fermi proposed to name them neutrinos.

13 First Detection of ν s

14 First Detection of ν s In 1934, Hans Bethe and Rudolf Peierls showed that the cross section between ν and matter should be so small that a ν go through the Earth without deviation

15 First Detection of ν s In 1934, Hans Bethe and Rudolf Peierls showed that the cross section between ν and matter should be so small that a ν go through the Earth without deviation In 1953 Frederick Reines and Clyde Cowan place a neutrino detector near a nuclear plant

16 First Detection of ν s In 1934, Hans Bethe and Rudolf Peierls showed that the cross section between ν and matter should be so small that a ν go through the Earth without deviation In 1953 Frederick Reines and Clyde Cowan place a neutrino detector near a nuclear plant 400 litters of water and cadmium chloride. ν e + p e + + n e + annihilates e of the surrounding material giving two simultaneous γ s. neutron captured by a cadmium nucleus with emission of γ s some 15 msec after

17 First Detection of ν s In 1934, Hans Bethe and Rudolf Peierls showed that the cross section between ν and matter should be so small that a ν go through the Earth without deviation In 1953 Frederick Reines and Clyde Cowan place a neutrino detector near a nuclear plant 400 litters of water and cadmium chloride. ν e + p e + + n e + annihilates e of the surrounding material giving two simultaneous γ s. neutron captured by a cadmium nucleus with emission of γ s some 15 msec after The neutrino was there. Its tag was clearly visible

18 Neutrino Helicity

19 Neutrino Helicity The neutrino helicity was measured in 1957 in a experiment by Goldhaber et al. Using the electron capture reaction e Eu ν Sm 152 Sm + γ with J( 152 Eu) = J( 152 Sm) = 0 and L(e ) = 0

20 Neutrino Helicity The neutrino helicity was measured in 1957 in a experiment by Goldhaber et al. Using the electron capture reaction e Eu ν Sm 152 Sm + γ Angular momentum conservation with J( 152 Eu) = J( 152 Sm) = 0 and L(e ) = 0 J z (e ) = J z (ν) + J z (Sm ) = = J z (ν) + J z (γ) + 1 J z (ν) = 1 2 J z(γ)

21 Neutrino Helicity The neutrino helicity was measured in 1957 in a experiment by Goldhaber et al. Using the electron capture reaction e Eu ν Sm 152 Sm + γ Angular momentum conservation with J( 152 Eu) = J( 152 Sm) = 0 and L(e ) = 0 J z (e ) = J z (ν) + J z (Sm ) = = J z (ν) + J z (γ) + 1 J z (ν) = 1 2 J z(γ) Nuclei are heavy p( 152 Eu) p( 152 Sm) p( 152 Sm ) = 0 So momentum conservation p(ν) = p(γ) ν helicity= γ helicity

22 Neutrino Helicity The neutrino helicity was measured in 1957 in a experiment by Goldhaber et al. Using the electron capture reaction e Eu ν Sm 152 Sm + γ Angular momentum conservation with J( 152 Eu) = J( 152 Sm) = 0 and L(e ) = 0 J z (e ) = J z (ν) + J z (Sm ) = = J z (ν) + J z (γ) + 1 J z (ν) = 1 2 J z(γ) Nuclei are heavy p( 152 Eu) p( 152 Sm) p( 152 Sm ) = 0 So momentum conservation p(ν) = p(γ) ν helicity= γ helicity Goldhaber et al found γ had negative helicity ν has helicity 1

23 The Other Flavours ν coming out of a nuclear reactor is ν e because it is emitted together with an e Question: Is it different from the muon type neutrino ν µ that could be associated to the muon? Or is this difference a theoretical arbitrary convention?

24 The Other Flavours ν coming out of a nuclear reactor is ν e because it is emitted together with an e Question: Is it different from the muon type neutrino ν µ that could be associated to the muon? Or is this difference a theoretical arbitrary convention? In 1959 M. Schwartz thought of producing an intense ν beam from π s decay (produced when a proton beam of GeV energy hits matter) Schwartz, Lederman, Steinberger and Gaillard built a spark chamber (a 10 tons of neon gas) to detect ν µ

25 The Other Flavours ν coming out of a nuclear reactor is ν e because it is emitted together with an e Question: Is it different from the muon type neutrino ν µ that could be associated to the muon? Or is this difference a theoretical arbitrary convention? In 1959 M. Schwartz thought of producing an intense ν beam from π s decay (produced when a proton beam of GeV energy hits matter) Schwartz, Lederman, Steinberger and Gaillard built a spark chamber (a 10 tons of neon gas) to detect ν µ They observe 40 ν interactions: in 6 an e comes out and in 34 a µ comes out. If ν µ ν e equal numbers of µ and e

26 The Other Flavours ν coming out of a nuclear reactor is ν e because it is emitted together with an e Question: Is it different from the muon type neutrino ν µ that could be associated to the muon? Or is this difference a theoretical arbitrary convention? In 1959 M. Schwartz thought of producing an intense ν beam from π s decay (produced when a proton beam of GeV energy hits matter) Schwartz, Lederman, Steinberger and Gaillard built a spark chamber (a 10 tons of neon gas) to detect ν µ They observe 40 ν interactions: in 6 an e comes out and in 34 a µ comes out. If ν µ ν e equal numbers of µ and e Conclusion: ν µ is a different particle

27 The Other Flavours ν coming out of a nuclear reactor is ν e because it is emitted together with an e Question: Is it different from the muon type neutrino ν µ that could be associated to the muon? Or is this difference a theoretical arbitrary convention? In 1959 M. Schwartz thought of producing an intense ν beam from π s decay (produced when a proton beam of GeV energy hits matter) Schwartz, Lederman, Steinberger and Gaillard built a spark chamber (a 10 tons of neon gas) to detect ν µ They observe 40 ν interactions: in 6 an e comes out and in 34 a µ comes out. If ν µ ν e equal numbers of µ and e Conclusion: ν µ is a different particle In 1977 Martin Perl discovers the particle tau the third lepton family. The ν τ was observed by DONUT experiment at FNAL in 1998 (officially in Dec. 2000).

28 Sources of ν s The Sun ν e = ν/cm 2 s E ν MeV Φ Earth ν The Big Bang ρ ν = 330/cm 3 E ν = ev Human Body Φ ν = ν/day Atmospheric ν s ν e, ν µ, ν e, ν µ Φ ν 1ν/cm 2 s SN1987 E ν MeV p Nuclear Reactors E ν few MeV ν e Earth s radioactivity Φ ν ν/cm 2 s Accelerators E ν GeV

29 ν in the SM

30 ν in the SM The SM is a gauge theory based on the symmetry group SU(3) C SU(2) L U(1) Y SU(3) C U(1) EM

31 ν in the SM The SM is a gauge theory based on the symmetry group 3 Generations of Fermions: SU(3) C SU(2) L U(1) Y SU(3) C U(1) EM (1, 2, 1 2 ) (3, 2, 1 6 ) (1, 1, 1) (3, 1, 2 3 ) (3, 1, 1 3 ) L L Q i L E R UR i DR i ( ( νe u i e )L d )L i e R u i R d i R ( ) ( νµµ c i L s )L i µ R c i R s i R ( ( ντ t i τ )L b )L i τ R t i R b i R

32 ν in the SM The SM is a gauge theory based on the symmetry group 3 Generations of Fermions: SU(3) C SU(2) L U(1) Y SU(3) C U(1) EM (1, 2, 1 2 ) (3, 2, 1 6 ) (1, 1, 1) (3, 1, 2 3 ) (3, 1, 1 3 ) L L Q i L E R UR i DR i ( ( νe u i e )L d )L i e R u i R d i R ( ) ( νµµ c i L s )L i µ R c i R s i R ( ( ντ t i τ )L b )L i τ R t i R b i R Spin-0 particle φ: (1, 2, 1 2 ) ( ) ( φ φ = + SSB 1 φ v + h )

33 ν in the SM The SM is a gauge theory based on the symmetry group SU(3) C SU(2) L U(1) Y SU(3) C U(1) EM 3 Generations of Fermions: (1, 2, 1 2 ) (3, 2, 1 6 ) (1, 1, 1) (3, 1, 2 3 ) (3, 1, 1 3 ) L L Q i L E R UR i DR i ( ( νe u i e )L d )L i e R u i R d i R ( ) ( νµµ c i L s )L i µ R c i R s i R ( ( ντ t i τ )L b )L i τ R t i R b i R Spin-0 particle φ: (1, 2, 1 2 ) ( ) ( φ φ = + SSB 1 φ v + h ) Q EM = T L3 + Y ν s are T L3 = 1 2 components lepton doublet L L ν s have no strong or EM interactions No ν R (they are singlets of gauge group)

34 SM Fermion Lagrangian L = 3 3 k=1 i,j=1 3 3 k=1 i,j=1 3 3 k=1 i,j=1 Q i L γµ ( i µ g s λ a,ij +U i R,k γµ ( i µ g s λ a,ij +D i R,k γµ ( i µ g s λ a,ij 2 Ga µ g τ a 2 δ ij W a µ g Y 2 δ ijb µ ) Q j L,k 2 Ga µ g Y 2 δ ijb µ ) U j R,k 2 Ga µ g Y 2 δ ijb µ ) D j R,k 3 +L L,k γ ( µ i µ g τ i 2 Wµ i g Y 2 B µ) LL,k +E R,k γ ( µ i µ g Y 2 B µ) ER,k k=1 3 k,k =1 ( ) λ u kk Q L,k (iτ 2 )φu R,k + λ d kk Q L,k φd R,k +λ l kk L L,k φe R,k + h.c.

35 SM Fermion Lagrangian L = 3 3 k=1 i,j=1 3 3 k=1 i,j=1 3 3 k=1 i,j=1 Q i L γµ ( i µ g s λ a,ij +U i R,k γµ ( i µ g s λ a,ij +D i R,k γµ ( i µ g s λ a,ij 2 Ga µ g τ a 2 δ ij W a µ g Y 2 δ ijb µ ) Q j L,k 2 Ga µ g Y 2 δ ijb µ ) U j R,k 2 Ga µ g Y 2 δ ijb µ ) D j R,k 3 +L L,k γ ( µ i µ g τ i 2 Wµ i g Y 2 B µ) LL,k +E R,k γ ( µ i µ g Y 2 B µ) ER,k k=1 3 k,k =1 ( ) λ u kk Q L,k (iτ 2 )φu R,k + λ d kk Q L,k φd R,k +λ l kk L L,k φe R,k + h.c. Invariant under global rotations q i e iαb/3 q i l i e iα L /3 i l i ν i e iα L /3 i ν i

36 SM Fermion Lagrangian L = 3 3 k=1 i,j=1 3 3 k=1 i,j=1 3 3 k=1 i,j=1 Q i L γµ ( i µ g s λ a,ij +U i R,k γµ ( i µ g s λ a,ij +D i R,k γµ ( i µ g s λ a,ij 2 Ga µ g τ a 2 δ ij W a µ g Y 2 δ ijb µ ) Q j L,k 2 Ga µ g Y 2 δ ijb µ ) U j R,k 2 Ga µ g Y 2 δ ijb µ ) D j R,k 3 +L L,k γ ( µ i µ g τ i 2 Wµ i g Y 2 B µ) LL,k +E R,k γ ( µ i µ g Y 2 B µ) ER,k k=1 3 k,k =1 ( ) λ u kk Q L,k (iτ 2 )φu R,k + λ d kk Q L,k φd R,k +λ l kk L L,k φe R,k + h.c. Invariant under global rotations q i e iαb/3 q i l i e iα L /3 i l i ν i e iα L /3 i ν i Accidental ( not imposed) global symmetry: B L e L µ L τ Each lepton flavour, L i, is conserved Total lepton number L = L e + L µ + L τ is conserved

37 Number of Neutrinos The counting of light active left-handed neutrinos is based on the family structure of the SM assuming a universal diagonal NC coupling: Z ν j µ Z = α ν αl γ µ ν αl ν

38 Number of Neutrinos The counting of light active left-handed neutrinos is based on the family structure of the SM assuming a universal diagonal NC coupling: Z ν j µ Z = α ν αl γ µ ν αl ν For m νi < m Z /2 one can use the total Z-width Γ Z to extract N ν N! ν = Γ inv 1 (Γ Z Γ h 3Γ l ) Γ ν [ Γ ν ] = Γ l 12πR hl Γ ν σh 0 R hl 3 m2 Z Γ inv = the invisible width Γ h = the total hadronic width Γ l = width to charged lepton

39 Number of Neutrinos The counting of light active left-handed neutrinos is based on the family structure of the SM assuming a universal diagonal NC coupling: Z ν j µ Z = α ν αl γ µ ν αl ν For m νi < m Z /2 one can use the total Z-width Γ Z to extract N ν N! ν = Γ inv 1 (Γ Z Γ h 3Γ l ) Γ ν [ Γ ν ] = Γ l 12πR hl Γ ν σh 0 R hl 3 m2 Z Γ inv = the invisible width Γ h = the total hadronic width Γ l = width to charged lepton Leads N ν = ±

40 Fermon Masses in SM A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant)

41 Fermon Masses in SM A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant) In the Standard Model mass comes from spontaneous symmetry breaking via Yukawa interaction of the left-handed doublet L L with the right-handed singlet E R : L l Y = λ l ijl Li E Rj φ + h.c. φ = the scalar doublet

42 Fermon Masses in SM A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant) In the Standard Model mass comes from spontaneous symmetry breaking via Yukawa interaction of the left-handed doublet L L with the right-handed singlet E R : L l Y = λ l ijl Li E Rj φ + h.c. φ = the scalar doublet After spontaneous symmetry breaking φ SSB 0 v+h 2 Ll mass = L L M l E R + h.c. M l = 1 2 λ l v Dirac mass matrix for charged leptons

43 Fermon Masses in SM A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant) In the Standard Model mass comes from spontaneous symmetry breaking via Yukawa interaction of the left-handed doublet L L with the right-handed singlet E R : L l Y = λ l ijl Li E Rj φ + h.c. φ = the scalar doublet After spontaneous symmetry breaking φ SSB 0 v+h 2 Ll mass = L L M l E R + h.c. M l = 1 2 λ l v Dirac mass matrix for charged leptons ν s do not participate in QED or QCD and only ν L is relevant for weak interactions there is no dynamical reason for introducing ν R, so

44 Fermon Masses in SM A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant) In the Standard Model mass comes from spontaneous symmetry breaking via Yukawa interaction of the left-handed doublet L L with the right-handed singlet E R : L l Y = λ l ijl Li E Rj φ + h.c. φ = the scalar doublet After spontaneous symmetry breaking φ SSB 0 v+h 2 Ll mass = L L M l E R + h.c. M l = 1 2 λ l v Dirac mass matrix for charged leptons ν s do not participate in QED or QCD and only ν L is relevant for weak interactions there is no dynamical reason for introducing ν R, so In SM Neutrinos are Strictly Massless

45 Helicity versus Chirality

46 Helicity versus Chirality We define the chiral projections P R,L = 1 ± γ 5 2 ψ = ψ L + ψ R ψ L = 1 γ 5 2 ψ ψ R = 1 + γ 5 2 ψ

47 Helicity versus Chirality We define the chiral projections P R,L = 1 ± γ 5 2 ψ = ψ L + ψ R ψ L = 1 γ 5 2 In the SM the neutrino interaction terms L int = i g 2 [j + µ W µ + j µ W + µ ] + ψ ψ R = 1 + γ 5 2 i g 2 cos θw j Z µ Z µ ψ j µ = l α γ µ P L ν α α = e, µ, τ j + µ = j µ j Z µ = ν α γ µ P L ν α

48 Helicity versus Chirality We define the chiral projections P R,L = 1 ± γ 5 2 ψ = ψ L + ψ R ψ L = 1 γ 5 2 In the SM the neutrino interaction terms L int = i g 2 [j + µ W µ + j µ W + µ ] + ψ ψ R = 1 + γ 5 2 i g 2 cos θw j Z µ Z µ ψ j µ = l α γ µ P L ν α α = e, µ, τ j + µ = j µ j Z µ = ν α γ µ P L ν α ν L interact and ν R do not interact chirality states are physical states for weak interactions

49 Helicity versus Chirality We define the chiral projections P R,L = 1 ± γ 5 2 ψ = ψ L + ψ R ψ L = 1 γ 5 2 In the SM the neutrino interaction terms L int = i g 2 [j + µ W µ + j µ W + µ ] + ψ ψ R = 1 + γ 5 2 i g 2 cos θw j Z µ Z µ ψ j µ = l α γ µ P L ν α α = e, µ, τ j + µ = j µ j Z µ = ν α γ µ P L ν α ν L interact and ν R do not interact chirality states are physical states for weak interactions However what Goldhaber measured was the helicity not the chirality of ν

50 Helicity versus Chirality

51 Helicity versus Chirality The Lagrangian of a massive free fermion ψ is L = ψ(x)(iγ m)ψ(x)

52 Helicity versus Chirality The Lagrangian of a massive free fermion ψ is L = ψ(x)(iγ m)ψ(x) The Equation of Motion is: i t ψ = Hψ = γ0 ( γ. p + m) ψ

53 Helicity versus Chirality The Lagrangian of a massive free fermion ψ is L = ψ(x)(iγ m)ψ(x) The Equation of Motion is: i t ψ = Hψ = γ0 ( γ. p + m) ψ In momentum space this equation has 4 possible solutions (γ p m)u s ( p) = 0 (γ p + m)v s ( p) = 0 s = ± 1 2 and u s( p) and v s ( p) are the four component Dirac spinors.

54 Helicity versus Chirality The Lagrangian of a massive free fermion ψ is L = ψ(x)(iγ m)ψ(x) The Equation of Motion is: i t ψ = Hψ = γ0 ( γ. p + m) ψ In momentum space this equation has 4 possible solutions (γ p m)u s ( p) = 0 (γ p + m)v s ( p) = 0 s = ± 1 2 and u s( p) and v s ( p) are the four component Dirac spinors. For this free fermion [H, J]=0 and [ p, J. p]=0 with J = L + σ 2 (σ i = γ 0 γ 5 γ i )

55 Helicity versus Chirality The Lagrangian of a massive free fermion ψ is L = ψ(x)(iγ m)ψ(x) The Equation of Motion is: i t ψ = Hψ = γ0 ( γ. p + m) ψ In momentum space this equation has 4 possible solutions (γ p m)u s ( p) = 0 (γ p + m)v s ( p) = 0 s = ± 1 2 and u s( p) and v s ( p) are the four component Dirac spinors. For this free fermion [H, J]=0 and [ p, J. p]=0 with J = L + σ 2 (σ i = γ 0 γ 5 γ i ) we can chose u s ( p) and v s ( p) to be eigenstates also of the helicity projector P ± = 1 ± 2 J p p 2 = 1 ± σ p p 2

56 Helicity versus Chirality The Lagrangian of a massive free fermion ψ is L = ψ(x)(iγ m)ψ(x) The Equation of Motion is: i t ψ = Hψ = γ0 ( γ. p + m) ψ In momentum space this equation has 4 possible solutions (γ p m)u s ( p) = 0 (γ p + m)v s ( p) = 0 s = ± 1 2 and u s( p) and v s ( p) are the four component Dirac spinors. For this free fermion [H, J]=0 and [ p, J. p]=0 with J = L + σ 2 (σ i = γ 0 γ 5 γ i ) we can chose u s ( p) and v s ( p) to be eigenstates also of the helicity projector P ± = 1 ± 2 J p p 2 = 1 ± σ p p 2 Only for massless fermions Helicity and chirality states are the same.

57 Dirac versus Majorana Neutrinos In the SM neutral bosons can be of two type: Their own antiparticle such as γ, π 0... Different from their antiparticle such as K 0, K 0... In the SM ν are the only neutral fermions

58 Dirac versus Majorana Neutrinos In the SM neutral bosons can be of two type: Their own antiparticle such as γ, π 0... Different from their antiparticle such as K 0, K 0... In the SM ν are the only neutral fermions OPEN QUESTION: are neutrino and antineutrino the same or different particles?

59 Dirac versus Majorana Neutrinos In the SM neutral bosons can be of two type: Their own antiparticle such as γ, π 0... Different from their antiparticle such as K 0, K 0... In the SM ν are the only neutral fermions OPEN QUESTION: are neutrino and antineutrino the same or different particles? ANSWER 1: ν different from anti-ν ν is a Dirac particle (like e)

60 Dirac versus Majorana Neutrinos In the SM neutral bosons can be of two type: Their own antiparticle such as γ, π 0... Different from their antiparticle such as K 0, K 0... In the SM ν are the only neutral fermions OPEN QUESTION: are neutrino and antineutrino the same or different particles? ANSWER 1: ν different from anti-ν ν is a Dirac particle (like e) It is described by a Dirac field ν(x) = X s, p ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi

61 Dirac versus Majorana Neutrinos In the SM neutral bosons can be of two type: Their own antiparticle such as γ, π 0... Different from their antiparticle such as K 0, K 0... In the SM ν are the only neutral fermions OPEN QUESTION: are neutrino and antineutrino the same or different particles? ANSWER 1: ν different from anti-ν ν is a Dirac particle (like e) It is described by a Dirac field ν(x) = X s, p ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi And the charged conjugate neutrino field the antineutrino field ν C = C ν C 1 = η C X s, p h b s ( p)u s ( p)e ipx + a s( p)v s ( p)e ipxi = η C C ν T (C = iγ 2 γ 0 ) which contain two sets of creation annihilation operators

62 Dirac versus Majorana Neutrinos In the SM neutral bosons can be of two type: Their own antiparticle such as γ, π 0... Different from their antiparticle such as K 0, K 0... In the SM ν are the only neutral fermions OPEN QUESTION: are neutrino and antineutrino the same or different particles? ANSWER 1: ν different from anti-ν ν is a Dirac particle (like e) It is described by a Dirac field ν(x) = X s, p ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi And the charged conjugate neutrino field the antineutrino field ν C = C ν C 1 = η C X s, p h b s ( p)u s ( p)e ipx + a s( p)v s ( p)e ipxi = η C C ν T (C = iγ 2 γ 0 ) which contain two sets of creation annihilation operators These two fields can rewritten in terms of 4 chiral fields ν L, ν R, (ν L ) C, (ν R ) C with ν = ν L + ν R and ν C = (ν L ) C + (ν R ) C

63 Dirac versus Majorana Neutrinos ANSWER 2: ν same as anti-ν ν is a Majorana particle : ν M = ν C M

64 ANSWER 2: ν same as anti-ν η C X s, p Dirac versus Majorana Neutrinos h b s ( p)u s ( p)e ipx + a s( p)v s ( p)e ipxi = X s, p ν is a Majorana particle : ν M = ν C M ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi

65 ANSWER 2: ν same as anti-ν η C X s, p Dirac versus Majorana Neutrinos h b s ( p)u s ( p)e ipx + a s( p)v s ( p)e ipxi = X s, p So we can rewrite the field ν M = s, p ν is a Majorana particle : ν M = ν C M ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi [ as ( p)u s ( p)e ipx + η Ca s( p)v s ( p)e ipx] which contains only one set of creation annihilation operators

66 ANSWER 2: ν same as anti-ν η C X s, p Dirac versus Majorana Neutrinos h b s ( p)u s ( p)e ipx + a s( p)v s ( p)e ipxi = X s, p So we can rewrite the field ν M = s, p ν is a Majorana particle : ν M = ν C M ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi [ as ( p)u s ( p)e ipx + η Ca s( p)v s ( p)e ipx] which contains only one set of creation annihilation operators A Majorana particle can be described with only 2 independent chiral fields: ν L and (ν L ) C which verify ν L = (ν R ) C (ν L ) C = ν R

67 ANSWER 2: ν same as anti-ν η C X s, p Dirac versus Majorana Neutrinos h b s ( p)u s ( p)e ipx + a s( p)v s ( p)e ipxi = X s, p So we can rewrite the field ν M = s, p ν is a Majorana particle : ν M = ν C M ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi [ as ( p)u s ( p)e ipx + η Ca s( p)v s ( p)e ipx] which contains only one set of creation annihilation operators A Majorana particle can be described with only 2 independent chiral fields: ν L and (ν L ) C which verify ν L = (ν R ) C (ν L ) C = ν R In the SM the interaction term for neutrinos L int = i g 2 [( l α γ µ P L ν α )W µ + ( ν α γ µ P L l α )W + µ ] + i g 2 cos θw ( ν α γ µ P L ν α )Z µ Only involves two chiral fields P L ν = ν L and ν P R = η C (ν L ) C T C

68 ANSWER 2: ν same as anti-ν η C X s, p Dirac versus Majorana Neutrinos h b s ( p)u s ( p)e ipx + a s( p)v s ( p)e ipxi = X s, p So we can rewrite the field ν M = s, p ν is a Majorana particle : ν M = ν C M ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi [ as ( p)u s ( p)e ipx + η Ca s( p)v s ( p)e ipx] which contains only one set of creation annihilation operators A Majorana particle can be described with only 2 independent chiral fields: ν L and (ν L ) C which verify ν L = (ν R ) C (ν L ) C = ν R In the SM the interaction term for neutrinos L int = i g 2 [( l α γ µ P L ν α )W µ + ( ν α γ µ P L l α )W + µ ] + i g 2 cos θw ( ν α γ µ P L ν α )Z µ Only involves two chiral fields P L ν = ν L and ν P R = η C (ν L ) C T C Weak interaction cannot distinguish if neutrinos are Dirac or Majorana

69 ANSWER 2: ν same as anti-ν η C X s, p Dirac versus Majorana Neutrinos h b s ( p)u s ( p)e ipx + a s( p)v s ( p)e ipxi = X s, p So we can rewrite the field ν M = s, p ν is a Majorana particle : ν M = ν C M ha s ( p)u s ( p)e ipx + b s( p)v s ( p)e ipxi [ as ( p)u s ( p)e ipx + η Ca s( p)v s ( p)e ipx] which contains only one set of creation annihilation operators A Majorana particle can be described with only 2 independent chiral fields: ν L and (ν L ) C which verify ν L = (ν R ) C (ν L ) C = ν R In the SM the interaction term for neutrinos L int = i g 2 [( l α γ µ P L ν α )W µ + ( ν α γ µ P L l α )W + µ ] + i g 2 cos θw ( ν α γ µ P L ν α )Z µ Only involves two chiral fields P L ν = ν L and ν P R = η C (ν L ) C T C Weak interaction cannot distinguish if neutrinos are Dirac or Majorana The difference arises from the mass term

70 ν Mass Terms A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant)

71 ν Mass Terms A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant) In the Standard Model mass comes from spontaneous symmetry breaking via Yukawa interaction of the left-handed doublet L L with the right-handed singlet E R : L l Y = λ l ijl Li E Rj φ + h.c. φ = the scalar doublet

72 ν Mass Terms A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant) In the Standard Model mass comes from spontaneous symmetry breaking via Yukawa interaction of the left-handed doublet L L with the right-handed singlet E R : L l Y = λ l ijl Li E Rj φ + h.c. φ = the scalar doublet After spontaneous symmetry breaking φ SSB 0 v+h 2 Ll mass = L L M (l) E R + h.c. M l = 1 2 λ l v Dirac mass matrix for charged leptons

73 ν Mass Terms A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant) In the Standard Model mass comes from spontaneous symmetry breaking via Yukawa interaction of the left-handed doublet L L with the right-handed singlet E R : L l Y = λ l ijl Li E Rj φ + h.c. φ = the scalar doublet After spontaneous symmetry breaking φ SSB 0 v+h 2 Ll mass = L L M (l) E R + h.c. M l = 1 2 λ l v Dirac mass matrix for charged leptons ν s do not participate in QED or QCD and only ν L is relevant for weak interactions there is no dynamical reason for introducing ν R, so

74 ν Mass Terms A fermion mass can be seen as at a Left-Right transition m f f L f R + h.c. (this is not SU(2) L gauge invariant) In the Standard Model mass comes from spontaneous symmetry breaking via Yukawa interaction of the left-handed doublet L L with the right-handed singlet E R : L l Y = λ l ijl Li E Rj φ + h.c. φ = the scalar doublet After spontaneous symmetry breaking φ SSB 0 v+h 2 Ll mass = L L M (l) E R + h.c. M l = 1 2 λ l v Dirac mass matrix for charged leptons ν s do not participate in QED or QCD and only ν L is relevant for weak interactions there is no dynamical reason for introducing ν R, so How can we generate a mass for the neutrino?

75 ν Mass Terms: Dirac Mass

76 ν Mass Terms: Dirac Mass OPTION 1: One introduces ν R which can couple to the lepton doublet by Yukawa interaction L (ν) Y = λν ijν Ri L Lj φ + h.c. ( φ = iτ 2 φ )

77 ν Mass Terms: Dirac Mass OPTION 1: One introduces ν R which can couple to the lepton doublet by Yukawa interaction L (ν) Y = λν ijν Ri L Lj φ + h.c. ( φ = iτ 2 φ ) Under spontaneous symmetry-breaking L (ν) Y L(Dirac) mass L (Dirac) mass = ν R M ν Dν L + h.c. 1 2 (ν RM ν Dν L +(ν L ) c M ν D T (ν R ) c )+h.c. X k m k ν D k ν D k M ν D = 1 2 λ ν v =Dirac mass for neutrinos V ν R M D V ν = diag(m 1, m 2, m 3 )

78 ν Mass Terms: Dirac Mass OPTION 1: One introduces ν R which can couple to the lepton doublet by Yukawa interaction L (ν) Y = λν ijν Ri L Lj φ + h.c. ( φ = iτ 2 φ ) Under spontaneous symmetry-breaking L (ν) Y L(Dirac) mass L (Dirac) mass = ν R M ν Dν L + h.c. 1 2 (ν RM ν Dν L +(ν L ) c M ν D T (ν R ) c )+h.c. X k m k ν D k ν D k M ν D = 1 2 λ ν v =Dirac mass for neutrinos V ν R M D V ν = diag(m 1, m 2, m 3 ) L (Dirac) mass involves the four chiral fields ν L, ν R, (ν L ) C, (ν R ) C

79 ν Mass Terms: Dirac Mass OPTION 1: One introduces ν R which can couple to the lepton doublet by Yukawa interaction L (ν) Y = λν ijν Ri L Lj φ + h.c. ( φ = iτ 2 φ ) Under spontaneous symmetry-breaking L (ν) Y L(Dirac) mass L (Dirac) mass = ν R M ν Dν L + h.c. 1 2 (ν RM ν Dν L +(ν L ) c M ν D T (ν R ) c )+h.c. X k m k ν D k ν D k M ν D = 1 2 λ ν v =Dirac mass for neutrinos V ν R M D V ν = diag(m 1, m 2, m 3 ) L (Dirac) mass involves the four chiral fields ν L, ν R, (ν L ) C, (ν R ) C The eigenstates of M ν D are Dirac particles (same as quarks and charged leptons) ν D = V ν ν L + V ν R ν R

80 ν Mass Terms: Dirac Mass OPTION 1: One introduces ν R which can couple to the lepton doublet by Yukawa interaction L (ν) Y = λν ijν Ri L Lj φ + h.c. ( φ = iτ 2 φ ) Under spontaneous symmetry-breaking L (ν) Y L(Dirac) mass L (Dirac) mass = ν R M ν Dν L + h.c. 1 2 (ν RM ν Dν L +(ν L ) c M ν D T (ν R ) c )+h.c. X k m k ν D k ν D k M ν D = 1 2 λ ν v =Dirac mass for neutrinos V ν R M D V ν = diag(m 1, m 2, m 3 ) L (Dirac) mass involves the four chiral fields ν L, ν R, (ν L ) C, (ν R ) C The eigenstates of M ν D are Dirac particles (same as quarks and charged leptons) ν D = V ν ν L + VR ν ν R Total Lepton number is conserved by construction (not accidentally): U(1) L ν = e iα ν and U(1) L ν = e iα ν U(1) L ν C = e iα ν C and U(1) L ν C = e iα ν C

81 ν Mass Terms: Majorana Mass

82 ν Mass Terms: Majorana Mass OPTION 2: One does not introduce ν R but uses that the field (ν L ) c is right-handed, so that one can write a Lorentz-invariant mass term L (Maj) mass = 1 2 νc L M ν M ν L + h.c. 1 2 m k ν M i k ν M i M ν M =Majorana mass for neutrinos is symmetric V ν T M M V ν = diag(m 1, m 2, m 3 )

83 ν Mass Terms: Majorana Mass OPTION 2: One does not introduce ν R but uses that the field (ν L ) c is right-handed, so that one can write a Lorentz-invariant mass term L (Maj) mass = 1 2 νc L M ν M ν L + h.c. 1 2 m k ν M i k ν M i M ν M =Majorana mass for neutrinos is symmetric V ν T M M V ν = diag(m 1, m 2, m 3 ) But under any U(1) symmetry U(1) ν c = e iα ν c and U(1) ν = e iα ν

84 ν Mass Terms: Majorana Mass OPTION 2: One does not introduce ν R but uses that the field (ν L ) c is right-handed, so that one can write a Lorentz-invariant mass term L (Maj) mass = 1 2 νc L M ν M ν L + h.c. 1 2 m k ν M i k ν M i M ν M =Majorana mass for neutrinos is symmetric V ν T M M V ν = diag(m 1, m 2, m 3 ) But under any U(1) symmetry U(1) ν c = e iα ν c and U(1) ν = e iα ν it can only appear for particles without electric charge Total Lepton Number is not conserved The eigenstates of MM ν are Majorana particles ν M = V ν ν L + (V ν ν L ) c (verify ν M c i = νi M )

85 ν Mass Terms: Majorana Mass OPTION 2: One does not introduce ν R but uses that the field (ν L ) c is right-handed, so that one can write a Lorentz-invariant mass term L (Maj) mass = 1 2 νc L M ν M ν L + h.c. 1 2 m k ν M i k ν M i M ν M =Majorana mass for neutrinos is symmetric V ν T M M V ν = diag(m 1, m 2, m 3 ) But under any U(1) symmetry U(1) ν c = e iα ν c and U(1) ν = e iα ν it can only appear for particles without electric charge Total Lepton Number is not conserved The eigenstates of MM ν are Majorana particles ν M = V ν ν L + (V ν ν L ) c (verify ν M c i = νi M ) But SU(2) L gauge invariance is broken!!!

86 General SU(2) L invariant ν Mass Terms

87 General SU(2) L invariant ν Mass Terms OPTION 3: Introduce ν Ri (i = 1, m) and write all Lorentz and SU(2) L invariant mass term L (ν) Y = λν ijν R,i L L,j φ 1 2 ν R,iM ν N,ijν c R,j + h.c.

88 General SU(2) L invariant ν Mass Terms OPTION 3: Introduce ν Ri (i = 1, m) and write all Lorentz and SU(2) L invariant mass term L (ν) Y = λν ijν R,i L L,j φ 1 2 ν R,iM ν N,ijν c R,j + h.c. After spontaneous symmetry-breaking L (ν) mass = ν R M D ν L 1 2 ν RM N ν c R + h.c. 1 2 νc M ν ν + h.c. with ν = ( νl ν c R ) and M ν = ( 0 M T ) D M D M N

89 General SU(2) L invariant ν Mass Terms OPTION 3: Introduce ν Ri (i = 1, m) and write all Lorentz and SU(2) L invariant mass term L (ν) Y = λν ijν R,i L L,j φ 1 2 ν R,iM ν N,ijν c R,j + h.c. After spontaneous symmetry-breaking L (ν) mass = ν R M D ν L 1 2 ν RM N ν c R + h.c. 1 2 νc M ν ν + h.c. with ν = ( νl ν c R ) and M ν = ( 0 M T D M D M N ) L (ν) mass = k 1 2 m kν M k νm k where V ν T M ν V ν = diag(m 1, m 2,..., m 3+m )

90 General SU(2) L invariant ν Mass Terms OPTION 3: Introduce ν Ri (i = 1, m) and write all Lorentz and SU(2) L invariant mass term L (ν) Y = λν ijν R,i L L,j φ 1 2 ν R,iM ν N,ijν c R,j + h.c. After spontaneous symmetry-breaking L (ν) mass = ν R M D ν L 1 2 ν RM N ν c R + h.c. 1 2 νc M ν ν + h.c. with ν = ( νl ν c R ) and M ν = ( 0 M T D M D M N ) L (ν) mass = k 1 2 m kν M k νm k where V ν T M ν V ν = diag(m 1, m 2,..., m 3+m ) In general if M N 0 3+m Majorana neutrino states ν M = V ν ν L + (V ν ν L ) c (verify ν M c i = ν M i ) Total Lepton Number is not conserved

91 The See-Saw Mechanism

92 The See-Saw Mechanism A particular realization of OPTION 3: Add m ν Ri so L (ν) mass = ν R M D ν L 1 2 ν RM N ν c R + h.c. 1 2 νc M ν ν + h.c. with ν = ( νl ν c R ) and M ν = ( 0 M T ) D M D M N

93 The See-Saw Mechanism A particular realization of OPTION 3: Add m ν Ri so L (ν) mass = ν R M D ν L 1 2 ν RM N ν c R + h.c. 1 2 νc M ν ν + h.c. with ν = Assume M N m D ( νl ν c R ) and M ν = ( 0 M T D M D M N )

94 The See-Saw Mechanism A particular realization of OPTION 3: Add m ν Ri so L (ν) mass = ν R M D ν L 1 2 ν RM N ν c R + h.c. 1 2 νc M ν ν + h.c. with ν = Assume M N m D ( νl ν c R ) and M ν = ( 0 M T D M D M N 3 light neutrinos ν s of mass m νl MD T M 1 N M D m Heavy ν s of mass m νh M N )

95 The See-Saw Mechanism A particular realization of OPTION 3: Add m ν Ri so L (ν) mass = ν R M D ν L 1 2 ν RM N ν c R + h.c. 1 2 νc M ν ν + h.c. with ν = Assume M N m D ( νl ν c R ) and M ν = ( 0 M T D M D M N 3 light neutrinos ν s of mass m νl MD T M 1 N M D m Heavy ν s of mass m νh M N The heavier ν H the lighter ν l See-Saw Mechanism Natural explanation to m ν m l, m q )

96 The See-Saw Mechanism A particular realization of OPTION 3: Add m ν Ri so L (ν) mass = ν R M D ν L 1 2 ν RM N ν c R + h.c. 1 2 νc M ν ν + h.c. with ν = Assume M N m D ( νl ν c R ) and M ν = ( 0 M T D M D M N 3 light neutrinos ν s of mass m νl MD T M 1 N M D m Heavy ν s of mass m νh M N The heavier ν H the lighter ν l See-Saw Mechanism Natural explanation to m ν m l, m q Arises in many extensions of the SM: SO(10) GUTS, Left-right... )

97 Summary I

98 Summary I In the SM: Accidental global symmetry: B L e L µ L τ m ν 0 neutrinos are left-handed ( helicity -1): m ν = 0 chirality helicity No distinction between Majorana or Dirac Neutrinos

99 Summary I In the SM: Accidental global symmetry: B L e L µ L τ m ν 0 neutrinos are left-handed ( helicity -1): m ν = 0 chirality helicity No distinction between Majorana or Dirac Neutrinos If m ν 0 Need to extend SM different ways of adding m ν to the SM breaking total lepton number (L = L e + L µ + L τ ) Majorana ν: ν = ν C conserving total lepton number Dirac ν: ν ν C

100 Summary I In the SM: Accidental global symmetry: B L e L µ L τ m ν 0 neutrinos are left-handed ( helicity -1): m ν = 0 chirality helicity No distinction between Majorana or Dirac Neutrinos If m ν 0 Need to extend SM different ways of adding m ν to the SM breaking total lepton number (L = L e + L µ + L τ ) Majorana ν: ν = ν C conserving total lepton number Dirac ν: ν ν C Question: How to search for m ν? Answer:

101 Summary I In the SM: Accidental global symmetry: B L e L µ L τ m ν 0 neutrinos are left-handed ( helicity -1): m ν = 0 chirality helicity No distinction between Majorana or Dirac Neutrinos If m ν 0 Need to extend SM different ways of adding m ν to the SM breaking total lepton number (L = L e + L µ + L τ ) Majorana ν: ν = ν C conserving total lepton number Dirac ν: ν ν C Question: How to search for m ν? Answer: Tomorrow...