Introduction to Neutrino Physics. TRAN Minh Tâm

Transcription

1 Introduction to Neutrino Physics TRAN Minh Tâm LPHE/IPEP/SB/EPFL This first lecture is a phenomenological introduction to the following lessons which will go into details of the most recent experimental situation of neutrino oscillations and neutrino mass, including the future plans of neutrino physics. 1

2 Notations Metric Feynman s metric) g µν = g µν g 00 = 1 g ii = 1 i = 1,, 3) g ij = 0 i j) Scalar product x µ = t, x) x µ = t, x) p µ = E, p) p µ = E, p) Dirac Matrices A B = A µ B µ = g µν A ν B µ = A 0 B 0 A B [ µ, ν ] + = µ ν + ν µ = g µν Standard representation : ) ) I = j σj = σ 0 I σ j 0 j = Pauli matrices ) ) ) i 1 0 σ 1 = σ 1 0 = σ i 0 3 = 0 1 ) 0 I 5 = 5 i = I 0 σ µν i [µ, ν ] [ µ, ν ] + = g µ ν [ µ, 5] + = µ µ = 0 Transposed j ) T = j for j = 0,, 5 Hermitian conjugates j ) T = j for j = 1, 3 0 ) = 0 5 ) = 5 j ) = j 0 µ ) 0 = µ j = 1,, 3

3 Dirac equation i µ µ m) ψx) = 0 We can expand the 4 component spinor into ψx) = 1 π) 3/ [ up) e ipx + vp) e ipx] Putting back in the Dirac equation, we get in momentum space : µ p µ m) up) = 0 µ p µ + m) vp) = 0 up) for the particles, vp) for the antiparticles. Adjoint spinors : For the adjoint spinors : ψ ψ 0 u u 0 v v 0 ψx) i µ µ + m) = 0 up) µ p µ m) = 0 v µ p µ + m) = 0 Charge conjugaison Consider the Dirac equation for a charged fermion in an external field A µ : i µ µ e µ A µ m) ψx) = 0 1) The charge conjugate spinor ψ c is such that i µ µ + e µ A µ m) ψ c x) = 0 ) Taking the complex conjugate of 1) : [ µ i µ + ea µ ) + m] ψ x) = 0 [i µ + ea µ ) µ + m] ψ x) = 0 If we can find a matrix such that C 0 ) µ ) C 0) 1 = µ 3) 3

4 we then recover equation 1) with ψ c C 0 ψ = C ψ T From 3) and the properties of the matrices page ), one gets C = i 0 and C 1 µ C = µ ) T one also has C = C 1 = C = C T Why can we NOT have neutrino mass in Standard Model? We can always write ψ = ψ L + ψ R 4) ψ R = ) ψ ψ L = 1 1 5) ψ 5) As 5 = 1, 5 = 5, 5 µ = µ 5 5 ψ R = ψ R 5 ψ L = ψ L ψ R = ψ R 0, ψ R = ψ ) 0 = ψ 1 1 5) 6) Thus, the mass term which is proportionnal to ψ ψ, is ψ R ψ R = 1 4 ψ 1 5) ) ψ = 0 as is ψ L ψ L = 0 7) In Standard Model, neutrinos are massless and lepton numbers are separately conserved. Neutrinos only appear as left handed : consider for example the Gauge Boson - Lepton interactions L GB L = M W G )1 [νe µ 1 5 )ew µ + e µ 1 5 )ν e W µ + muon terms + tau terms ] in which neutrinos appear as 1 5 )ν e and antineutrinos as ) ν e. In the following, we will give 3 arguments for neutrinos not being massive in Standard Model. 4

5 a) A standard mass term in the Lagrangian also called Dirac mass term) has the form : = m D ψ ψ = m D ψr + ψ L ) ψr + ψ L ) = m D ψr + ψ L ) ψr + ψ L ) = m D [ ψr ψ L + ψ L ψ R ] This requires the existence of ψ R, that is the existence of right handed neutrinos which do not occur in Standard Model. Of course, you may add them in the Theory, but they do not interact, as they do not appear anywhere in any interaction term : they are sterile and do not couple to the Z 0 as shown by the results obtained at LEP. 8) N = N = 3 N = 4 σ nb) Energy GeV) b) Consider the chirality transformation ψ 5 ψ Then : ψ 5 ψ) = ψ 5 0 = ψ 5 0 = ψ 5 Putting this in L GB L, we see that L GB L is invariant under this chirality transformation ; but mψ ψ mψ 5 5 ψ = mψ ψ 9) i.e. changes sign : is not invariant. 5

6 c) Suppose that you produce in Standard Model a left handed neutrino. If it is massless, its helicity is exactly λ = 1/. If it is massive but moving fast, its helicity is close to λ = 1/. Now, since it is massive, its speed is v < c. Sitting on a reference frame moving along the direction of v, but faster than the particle, you will see the neutrino moving backward, but with the same spin, i.e. its helicity will appear to you to be close to λ = + 1/ : its is then meaningless to say that neutrinos are only left-handed. In Standard Model, there is no inconsistency : all ν masses are zero. Conclusions 1 : To accomodate a mass, one must Introduce sterile ν lr with l = e, µ, τ The Dirac mass term is m D ψ ψ This mass term does not violate lepton number, as we stay within one generation. There would exist neutrinos of both helicities. This is not the only way to have a mass... In general, the mass term is quadratic in ψ and Lorentz invariant. 6

7 The Majorana mass term Recall that under Charge conjugation ψ ψ c = C ψ T with C = i 0 10) ψ ψ c = C ψ T = ψ T) C 0 = ψ ) C 0 ψ c = ψ T 0) T C 0 ψ c = ψ 0) C 0 as 0 = 0) = 0 ) = 0 ) T ψ ψ c = ψ T C 11) Now, claim that ψ c ψ is Lorentz invariant. Let us take a simple pure Lorentz transformation. Recall that under a pure Lorentz transformation of speed v along the third axis Oz : ψ ψ = exp ω 0 3 ) ψ 1) where cosh ω = = 1 1 β, tanh ω = v/c see Bjorken and Drell Relativistic Quantum Mechanics pp ) For the adjoint of the charge conjugate : ψ c ψ c = ψ ) T C [ ψ ) T ω )] = ψ T exp 0 3 T ω = ψ T exp So, under Lorentz transformation : ω ψ c ψ ψ T exp ) 3 T ) ) 0 T C exp 3 ) T 0 ) T ) ω 0 3 ) ψ But expα) = lim 1 α ) N, N N in the development of the left exponential, there is the term 3) T ) 0 T. 7

8 As 0) T = C 1 0 C, one has 3 ) T 0 ) T C = 3 ) T C 1 0 C C = }{{} C = C 1 3 ) T C 1 0 C 1) C }{{} = 1 3 ) T 0 ) T C = 3 ) T C 1 0 = }{{} C 1 µ C = µ ) T C 1 3 CC 1 0 = C 3 0 Finally : ψ c ψ ψ T C exp = ψ T C exp ω 3 0 ) exp ω 0 3 ) ω 0 3 ) ψ = exp ω 0 3 ) ψ c ψ ψ T C ψ = ψ c ψ 13) ψ ψ c ψ is Lorentz invariant as is ψ ψ c Majorana type mass term m M [ψ c ψ + ψ ψ c ] What happens to lepton numbers with this kind of mass term? Recall that ψ νe annihilates one ν e or creates one ν e, destroys one unit of electronic lepton number : L e = 1 Similarly, for ψ νe, it s the reversed adds one unit of electronic lepton number : L e = +1 For ψ c ν e, the rôle of neutrinos and antineutrinos is inverted, ψ c ν e adds one unit of electronic lepton number : L e = +1 ψ νe ψ c ν e adds units of L e With Majorana mass term, L e,... are NOT conserved 8

9 Majorana-type mass term with just ψ L It is possible to construct a Majorana-type mass term using just lefthanded fields or just right handed ones). For we have that ψ L ) c = C ψ L T = C 1 1 5) ψ T = C [ ψ 1 1 5) 0] T 5 = 5) = C 1 0 T 1 T 5 ) ψ = C T 5 )0 ψ as C T 5 = 5 C ψ L ) c = 1 ) C ψ T = ψ c ) R 14) m [ M and so ψ L ) c ψ L + ψ L ψ L ) c] does not vanish. Since ψ L ψ L = ψ R ψ R = 0, we can rewrite the above equation as : m M [ψ L ) c + ψ L ] [ψ L ) c + ψ L ] 15) Define ψ M L = ψ L + ψ L ) c, the mass term becomes : m M ψ M L ψm L and has the Dirac form for a mass term. Under Charge conjugation : ψ M L ψ M L ) c = ψl ) c + ψ L = ψ M L Majorana field is equal to its own C-conjugate : the corresponding Majorana particles are identical to their antiparticles. The Majorana mass term is then : m M ) ψ M c L ψ M L 16) Conclusion : to extend Standard model to accomodate neutrino masses : Include Majorana type mass terms involving ν e L, ν µ L, ν τ L. Neutrinos would be the same as antineutrinos ν ν experimental consequences. Lepton number would be violated. 9

10 Another way to write the mass terms of the Lagrangian for neutrinos We have seen that the Dirac mass term is : As ν L ν R = ν R ) c ν L ) c, L D = m D ν L ν R + h.c. ν c L = C ν L T, ν R ) c = ν T R C, ν R ) c ν L ) c = ν T R C C ν L T = ν L T ν T R = ν L ν R ), one can write : L D = m D ν L ν R + ν R ) c ν L ) c) + h.c. For the Majorana mass term, including also right-handed neutrinos : L M = ml M ν L) c ν L + h.c. mr M ν R) c ν R + h.c. Then the combined L D + L M can be written in matrix form : L D + L M = 1 ν L, ν R ) c) m L M m ) D νl ) c ) + h.c. 17) m D m R M ν R Present neutrino mass limits ν e : m ν e < 3 ev/c ν µ : m ν µ < 190 kev/c ν τ : m ν τ < 18. MeV/c conservative evaluation of Particle Data group based on end point of β spectrum fron H 3 decay and neutrino arrival time from SN1987A Particle Data Group from π + µ + ν µ Uncertainties from pion mass Particle Data Group from τ π π + ν τ 939 events) τ 3 π π + π 0 ) ν τ 5 events) 10

11 Neutrino Mass if it is a Majorana particle There are modes of β decays in which two electrons are simultaneously emitted : these are A, Z) A, Z + ) + e + ν e Double beta decay A, Z) A, Z + ) + e Neutrinoless double beta decay The two decays differ by the spectra of their emitted electrons. The usual double beta decay is a rare, second order transition inside the nucleus ; the life time of the nuclei which undergo such a transition is of the order of years. e ν ν e e e ν ν 0ν The neutrinoless double beta is more subtle : n e - p n Dirac : Majorana : n n Dirac : ν n e p Majorana : ν n e p, p e - ν must have negative helicity p e ν p e ν, ν has mostly positive helicity The ν e is mainly of positive helicity, but to be reabsorbed, the neutrino must have negative helicity. The flip in helicity is possible as long as m νe 0, but its amplitude is proportionnal to m ν /E ν, which is much less than 1. No neutrinoless double beta decay has been detected. From the Heidelberg-Moscow limit : Half life Ge 76 ) neutrinoless = ) 10 5 years, 95% C.L. m ν = ) ev/c, 95% C.L. 11

12 Neutrino Oscillations We can assume that the mass eigenstates, i.e. the states whose wave function has the usual stationary time dependence e iet/, are linear combinations of ν e, ν µ, ν τ which are produced by the weak interaction. Now writing the mass eigenstates as ν 1, ν, ν 3 : ν e U 11 U 1 U 13 ν 1 ν µ = U 1 U U 3 ν ν τ U 31 U 3 U 33 ν 3 The matrix U is unitary U U = 1. Useful notation : i = 1,, 3 and α = e, µ, τ U has the elements U α i and we can write : 18) ν α = U α i ν i 19) Assume that, via weak interaction, we produce ν e, ν µ or ν τ at x = 0 and t = 0. Assume also that the neutrino is produced with a well-defined momentum and propagates in the x-direction strictly we should deal with wave packets, but for the cases of interest, it is okay to take a definite value of the momentum). The mass eigenstates develop in time as take = 1) : ν i p, t) = e i E it ν i p) 0) Now, we always assume that m i << E i, then : E i = p + m i p + m i p Hence, if we begin with then, at time t, it has evolved into c = 1) ν i p, t) = e i p t e i m i t/ E ν i p) 1) ν α = U α i ν i p) at t = 0 ν α p, t) = e i p t U αi e i m i t/ E ν i p) ) The probability to find ν β at time t is P ν α ν β, t) = ν β ν α t) 1

13 Mixing of two flavors Let us take να ν β ) = cos θ sin θ sin θ cos θ Then, for β α ) m P ν α ν β, L) = sin θ sin 4 E L L being the distance to the detector. The probability for ν α to remain ν α is ) ν1 ν ) 1.7 m = sin θ sin ev ) )Lkm) E GeV ) 3) P ν α ν α, x = L) = 1 P ν α ν β, x = L) 4) The CPT theorem implies that P ν α ν β ) = P ν β ν α ) Pratical issues of seeing oscillations 1) Rôle of the mixing angle : Clearly, the whole effect depends on the mixing angle θ and disappears if the mixing is very weak θ 0). The mixing angle will show up in the amplitude of the oscillation probability. ) Range of m : The wave length of the oscillations is Λ km) = π E GeV ) 1.7 m ev ) One has to locate the detector in such a way that the experiment fits to the desired maximum or minimum) oscillation probability. Consequences of seeing oscillations 1) Oscillating neutrinos implies that at least one of the neutrinos is massive. ) If a neutrino is massive, an extension of the Standard Model is necessary! 13

14 The question of coherence In reality, the produced neutrino forms a wave packet with mean momentum p. Let the packet have length δ along the propagation direction. The mass eigenstates have slightly different speeds : β i = p E = i m i 1 1 mi ) 5) E i E i E Hence, after a flight time t, the packets are separated by a distance t β L m E For the wave packets to interfere, they must overlap. Hence we need < δ i.e. L m E < δ 6) πl that is : Λ < δ E In other words, the maximum number of oscillations before incoherence is Examples N max δ E π 10 1 δ meter) E MeV ) 7) a) Neutrino from β-decay : expect δ c τ where τ = decay life-time 10 3 to 1 second, E few MeV. δ and N max is huge. b) Neutrino from π decay : expect δ c Again N max is huge. Neutrino Oscillations : where do we stand? See Neutrino Oscillations, an overview http ://lphe.epfl.ch/ mtran/seminaires/neutrino.pdf 14